WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae

Detailed explanations in West Bengal Board Class 9 Life Science Book Solutions Chapter 1.3C Classification of Kingdom Plantae offer valuable context and analysis.

WBBSE Class 9 Science Chapter 1.3C Question Answer – Classification of Kingdom Plantae

Very Short Questions and Answers : (1 mark for each question)

Question 1.
Members of which group of plants are not divisible into roots, stems and leaves?
Answer:
Members of thallophyta are not divisible into roots, stems and leaves.

Question 2.
Name an alga which has spirally coiled chloroplast In its cells.
Answer:
Spirogyra is an alga, has spirally coiled chloroplast in its cells.

WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae

Question 3.
Name a filamentous alga having prokaryotic cells.
Answer:
Nostoc is a filamentous alga, having prokaryotic cells.

Question 4.
Name a colony-forming alga.
Answer:
Volvox is a colony-forming alga.

Question 5.
Name a unicellular alga, which is now treated as a protist.
Answer:
Chlamydomonas is a unicellular alga, which is now treated as a protist.

Question 6.
What is the fine, branched, colourless, filamentous structure of a fungus called?
Answer:
The fine, branched, colourless, filamentous structure of a fungus is called mycelium.

WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae

Question 7.
What are the filaments of a fungal mycelium known as?
Answer:
The filaments of a fungal mycelium are known as hyphae.

Question 8.
What is the name of the fungal component of a lichen?
Answer:
The fungal component of a lichen is called mycobiont.

Question 9.
What is the name of the algal component of a lichen?
Answer:
The algal component of a lichen is called phycobiont.

Question 10.
Which type of bryophytes form velvety thallus on moist soil?
Answer:
Liverworts form velvety thallus on moist soil.

Question 11.
What is the name of the erect portion of moss?
Answer:
The erect portion of moss is called gametophore.

WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae

Question 12.
What is the name of the stem-like portion of a moss?
Answer:
The stem-like portion of a moss is called caulid.

Question 13.
What is the name of the fine root-like structure of a moss?
Answer:
The fine root-like structure of a moss is called rhizoid.

Question 14.
What does the rhizoid of a moss do?
Answer:
Rhizoid of a moss absorbs water from the soil and affix the plant body to the soil.

Question 15.
Which type of plants possess leaves, stems and roots but do not develop flower and fruit?
Answer:
Ferns possess leaves, stems and roots but do not develop flower and fruit.

Question 16.
Which plants possess ramenta on their stems and the petioles of leaves?
Answer:
Ferns posses ramenta on their stems and the petioles of leaves.

Question 17.
Name a plant in which aerial part is comprised of pileus and stipe.
Answer:
Aerial part of Agaricus is comprised of pileus and stipe.

Question 18.
What are the wart-like structures below the leaflets of Dryopteris called ?
Answer:
The wart-like structures below the leaflets of Dryopteris are called sori.

WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae

Question 19.
Members of which division of plants are known as amphibious plants?
Answer:
Plants under division bryophyta are known as amphibious plants.

Question 20.
In which division of plants, seeds develop without formation of fruits?
Answer:
In gymnosperms, seeds develop without formation of fruits.

Question 21.
Which type of plant body is seen amongst ferns?
Answer:
Sporophytic plant body is seen amongst ferns.

Question 22.
What is the name of the symbiotic association of fungi and the roots of higher vascular plants?
Answer:
The symbiotic association of fungi and the roots of higher vascular plants is known as mycorrhiza.

Question 23.
Which division of non-flowering plants possess vascular system?
Answer:
Plants under division pteridophyta are nonflowering plants, which possess vascular system.

Question 24.
In which group of plants usually many embryos are formed in a single seed?
Answer:
In gymnosperms, usually many embryos are formed in a single seed.

WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae

Question 25.
Which group of plants always show tap root system?
Answer:
Dicotyledonous plants always show tap root system.

Short Questions and Answers : (2 marks for each question)

Question 1.
What is meant by algae?
Answer:
The simplest form of autotrophic, aquatic plants, with thallophytic body, made up of mainly eukaryotic cells, without any tissue system and having unicellular sex organs, are called algae. Example-Spirogyra sp., Chara sp.

Question 2.
What is thallophyta?
Answer:
Thallophyta is a plant group, that includes those plants having flat thallus-like body, which are not divisible into root, stem and leaf. Example-Algae.

WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae

Question 3.
What are bryophytes?
Answer:
Bryophytes are simple non-flowering plants without true roots, stems, leaves and vascular tissues. They grow in moist soil but perform fertilisation in water and maintain gametophytic generation throughout the major part of their life cycle. Example-Pogonatum sp., Sphagnum sp.
WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae 1

Question 4.
What are pteridophytes?
Answer:
Pteridophytes are terrestrial non-flowering plants having true roots, stems, leaves and vascular tissues. They maintain sporophytic generation throughout the major part of their life cycle.
Example — Marsilea sp., Selaginella sp.
WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae 2
Question 5.
What are gymnospenns?
Answer:
The group of plants, in which no fruit is produced, so that the seeds remain exposed and usually many embryos form inside a seed, are called gymnosperms or naked-seeded plants, where endosperm forms before fertilisation. Example — Cycas sp., Pinas sp.

WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae 3

Question 6.
What are anglesperms?
Answer:
The flowering plants, in which fruits develop from flowers and seeds remain within the fruits, are called angiosperms. Here, the endosperm forms after fertilisation. Example — Solanum tuberosum (potato), Malus domestica (apple).

WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae

Question 7.
What is meant by dicotyledonous plants?
Answer:
The angiospermic plants, in which seeds have two cotyledons, show tap root system and have reticulate venation in leaves, are called dicotyledonous plants. Example-Mangifera indica (mango), Pisum sativum (pea).
WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae 4

Question 8.
What is meant by monocotyledonous plants?
Answer:
The angiospermic plants, in which seeds have only one cotyledon, show adventitious root system and have parallel venation in leaves, are called monocotyledonous plants. Example-Cocos nucifera (coconut), Triticum aestivum (wheat).

Question 9.
What are rhizoids?
Answer:
Rhizoids are branched, colourless, very fine thread-like structures growing from the base of the gametophores or the ventral side of the thallus of mosses, which help to absorb water and minerals from soil and affix the plant with the subtratum.

Question 10.
What is a gametophore?
Answer:
The gametophore is a structure, present in the gametophytic generation of mosses and ferns, which bears the male and female sex organs, known as antheridium and archegonium respectively.

WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae

Question 11.
What are ramenta? What is the function of ramenta?
Answer:
(i) Ramenta: Some species of ferns have several thin, brown scale or hair-like projections upon the young shoots and the petioles of leaves. These are called ramenta.
(ii) Function: Ramenta prevent the ferns from drying out by reducing the rate of transpiration in arid condition.

WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae 5

Question 12.
What is circinate vernation?
Answer:
The manner in which immature leaves of ferns are tightly curled so that the tender growing tip of the frond remains protected within a coiled structure, is called circinate vernation. This is commonly seen in Dryopteris sp.
WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae 6

Long Questions and Answers : (5 marks for each question)

Question 1.
Classify plant kingdoms with the help of a chart and example.
Answer:
Classification of plant kingdom with examples
WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae 7

WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae

Question 2.
Mention the similarities between algae and fungi. Discuss their differences.
Answer:
Similarities between algae and fungi
The similarities between algae and fungi are as follows-

  • Both are eukaryotic, unicellular or multicellular.
  • Both forms are thallophytic, that is their body is not differentiated into roots, stems and leaves.
  • Both algae and fungi have distinct cell wall outside their cells.
  • Both algae and fungi are devoid of vascular tissues.

Differences between algae and fungi

Features Algae Fungi
1. Habitat Mostly aquatic, few grow on moist soil Mostly terrestrial
2. Nutrition Mostly autotrophic All are heterotrophic, mostly saprophytic, some are parasitic
3. Photosynthetic pigments Chlorophyll and other pigments Photosynthetic pigments absent
4. Main component of cell wall Main component is cellulose Main component is chitin
5. Stored food Stored food is starch and other polysaccharides Stored food is glycogen
6. Vacuole Centrally large and few vacuoles present Small and many vacuoles present

Question 3.
Mention the similarities between bryophytes and pteridophytes. Discuss the differences between them.
Answer:
Similarities between bryophytes and pteridophytes
The similarities between bryophytes and pteridophytes are as follows-

  • Both bryophytes and pteridophytes are non-flowering plants.
  • Both show distinct gametophytic and sporophytic generations in their respective life cycles.
  • Both require water for fertilisation.
  • Both carry out asexual reproduction by means of spores.
  • The structure of male and female gametangia are alike in both bryophytes and pteridophytes. Reproductive cells are externally covered by sterile cells.

Differences between bryophytes and pteridophytes

Features Bryophytes Pteridophytes
1. Body Body differentiated into stem and leaves, true roots absent Body differentiated into true roots, stem and leaves
2. Main plant type Main plant type is gametophytic Main plant type is sporophytic
3. Vascular tissue True vascular tissues are absent Xylem and phloem are present
4. Habitat Moist soil and rocky surfaces Cool, shady but dry surfaces
5. Roots True root system absent, root-like rhizoids are present True root system is present
6. Sperms Sperms biflagellate Sperms multi flagellate

WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae

Question 4.
Mention the similarities between gymnosperms and angiosperms. Write the differences between them.
Answer:
Similarities between gymnosperms and angiosperms
Similarities between gymnosperms and angiosperms are as follows—

  • Both gymnosperms and angiosperms have their bodies differentiated into true roots, stems and leaves.
  • Both the types of plants develop seeds after fertilisation.
  • In both forms, the ovules remain covered by integument.
  • Both gymnosperms and angiosperms have well-organised vascular bundle.

Differences between gymnosperms and angiosperms:

Features Gymnosperms Angiosperms
1. Nature Perennial and evergreen Annual, biennial or perennial, evergreen or deciduous
2. Vascular tissue

 

Xylem and phloem are devoid of trachea and companion cells respectively We developed xylem and phloem present
3. Flower and fruit Flowers lack ovary, therefore no fruit formation occurs Flowers have ovary, which matures into fruit
4. Seed Seeds remain exposed Seeds remain inside the fruit
5. Embryo More than one embryo originates in ovule, but only one matures Single embryo is formed in an ovule
6. Endosperm Endosperm forms before fertilisation Endosperm forms after fertilisation

Question 5.
Mention the differences between monocotyledonous and dicotyledonous plants. Write the scientific names of two monocot and two dicot plants.
Differences between monocotyledonous and dicotyledonous plants
Answer:

1. Nature Annual, mostly herbs, rarely shrubs and trees Perennial, majority are trees, but several herbs and shrubs are also present
2. Cotyledons Seed has single cotyledon Seed has two cotyledons
3. Leaves Isobilateral, with parallel venation Dorsiventral, with reticulate venation
4. Stem Generally unbranched and less woody Well-branched and mostly woody
5. Root Adventitious roots present Taproot system present
6. Vascular bundle Numerous and scattered vascular bundles in stem Few vascular bundles in stem, arranged in circles or rings

Scientific name of monocot and dicot plants
The scientific names of two inonocot plants are Oryza sativa (rice) and Zea mays (maize) and the scientific names of two dicot plants are Mangifera indica (mango) and Pisum sativum (pea)

Question 6.
Mention the main characteristic features of algae, Give two examples.
Answer:
Characteristic features of algae
WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae 8
The main characteristic features of algae are as follows—

  • Algae are unicellular or multicellular aquatic thallophytes.
  • These are autotrophic, i.e. they can synthesise their own food within the body, due to the presence of chlorophyll.
  • Algal cell wall is mainly composed of cellulose and pectin.
  • Algal cells have variously shaped chloroplasts, which contain photosynthetic pigments and stored food like starch.
  • Sub-cellular micro compartments, called pyrenoids, are present in the chloroplasts.
  • Algae perform vegetative, asexual and sexual reproduction.

Examples of algae
Two examples of algae are Spirogyra maxima and Volvox globator.

WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae

Question 7.
Mention the main characteristic features of bryophytes. Give two examples. Characteristic features’ of bryophytes
Answer:
The main characteristic features of bryophytes are as follows:
WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae 9

  • Bryophytes are green, photosynthetic and non-vascular plants.
  • Plants are either thallus-like or have indistinguishable stem-like caulid, leaf-like phyllids and rhizoids in place of roots.
  • The plant body of bryophytes is gametophytic.
  • Sporophyte depends upon the gametophyte for survival.
  • Both male and female reproductive organs are multicellular.
  • Alternation of generation is distinct and unequal, where gametophytic generation dominates over the sporophytic generation.

Examples of bryophytes
Two examples of bryophytes are Riccia fluitans and Funaria hygrometrica.

Question 8.
Mention the main characteristic features of pteridophytes. Give two examples.
Answer:
Characteristic features of pteridophytes:
WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae 10
The main characteristic features of pteridophytes are as follows-

  • Pteridophytes are muiticellular, non-flowering plants with vascular tissue system.
  • Vascular system is composed of xylem and Marsilea minuta phloem.
  • Vegetative body is differentiated into roots, stems and leaves.
  • The independent plant body is sporophytic.
  • Sporophylls or spore bearing leaves either arrange themselves to form cones or remain free as compound leaves.
  • Alternation of generation is distinct and unequal, where sporophytic generation dominates over the gametophytic generation.

Examples of pteridophytes:
Two examples of pteridophytes are Dryopteris filix-mas and Marsilea minuta.

Question 9.
Mention the main characteristic features of gymnosperms. Give two examples.
Answer:
Characteristic features of gymnosperms:
WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae 11
The main characteristic features of gymnosperms are as follows-

  • Gymnosperms are perennial sporophytic trees.
  • Body differentiated into roots, stems and leaves.
  • Leaves are of two types-dry brown scale leaves and green foliage leaves. Foliage leaves are either needle-like or compound.
  • Vascular system present but xylem and phloem are devoid of trachea and companion cells respectively.
  • Fruit absent, that is why seeds remain exposed or naked.
  • Seeds bear more than one embryo, a feature typically known as polyembryony.
  • In gymnosperms, endosperm develops before fertilisation.

Examples of gymnosperms:
Two examples of gymnosperms are Pinus roxburghii and Gnetum ula.

WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae

Question 10.
What is a sorus? What is cone or strobilus? Give the differences between cryptogams and phanerogams.
Answer:
Sorus
The small, kidney-shaped, wart-like structure which grows on the ventral surface of the leaflets of ferns, to hold the sporangia within it, is called sorus.
WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae 12

Cone or Strobilus
In most gymnosperms, a number of male or female reproductive organs or sporangia-bearing, scale-like structures arrange closely around a central axis to form a cone-shaped or oval structure. This is called cone or strobilus.

Differences between cryptogams and phanerogams:

Features Cryptogams Phanerogams
1. Seeds Do not produce seeds Produce seeds
2. Vascular system May or may not be present Always present
3. Need for water External water serves as medium External water is not required
4. Evolutionary status These are primitive type These are advanced type

Question 11.
Mention the main characteristic features of angiosperms. Give two examples.
Answer:
Characteristic features of angiosperms
WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae 13
The main characteristic features of angiosperms are as follows-

  • Angiosperms are either herbs, shrubs or trees.
  • Their life span varies from annual, biennial to perennial.
  • Vascular tissue system is highly developed, composed of all components of xylem and phloem.
  • Flowers develop ovary, which matures into fruit. Seeds remain protected inside the fruit.
  • Seeds may have one or two cotyledons.
  • In angiosperms, endosperm develops after fertilisation.

Examples of angiosperms
Two examples of angiosperms are Cocos nucifera and Mangifera indica.

WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae

Question 12.
Mention the main characteristic features of monocotyledonous plants. Give two examples.
Answer:
Characteristic features of monocotyledonous plants
WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae 14
The main characteristic features of monocotyledonous plants are as follows-

  • Monocots are mostly herbs, a few are trees.
  • Their life span varies from annual to perennial.
  • Their seeds have a single cotyledon.
  • Their roots are adventitious.
  • Stems are unbranched and leaves are isobilateral with parallel venation.
  • Fruit coat remains fixed with the seed coat.

Examples of monocotyledonous plants
Two examples of monocotyledonous plants are Oryza sativa and Zea mays.

WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae

Question 13.
Mention the main characteristic features of dicotyledonous plants. Give two examples.
Answer:
Characteristic features of dicotyledonous plants
WBBSE Class 9 Life Science Solutions Chapter 1.3C Classification of Kingdom Plantae 15
The main characteristic features of dicotyledonous plants are as follows-

  • Dicots are mostly trees, some are herbs and shrubs.
  • Their life span varies from annual, biennial to perennial.
  • Their seeds have two cotyledons.
  • Well developed tap root system is present in these type of plants.
  • Stems are branched and leaves are dorsiventral with reticulate venation.
  • Fruit coat and seed coat remain separate.

Examples of dicotyledonous plants
Two examples of dicotyledonous plants are Artocarpus heterophyllus and Pisum sativum.

WBBSE Class 9 Life Science Solutions Chapter 1.3B Five Kingdoms of Life

Detailed explanations in West Bengal Board Class 9 Life Science Book Solutions Chapter 1.3B Five Kingdoms of Life offer valuable context and analysis.

WBBSE Class 9 Science Chapter 1.3B Question Answer – Five Kingdoms of Life

Very Short Questions and Answers : (1 mark for each question)

Question 1.
Which type of cells are found among the members of kingdom Monera?
Answer:
Members of kingdom Monera have prokaryotic cells.

Question 2.
The cell wall of the members of kingdom Monera is made up of what?
Answer:
Cell wall of the members of kingdom Monera is made up of polysaccharides and amino acids.

WBBSE Class 9 Life Science Solutions Chapter 1.3B Five Kingdoms of Life

Question 3.
Mention the nature of cells, found among the members of kingdom Protista.
Answer:
Members of kingdom Protista have eukaryotic cells.

Question 4.
Which cell wall component is found among the members of kingdom Fungi?
Answer:
Chitin is the cell wall component found amongst the members of kingdom Fungi.

Question 5.
Which major cell wall component is found among the members of kingdom Plantae?
Answer:
Cellulose is the major cell wall component, found amongst the members of kingdom Plantae.

Question 6.
Which type of nutrition is performed by the members of kingdom Plantae?
Answer:
Members of kingdom Plantae perform autotrophic mode of nutrition.

Question 7.
Which type of nutrition is performed by the members of kingdom Fungi?
Answer:
Saprophytic or parasitic nutrition is commonly performed by the members of kingdom Fungi.

Question 8.
Which type of nutrition is performed by the members of kingdom Animalia?
Answer:
Members of kingdom Animalia perform heterotrophic mode of nutrition.

WBBSE Class 9 Life Science Solutions Chapter 1.3B Five Kingdoms of Life

Question 9.
Which members of living world perform the role of producers in any ecosystem?
Answer:
Members of kingdom Plantae perform the role of producers in any ecosystem.

Question 10.
Which members of living world perform the role of consumers in any ecosystem?
Answer:
Members of kingdom Animalia perform the role of consumers in any ecosystem.

Question 11.
Which members of living world perform the role of decomposers in any ecosystem?
Answer:
Members of kingdom Fungi perform the role of decomposers in any ecosystem.

Question 12.
Which members of kingdom Protista typically possess two flagella?
Answer:
Among the members of kingdom Protista, dinoflagellates typically possess two flagella.

Question 13.
Members of which kingdom perform the role of producers, consumers as well as decomposers in any ecosystem?
Answer:
Members of kingdom Protista perform the role of producers, consumers as well as decomposers in any ecosystem.

Question 14.
Members of which kingdom/kingdoms cause maximum number of human diseases?
Answer:
Members of kingdom Monera, Protista and Fungi cause maximum number of human diseases.

Short Questions and Answers : (2 marks for each question)

Question 1.
Name the five kingdoms of the living world, as proposed by Whittaker.
Answer:
The five kingdoms of the living world, as proposed by Whittaker are – Monera, Protista, Fungi, Plantae and Animalia.

Question 2.
What is meant by Monera?
Answer:
The word Monera has been derived from a Greek word ‘monos’, which means ‘single’. Monera is a kingdom of the living world, which is comprised of unicellular, microscopic and very primitive type of prokaryotic organisms. Example-Bacteria, blue-green algae etc. belong to this group.

WBBSE Class 9 Life Science Solutions Chapter 1.3B Five Kingdoms of Life

Question 3.
What is meant by Protista?
Answer:
The word Protista has been derived from a Greek word ‘protistos’, which means ‘first of all.’ Protista is a kingdom of the living world, which is comprised of unicellular, eukaryotic, microscopic organisms. Example-Amoeba proteus, Euglena viridis etc.

Question 4.
What is meant by Fungi?
Answer:
The word Fungi has been derived from a Latin word ‘fungour’, which means ‘growing fast’. Fungi is a kingdom of the living world, which comprises eukaryotic, non-chlorophyllous, spore-bearing organisms, having cells with chitin-rich cell wall and performing both sexual and asexual reproduction. Example – Mucor mucedo, Aspergillus niger.

Question 5.
What is meant by eukaryotic cells?
Answer:
The cells, which possess well-organised nucleus and membrane-bound cell organelles, are known as eukaryotic cells. Example-The cells of all higher plant and animal cells.

Question 6.
What is meant by prokaryotic cells?
Answer:
The cells, which do not possess true nucleus and membrane-bound cell organelles, are called prokaryotic cells. Example-Bacteria, Mycoplasma, blue-green algae etc.

WBBSE Class 9 Life Science Solutions Chapter 1.3B Five Kingdoms of Life

Question 7.
Mention two disadvantages of five kingdom classification.
Answer:
Two disadvantages of five kingdom classification are as follows-

  • Different members Algae have been placed under three different kingdoms-blue-green algae under Monera, Euglena under Protista and brown algae under Plantae, which is ambiguous.
  • Virus has not been included under any of the five kingdoms.

Question 8.
Which type of organisms are placed under kingdom Plantae?
Answer:
According to five kingdom classification of the living world, all chlorophyllous, photosynthetic, eukaryotic, multicellular organisms with cellulose containing cell wall, are included under kingdom Plantae.

Long Questions and Answers : (5 marks for each question)

Question 1.
Explain the basis of five kingdom classification.
Answer:
Basis of five kingdom classification
The characteristic features, based on which Whittaker proposed his five kingdom classification, are discussed below.
1. Complexity of cellular structure: On the basis of this feature, the living world is divided into two main divisions-

  • Prokaryotes: Primitive forms without true nucleus.
  • Eukaryotes: More complex forms along with true nucleus and membrane-bound cell organelles.

2. Complexity of body structure: Based on the number of cells present in a living body, organisms are classified into two groups-

  • Unicellular: Body is made up of single cell.
  • Multicellular: Body is made up of many cells. In case of multicellular forms, the nature of tissue and tissue systems are also taken into consideration.

WBBSE Class 9 Life Science Solutions Chapter 1.3B Five Kingdoms of Life

3. Mode of nutrition: Living forms are classified into two main types, on the basis of their modes of nutrition. These are-

  • Autotrophs: They are photosynthetic forms capable of preparing their own food.
  • Heterotrophs: These are organisms which depend directly or indirectly on autotrophs for food.

WBBSE Class 9 Life Science Solutions Chapter 1.3B Five Kingdoms of Life 1

4. Role in ecosystem: Based on their roles in ecosystem, living forms are classified into three types-

  • Producers: They synthesise their own food within their body. Mainly photosynthetic green plants belong to this group.
  • Consumers: They depend directly or indirectly on producers for food. These include all animals.
  • Decomposers: They take part in biodegradation of organic matters. Various bacteria and fungi belong to this group.

5. Phylogenetic relation: Besides morphological, anatomical, biochemical and genetic features, the evolutionary trend of different groups are equally emphasized in this type of classification.

WBBSE Class 9 Life Science Solutions Chapter 1.3B Five Kingdoms of Life

Question 2.
Compare the features of the five different kingdoms.
Answer:
Comparative features of the five kingdoms
WBBSE Class 9 Life Science Solutions Chapter 1.3B Five Kingdoms of Life 2

Question 3.
Mention three identifying features of kingdom Monera. Give two examples of Monera.
Answer:
Identifying features of kingdom Monera
Three identifying features of kingdom Monera are as follows-
1. Ceil and cellular organisation : Members of this kingdom are unicellular, microscopic organisms. Their cells do not contain any membrane-bound organelle and membrane-bound true. nucleus is absent (i.e. prokaryotic). Genetic material is represented by a naked circular DNA. Ribosomes and chromatophores are present. Cell wall is composed of polysaccharides and amino acids.

2. Mode of nutrition: Nutritionally these are heterotrophic (parasitic, saprophytic, symbiotic) or autotrophic. Autotrophs are either photosynthetic or chemosynthetic. Among heterotrophs, saprophytes obtain nutrients from dead and decaying organic matters. Symbionts obtain nutrition by the help of other organisms.

3. Role in ecosystem: Photosynthetic and chemosynthetic forms play the role of producers. Heterotrophs act as consumers and saprophytes act as decomposers.
WBBSE Class 9 Life Science Solutions Chapter 1.3B Five Kingdoms of Life 3

Two examples of Monera:
Two examples of Monera are Mycobacterium tuberculosis and Escherichia coli.

WBBSE Class 9 Life Science Solutions Chapter 1.3B Five Kingdoms of Life

Question 4.
Mention three identifying features of kingdom Protista. Give two examples of Protista.
Answer:
Identifying features of kingdom Protista:
Three identifying features of kingdom Protista are as follows-

1. Cell and cellular organisation: These are unicellular, microscopic organisms. Cells contain membrane-bound cell organelles and true nucleus (i.e. eukaryotic). Nucleus contains chromosomes, which contain genetic material.

2. Mode of nutrition: These are heterotrophic (parasitic, symbiotic) or autotrophic in nature. Parasites obtain nutrients from the body of the host. Symbionts obtain nutrition by the help of other organisms.

3. Role in ecosystem: Photosynthetic forms (autotrophs) play the role of producers. Heterotrophs act as consumers.

WBBSE Class 9 Life Science Solutions Chapter 1.3B Five Kingdoms of Life 4

Two examples of Protista
Two examples of Protista are Amoeba proteus and Euglena viridis.

Question 5.
Mention three Identifying features of kingdom Fungi. Give two examples of Fungi.
Answer:
Identifying features of kingdom Fungi
Three identifying features of kingdom Fungi are as follows-

1. Cell and celiular organisation: These are unicellular or multicellular eukaryotic organisms. Cell wall typically contains chitin. Glycogen is the distinctive stored food. Multicellular forms often develop hyphae or mycelium.

2. Mode of nutrition: These are mostly saprophytic. They collect nutrition from dead and decaying organic matters. Some are parasitic, which draw nutrition from living plants and animals. Few symbiotic forms obtain nutrition with the help of their photosynthetic partners.

3. Role in ecosystem: All saprophytic forms act as decomposers. Parasitic forms play the role of consumers.
WBBSE Class 9 Life Science Solutions Chapter 1.3B Five Kingdoms of Life 5
Two examples of Fungi
Two examples of Fungi are Agaricus bisporus and Penicillium notatum.

WBBSE Class 9 Life Science Solutions Chapter 1.3B Five Kingdoms of Life

Question 6.
Mention three identifying features of kingdom Plantae. Give two examples of Plantae.
Answer:
Identifying features of kingdom Plantae
Three identifying features of kingdom Plantae are as follows-

1. Cell and cellular organisation: These are multicellular, organisms. The cells contain membrane-bound organelles and nucleus is well-organised (i.e. eukaryotic). Cell wall is typically rich in cellulose and other polysaccharides. Plastids and large vacuole present in cells. Tissue and tissue systems develop in the body.

2. Mode of nutrition: All members are autotrophs, which perform nutrition by the process of photosynthesis.

3. Role in ecosystem: All members act as producers of the ecosystem.

WBBSE Class 9 Life Science Solutions Chapter 1.3B Five Kingdoms of Life 6
Two examples of Plantae
Two examples of Plantae are sunflower (Helianthus annus) and Riccia glauca.

Question 7.
Mention three identifying features of kingdom Animalia. Give two examples of Animalia.
Answer:
Identifying features of kingdom Animalia
Three identifying features of kingdom Animalia are as follows-

1. Cell and cellular organisation: These are multicellular, eukaryotic organisms. Cells have centrioles. Cell wall and plastids are absent. Body with definite shape and size. Body has well organised tissues, tissue systems, organs and organ systems.

2. Mode of nutrition: All members are heterotrophs, which perform holozoic mode of nutrition.

3. Role in ecosystem: All members act as consumers of the ecosystem.
WBBSE Class 9 Life Science Solutions Chapter 1.3B Five Kingdoms of Life 7

Two examples of Animalia
Two examples of Animalia are starfish (Asterias rubens) and pigeon (Columba livia).

WBBSE Class 9 Life Science Solutions Chapter 1.3B Five Kingdoms of Life

Question 8.
Mention the advantages of five kingdom classification. According to five kingdom classification of the living world, which type of organisms are placed under kingdom Animalia?
Answer:
Advantages of five kingdom classification
The advantages of five kingdom classification are as follows-

  • All unicellular prokaryotes are separated from eukaryotic unicellular protists and placed under kingdom Monera. So this classification is justified.
  • Isolation of non-photosynthetic fungi from photosynthetic green plants is also justified because the mode of nutrition of fungi is different from that of green plants.
  • Separating Protozoa from Animalia, makes the classification system more accurate.
  • This classification rightly emphasises on the phylogeny of the living world.

WBBSE Class 9 Life Science Solutions Chapter 1.3B Five Kingdoms of Life 8
Organisms under kingdom Animalia:
According to five kingdom classification of the living world, all heterotrophic, multicellular organisms, having cells without cell wall and holozoic mode of nutrition, are included under kingdom Animalia. It comprises all the consumers of different ecosystem.

WBBSE Class 9 Life Science Solutions Chapter 1.3A Classification of Diversity of Life: Taxonomy

Detailed explanations in West Bengal Board Class 9 Life Science Book Solutions Chapter 1.3A Classification of Diversity of Life: Taxonomy offer valuable context and analysis.

WBBSE Class 9 Science Chapter 1.3A Question Answer – Classification of Diversity of Life: Taxonomy

Very Short Questions and Answers : (1 mark for each question)

Question 1.
Who introduced the concept of species?
Answer:
The concept of species was introduced by John Ray.

WBBSE Class 9 Life Science Solutions Chapter 1.3A Classification of Diversity of Life: Taxonomy

Question 2.
Name the scientist who gave the modern definition of species.
Answer:
Scientist Ernst Mayr (1966) gave the modern definition of species.

Question 3.
Name the systematic framework of classification with fixed number of rank.
Answer:
The systematic framework of classification with fixed number of ranks, is called hierarchy.

Question 4.
What is the scientific name of India’s national fruit?
Answer:
The scientific name of India’s national fruit (mango) is Mangifera indica.

Question 5.
What is the full form of ICZN?
Answer:
Full form of ICZN is International Code of Zoological Nomenclature.

Question 6.
Before which year, the published scientific names of animals were considered invalid?
Answer:
Before the year 1758 , the published scientific names of animals were considered invalid.

Question 7.
Name an aquatic animal having same common name and generic name.
Answer:
Octopus is an aquatic animal, having same common and generic name.

WBBSE Class 9 Life Science Solutions Chapter 1.3A Classification of Diversity of Life: Taxonomy

Question 8.
Give an example of a fish having same generic and specific name.
Answer:
Scientific name of catla fish (Catla catla) has same generic and specific name.

Question 9.
Name the branch of biology that deals with identification, classification and nomenclature of organisms.
Answer:
The branch of biology that deals with identification, classification and nomenclature of organisms, is known as taxonomy.

Question 10.
In which edition of ‘Systema Naturae’ was the binomial nomenclature of animals accepted by ICZN?
Answer:
In the 10th edition of ‘Systema Naturae’, the binomial nomenclature of animals was accepted by ICZN.

Question 11.
Which year’s edition of ‘Species Plantarum’ was accepted by ICBN for binomial nomenclature of plants?
Answer:
The edition of ‘Species Plantarum’ published in the year 1753, was accepted by ICBN for binomial nomenclature of plants.

Question 12.
Define characterisation.
Answer:
Characterisation is an important feature of taxonomy which involves the process of listing the identifying features of collected specimen.

Question 13.
Name a terrestrial animal, having same common and scientific name.
Answer:
Gorilla is a terrestrial animal, which has same common name and scientific name (Gorilla gorilla).

WBBSE Class 9 Life Science Solutions Chapter 1.3A Classification of Diversity of Life: Taxonomy

Question 14
Which type of organisms are mostly known by their scientific names?
Answer:
Most of the protozoa are known by their scientific names, such as Amoeba, Paramoecium etc.

Short Questions and Answers : (2 marks for each question)

Question 1.
How did the word ‘taxonomy’ originate?
Answer:
The word ‘taxonomy’ originated from two Greek words-‘taxis’ meaning ‘rank’ and ‘nomos’, meaning ‘law’ or ‘custom’. Therefore, the word ‘taxonomy’ literally means the ‘custom of ranking.’

Question 2.
What are the three basic features of Linnaean taxonomy?
Answer:
Three basic features of Linnaean taxonomy are-

  • Characterisation of living organisms based on certain observable features
  • Naming the organisms according to binomial nomenclature and
  • Placing the individuals into seven major levels of hierarchical ranks.

Question 3.
Mention the elements of discussion under taxonomy.
Answer:
The main elements of discussion under taxonomy are-

  • identification
  • nomenclature
  • classification and
  • documentation.

WBBSE Class 9 Life Science Solutions Chapter 1.3A Classification of Diversity of Life: Taxonomy

Question 4.
What is meant by identification?
Answer:
Identification is a taxonomic process of separating out any organism from a cluster of other forms, on the basis of its distinctive features.

Question 5.
What does spot identification mean?
Answer:
Spot identification is the process of separating out any plant or animal instantly from a cluster of other forms, on the basis of some external features.

Question 6.
What is meant by categorisation or grouping?
Answer:
The taxonomic process by which any organism is placed under a taxon or rank, based on its similarities with the other members of that group, is called grouping or categorisation.

Question 7.
What is meant by nomenclature?
Answer:
Nomenclature is a taxonomic process of providing a scientific name to a properly identified plant or animal, according to the internationally accepted rules in order to distinguish it from others.

Question 8.
What is classification?
Answer:
Classification is a taxonomic process of placing or grouping organisms into proper taxonomic categories on the basis of similarities in characters.

Question 9.
What is meant by documentation?
Answer:
Documentation is the scientific method of preservation of the type specimens, their description and all relevant data in museum or in laboratories, after the completion of taxonomic processes like identification, nomenclature and classification of a specimen.
WBBSE Class 9 Life Science Solutions Chapter 1.3A Classification of Diversity of Life Taxonomy 1

Question 10.
What are taxonomic categories?
Answer:
Taxonomic categories are the units in which plants or animals are grouped during classification. Species, genus, family, order, class, phylum etc. are the examples of different taxonomic categories.

Question 11.
What is taxon?
Answer:
Taxon is the basic unit of hierarchy, which is a rank, given to a group of organisms with similar features, during classification.

WBBSE Class 9 Life Science Solutions Chapter 1.3A Classification of Diversity of Life: Taxonomy

Question 12
What is systematics?
Answer:
According to Simpson (1961), systematics is the scientific study of the kinds and diversities of organisms and relationship among them. The terms, systematics and taxonomy, often complement each other.

Question 13.
Mention the main objective of systematics.
Answer:
The main objective of systematics is to assort different organisms on the basis of their similarities and dissimilarities, so as to determine the inter-relationship amongst them and to point out their origin and trend of evolution.

Question 14.
Mention two problems of using common name of organisms with example.
Answer:
Two problems of using common names are:

  • Common names vary with regional languages, due to which an organism cannot be recognised universally. For example, crow is called ‘kak’ in bengali, ‘kawa’ in hindi, ‘corneille’ in french, ‘karasu’ in japanese etc.
  • Common name may create identity crisis. For example, silver fish is an insect, but the name apparently indicates a fish.

Question 15.
What is Linnaean hierarchy?
Answer:
The five-category system of hierarchical classification system proposed by Linnaeus in the 10th edition of his famous book, ‘Systema Naturae,’ is known as Linnaean hierarchy. He classified the living organisms into five categories, which was later modified into seven categories.
WBBSE Class 9 Life Science Solutions Chapter 1.3A Classification of Diversity of Life Taxonomy 2

Question 16.
What is binomial nomenclature?
Answer:
Binomial nomenclature is the method of providing a scientific name to an organism. The name should contain two distinct epithets, the first one being genus and the second one is species. For example, Panthera tigris is the binomial nomenclature of tiger.

Question 17.
What is trinomial nomenclature?
Answer:
Trinomial nomenclature is the method of providing a scientific name to an organism where the name should contain three distinct epithets, genus, species and sub-species. For example, Homo sapiens sapiens is the trinomial nomenclature of human.

Question 18.
Mention the seven categories of modified Linnaean hierarchy.
Answer:
Seven categories of modified Linnaean hierarchy, from highest to lowest rank are
Kingdom → Phylum → Class → Order → Family → Genus → Species.

WBBSE Class 9 Life Science Solutions Chapter 1.3A Classification of Diversity of Life: Taxonomy

Question 19.
What is meant by structure of hierarchy? Represent the structure of Linnaean hierarchy.
Answer:

  • Structure of hierarchy: The arrangement of all categories of hierarchy from the base to the top collectively forms the structure of hierarchy.
  • Structure of Linnaean hierarchy: The structure of Linnaean hierarchy can be respresented as follows-

WBBSE Class 9 Life Science Solutions Chapter 1.3A Classification of Diversity of Life Taxonomy 3

Question 20.
Mention the importance of classification.
Answer:
The importance of classification are as follows-

  • By means of classification, an organism can be sorted out easily and quickly from a cluster of organisms.
  • Classification helps to determine the inter-relationship amongst different groups of organisms.

Long Questions and Answers : (5 marks for each question)

Question 1.
Write down the brief history of the development of taxonomy in the pre-Linnaean period.
Answer:
Pre-Linnaean development of taxonomic science
Taxonomy is one of the oldest branches of biology, which has developed gradually since ancient time. The developmental phases of taxonomy, according to the ages, have been discussed below.
WBBSE Class 9 Life Science Solutions Chapter 1.3A Classification of Diversity of Life Taxonomy 4
(1) Ancient age: The concept of taxonomy has developed in India about 2500 years back. The glimpse of classification was noticed in Rig veda, Manusamhita and Agnipuran. Manu, an ancient sage, classified plants into oshadhi, banaspati, briksha, gulma, trina and balli. In ‘Briksha ayurveda’, Rishi Parashar (1500 BC) classified plants on the basis of morphological features.

Susruta, the famous ancient Indian medic, classified medicinal plants into 37 genera. Greek philosopher, Aristotle (384-322 BC) classified animal kingdom in two divisionsAnaima (having colourless blood) and Enaima (having red blood). Theophrastus (371-287 BC), the disciple of Aristotle, described 480 species of plants and classified those into trees, shrubs and herbs, in his famous book, ‘The Historia Plantarum’

WBBSE Class 9 Life Science Solutions Chapter 1.3A Classification of Diversity of Life: Taxonomy

(2) Roman age: Famous Roman naturalist Pliny the Elder (23-79 AD), classified plants in his book, ‘Naturalis Historia’ on the basis of their medicinal qualities. Roman physician Discordies, described about 600 medicinal plants in his famous book ‘The Materia Medica.’

(3) Middle age: Italian botanist Caesalpino (1583), in his famous book ‘The Plantis’, classified 1500 plants into trees, shrubs and herbs. In his classification, he put emphasis on the character of fruits and seeds. For this great work, Caesalpino is regarded as the pioneer of plant taxonomy.

Gaspard Bauhin (1596) classified 6000 species of plants in his famous book ‘Pinax’. He was the first person who tried to show the difference between genus and species. Famous scientist John Ray (1627-1705), classified plants based on the basis of the number of cotyledon of seeds in his book ‘Methodum Plantarum’.

Question 2.
Write down the brief history of the development of taxonomy in pre-Linnaean period.
Answer:
Unnaeus’s role in developing modern taxonomy:

Modern taxonomy was set into light after the work of great Swedish scientist Carolus Linnaeus. He was the pioneer of scientific classification of plants and animals. In 1751, he first suggested the concept of binomial nomenclature in his famous book Philosophia Botanica

He studied numerous plants and animals, which he described in the latest edition of his two famous books, ‘Species Plantarum’ (1753) and ‘Systema Naturae’ (1758), respectively.

WBBSE Class 9 Life Science Solutions Chapter 1.3A Classification of Diversity of Life Taxonomy 5

Taxonomists of post-Linnaean period:
Augustin de Candolle (1799), George Bentham (1862-1883), Joseph Dalton Hooker (1855), Adolph Engler, Karl Prantl (1887-1915), John Hutchinson (1926), Ernst Mayr (1963) are some of t

he renowned scientists who have enriched the taxonomic science to a great extent in the post-Linnaean period.

In last twenty years, there was immense development noticed in the field of systematic classification of plant kingdom based on phylogenetic analysis. Takhtajan (1997), Cronquist (1981) etc. played promising role in this aspect.

Question 3
What is meant by taxonomic hierarchy? How did Linnaean hierarchy modify in course of time?
Answer:
Taxonomic heirarchy
Taxonomic hierarchy is the sequential arrangement of various categories or units of classification comprising of different taxa, from larger to smaller ranks. This includes seven categories.

Modification of Linnaean hierarchy
In course of time, newer variants of organisms have been discovered. Modern taxonomists faced problems in positioning those variants in the prevailing seven-step format. Consequently, Simpson (1961) suggested a
21-step hierarchy for animal classification.

In this format, a new step ‘tribe’ was being inserted in between genus and sub-family. Besides this, some intermediate steps were also inserted in between the main steps prefixed with ‘supra,’ ‘sub’ and ‘infra’. Later, ICBN proposed a 24-step hierarchy structure.

Here, few more new steps were introduced, namely ‘section’ and ‘series’ between genus and species and ‘variety, ‘form’ and ‘clone’ after species. However, the intermediate steps, prefixed with ‘supra’ and ‘infra’ are discarded. Thus, the old Linnaean hierarchy is modernised in course of time.

WBBSE Class 9 Life Science Solutions Chapter 1.3A Classification of Diversity of Life: Taxonomy

Question 4.
Describe every category of taxonomic hierarchy with example.
Answer:
Categories of taxonomic hierarchy
Taxonomic hierarchy includes seven categories or ranks, which are described below.
1. Kingdom: It is the topmost category of taxonomic hierarchy. All different phyla or divisions constitute a kingdom. American scientist, R H Whittaker (1969) classified the whole living world into five different kingdoms-Monera, Protista, Fungi, Plantae and Animalia. Example-Human belongs to kingdom ‘Animalia’ and mango tree belongs to kingdom ‘Plantae.’

2 Phylum/Division: Related classes of animals constitute a phylum. In case of plants, this category is known as division. Example-In case of human, phylum is ‘Chordata’ and for mango tree, division is ‘Angiospermae.’

3 Class: A class consists of several related orders.
Example-Human belongs to class ‘Mammalia’ and mango tree belongs to class ‘Dicotyledoneae.’

WBBSE Class 9 Life Science Solutions Chapter 1.3A Classification of Diversity of Life Taxonomy 6

4. Order: Several related families add up to create an order. Example-Human belongs to the order ‘Primates’ and mango tree belongs to the order ‘Sapindales’.

5. Family: All related genus together comprise a family. Example-Family of human is ‘Hominidae’ and for mango tree, it is ‘Anacardiaceae’.

6 Genus: Several related species constitute a genus. Example-Genus of human and mango tree are Homo and Mangifera respectively.

7 Species: It is the lowest category of taxonomic hierarchy. The taxon under this category include the group of naturally interbreeding organisms, which remain reproductively isolated from other related groups.
Example-Species of human and mango tree are sapiens and indica respectively.

Question 5.
What are the advantages of hierarchical classification? Mention the systematic position of mango tree according to Linnaean hierarchy.
Answer:
Advantages of hierarchical classification
The advantages of hierarchical classification has been discussed below.
1. Inter-relation amongst groups: Hierarchical system of classification provides ready information necessary to determine inter-relationships amongst two groups of plants or two groups of animals.

2. Flawless grouping: Before this system of classification, characteristic features of the sample specimens were studied vividly. Emergence of this system of classification helps us to study the characteristics of the sample specimens more specifically. Therefore, the grouping becomes flawless.

3. Easy identification of taxon: From a hierarchical classification, any taxon can be easily identified.

WBBSE Class 9 Life Science Solutions Chapter 1.3A Classification of Diversity of Life: Taxonomy

4. Other advantages: In this system of classification, repetition of characteristic features does not occur amongst the taxa of higher and lower categories. Only additional similarities are added to the lower categories.

Systematic position of mango tree
The systematic position of mango tree has been represented below.
WBBSE Class 9 Life Science Solutions Chapter 1.3A Classification of Diversity of Life Taxonomy 7

Question 6.
Mention the rules of binomial nomenclature.
Answer:
Rules of binomial nomendature

The basic principles of binomial nomenclature was initially proposed by Linnaeus. Later in the 19th and 20th century, several international bodies of taxonomy, namely ICBN, ICZN, ICBacN, ICVN and ICNCPP reviewed the prevailing rules and published the final edited version of the rules of binomial nomenclature. The most important rules are as follows—

1. Every organism must have a definite scientific name in Latin or Greek, comprising of two epithets – a generic name and a specific name.

2. Each of the epithets must not be composed of less than three or more than twelve roman alphabets. The generic name is a noun, which should be written first, starting with a capital letter.

3. A generic name cannot be repeated in any other organism of the same or different kingdom.

4. Specific name should be written after generic name. It should be written in small letters.

5. A specific name may be repeated under different genus.

6. The name of the discoverer may be mentioned in full or in abbreviation, after the specific name.

7. A scientific name should be printed in italics. If handwritten, each epithet should be separately underlined. For example, scientific name of mango tree should be written as Mangifera indica L. (Here L. stands for Linnaeus, the discoverer).

8. Any plant name, used before publication of ‘Species Plantarum’ (1753) and animal name, used before publication of 10th edition of ‘Systema Naturae’ (1758) must be discarded.

9. If a single organism is assigned two different names by two separate discoverers in different times, the former will be accepted and the latter will be discarded. This is known as Law of priority.

10. Detailed description with photograph of the sample used for nomenclature, must be published in any internationally accepted science journal. The type specimen, used for the nomenclature, must be preserved in any museum competent for such job.

WBBSE Class 9 Life Science Solutions Chapter 1.3A Classification of Diversity of Life: Taxonomy

Question 7.
Mention the systematic position of human according to taxonomic hierarchy and the important characteristic features of different taxa.
Answer:
Systematic position of human and characteristic features of different taxa

MSBM Taxon Characteristic features
1. Kingdom Animalia Multicellular, cells without plastids and cell wall, heterotrophic, capable of locomotion
2. Phylum Chordata Notochord present at any stage of the life cycle, dorsal tubular nerve cord present, pharyngeal gill slits and post anal tail present
3. Class Mammalia Body covered with hair, sweat glands and sebaceous glands present in skin, mammary glands and external pinna present
4. Order Primata or Primates Palm and foot can grip any object, all digits have nails, vision is strong but sense of smell (olfactory sense) limited
5. Family Hominidae Body erect, capable of bipedal movement, hands shorter than legs
6. Genus Homo Brain is advanced and structurally complex, structure of palm is well developed (to use and prepare tools), capable of talking
7. Species sapiens High forehead, distinct lower jaw-line, culturally advanced

 

WBBSE Class 9 Life Science Solutions Chapter 1.2 Biology and its Branches

Detailed explanations in West Bengal Board Class 9 Life Science Book Solutions Chapter 1.2 Biology and its Branches offer valuable context and analysis.

WBBSE Class 9 Science Chapter 1.2 Question Answer – Biology and its Branches

Very Short Questions and Answers : (1 mark for each question)

Question 1.
What is anatomy?
Answer:
Anatomy is the study of the structure and location of internal organs and organ systems of animals and plants.

Question 2.
What is genetics?
Answer:
Genetics is the study of genes and hereditary features of organisms.

WBBSE Class 9 Life Science Solutions Chapter 1.2 Biology and its Branches

Question 3.
What is embryology?
Answer:
Embryology is the study of structure and development of embryos of living organisms.

Question 4.
What is evolutionary biology?
Answer:
Evolutionary biology is the study of origin of life and trends of changes in different forms of life with time.

Question 5.
What is biogeography?
Answer:
Biogeography is the study of distribution of plants and animals in different geographical regions.

Question 6.
What is cybernetics?
Answer:
Cybernetics is the science that deals with the technique of communications and control systems in both machines and living organisms.

Question 7.
What do you mean by hybridization?
Answer:
Hybridization is the process of crossing two genetically different individuals to produce an offspring with a different, often preferred, set of characters.

WBBSE Class 9 Life Science Solutions Chapter 1.2 Biology and its Branches

Question 8.
Mention any three branches of applied biology.
Answer:
Three branches of applied biology are — sericulture, apiculture and animal husbandry.

Question 9.
What is agriculture?
Answer:
Agriculture is the scientific practice of cultivation of commercially important crops.

Question 10.
What is horticulture?
Answer:
Horticulture is the scientific practice of production of flowers and fruits.

Question 11.
What is apiculture?
Answer:
Apiculture is the scientific practice of rearing honeybees in artificial hives and extracting honey from it.

Question 12.
What is sericulture?
Answer:
Sericulture is the scientific practice of rearing silkworms and extracting silk fibres from cocoons for economic purpose.

Question 13.
What is animal husbandry?
Answer:
Animal husbandry is the practice of breeding and rearing domestic animals like cattles, pigs, poultry birds etc. for production of milk, meat and eggs.

Question 14.
What is pharmacy?
Answer:
Pharmacy is the study of principles and practices of developing drug for therapeutic use in man, animals and plants.

WBBSE Class 9 Life Science Solutions Chapter 1.2 Biology and its Branches

Question 15.
What is silviculture?
Answer:
Silviculture is the study of principles and practices of cultivation of trees for wood and use of timber and its by-products.

Question 16.
What is entomology?
Answer:
Entomology is the study of different insects which are useful and harmful to man, animals and plants.

Short Questions and Answers : (2 marks for each question)

Question 1.
What is biology?
Answer:
The branch of science that deals with the study, investigations, experiments, observations and discussions on the origin, development, structural features and physiological activities of different living organisms, is known as biology

Question 2.
What do you mean by applied branches of biology?
Answer:
Applied branches of biology are those branches of biological science, in which, knowledge obtained from biology is applied for the welfare of human, animals and plants.

Question 3
What is zoology?
Answer:
The branch of biological science that deals with the study of morphology, anatomy, physiology, taxonomy, geographical distribution and conservation of animals, is known as zoology.

Question 4.
What is botany?
Answer:
The branch of biological science that deals with the study of morphology, anatomy, physiology, taxonomy, geographical distribution and medinical uses and conservation of plants, is known as botany.

Question 5.
What is physiology?
Answer:
The branch of biology that deals with the study of different activities and reactions related to various metabolic functions of living organisms, is called physiology.

Question 6.
What is taxonomy?
Answer:
Taxonomy is the oldest branch of biology, that deals with the identification, nomenclature and classification of living organisms on the basis of certain principles and processes.

Question 7.
What is microbiology?
Answer:
Microbiology is a branch of biological science, that deals with the study of microorganisms like virus, bacteria and all other microscopic organisms which are useful and harmful to human, animals and plants.
WBBSE Class 9 Life Science Solutions Chapter 1.2 Biology and its Branches 1

Question 8.
What is biochemistry?
Answer:
Biochemistry is a branch of science, developed by the combination of biology and chemistry. in order to study the nature of various chemicals, their classification and reaction processes occurring in the living organisms.

Question 9.
What is biophysics?
Answer:
Biophysics is an interdisciplinary branch of science, that is developed by (he combination of biology and physics. Ir deals with the study and explaination of the critical structure of various organs and the physical activities related to movement and locomotion, food capturing etc. of living bodies under the light of physics. Various instruments have been invented with the help of biophysics, such as endoscope, ultrasonograph, microscope etc.
WBBSE Class 9 Life Science Solutions Chapter 1.2 Biology and its Branches 2

Question 10.
What is molecular biology?
Answer:
The branch of biology that deals with the study of molecular structures of different intracellular biomolecules, their functions and applications, is known as molecular biology.

Question 11.
What is statistics?
Answer:
The branch of science, that deals with the collection, analysis and interpretation of data and finally presentation and organisation of the collected data, is known as statistics.

Question 12.
What is immunology?
Answer:
Immunology is a branch of science, that deals with the study of structure and function of body’s immune system and the nature of reactions occuring within the body between different antigens and antibodies.

Question 13.
What is astrobiology or exobiology?
Answer:
Astrobiology or exobiology is a combination of biology and space science, which is concerned with the study of the effect of hostile environment of outer space on various life activities of organisms on earth and different biochemical reactions.

Question 14.
What are the objectives of studying exobiology?
Answer:
The most important objective of studying exobiology is to find those organisms, that can withstand extreme conditions in space. Study of these organisms gives us an idea of those life forms, which came into being in extremely hostile environment of primitive earth. Study of this subject also gives us an idea of the reactions, which took place during abiogenesis of life on earth.

Question 15
Mention the different branches of cinical biology.
Answer:
The different branches of clinical biology are-immunology, serology, haematology, oncology, radiation biology, IVF (In vitro Fertilisation) technology, veterinary science etc.

Question 16
What is serology?
Answer:
Serology is the scientific study or diagnostic examination of blood and serum, especially with respect to the response of the immune system to pathogens or introduced substances.

Question 17
What is meant by pearl culture? What is lac culture?
Answer:

  • Pearl culture: Pearl culture is a branch of applied biology that deals with the methods and techniques of rearing pearl oysters and extraction of pearl from them.
  • Lac culture: Lac culture is a branch of applied biology that deals with the methods of rearing lac insects and extraction of shellac (a resin secreted by the female lac bug) from them.

Question 18.
What is meant by bioenergetics? What is thanatology?
Answer:

  • Bioenergetics: The branch of biology that deals with the transformation of energy required for the synthesis and breaking up of biomolecules, is known as bioenergetics.
  • Thanatology: Thanatology is the branch of science, that is concerned with the study of changes in a body after death and finding the cause of death.

Question 19.
What is bionics?
Answer:
Bionics is a new branch of science developed by the combination of biology and electronics to study the mechanical systems that function like living organisms or part of living organisms.

Question 20.
What is gerontology?
Answer:
Gerontology is a branch of science, that deals with ageing process. It is also known as geriatric studies. It mainly deals with the physical activities and problems faced during old age, senescence process of living cells, developing anti-oxidant medications etc.

Question 21.
What is biotechnology?
Answer:
Biotechnology is a broad discipline of bioscience, which involves the exploitation of biological processes for industrial and other purposes, especially the genetic manipulation of microorganisms for the production of antibiotics, hormones, etc.

Question 22.
Write two objectives of studying biology.
Answer:
Two objectives of studying biology are:

  • To understand the relationship of living organisms with primary components of the environment and their interdependence.
  • To gain knowledge about our local flora and fauna and developing healthy interaction with them.

Long Questions and Answers : (5 marks for each question)

Question 1.
Give a brief idea wa different brauches of classical biology.
Answer:
Branches of classical biology
Different branches of classical biology are discussed below.

  • Morphology: This Is the branch of classical biology, that deals with the study of morphs I.e. size.
    shape and external features of living organisms.
  • Cytology: This branch of biology deals with the study of structure and function of cells.
  • Histology: The study of structure, distribution and function of tissues of living organisms, is called histology.
  • Anaty: [his branch of biology deals with the structures of different Internal organs of plants and animals.
  • Pyslolog: This is the branch of classical biology, that involves the study of various metabolic reactions and the functional activities of life.
  • Genetics: This branch of biology deals with the study of heredity and variations of inherited features.
  • Evolutionary biolo: This is the study of origin of life and gradual complexities of organisms.
  • Ethology: This branch of biology is concerned with the study of behaviour of organisms.
  • Ecology: This branch of classical biology deals with the study of interrelation between organisms and the environment.
  • Taxonomy: This is the oldest branch of classical biology that deals with the theoritical study of classification of living organisms including its basic principles and processes. It involves identification, nomenclature and classification of organisms.
  • Immunology: This branch of science deals with the study of immune system of our body and the reactions of antigens and antibodies occuring in our body.
  • Embryology: This branch of science deals with the structure and development of embryo of living organisms.
  • Pathology: This branch of science deals with the study of different diseases, their causes and preventive measures.
  • Paleontology: This branch of classical biology deals with the study of extinct organisms, through their fossils and, thus, knowing their geological history.

Question 2.
Give a brief idea on different branches of applied biology.
Answer:
Branches of applied biology
Different branches of applied biology are discussed below.
1. Agriculture: This branch of applied biology deals with the practice of cultivation of crops and vegetables.

2. Horticulture: This is a branch of applied biology, which deals with the scientific practice of cultivation of flowers and fruits.

3. Forestry: This branch of applied biology deals with the methods, which help to reduce deforestation, such as afforestation and conservation of forest.

4. Pharmacognosy: This branch of science deals with the rearing and conservation of medicinal herbs and study of crude drugs and herbal medicines.
WBBSE Class 9 Life Science Solutions Chapter 1.2 Biology and its Branches 3
5. Sericulture: This branch of science deals with the scientific practice of rearing silkworm and extracting silk fibres from cocoons, for economic interest.

6. Pisciculture: This branch of applied biology deals with the scientific practice of breeding, rearing and capturing fishes for economic purpose.

7. Apiculture: This branch of science deals with the scientific practice of rearing honeybees in artificial hives and extracting honey from them.

WBBSE Class 9 Life Science Solutions Chapter 1.2 Biology and its Branches 4

8. Pearl Culture: This is a branch of applied biology, that deals with the methods of rearing pearl oysters and extraction of pearl from them.

9. Lac culture: This branch of applied biology deals with the methods of rearing lac insects and extraction of shellac from them.

10. Poultry farming: This branch of applied biology deals with the breeding and rearing of poultry birds, for the production of eggs and meat.

11. Dairy farming: This branch of applied biology deals with the rearing of cattle for the production of milk.

12. Animal husbandly : This branch of applied biology deals with the practice of breeding of farm animals (cattle, pigs, horses etc.) in order to develop their genetic qualities for human profit.

13. Plant breedhg This branch of applied biology deals with various methods for production of improved variety of plants.

14. Molecular biology : The branch of science that deals with the study of different biomolecules and their activities within our body, is known as molecular biology.

15. Biochemistry : This branch of science deals with the study of chemical processes occurring inside living organisms.

16. Microbiology: It is the study of microorganisms that are useful or harmful to man, plants and animals.

Question 3.
Mention the relationship of biology with other branches of science.
Answer:
Relationship of biology with other branches of science:
In course of time, biological science has developed further by sharing its concepts with other fields of science. This gave rise to some new and modern branches of science subjects. Some of these are discussed below.

1. Biophysics: This branch of science has been developed by the combination of biology and physics. It deals with the study and explanation of the critical structures of various organs and physical activities related to movement, locomotion, food capturing etc. of living bodies, under the light of physics.

2. Biochemistry: This is a branch of science, developed by the combination of biology and chemistry. It deals with the study of nature of various chemicals, their classification and reaction processes occurring in the living organisms.

3. Biotechnology: This is a modern branch of science, developed for producing improved variety of plants or animals for human benefit, by means of genetic manipulation.

4. Biometry: This is a branch of science, developed by combination of biology, mathematics and statistics. This subject helps to analyse different events and facts of biology, under the light of statistics.

5. Bionics: This branch of science has developed by the combination of biology’ and electronics. This is the study of mechanical systems which function like living organisms or parts of living organisms.

6. Biogeography: This branch of science has developed by the combination of biology and geography. It deals with the study of the distribution of flora and fauna in different geographical regions.

7. Anthropology: This branch of science has been developed for the study of humankind, under the light of geography, sociology and palaeontology.

8. Palaeontology: This branch of science is developed by the combination of biology and geology. It deals with the study of the fossils of primitive life forms.

9. Bioinformatics: This branch of science has been developed by the combination of biology and computer science. It deals with the analysis of complex biological data, especially genetic codes.

10. Astrobiology or exobiology: This is a modern branch of science, developed by the combination of biology and space science to study the effect of environment of outer space on living organism of earth.

Question 4.
Explain the applications of biology in various fields.
Answer:
Applications of biology:
Some important applications of biology are discussed below.

1. Development in agriculture: The ever-growing population of our country led to the problem of food scarcity, thereby increasing the demand for food. Application of biology has solved this problem to a great extent. Biology is applied at its best in the field of agriculture, to increase the productivity and quality of crops. Various fields of application of biology in agriculture are —

  • Production of high-yielding varieties of crops: By using the knowledge of genetics and applying the techniques of hybridization, high-yielding varieties of paddy, wheat, maize and several other crops have been successfully produced.
  • Production of seedless fruits: Concept of genetics and horticulture are combined to produce seedless varieties of fruits and to improve the productivity and size of fruits.
  • Pest control: Concept of biotechnology has been applied to develop certain pest resistant varieties of crops and vegetables. Several pests are now controlled biologically by using predators and parasites.
  • Storing of crops and food: Inventions in the field of biology help us to protect and preserve crops and food for a longer period of time. ‘

2. Fish production: Knowledge of biology has been applied in induced breeding technique and composite fish culture, to increase fish production and to meet the growing demand.

3. Poultry and dairy: Knowledge of biology has been used for the advancement of poultry and dairy farming techniques. By this process, the production of eggs, meat and milk has been increased.

4. Preparation of medicines: Herbal medicines are used since ancient times. Study of biology has revealed several medicinally important plants and animal products, which are used for the benefit of human and animals. Knowledge of microbiology has helped to produce a number of antibiotics, such as penicillin, streptomycin, neomycin, tetracycline etc.
WBBSE Class 9 Life Science Solutions Chapter 1.2 Biology and its Branches 5
5. Application of biology in space: Chlorella is the most photosynthetically efficient green alga. It is cultured in space stations, to maintain a continuous supply of oxygen and removal of carbon dioxide. Many microbes and other organisms are studied in space laboratories to know the effect of extreme conditions of space on life forms.

6. Industrial production: Industries like sericulture, lac culture, pearl culture are different wings of biology. The concepts of biology are utilised to increase the production of cosmetics, medicines, textile, paper, plastic, rubber, leather, wood, tea, coffee, bakery and brewery industries.

7. Controlling environmental pollution: Biological notions help us to control environmental pollution. Afforestation, social forestry and conservation of forest are some of the positive steps, undertaken globally for pollution control.

WBBSE Class 9 Life Science Solutions Chapter 1.2 Biology and its Branches 6

8. Population control: Population explosion is the burning problem of present world. By applying different concepts of biology and medicine, scientists have invented different modern and effective birth contol techniques for controlling population.

WBBSE Class 9 Life Science Solutions Chapter 1.1 Basic Properties of Life

Detailed explanations in West Bengal Board Class 9 Life Science Book Solutions Chapter 1.1 Basic Properties of Life offer valuable context and analysis.

WBBSE Class 9 Science Chapter 1.1 Question Answer – Basic Properties of Life

Very Short Questions and Answers : (1 mark for each question)

Question 1
Mention five common signs of life.
Answer:
Reproduction, growth, metabolism, response to stimuli and adaptability are the five common signs of life.

Question 2.
What does reproduction mean?
Answer:
Reproduction is the biological process by which new individuals or offsprings are produced from their parents.

WBBSE Class 9 Life Science Solutions Chapter 1.1 Basic Properties of Life

Question 3.
Define excitability.
Answer:
Excitability is defined as a response of a living organism to any external and internal stimulus.

Question 4.
What is meant by response?
Answer:
The reaction shown by a living organism towards any impulse, is called response.

Question 5.
What does metabolism mean?
Answer:
A set of complex physiological and chemical processes necessary for sustaining life, involving formation and breakdown of various nutrients within the body is called metabolism.

Question 6.
What is growth?
Answer:
Growth is the irreversible increase in dry mass and size of a living body.

WBBSE Class 9 Life Science Solutions Chapter 1.1 Basic Properties of Life

Question 7.
What is meant by biopoiesis?
Answer:
Biopoiesis is the concept of the origin of living organisms from non-living matters.

Question 8.
Who first suggested that living organisms originated from non-living matters?
Answer:
Philosopher Aristotle first suggested that living organisms originated from non-living matters.

Question 9.
From where did primary organic micromolecules originate from inorganic compounds?
Answer:
Primary organic micromolecules originated in the primitive atmosphere from simple inorganic compounds.

Question 10.
How was the atmosphere on earth before the origin of life?
Answer:
Before the origin of life, earth’s atmosphere was reducing in nature.

Question 11.
Why the atmosphere of primitive earth was reducing in nature?
Answer:
The atmosphere of primitive earth was reducing in nature because there was no free oxygen in the atmosphere.

WBBSE Class 9 Life Science Solutions Chapter 1.1 Basic Properties of Life

Question 12.
Name the two scientists who suggested that organic micromolecules originated on earth by chemical reactions amongst simple inorganic compounds.
Answer:
Scientists Alexander Oparin and JBS Haldane, first suggested that organic micromolecules originated on earth by chemical reactions amongst simple inorganic compounds.

Question 13.
How did primary organic micromolecules reach the earth’s surface?
Answer:
Primary organic micromolecules reached the earth’s surface through rainwater.

Question 14.
Who coined the term ‘primordial soup’?
Answer:
Scientist A1 Oparin coined the term ‘primordial soup’ to denote the primitive oceanic water rich in organic matter.

Question 15.
Which term was used by Haldane to indicate the boiling hot oceanic water of the primitive earth rich in organic matter?
Answer:
Haldane denoted the organic matter-rich boiling hot oceanic water of the primitive earth as ‘hot dilute soup’.

Question 16.
What are proteinoids?
Answer:
Proteinoids are tiny clusters of protein molecules formed abiotically from amino acids by cross-linkage.

WBBSE Class 9 Life Science Solutions Chapter 1.1 Basic Properties of Life

Question 17.
What are microspheres?
Answer:
Microspheres are spherical clusters of proteinoids, covered with thin film of phospholipids.

Question 18.
Name two primitive forms, which are considered as predecessors of living organisms on earth.
Answer:
Microspheres and coacervates are two primitive forms, which are considered as predecessors of living organisms on earth.

Question 19.
Which organic matters constitute a protobiont?
Answer:
A protobiont is constituted of organic matters like nucleic acids, lipids and various protein molecules.

Question 20.
What is meant by alpha (α) diversity?
Answer:
Alpha (α) diversity indicates the richness of species in a specific area or an ecosystem.

Question 21.
What is meant by beta (β) diversity?
Answer:
Beta (β) diversity is the measure of the comparative turnover of species in two different habitats of an ecosystem.

Question 22.
What is meant by gamma (γ) diversity?
Answer:
Gamma (γ) diversity is the overall biodiversity of a large geographical realm, consisting of many typical habitats and ecosystems.

Question 23.
What are the two main causes of biodiversity on earth?
Answer:
Genetic variation and natural selection are the two main causes of biodiversity on earth.

Question 24.
Which class of animals shows the highest level of diversity in India?
Answer:
In India, birds show the highest level of diversity, where about 12.6% of global bird species are found.

Question 25.
What is meant by biodiversity hotspot?
Answer:
Biodiversity hotspot is a geographical area that provides shelter to a very rich variety of endemic species that are facing the threat of extinction.

WBBSE Class 9 Life Science Solutions Chapter 1.1 Basic Properties of Life

Question 26.
How many biodiversity hotspots are there in India?
Answer:
In India there are three biodiversity hotspots-the Western Ghats, the Himalayas and the Indo-Burma regions.

Question 27.
How many biodiversity hotspots are there in the World?
Answer:
According to Conservation International (2004), there are 34 hotspots in the world.

Question 28.
Which type of diversity is also known as turnover diversity?
Answer:
Beta diversity is also known as turnover diversity.

Short Questions and Answers : (2 marks for each question)

Question 1.
What is life?
Answer:
Life is the condition that establishes the existence of a living organism on the basis of various characteristics such as growth, reproduction, metabolic activities, response to stimuli, adaptability, ageing, death etc.

Question 2.
What are non-living objects?
Answer:
Those objects which do not show any of the signs of life, such as growth, reproduction, metabolic activities, response to stimuli, ageing and senescence etc., are called non-living objects.
WBBSE Class 9 Life Science Solutions Chapter 1.1 Basic Properties of Life 1

WBBSE Class 9 Life Science Solutions Chapter 1.1 Basic Properties of Life

Question 3.
What is meant by life cycle?
Answer:
The cyclic course of events undergone by an organism, starting from its birth followed by growth, maturity, reproduction, ageing and finally death, is known as its life cycle.
WBBSE Class 9 Life Science Solutions Chapter 1.1 Basic Properties of Life 2

Question 4.
Define heredity.
Answer:
Heredity is the process of transfer of different characteristics from parents to their offsprings.

Question 5.
What is evolution?
Answer:
Evolution is the slow, steady and continuous emergence of advanced and complex forms of life from primitive and simpler forms. It also involves change in heritable characters of biological population from one generation to the next.

Question 6.
What is meant by mutation? What is mutability?
Answer:

  • Mutation: Mutation is a sudden, permanent and inheritable transformation in living organisms, caused due to any change in chromosome or gene either naturally or by any external factor.
  • Mutability: The capacity of a living organism to undergo mutation is called mutability.

Question 7.
What is meant by variation?
Answer:
The dissimilarities in external or internal features or behaviours found amongst the different members of a species in a population, is called variation.

WBBSE Class 9 Life Science Solutions Chapter 1.1 Basic Properties of Life

Question 8.
Give an example of response to stimuli shown by plant.
Answer:
The leaves of ‘touch me not’ (Mimosa pudica) plant droop when touched. This is an example of response to stimuli shown by a plant.
WBBSE Class 9 Life Science Solutions Chapter 1.1 Basic Properties of Life 3

Question 9.
What is meant by senescence?
Answer:
Senescence is the process of gradual and irreversible deterioration in the structural components of the body and decrease in the ability of a matured organism to deal with the stress of the environment.

Question 10.
what is meant by ageing? What is death?
Answer:

  • Ageing: Ageing is the process of growing older with time, that is genetically determined and environmentally modulated.
  • Death: Death is the ultimate disappearance of all signs of life from a living organism.

Question 11.
What does the ‘Ghemogeny of life’ mean?
Answer:
Chemogeny or chemosynthesis of life is the most modern concept, used to explain the biochemical origin of life on the earth. According to this concept, life originated naturally and spontaneously from simpler inorganic matters on earth, through various physical and chemical processes.

Question 12.
What is meant by ‘hoi dilute soup’?
Answer:
According to the theory of chemosynthesis, several organic molecules accumulated in the hot boiling water of the primitive ocean before the origin of life. Haldane denoted this hot broth of organic matter as ‘hot dilute soup!

Question 13.
What are coacervates?
Answer:
Coacervates are minute (1-100μ in diameter), spherical, bubble-shaped, colloidal matters, composed of a thin layer of organic molecules rich in lipids. Oparin (1924) first suggested this structure and named it coacervate.
WBBSE Class 9 Life Science Solutions Chapter 1.1 Basic Properties of Life 4

WBBSE Class 9 Life Science Solutions Chapter 1.1 Basic Properties of Life

Question 14.
Mention the living features of microspheres.
Answer:
The living features of microspheres are as follows —

  • They are surrounded by a double-layered membrane
  • They show osmotic activity.
  • Certain protein molecules present in the microspheres, have enzymatic activity
  • Microspheres may divide through a process similar to binary fission.

Question 15.
What are proiocells?
Answer:
In the primitive sea, nucleic acid molecules entered into protein coacervates to take the shape of primitive cells. These primitive structures are known as protocells, in which the first sign of life was noticed.

Question 16.
What are naked genes?
Answer:
According to the chemosynthetic theory, during the last phase of ceil formation, purines, pyrimidines and pentose sugars united to form nucleic acids. These nucleic acid molecules were capable of seif-replication. They were devoid of histone and other proteins and as they were not surrounded by nuclear membrane. They were called naked genes.

Question 17.
What is meant by biodiversity?
Answer:
Biodiversity refers to the variety of different types of living organisms on the basis of structure, shape and behaviour found on earth, that are inhabiting different habitats of various ecosystems.

Question 18.
What is meant by genetic diversity?
Answer:
The different types of variations and combinations in the genetic composition among the individuals of a species is considered as genetic diversity. Due to genetic diversity, different individuals of same species look and behave differently.

Question 19.
What is meant by species diversity?
Answer:
The diversity in number, richness and evenness in distribution of various species in a biotic community is regarded as species diversity. This is estimaed by determining the number of variety of a species and the number of species Inhabiting a specified area.

Question 20.
What is meant by ecological diversity?
Answer:
The variation of several ecosystems in a large geograpical area or throughout the planet. is known as ecological diversity. The diversity of organisms from a microhabitat to a huge hiome is included under ecological diversity.

Question 21.
What is meant by species?
Answer:
Species is a group of closeI related individuals, which are capable of naturally interbreeding to produce fertile offsprings. They remain reproductively isolated from other groups.

Question 22.
How do variation occur In a population?
Answer:
Almost every individual of any sexually reproducing population is slightly different from one another. There are three main reasons behind —

  • Formatilin of different types of ganletes I due to crossing over,
  • Random union of gametes and
  • Change in the genetic material.

WBBSE Class 9 Life Science Solutions Chapter 1.1 Basic Properties of Life

Question 23.
Give a brief note on the biodiversity of India.
Answer:
The biological diversity of India includes a huge variety of plant and animal species, their genetic variability and the organisation of species into different ecosystems. India is one of the seventeen megadiverse countries on earth. In global perspective, India houses about 7.6% of all mammals, 12.6% of all birds, 6.2% of all reptiles, 4.4% of all amphibians, 11.7% of all fishes and 6.0% of all flora.

Question 24.
What are the reasons for the loss of biodiversity?
Answer:
The main reasons for the loss of biodiversity are-

  • Deforestation,
  • Habitat fragmentation,
  • Environmental pollution,
  • Over exploitation of resources by Illegal hunting and poaching,
  • Natural calamities,
  • Alteration of natural environment due to green house effect, ozone layer depletion etc.,
  • Genetic incompatibility,
  • Introduction of exotic species etc.

Question 25.
What are the contributing factors of biodiversity?
Answer:
The contributing factors of biodiversity are-

  • Variation
  • Mutation
  • Speciation
  • Adaptation
  • Geographical isolation
  • Change in climatic condition
  • Change in habitat etc.

Question 26.
Compare between coacervate and microsphere.
Answer:

features Coacervate Microsphere
1. Definition These are minute spherical, bubble-shapped structures originating in the ‘hot dilute soup’ These are spherical clusters of proteinoids covered by a thin film of lipid bilayer
2. Function Plays an important role as the precursor of protocell Plays an Important role as the precursor of the early or primitive cell
3. Structural component Structure is composed of protein, fat, carbohydrate and nucleic acid Structure is composed mainly of proteins formed from proteinoids


Long Questions and Answers : (5 marks for each question)

Question 1.
Explain the basic properties of lite.
Answer:
Basic properties of life
From biological point of view, there are five basic properties of life. They are —

(1) Reproduction: Reproduction is a biological process by which living organisms give birth to offsprings in their lifetime, in order to protect their races from becoming extinct.

WBBSE Class 9 Life Science Solutions Chapter 1.1 Basic Properties of Life

(2) Growth: Growth is the irreversible increase in dry mass and size of a living body. A living organism must exhibit growth and development in due course of time.

(3) Metabolism: Metabolism is a set of life sustaining chemical reactions involving formation and breakdown of various nutrients within the body. Different metabolic activities like nutrition, respiration, excretion etc. take place in a living organism.

(4) Response to stimuli: Response to external and internal stimulus is an important character of a living organism. For example, if a moving millipede is touched, it coils around.

(5) Adaptability: Adaptability is a property of life where any living organism tries to adjust itself, either structurally or behaviourally, according to the changing environment.

Question 2.
Mention the differences between living beings and non-living objects.
Answer:
Differences between living beings and non-living objects

Features Living beings Non-living objects
1. Reproduction Living beings maintain their race by reproduction Non-living objects do not reproduce
2. Response to stimuli Living beings respond to external or internal stimuli Sensitivity to stimulus is absent in non-living objects
3. Growth Living beings grow gradually with time Non-living objects never show any growth
4. Metabolism Living beings perform various metabolic activities like photosynthesis, nutrition, respiration etc. Non-living objects never perform any metabolic function
5. Life span Ali living forms have a definite life span Non-living objects do not have any specific duration of existence
6. Movement and locomotion Living beings spontaneously exhibit movement and locomotion Movement occurs as a result of external influence
7. Protoplasm Cells of living organisms have active protoplasm Protoplasm is absent
8. Adaptability Living organisms can adapt themselves according to changing environment Non-living objects do not show any adaptation
9. Life cycle Living organisms have life cycle Non living objects do not have life cycle
10. Senescence and death Living organisms show signs of senescence and death Non-living objects do not show signs of senescence and death

Question 3.
Briefly describe the origin of earth and gradual changes in the primitive atmosphere according to the theory of chemosynthesis of life.
Answer:
Origin of earth and gradual changes in the primitive atmosphere
According to the theory of chemosynthesis of life or the theory of abiogenesis about 5-6 billion years ago, the earth originated from a huge explosion known as the Big Bang in the cosmos. At that time, the earth was a mass of burning gas. It took several million years to cool down and condense.

The temperature of primitive earth was around 5000°C — 6000°C. At that high temperature, it was difficult for certain gases such as hydrogen, nitrogen and oxygen etc. to exist independently in the atmosphere. As a result, they were present in the form of metallic or non-metallic compounds.

Therefore, the primitive atmosphere was reducing in nature due to the absence of free oxygen. With gradual decrease in temperature, water vapour condensed to form huge clouds, which came down on the earth in the form of rain. Gradually, the earth started to cool down further and condensed to form solid matters. The rainwater accumulated in the lowlands to form the primitive ocean.

WBBSE Class 9 Life Science Solutions Chapter 1.1 Basic Properties of Life

Question 4.
Briefly describe chemogeny of life.
Answer:
Chemogeny of life
The process of synthesis of various organic compounds essential for the origin of life on the earth, was denoted as chemogeny by scientists Oparin and Haldane. This involves 3 steps. The steps of chemogeny are discussed below.

1. Origin of simple organic compounds: With gradual decrease in atmospheric tempertaure, the chemical components like hydrogen, methane, ammonia, water vapour etc., started reacting with each other to form simple organic molecules such as amino acids, simple sugars, fatty acids etc. The energy required to carry out these reactions was provided by lighting, ultraviolet ray and different cosmic rays. Through downpour, these compounds precipitated on earth and mixed in the primitive ocean.

2. Formation of complex organic compounds: The simple organic molecules condensed in the primitive ocean to form complex organic compounds such as proteins, carbohydrates, lipids, purines, pyrimidines etc. The boiling hot water of the primitive ocean containing the complex organic compounds was denoted as hot dilute soup by Haldane. Oparin termed the same as primordial soup.

3. Formation of coacervates: In the hot dilute soup, the organic matters, especially lipids, clustered to form a self-organised colloidal matter. Oparin named this bubble-shaped structures as coacervates. American scientist Sidney Fox (1957), denoted a similar substance as microsphere, which, according to him, was formed by unification of several globoid duster of proteins, called proteinoids, surrounded by a double-layered lipid membrane.

Question 5.
Explain biogeny or organic evolution of life.
Answer:
Biogeny or organic evolution of life:
According to Oparin, as the temperature of the atmosphere gradually started to cool down and simple organic compounds reacted with each other to form complex organic compounds by chemosynthesis, the boiling hot water of primitive ocean acted as the site of formation of lipid clusters, called coacervates. These coacervates absorbed nucleic acids and proteins from the primordial soup and transformed into a structure, capable of performing few metabolic activities which were named as protobionts.

WBBSE Class 9 Life Science Solutions Chapter 1.1 Basic Properties of Life

The protobionts were capable of synthesising some organic matters within it through various chemical reactions. Later, these protobionts transformed into self-replicating organic matters, which the scientists denoted as eubionts. These were the first living entity on earth. Later, cytoplasm formed within them and thus the earliest prokaryotic cells came into being.

WBBSE Class 9 Life Science Solutions Chapter 1.1 Basic Properties of Life 5

Question 6.
Classify different levels of biodiversity.
Answer:
Levels of biodiversity : Biodiversity has been classified into three levels-

  • Genetic diversity,
  • Species diversity and
  • Ecological diversity,

1. Genetic diversity: The various types of variations and combinations in the genetic composition among the individuals of a species is considered as genetic diversity, Due to genetic diversity, individuals of same species look and behave differently. Example -The diversity in the genes of different varieties of paddy, wheat, dogs, etc.

2. Species divesity : The diversity in number, richness and evenness in distribution of various species in a biotic community is regarded us species diversity. This is estimated by determining the number of variety of a species and the number of species inhabiting a specified area. Example –The diversity of species inhabiting the mangrove forest of Sundarban.

3. Ecological diversity: The variation of several ecosystems in a large geographical area is known as ecological diversity. The diversity of organisms from a microhabitat to a huge biome is included under ecological diversity. American ecologist Whittaker (1965), classified ecological diversity into three different sub-types. They are—

  • α diversity: α diversity indicates the richness of species in a specific area or an ecosystem.
  • β diversity: β diversity is the measure of comparative turnover of species in two different habitats of an ecosystem.
  • γ diversity: γ diversity is the overall biodiversity of a large geographical realm, consisting of many typical habitats and ecosystems.

WBBSE Class 9 Life Science Solutions Chapter 1.1 Basic Properties of Life

Question 7.
Describe the different types of variation along with their examples and significance.
Answer:
Types of variation
Variation is broadly classified into two types—

  • Continuous variation and
  • Discontinuous variation.

Continuous variation: Any characteristic of a species that changes gradually over a range of values is considered to exhibit continuous variation. It is represented by a line graph. Example — Variation in human height, skin colour and weight

Significance — Continuous ‘variation creates diversity amongst the members of same species. The characteristic features which undergo natural selection are ultimately transferred to the next generation.

Discontinuous variation: Any characteristic of a species having only a limited number of possible values is considered to exhibit discontinuous variation. It is represented by a bar graph. Example — Variation in human blood group and human eye colour.

Significance — Mutation is the prime reason behind discontinuous variation. Mutation leads to the development of new species at a comparatively faster rate. Discontinuous variation leads to species diversity.

WBBSE Class 9 Life Science Solutions Chapter 1.1 Basic Properties of Life 6
Question 8.
What are the important features of biodiversity?
Answer:
Important features of biodiversity
The important features of biodiversity are as follows –

  • Comparatively high level of biodiversity has been noticed in tropical and subtropical region than temperate region.
  • Rate of biodiversity is very high in tropical rain forests in terrestrial ecosystem and coral reefs in aquatic ecosystem.
  • Insects are the diverse of all living organisms (56.4% of total species).
  • Reduction of genetic diversity of a species gradually leads to its extinction.
  • Biodiversity is not evenly distributed in all parts of the world.

WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry

Comprehensive WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry can help students make connections between concepts.

Organic Chemistry Class 10 WBBSE Notes

Organic Chemistry : All carbon containing compounds except oxides of carbon, metal carbonates, bicarbonate, hydrogen cyanide and metallic cyanides are organic compounds and the chemistry of organic compounds is known as organic chemistry.

Organic compounds is a separate branch because organic compounds:

  • are large in number
  • show isomerism
  • are soluble in non-polar solvents
  • have complex structures
  • are bad conductors
  • show catenation
  • are made from small number of elements.

WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry

Role of organic compounds in life process : Organic compounds are intemately related to our daily life. In order to maintain the life process of our body the following process are required.

  • Intake of nutritional food for the growth the living body.
  • For repairing the depreciation of health.
  • Mobility
  • Reproduction

All these are fulfilled almost by all organic materials.

(i) The main ingredients in our food chart are classified of three types : carbohydrates (e.g. rice, wheat, sugar etc.), proteins (e.g. milk, fish, egg etc.) and fats (e.g. milk, oil, butter etc.)

(ii) These food materials are ultimately converted to single organic compounds in different metabolic process with the help of enzymes in our body.

(iii) Vitamin, an organic compound plays indirectly a very important role for maintaining the growth of our health.

(iv) For the translation and mobility of the animals, energy which is required is stored in our body. This energy is associated in an organic compound. This is known as ATP (adenosine triphosphate).

(v) Haemoglobin which is an organic compound an oxygen carrier supplies oxygen in the body.

Biomoleculcs: The molecules of compounds like carbohydrates, amino acids, proteins, vitamins, fats, RNA, DNA etc. which are essential for the continuation of life processes of biological species like plants and animals (including human beings) are known as biomolecules.

WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry

Some essential biomolecules are :

(i) Carbohydrates : These molecules are composed of carbon, hydrogen and oxygen. General formula of it is Cn(H2O)n. Carbohydrates. are the class of compounds that include polyhydric aldehydes, polyhydric ketones and large polymeric molecules that can be broken down to polyhydric aldehydes and ketones.

Carbohydrates which contain aldehyde groups are called aldose and the carbohydrate containing keto groups are called ketose.

Carbohydrates are mainly of three types :

(a) Monosaccharides : These are the simplest form and are not hydrolysed, e.g. glucose, fructose etc.

(b) Oligosaccharides: During hydrolysis they form fixed number (2-10) of more saccharides.
WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy 18

(c) Polysaccharides: These are the natural polymer of large number of mono saccharides. It has general formula (C6H10O5)n. e.g. starch, cellulose etc.
WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy 19

(ii) Proteins: Another class of compounds essential for living being are proteins. They contain carbon, hydrogen, oxygen and nitrogen. Proteins are consumed daily from our diet. During hydrolysis of proteins by acid, base or enzyme amino acids are produced.WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy 20
Proteins are of three types :
(a) Simple protein
(b) Conjugated protein
(c) Derived protein.

WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry

(a) Types of simple protein :

  • Albumin: This type of protein can be isolated from egg, milk, wheat, serum etc.
  • Globulin: Proteins that are isolated from egg yolk, tissues, belong to this class of protein.
  • Glutemin: This type of protein is present in wheat, rice etc.
  • Protamin: This type of protein is present in wheat, barli etc.

(b) Conjugated protein : Examples of this type of protein are : nucleo protein, chromoprotein, glycoprotein, phosphoprotein, lipoprotein, metalloprotein.

(c) Derived protein : This type of proteins are available by the partial hydrolysis of protein of very high molecular weight by acid, base or enzyme to simpler protein.

(ii) Fats: Fats are composed of carbon, hydrogen and oxygen. These are made of glycerol and fatty aids. Fats may be of animal or vegetable origin.

Animal fats are found in : milk, cheese, butter, eggs, meat and oily fish.

Vegetable fats are found in : walnut, almonds, groundnut and coconut, mustard seeds, seasame seed etc.

Fat is a concentrated source of energy.

(iv) Vitamins: Vitamins are organic compounds containing carbon, hydrogen, oxygen and sometimes nitrogen and sulphur. Vitamins are necessary to carry out many vital processes. Their absence or deficiency causes many ailments and impairs the healthy living oi human being.

(v) Amino acids: Proteins are made of amino acids. These are the essential for living beings. Amino acids are the class of compounds containing amino group and carboxylic acid group. 1

WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry

Amino acids are of three types :

Neutral amino acids : Amino acids containing the same number of amino and carboxylic acid groups are called neutral amino acids. These have one -NH2 group and one -COOH group. Total number  of amino acids in this group is 19. e.g. glycine (NH2CH2COOH), alanin [CH3CH(NH2)COOH] etc.

  • Basic amino acids : This type of amino acids have two -NH2 groups and one-COOH group, e.g. lysine, histidine etc.
  • Acidic amino acids : This type of amino acids have two -COOH groups ’• and one -NHZ group, e.g. aspartic acid, glutamic acid.

The essential amino acids are :

  • Valine
  • Lucine
  • Iso-lucine
  • Phenylalanine
  • Threonine
  • Methionine
  • Argenine
  • Lysine
  • Histidine
  • Tryptophan.                                                                                                              1

DNA (Deoxyribo nucleic acids): It is found in nucleus, mitochondria and chloroplast of the body cells. James Watson and Francis Crick were able to establish the structure of DNA in 1953. DNA molecule is double helical in nature. It contains three different chemical compositions.

  • Phosphoric acid
  • Deoxy ribose
  • Pyridine like base adenine and guanine, pyrimidine like base thymine and cytonine.

DNA has two strands of polynucleotides coiled around each other by hydrogen; bond in the form of double helix.

Function of DNA :

  • Transmission of herediatary characteristics.
  • Biosynthesis of proteins.

The genetic information for the cell is preserved in the sequence of this base in the DNA molecule. When a cell divides, DNA molecules replicate and make exact copies of themselves so that daughter cell will have identical DNA to that of parent cell.

RNA (Ribonucleic acid): RNA molecule is single stranded, containing 1 ribose, phosphoric acid and uracil as the nitrogenenous base. RNA plays a very Vital role in life process. RNA carries the message of DNA and acts accordingly.

WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry

Type of RNA : (a) mRNA (b) tRNA (c) rRNA. RNA(tRNA, mRNA) helps synthesis of polypeptides, the intermediates of proteins.

Bonding In organic compounds Electronic configuration of carbon shows that the number of electrons available in the outermost orbit is 4. The outermost electrons can only be capable to form covalent bond by sharing and pairing of its electrons with that of other element. So, valency of carbon becomes 4.

Catenation : The property by virtue of which carbon forms covalent linkage amongst them to form stable carbon chains is knoun as catenation. Catenation is an unique property of carbon. As a result of catenation, carbon can form innumerable chemical species with the supplement in various ways by the necessary addition of other elements like H, O, N, P, S, halogens etc.

Difference between organic and inorganic compounds :

Organic Compounds Inorganic Compounds
(i) Catenation property of carbon atoms and its effect : The catenation property of carbon atoms among themselves and other atoms gives rise to limitless number of organic compounds with strain chains and rings of carbon atoms. (i) Due to absence of catenetion property of its atoms, the number of inorganic compounds formed by the rest erf the elements in not so large.
(ii) Class similarity : Organic compounds can be easily classified into different groups with certain similar characteristic properties. (ii) In organic compounds cannot generally be classified into such groups with similar characteristics.
(iii) Thermal stability : Organic
compounds are in general, thermally unstable and decompose at higher temperature.
(iii) These are more or less thermally stable., These can tolerate comparatively higher temperatures without decomposition.
(iv) Bonding: Organic compounds are formed through covalent bonding in their molecules. (iv) These compounds are generally formed by eledrovalent or ionic bonding in their molecules.
(v) Melting and boiling points: As these compounds are covalent in nature, so they have comparatively lower melting and boiling points. (v) As these compounds are ionic in nature, so they have high melting and boiling points.
(vi) Solubility: These are soluble in organic solvents like alcohol, ether, benzene etc. but generally insoluble in water. (vi) These are soluble in ionising solvents like water but are generally insoluble in organic solvents.
(vii) Structural characteristics: They have widely varying structural forms, from simple chain structure to complex ring. (vii) They have mostly simple structure.
(viii) Isomerism: Different types of isomerism often occur in organic compounds. (viii) Isomerism is practically unknown in inorganic compounds.
(ix) Electrolytic character: These are generally non-electrolytes (ix) Inorganic compounds are general electrolytes.

Classification of organic compounds :
WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry 3

WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry

Functional group : The groups which are responsible to characterise the compounds are Known as functional group.

WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry 2

Homologous series : It is a group of compounds in which the various members have similar chemical properties, possess the same functional group and can be represented by the same general formula. The different members of a homologous series are called homologous and the phenomenon is called homology.

Constitutional or Structural Isomerism : The phenomenon in which two or more different compounds obviously having different physical and chemical properties possess the same molecular formula but different molecular structures due to the arrangements of the constitutional or structural isomerism.

Types of structural isomerism : Three types

(a) Chain isomerism; The compounds of the same class, with the same molecular formula, have chain isomerism due chain carbon atoms.

WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry 4

(b) Position isonierism : The compounds of the same dass, with the same molecular formula, have position isomerism due to the position of functional groups.

WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry 5

(c) Functional group isomerism : The compounds of the same class, with the same molecular formula, have functional group isomerism due to different functional groups.
WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry 6

Hydrocarbons : Hydrocarbons are organic compounds containing only carbon and hydrogen atoms in their molecules.

WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry

Types of hydrocarbons: Hydrocarbons are divided into two classes, namely

  • Saturated hydrocarbons
  • Unsaturated hydrocarbons.

Alkane : Open chain saturated hydrocarbons are known as alkanes. General formula of these compounds are : CnH2n+2 (n = integers)

Carbon atoms of these compounds are linked by single covalent bonds. Hy­drogen atoms of these compounds are also linked to one or more carbon atoms by a single covalent bonds.

Saturated hydrocarbon molecules do not contain any formal functional group and as such exhibit very little chemical activity. These are therefore known as paraffins.

Methane: Methane is the first member of the hydrocarbons of the alkane series. It is a colourless gas with molecular formula CH4.

Source:
WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry 7

(ii) Main source of methane is the natural gas. It is formed in nature in bio­logical decay of plants of swamps and marshes. Methane is called marsh gas. A little phosphine (PH3) and phosphorous dihydride (P2H4) accompany the methane gas and make it spontaneously inflammable. So, it often burns with flashes of flame over marshes. This is well known as ‘will-O-the-wisp’. It is just a natural phenomenon.

Chemical properties:
(i) Burning of methane: It burns in air or oxygen when ignited, and is oxidised to carbon dioxide and water with the evalution of heat.
CH4+2O2 = CO2 + H2O + 213 kcal/ molecule.

(ii) Substitution reaction : It undergoes substitution reaction. The first step of the reaction is :
WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry 8

Methane: A green house gas: Like carbon dioxide, water vapour etc. methane is known as a green house gas (contribution 19%) because it is capable to absorb visible light from sun-rays and keep the earth surface and its surrounding warm.

WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry

Uses of methane:

  • It is used as a domestic and industrial fuel by burning in gas-stoves. Almost 40% by volume of methane is present in coal gas.
  • It is used in the large-scale production of carbon black which is largely
    used for making printers’ ink, paints, typewriter, ribbon, motor tyres etc.
  • It is now used as a source of hydrogen for the synthesis of NH3.
  • It is used for the manufacture of methanol, chloromethane, chloroform etc.

Alkene: An alkene is an unsaturated aliphatic hydrocarbon with a double bond in-between two adjacent carbon atoms of its molecule.

General formula of alkene : CnH4 (where n is an integer)

Ethylene [or, Ethene]
The molecular formula of ethylene is C2H4 which can be written as CH2 = CH2.
It i colourless gas.

Source :
(i)
WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry 9
(ii) It is present in very small amount in coal gas.
(iii) Natural gas rich in methane when ignited at 1400°C — 1500°C at normal pressure, acetylene is obtained after cooling.
WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry 10

Chemical reaction :
(i) Addition reaction with hydrogen:
WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry 11
(ii) Addition reaction with bromine :
WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry 12
Uses of acetylene :

  • As an illuminant acetylene is used in acetylene lamp.
  • It is used as an oxyacetylene flame 2700°C, for cutting and welding iron plates and sheets.
  • It is used for the preparation of acetaldehyde, acetic acid, ethyl alcohol, acetone etc.
  • Acetylene is used for the preparation of weston, a solvent used in industry.
  • It is used for the preparation of synthetic rubber and plastics.
  • It is used as an anaesthatic agent.

Polymer : There are some organic molecules of low molecular weight, a large number of which join with one another repeatedly to produce a very big molecule of high molecular weight. The small molecules are known as monomers and the resulting large molecules are known as p tymers.

WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry

Some common polymers :
WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry 13

Hazards of using polymer materials in our environment :

  • Polythene, PVC, synthetic rubber after use are left anywhere else. Tyre used in the wheel of a vehicles are also thrown in the dustbin or these are burnt out. These materials have no natural decay in the soil by bacteria.
  • As a result there is a dumping of plastic materials and creates an environ­ment pollution.
  • It destroys the soils where it is dumping.
  • The plastic materials left in hills and mountain, is the cause of landslide.
  • Tyre prepared from synthetic rubber when left in air produces free radicals in presence of sunlight which is cancer producer.

Possible alternatives to avoid the hazards :

  • To get relief from these, limited uses are desirable
  • Now-a-day some reagents are added to plastic materials so that they would not create any problem or hazard to mankind.
  • Bags made to cloth, paper, jute should be used instead of polythene bags.

IUPAC (International Union of Pure and Applied Chemistry, 1957) nomenclature simple organic compounds :

WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry 14
WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry 15
Polymerisation: The process or technique through which monomer units combine to give a polymer is known as polymerisation.

WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry

Biodegradable polymers : The polymers that can be dissociated into small segments enzymes (produced by microorganisms) are called biodegradable polymers.

Example :
(i) PHBV (poly β-hydroxy butyrate co-β-hydroxy-valerate)
WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry 16

LPG (Liquified Petroleum Gas) : In refinning the crude petroleum, the gaseous products thus obtained the hydrocarbons of lower molecular weight after purification of these gaseous mixtures if there are cooled at hith pressure and low temperature, these are all condensed to liquidand. The liquid is known as Liquified Petroleum Gas (LPG).

Types of LPG :

  • Heavier grade LPG (80-90%) : C4-hydrocarbon (Butane, Isobutane, butene)
  • Lighter grade LPG (90%) : C3 hydrocarbon (Propane and propene, 10% C4-hydrocarbon (Butane and butene) 2% Ethane and ethylene.

Calorific value : Amount of heat energy produced by the complete com­bustion of one gram of the fuel is known as its calorific value.
Calorific value of LPG ; 29500 kcal/m3

CNG (Compressed Natural Gas):
It is compressed natural gas. It contains mainly methane. It has calorific value 21300 Btu/Ib

Uses of Ethanol:

  • Ethanol (C2H5OH) is used for drinking purpose.
  • Ethanol is widely used as a solvent.
  • Ethanol burns of give carbon dioxide and water and thus can be used as fuel.
  • Ethanol is used as antifreeze.

WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry

Physical properties of Ethanol :

  • It is colourless liquid having a sweet smell.
  • It is highly soluble in water.
  • It is volatile and inflammable.
  • It is neutral and it does not change the colour of litmus solution.

Reaction of Ethanol :

(i) Reaction with sodium (active metal): Ethanol reacts with sodium, to form sodium ethoxide and hydrogen.
2C2H5OH + 2Na → 2C2H5ONa + H2

(ii) Reaction of ethanol with cone. H2SO4 for dehydration :
Ethanol reacts with cone. H2SO4 at about 170°C to form ethylene.
WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry 18

Uses of acetic acid (CH3COOH)

  • It is used as an aid in cooking and for making pickles as also a preservative of fish, meat, vegetable etc. as vinegar (6-8% water solution of acetic acid)
  • It is used to prepare white paint.
  • It is used to concentrate the latex of leather.

Reactions of acetic acid :

(i) Reaction of acetic acid with NaHCO3 : During reaction of acetic acid with NaHCO3, sodium acetate, carbon dioxide and water are produced.
CH3COOH + NaHCO3 → CH3COONa + CO2↑+ H2O

(ii) Reaction of acetic acid with NaOH : Acetic acid reacts with NaOH to form sodium acetate and water.
CH3COOH + NaOH → CH3COONa + H2O

(iii) Reaction of acetic acid with C2H5OH : This reaction is called esterifi­cation. During this reaction ethyl acetate and water are produced.
WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry 17

Toxicity of Methanol : Methanol has a high toxicity in human. If as little as 10 mL of pure methanol is ingested, for example, it can break down into formic acid, which can cause permanent blindness by destruction of the opti nerve and 30 mL is potentially fatal, although the median lethal dose is typically 100 mL. Toxic effects take hours to start and effective antidotes can often prevent permanent damage.

WBBSE Class 10 Physical Science Notes Chapter 8.6 Organic Chemistry

Methanol is toxic by two mechanisms. First methanol (whether it enters the body by ingestion, inhalation or absorption through the skin) can be fatal due to its CNS depressant properties in the same manner as ethanol poisoing. Second, in a process of toxication, it is metabolized to formic acid (which is present as the formate ion) via formaldehyde in a process initiated by the enzyme alcohol dehydrogenase in the liver.

Methanol is converted to formal­dehyde via alcohol dehydrogenase (ADH) and formaldehyde is converted to formic acid (formate) via aldehyde dehydrogenase (ALDH). The conversion to formate via ALDH proceeds completely, with no detectable formaldehyde remaining.

Formate is toxic because it inhibits mitochondrial cytochrome C oxides, causing the symptoms of hypoxia at the cellular level, and also caus­ing metabolic acidosis, among a variety of other metabolic disturbances. So methanol is highly toxic and causes blindness insanity and even death. Ethanol is non-toxic but when drunk, it produces physiological effect and disturbs the brain activities.

Denatured spirit : Alcohol used for making beverages is highly taxed but the ethyl alcohol supplied to the chemical laboratories and industries is very cheap.

To make it unfit for drinking purposes, industrial alcohol is mixed with poisonous substance like methanol, pyridine etc. The process is known as denaturing and the product is called denatured alcohol. The ethylalcohol mixed with the methanol is called methylated spirit.

WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy

Comprehensive WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy can help students make connections between concepts.

Metallurgy Class 10 WBBSE Notes

Occurrence of metals : A few metals particularly noble metals (Au, Pt, Ag etc.) having least electropositive charader occur in free state. About 75% of the known elements are metals. Most of the metals are found in the combined state in nature as their compounds.

Minerals : The natural materials containing metals and their compounds associated with earthy and other impurities found under earth’s crust are called minarals.

WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy

Ores : The minerals from which the metals can be extracted conveniently and profitably. All ores are minerals but all minerals are not ores.

Metallurgy : It is a process of extracting metals from their ores.

Gangue or Matrix: They are the rocky and earthy impurities associated with the minerals.

Concentration of ore : It is the process of removal of gangue from the ore.

Flux: It is the substance which combines with gangue to form light and easiÍilusible material called slag.

WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy 1

WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy 2

Slag : It is the light and fusible material obtained by the combination of flux with gangue.

Calcination : It is the process of heating an ore in a limited supply of air or in absence of air at a temperature just below its melting point.

Roasting : It is the process of heating an ore in a limited supply of air or in absence of air at a temperature just below its melting point.

WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy

Extraction of metal : Different processes that are used for the extraction of pure metals from ores are called extraction of meials;

Alumimium (AI)

Symbol: Al; Valency : 3;
Atomic weight: 26-78
Atomic Number: 13,

Position in Periodic Table:
Period: 3, Group IIIA (13);
Electronic configuration : ls22s22p63s23p1

Occurrence: Aluminium is not found in pure state. It occurs to the extent of 8% in earth’s crust in many silicate rocks and clays. It is the third most abundant element in the earth’s crust.

Important aluminium minerals are :
WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy 3

  • Chief ore of aluminium : Chief ore of aluminium is bauxite (Al2O3-2H2O).
  • Mines : In India, large quantities of bauxite are available in Bihar, Jharkhand, Orissa, Andhra Pradesh, Madhya Pradesh, Rajasthan, Tamilnadu, Gujarat and Jammu-Kashmir. But large deposit of quality grade bauxite are found in Andhra Pradesh.
  • Extraction of aluminium : Initially alumina is prepared from bauxite then after mixing with cryolite and fluosper aluminium is extracted by electrolysis at about 950°C
  • The ore of aluminium (bauxite) is purified by Baeyer’s process, Hall’s process or Serpeck’s process depending upon the type of impurities present in it. Baeyer’s process or Hall’s process is used for ores containing oxides of iron as main impurity while Serpeck’s process is used for ores containing silica as main impurity.

WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy

Physical properties of Aluminium :

  • It is silvery white metal and is very light, (sp.gr. = 21)
  • It melts at 660°C and boils at 2270°C.
  • It is a good conductor of heat and electricity.
  • It is malleable and ductile.
  • Atomic radius of aluminium is 125 pm.
  • The ionisation enthalpies of Aluminium are 600 KJ mol-1, 2427 KJ mol-1 and 3568 KJ mol-1
  • Sum of the first three ionisation enthalpies being very high, aluminium forms covalent compounds.

Chemical Properties :

(i) Action on

  • Dry air has no action on aluminium. In moist air, it forms a thin film of oxide on its surface.
  • When heated strongly, it burns with a brilliant white light, producing aluminium oxide and little nitride.

WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy 4

(ii) Action of water : Pure water has no effect on the metal due to the presence film on the surface. Amalgamation with mercury prevents the protective oxide film. Amalgamated metal, therefore, can decompose even cold water. Commercial aluminium can be readily attacked by sea water. It decomposes boiling water, liberating hydrogen. 2Al + 6H2O = 2Al(OH)3 + 3H2

(iii) Action of chlorine : when aluminium is heated gently with chlorine, it forms aluminium chloride.
2Al + 3Cl2 = 2AlCl3
A mixture of aluminium powder and iodine catches fire when a drop of water is added on it.

(iv) Action of acids: (a) With dilute sulphuric acid, hydrogen gas is evolved at a slow rate. In hot concentrated sulphuric acid, metal dissolves with the evolution of
2Al + 3H2SO4 (dilute) = Al2(SO4)3 + 3H2
2Al + 6H2SO4 (cone) = Al2(SO4)3 + 3SO2 + 6H2O

(b) With dilute nitric acid, aluminium nitrate and ammonium nitrate are produced.
Action with carbon Sulphur :
8Al + 3OHNO3 = 8Al(NO3)3 + 9H2O + 3NH4NO3

WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy

(v) Action with carbon and sulphur :
Aluminium reacts with boiling caustic soda or caustic potash forming sodium aluminate or potassium aluminate respectively with the evolution of hydrogen gas.
WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy 6

(vi) Action of alkalls: Aluminium reacts with boiling caustic soda or caustic potash forming sodium aluminate or potassium aluminate respectively with the evolution of hydrogen gas.
2Al + 2NaOH + 2H2O = 2NaAlO2 + 3H2
2M + 2KOH + 2H2O = 2KAlO2 + 3H2
That is why, caustic soda is not boiled in aluminium pot.

(vii) Displacement reactions: Aluminium is strongly electropositve. Hence, it readily displaces less electropositive metals, such as-murcury, silver, gold etc.
2Al + 3HgCl2 = 2AlCl3 + 3Hg.

(viii) Peducing property: Due to its strong affinity of oxgen it acts as a strong reducing agent. It readily reduces certain metallic oxides to the metal.
WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy 5
These reactions are highly exothermic and they can be utlised for welding (Thermite welding) of metals.

Uses of Aluminium:
At present, use of aluminium is just after iron.

  • Being a light material as also stable and resistance to water vapour it is used to prepare external structures of air-ships and motor cars.
  • It is used in electric wire and electric equipments as it is a very good conductor of electricity.
  • It forms many useful alloys used for different purposes. Alloys like dural­umin, magnelium bronze etc. and also in flash bub used in photography aluminium metals are used.
  • Aluminium is used in thermit process. Due to its strong affinity for oxygen, it is used for the extraction of metals like Cr and Mn from their ores.
  • Mixed with linseed oil, powdered aluminium is used as paint.
  • Ammonal, a mixture of ammonium nitrate and aluminium powder, is used in explosive bombs.
  • Aluminium power has the use in fireworks.
  • Aluminum is used in the preparation of different utensils, chair, table and the packing over a cigarette, coffee, chocolate etc.

Artificial gem are produced by dropping powdered alumina mixed with little of other metallic oxides to obtain desired colour (e.g. 2-5% chromic oxide for artificial ruby). Sapphire is similary produced from alumina with addition of a little magnetic iron oxide and titanium oxide.

WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy

Foods having sour taste should not be preserved in aluminium foil, since fatty acid present in those types of foods react with aluminium foil producing salts which are very harmful to health.

Magnesium (Mg)

Symbol: Mg ; Valency: 2 ; Atomic weight: 24
Atomic Number: 12 :
Position in Periodic Table :
Period: 3; Group: IIA (2) :
Electronic Configuration : Is22s22p63s2
Occurrence: Magnesium is not found in free state in nature.

Important magnesium minerals are :

Nature Minerals
1. As carbonates Magnesite : MgCO3
Dolomite : MgCO, CaCO
2. As chloride Carnallite : MgCl2 . KCl . 6H2O
3. As sulphate Kieserte : MgSO4 . H2O
Kainite : MgSO4 – KCl- 3H2O
Epsom Salt : MgSO4– 7H2O
4. As silicate Asbestos: CaMg3 (SiO3)4

Chief ore: Magnesium is extracted from the ores magnesite and carnallite. Magnesium metal is also found in sea-water, chlorophyll and also in sping.

Mines : The chief ore of magnesium is carnallite. But carnallite is not available in India. It is imported from Germany, where it exists as deposits at stassfurt. Other important ores of magnesium are magnesite and dolomite which are abundantly available in Tamilnadu, Karnataka and in many parts of Bihar, Jharkhand, Orissa, West Bengal and Sikkim.

Extraction: Magnesium is prepared by:

  • Electrolytic reduction of the fused magnesium chloride.
  • Chemical reduction of MgO by carbon or ferro-silicon.

Physical properties of magnesium:

  • Magnesium is a light, white metal with a bright lustre which tarnishes in moist air due to the deposition of a film of oxide.
  • Soft, malleable, and ductile, it can be readily drawn into a wire of ribbon.

Chemical properties:
(i) Action with air: Mg does not react with dry air. If magnesium metal is left in moist air it gradually becomes dull and pale due to the formation of a layer of MgO. Magnesium metal when heated in presence of air it burns with bright white flame. Magnesium oxide and magnesium nitride are thus formed.
2Mg + O2 = 2MgO ; 3Mg + N2 = Mg3N2

(ii) Reaction with water (a) At ordinary temperature Mg metal does not react with water, (b) Heated Mg burns in presence of water vapour magnesium oxide and hydrogen are formed.
Mg + H2O = MgO + H2

WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy

(iii) Reaction with dilute nitric acid: It dissolves in dilute nitric acid, liberating hydrogen, but not in alkalis.
Mg + 2HNO3 = Mg(NO3)2 + H2

(iv) Mg acts as reducing agent. It is a powerful reducing agent; the heated metal reduces such stable oxides as silica, sodium oxide and potassium oxide. Ignited magnesium continues to burn in CO2 depositing carbon.
WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy 7

(v) Reaction with chlorine Heated magnesium burns in chlorine and form magnesium chloride.
Mg + Cl2 = MgCl2

Uses of magnesium : It is used

  • In flash light powders (magnesium powder mixed with potassium chlorate or barium peroxide) for photography, in military star shell, in light signals and in fire works.
  • for the preparation of Grignerd reagent in organic chemistry and for reducing agent.
  • In the production of light alloys, such as-duralumin, magnelium in the construction of airships and motor cars.
  • In incendiary explosive (goop is an incendiary containing magnesium powder).
  • In the extraction boron and silicon.

Zine (Zn)

Symbol : Zn ; Valency : 2; Atomic weight: 65.3
Atomic Number : 30;
Position in Periodic Table;
Period: 4, Group : IIB (12);
Electronic Configuration : ls22s22p63s23p63d104s2

Occurrence : Zinc does not occur free in nature. In the combined state, it occurs in the form of following minerals.

Nature Minerals
1. As sulphide Zinc blende : ZnS
2. As oxide Franklinite : ZnC Zincite : ZnO,Fe2O3
3. As carbonate Calamine : ZnCO3
4. As silicate Willemite : Zn2SiO4

Chief ore: The most important source of zinc is zinc hiende, usually containing about 6% Zn.

Mines: Zinc blende is obtained from zawar at Udaipur (Rajasthan) in India. Calamine is obtained at smithsonite in the U.S.A.

WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy

Extraction: Zinc is extracted mostly from zinc blende. The zinc blende which invariably contains so me galena in concentrated by oil floating process. Zinc is obtained from the zinc concentrate either. (a) by the carbon reduction process (b) by the electrolytic process.

  • Physical properties of zinc:
  • It is a bluish white solid which melts at 420°C and boils at 920°C.
  • Its density is 7-14 g/c.c.
  • It is a good conductor of electricity.

Chemical properties of zinc :

(i) Action of air: It burns in air with a green flame, if strongly heated, forming white clouds of zinc oxide which settles in wooly flocks (philosopher’s wool) 2Zn + O2 = ZnO

(ii) Action of water Pure zinc has no action of water. Impure zinc decomposes boiling water. Zn-Cu couple decomposes water producing hydrogen gas.
Zn + 2H2O = ZnO + H2

(iii) Action of chlorine and sulphur Zinc reacts with chlorine forming zinc choride with sulphur, zinc sulphide is formed.
Zn + Cl2 = ZnCl2
Zn + S = ZnS

(iv) Reactions with acids :
(a) With HCl : Zn + 2HCl (dilute) = ZnCl2 + H2
(b) With H2SO3 :
(i) Zn + H2SO4 (dilute) = ZnSO4 + H2
(ii) Zn + 2H2SO4 (cone.) = ZnSO4 + SO2 + 2H2OWBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy 9
Action with alkalis:
Zinc dissolves in hot solutions of caustic soda and potash, yielding hydrogen and zincate.
WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy 8

(vi) Displacement of less electropositive metals :
When zinc is added to the salt solution of a less electropositive metal like copper, silver, the metal is displaced.
CuSO4 + Zn = ZnSO4 +Cu ↓
2AgNO3 + Zn = Zn(NO3)2 + 2Ag ↓

WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy

Uses of zinc : It is used:

  • For galvanisation of iron sheets because zinc is resistant to atmospheric corrosion due to the formation of basic carbonate on its surface.
  • In the extraction of silver and gold by cyanide process.
  • as a cathode in dry cells.
  • zinc is used for the preparation white paint zinc oxide (ZnO).
  • for the preparation of hydrogen in the laboratory.
  • in the preparation of alloys such as brass, german silver, etc.

Iron (Fe):

Symbol: Fe ; Valency : 2, 3; Atomic weight: 55-85
Atomic number: 26 ;
Position in Periodic Table :
Period : 4, Group VIII(8);
Electronic Configuration : 1s2 2s22p63s23p63d64s2

Occurrence : Iron is not obtained in free state. Next to aluminium, iron is the most abundant metal in the earth’s crust. Compounds of iron occur in the soil, in green plants and in haemoglobin (0-336% Fe) the red colouring matter of blood.

Meteorites sometimes consist of iron with (3 -30)% of Ni; because of its nickel content meteoric iron does not rust readily.

The chief minerals of iron are given below :

Nature Minerals
1. As oxide
Brown Haematite: Magnetite: Fe3O4
Red Haematite: Fe2O3
2Fe2O3. 3H2O (Limonite)
2. As sulphide Iron pyrites: FeS2
3. As carbonate (Siderite) Spathic iron ore: FeCO3
  • Chief ore : Iron is extracted mainly from the ore Haematite an<-’ magnetite may be used.
  • Extraction : Iron is extracted from its oxide and carbonate ores (burnt pyrites which is chiefly ferric oxide, is sometimes used) by reduction with carbon in a blast furnace.
  • Mines : Rich deposits of good quality haematite are available in many parts of India, such as in Orissa, Bihar, West Bengal, Tamilnadu, Karnataka, Andhra Pradesh and Madhya Pradesh.

WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy

Physical properties of iron :

  • Pure iron is silvery white lustrous metal resembling platinum in appearance. It can take high polish.
  • It is malleable and ductile and good conductor of heat and electricity.
  • It is relatively soft and possesses high tensile strength.
  • It is the most magnetic of all metals but loses this property above 760°C.
  • Iron melts at 1525°C, boils at 2450°C and its density is 7-86 g cm-3.

Chemical properties :

(i) Action of water : Iron decomposes steam to liberate hydrogen. This reaction is utilized for the manufacture of hydrogen. 3Fe + 4H2O = Fe3O4 + 4H2

(ii) Action of moisture : When exposed to moist air, iron gets covered with a reddish yellow film. This is known as rust and the phenomenon is called rusting. The composition of rust is 2Fe2O3. 32O. Rusting is formed when iron reacts with oxygen and water vapour.

(iii) Action of water : Iron decomposes steam to liberate hydrogen. This reaction is utilized for the manufacture of hydrogen. 3Fe + 4H2O = Fe3O4 + 4H2

(iv) Action of halogens and sulphur
WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy 17

WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy

(v) Displacement of less electropositive metals: It displaces less electropos­itive metals like copper from their salts. CuSO4 + Fe = FeSO4 + Cu4

(vi) Reaction with alkalis : It has no reaction with alkalis.

(vii) Action of acids :
(a) Fe + H2SO4 (dilute) = FeSO4 + H↑
(b) Fe + 2HCl (dilute) = FeCl2 + H2
(c) 4Fe + 10HNO3 (dilute) = 4Fe(NO3)2 + NH4NO3 + 3H2O
(d) (i) Fe + 2H2SO4 (cone.) = FeSO4 + SO2 + 2H2O
(ii) 2FeSO4 + 2H2SO4 (cone.) = Fe2(SO4)3 + 2H2O + SO2

(viii) Passivity of iron: Fairly strong nitric acid gives ferric nitrate and a mixture of oxides of nitrogen. However, cone., nitric acid makes iron passive i.e. it becomes chemically inert and does not react further.

Uses of iron :

  • Iron is widely used for making beams, rafters, rods for building and for con­structing ships, submarines, rails and rail-wagons.
  • It is also used for preparing nuts, bolts, screws, wires, covers of drains, rain water pipes and pipes for conveying drinking water etc.
  • It is also used to make ordinary steel and alloy-steels.

Iron is classified according to quantity of carbon present:

(i) Cast iron : (2-4)%
carbon and small amount of silica, Mn, R S etc are present in iron.
use: (t is used for the manufacture of railings, drain pipes, radiators, gutters.
(ii) Wrought iron :
Smallest quantity of carbon is present in it. Iron of this type contains (01- 015)% of carbon.
use: It is used for the manufacture of nails, chair, anchors, horse shoes.
Steel : In steel (0-15- 1-5)% and small quantities of silicon, Mn are mixed with iron.
use: It is used for the manufacture of car wheel springs razor blades, ma­chine parts, ships, opera­tion equipments, building structures and arms.

Copper (Cu)

Symbol: Cu; valency : 1, 2; Atomic weight: 63-5
Atomic number : 29 ;
Position in Periodic Table :
Period : 4, Group IB(ll) ;
Electronic Configuration : ls22s22p63s23p63d104s1

Occurrence : Copper occurs in minute quantities in plants, particularly in green peas. It is present in the haemocyanin of blood cuttlefish, which acts like haemoglobin as an oxygen carrier. Copper helps in the formation of haemoglobin in the body. Lower organisms are very sensitive to copper salts and hence their use as fungicide, copper exists in small amount in free state in nature. The chief minerals are :

Nature Minerals
(i) As oxide Cuprite : Cu2O
Malaconite : CuO
(ii) As sulphide Chalcocite (copper glance): Cu2S
Chalcopyrite (copper pyrites): Cu2SFe2S3
(iii) As carbonate Malachite : CuCO3Cu(OH)2
Azurite : 2CuCO3 Cu(OH)2
(iv) As chloride Attacamite : CuCl2 . 3Cu(OH)2

Chief ore: Copper is mainly extracted from copper pyrities.

Mines: The chief copper-producing regions are : Montana, Ariz una, Canada, Japan, Russia, Germany, Cango, Spain and Chile. India is not rich in copper ore. In India, copper pyrites is available in small amounts in Singhbhum, Bihar, Tamilnadu, Karnataka, Uttar Pradesh, Sikkim, Orissa.

Extraction of copper from copper pyrities :

  • crushing
  • concentration by froth floatation
  • roasting
  • smelting with coke and sand in a blast furnace
  • best merization of molten matte obtained from blast furnace to get blister copper.
  • purification by electrolytic process.

Physical properties of copper :

  • It is a reddish brown metal.
  • It is highly malleable and ductile.
  • Copper melts at 1083°C and boils at 2323°C
  • Copper is a quite heavy metal (density 8.94g/cc )
  • It is an extremely good conductor of heat and electricity.
  • It readily forms alloys.

WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy

Chemical properties of copper :

(i) Action of air: (a) Dry air has no effect on copper metal. However, when it is exposed to damp air for a long time, a green coating of basic copper carbonate. CaCO.Cu(OH)2 is formed on its surface.
WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy 11

(b) When heated with air, copper forms cupric and cuprous oxides.
WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy 12

(ii) Action of Water (and even stearia) has no action on copper. How ever, at very 1 temperatures (at white heat), it is attacked by steam when slight oxidation takes place.

(iii) Action of acids on copper:
WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy 13

(iv) Reaction with chlorine and sulphur:
WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy 14

(v) Reducing properties:
WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy 15

Displacement reactions: Copper displaces less electropositive metal like silver, gold, platinum and mercury from their salts.
Cu + 2AgNO3 = Cu(NO3)2 + 2Ag
3Cu + 2AuCl3 = 2Au + 3CuCl2

(vi) Action with cyanides: It also dissolves in an aqueous solution of potassium cyanide yielding the complex, [Cu(CN)4]2-

(vii) Formation of colloidal copper : On striking an ore between two copper electrodes immersed in water, a dark red colloidal solution of copper is obtained.

WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy

Uses of copper :

  • Due to its very high electrical conductivity it is largely used for telegraph and telephone, and also for electrical machinary.
  • Copper is not attacked by air or water. So, it can be used for making household utensils, and water pipes.
  • It is used in electroplating and electrotyping and in coinage.
  • It is used in making a number of useful alloys.
    e.g. Brass: Cu (60-80)%, Zn (40-20%); Bronze: Cu (75-90)%, Sn (25-10)%

Alloy : It is homogeneous mixture (or, sometimes, a heterogeneous combination) of two or more metals, behaving as single metal in most of its physical properties.

Alloy is formed not only combination metals alone but also many non-metals viz. C, Si, S, P are added to prepare alloys.

Amalgam : If one of the component metals of an alloy be mercury, it is termed as amalgam, e.g. sodium amalgam (Na/Hg), aluminium amalgam (Al/Hg) etc.

Advantages of using alloys over pure metals :

  • Increase of hardness: as in gold for ornaments, high speed steel etc. N.B. Pure gold is somewhat soft. It is made harder by alloying with a little amount of Cu or Ag.
  • Increase in resistance to corrosion: as in stainless steel.
  • Increase in strength: as in steel, duralumin etc.
  • Increase of resistance to electricity: as in nichrome.
  • Increase in fusibility or decrease in melting point : as in solders, fuse wires etc.
  • Betterment of appearance . as in aluminium bronze, stainless steel etc.
  • Improvement of casting property : as in type metal.
  • Change in chemical activity : as in sodium amalgam, zinc amalgam etc.

WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy

Some important alloys :

WBBSE Class 10 Physical Science Notes Chapter 8.5 Metallurgy 16

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus

Comprehensive WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus can help students make connections between concepts.

Atomic Nucleus Class 10 WBBSE Notes

Atom and its constituents:

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus 1

Inventor Sub-atomic particles
J.J. Thomson (1897) Electron
E. Rutherford (1911) Proton
J. Chadwick (1932) Neutron
Anderson (1932) Positron
Yukawa (1935) π – meson
Name given by Fermy (1956) Neutrino

Nucleus : From Rutherford’s experiment on scattering of α-particles, it was established that the positive charge of an atom and almost whole of its mass must be concentrated in a small region (10-12 -10-13cm) & (while dimensions of atoms 10-8cm)
Atomic number: The number of protons in the nucleus is called the atomic number of atom.
Mass number: The sum of the number of protons and neutrons is called mass number.
Nucleons : The protons and neutrons are collectively known as nucleons.
Nuclear reactions : Nuclear reactions are the reactions in which nucleus of an atom undergoes a change.
Nuclear chemistry: The branch of chemistry which deals with the phenomenon of nucleus of atoms is known as nuclear chemistry.

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus

Type of species :

(i) Isotopes: The atoms of an element having same atomic number but different mass number are called isotopes.

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus 17

(ii) Isobars : The nuclides of different chemical elements having same mass number but different atomic numbers are called isobars.

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus 3

(iii) Isotones: The nuclides of different chemical elements having the same number of neutrons but different atomic numbers are called isotones.

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus 4

(iv) Nuclear isomers : The nuclear species having same atomic number and same mass number but different radioactive properties are called isomers or nuclear isomens.

Examples:
WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus 5

(v) Isodiaphers : Atoms having the same isotopic number or isotopic excess (number of neutrons-number of protons) are called isodiaphers.

Examples : \(\quad{ }_{92}^{238}\) U and \({ }_{90}^{234}\) Th

(vi) Isoster : Molecules or ions with same number of atoms and also the same number of electrons are said to form isosteric group or more simply isosters.

Examples : In N2 O and CO2;
number of electrons = 22. So, they are isosters.

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus

Radioactive radiations :

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus 7

Characteristics of alpha (α), beta (β) and gamma (γ) – rays.

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus 8

Stability of Nucleus : The stability of the nucleus is decided by the following factors:

(a) Nuclear forces: The forces which held the nucleons together within the small nucleus are called nuclear forces.
These force exist among p-p, p-n and n-n.

Nuclear force has been explained by yukawa by the discovery of a new fundamental particle called π-meson
These may be denoted as π+, π and π0

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus 9

The mesons keep on exchanging among the nucleons very rapidly (about 1024 transfers per second) and hold the nucleons together.

All nuclei with 84 or more protons i.e. atomic number ≥ 84 are unstable.

(b) Mass defect : Δm, the difference between the experimental and calculated masses of the nucleus is called the mass defect.

(Experimental mass of nucleus) – (mass of proton + mass of neutron) = mass defect.

(c) Binding energy: Atomic nucleus is made of protons and neutrons closely in a small volume. Although there exist intensive repulsive forces between the component protons, the nucleus is not slit apart. This is so because the nucleons are bound to one another by very powerful forces. The energy that binds the nucleons together in the nucleus is called the nuclear binding energy.
The binding energy of a nucleus can be calculated from its mass defect by using Einstein’s equation,
Δe = Δ m × c2 or Binding energy = Δm × 931 MeV

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus

By plotting the binding energy per nuclear against the mass number, we get the graph shown in figure below.

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus 10

(d) Packing fraction : Packing fraction was proposed by Aston and defined as the difference of actual isotopic mass and the mass number.

Packing fraction =\(\frac{\text { (acutal isotopic mass-mass number) }}{\text { mass number }} 10^4\)

The value of packing fraction may be negative, positive or zero. A negative packing fraction indicates less stability of nucleus. In general, for lower packing fraction the binding energy will be greater and the nucleus will be more stable.

(e) Neutron-Proton ratio (N / P ratio) : The stability of a nucleus seems depend on the neutron to proton (n / p) in the nucleus.

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus 11

(i) In lower elements (upto z=20 ), the stable nuclei have about equal number of protons and neutrons i.e. \(\frac{n}{p}\) = 1
(ii) For higher elements to be stable, there must be more neutrons than protons i.e. \(\frac{n}{p}\) >1.
(iii) The shaded portion in figure represents the region or belt of stability. The elements whose n / p ratios lie inside the belt are stable.
(iv) A nucleus whose n / p lies above or below the stability belt is radioactive or unstable on account of unfavourable n / p ratio. It emits α or β-particles so as to move into the stability range.

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus

Know more :
(a) If \(\frac{n}{p}\) > 1 to 1.5 i.e. it places above the stability belt. Such nuclei emit- β particle in order to lower its \(\frac{n}{p}\) ratio

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus 12

(b) If \(\frac{n}{p}\) < 1 to 1.5 i.e. places below its stability belt, such nuclei either emit positrons or undergo electron capture.

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus 13

(c) The even odd nature of the number of protons and neutrons : The number of stable nuclides is maximum when both p and n are even number. However, the number of stable nuclides in which either the z(p) or n is odd is about one third of those where both are even.

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus 14

The magic numbers : Magic numbers are the numbers 2, 8, 20,50, 82 and 126. Nuclides having magic number of either proton or neutrons or both are more stable. Nuclides with atomic numbers equal to magic numbers have more number of stable isotopes than their neighbours. Nuclides having even number of protons and neutrons are more stable than nucli containing odd number of these particles.

Radioactivity : It is a process in which nucleus of certain elements undergo spontaneous disintegration without excitation by any external means. This phenomenon was first discovered by Henry Becquerel (1866). However, the term radioactivity was proposed by Madam Curie.

Group Displacement Law (Fajans and Soddy in 1913) : We know that an α-emission decreases the atomic number of the parent by 2 and β emission increases the atomic number by 1 .
Thus ‘In an α emission, the parent element will be displaced to a group two places to the left and in β emission, it will be displaced to a group one place to the right’.

Disintegration Series : The whole series of elements starting with the parent radioactive element to the stable end product is called a radioactive disintegration series.

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus 15

Radioactive Decay: If N be the number of undecay atoms of an isotope present in a sample of the isotope, at time t, then

\(\frac{-d N}{d t} \alpha N\)
or, \(\frac{-d N}{d t}=\lambda N\)

where, \(\frac{-d N}{d t}\) tmeans the rate of decrease in the number of radioactive atoms in the sample and λ is the proportionality constant. This is known as disintegration constant.

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus

Disintegration constant : It may be defined as the proportion of atoms of an isotope decaying per second.

Various forms of equation for radioactive decay are :

(i) Nt = N0 e-λt
(ii) λ = \(\frac{2 \cdot 303}{t}\) log \(\frac{N_0}{N_t}\)
(N0= No. of atoms initially
Nt = No. of atoms after time t)

Half-Life period : Half-life period of a radioactive isotope is the time required for one-half of the isotope of decay.

t1/2 = \(\frac{0.693}{\lambda}\)

The value of λ can be found experimentally be finding the number of disintegrations per second with the help of a Geiger-Muller Counter.

The unit of half life period : time-1

Average life : The statistical average of the lives of all atoms present at any time is called the ‘average life’.

τ = 1.44 × t1/2
(τ = Average life
τ = \(\frac{1}{\lambda}\))

Activity of a Radioactive Substance :

  • Higher is the activity of a substance, faster will be its disintegration and vice-versa.
  • The greater the half-life of the substance, lesser is its activity and vice-versa.
  • The activity of a radioactive sample is usually determined experimentally with the help of a Gieger Muller counter.

Radioactive Equilibrium : Radioactive change being an irreversible process shows equilibrium when a daughter element disintegrates at the same rate at which it is formed from parent element.

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus 16

Unit of Radioactivity :

  • Curie : If a radioactive substance disintegrates at the rate of 3.7 × 1010 disintegrations per second, its activity is said to be 1 curie.
  • Rutherford : If a radioactive substance has 106 disintegrations per second, it is said to have an activity of one Rutherford.
  • Becqurel (SI unit) : If a radioactive substance has 1 disintegration per second, it is said to have an activity of one Becquerel.

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus

Measurement of Radioactivity : The instruments used are

  • Geiger-Muller counter
  • Wilson cloud
  • Electroscope

Natural transmutation : It is a process in which elements such as radium undergoes transmutation on their own.

Artificial transmutation: It is the process in which a stable nuclei is changed into another by artificial methods i.e. by bombarding the target nuclide with projectiles such as α-particles, neutrons etc.
The first artificial transmutation was carried out by Rutherford (1919) by bombarding nitrogen with α-particles.

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus 18

Artifical radioactivity : It is the phenomenon in which artificial transmutation of a stable nucleus leads to the formation of a radioactive nuclide.

Contributions of Artificial Transmutation : The following are the important contributions of artificial transmutation :
(i) Discovery of neutron
(ii) Artificial radioactivity
(iii) Nuclear fission
(iv) Nuclear fusion

Alchemy : The process of transforming one element into other is known as alchemy and the person involved in such experiments is called alchemist.
Nuclear reaction: It is a reaction in which the number of protons in the nucleus of an element changes to form a new element.

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus

Difference between Nuclear and Chemical Reactions :

Nuclear reactions Chemical reactions
(i) Proceed by distribution of nuclear particles. (i) Proceed by the rearrangement of extra nuclear electrons.
(ii) One element may be converted into another. (ii) No new element can be produced.
(iii) Often accompanied by release or absorption of enormous amount of energy. (iii) Accormpanied by release or absorption of relatively small amount of energy.
(iv) Rate of reaction is unaffected by external factors such as concentration, temperature, pressure and catalyst. (iv) Rate of reaction is influenced by external factor.

Types of Nuclear reactions-

A. On the basis of mechainism

(i) Projectile capture reaction : The bombarding particle is absorbed with or without the emission of γ radiations
Example:
WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus 19
(ii) Particle-particle reactions : Majority of nuclear reactions come under this category

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus 20

B. On the basis of bombarding particle

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus 21

Nuclear Fission : The process of artificial transmutation in which heavy nucleus is broken down into two lighter nuclei of nearly comparable masses with release of large amount of energy is termed as nuclear fission.
(i) The first nuclear fission to be discovered was that of \({ }_{92}^{225} \mathrm{U}\).
(ii) Over 200 different isotopes of 35 different elements have been found among the fission products of \({ }_{92}^{225} \mathrm{U}\). Most of them are radioactive.
(iii) During fission, there is always lose of mass which is converted into energy according to Einstein equation.
E = Δ mc2(Δ m = mass defect)

(iv) It has been found that at the moment, the energy available from 1 kg of uranium is equivalent to that available from 20,000 kg of coal

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus 25

Application of Nuclear fission –

(i) Nuclear reactor: A nuclear fission reactor is a device that permits a controlled chain nuclear fission, control rods made of elements such as boron and cadmium, absorb additional neutrons and can therefore, show the chain reaction.
(ii) Atom bomb : It is based on uncontrolled chain reaction. In the atom bomb fissionable material ( 235U or 239Pu) is taken in parts in such a way that each is in sub-critical stage. At the time of explosion these pieces one driven together rapidly by using explosives like TNT lying behind each piece of fissonable material make one large piece of fissonable material. At this instant, the overcritical stage is achieved and a fast chain reaction is set up. This result is a violent explosion with the release of tremendous amount of energy.

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus 22

Nuclear fusion : A nuclear reaction in which two lighter nuclei are fused together to form a havier nuclei is called nuclear fusion.
The process of fusion can take place at extremely high temperature only (>106 K). Such reactions are known as thermo-nuclear reactions.
WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus 23

Application of Nuclearfusion :
(i) Energy of sun : Every second, the sun loses 4.3 × 109 kg (4,20,000 tons) of mass by the fusion reactions. This mass is converted to energy.
(ii) Hydrogen bomb : Hydrogen bomb is based on the fusion of hydrogen nuclei into heavier ones by the thermonuclear reactions with release of enormous energy. In hydrogen bomb, a mixture of deuterium oxide (D2O) and tritium oxide (T2O) is enclosed in a space surrounding an atomic bomb. The temperature produced by the explosion of atomic bomb initiates the fusion reaction between \({ }_1^3 \mathrm{H}\) Hand \({ }_1^2 \mathrm{H}\) Hreleasing huge amount of energy.

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus

Difference between Nuclear Fission and Nuclear Fusion :

Nuclear fission Nuclear fusion
(i) A bigger nucleus splits into smaller nuclei. (i) Lighter nuclei fuse together to form the heavier nucleus.
(ii) It does not require high temperature. (ii) Extremely high temperature is required for fusion to take place.
(iii) A chain reaction sets in. (iii) It is not a chain reaction.
(iv) It can be controlled and energy released can be used for peaceful purposes. (iv) It cannot be controlled and energy released cannot be used properly.

Application of Redioactivity

WBBSE Class 10 Physical Science Notes Chapter 7 Atomic Nucleus 24

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Comprehensive WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry can help students make connections between concepts.

Inorganic Chemistry in the Laboratory and in Industry Class 10 WBBSE Notes

Discovery: In 1774 Priestly prepared the ammonia gas by heating a mixture of ammonium chloride and calcium hydroxide.

Occurrence: Traces of ammonia occur in the atmosphere. Ammonia is a product of the decomposition of organic matter containing nitrogen. The stable manure e.g. urea [CO(NH2)2], derived from the urine of animals. The urea is converted by the action of bacteria into ammonium carbonate which slowly decomposes, yielding ammonia and hence its smell is a stable and its presence in traces in air.

  • Molecular formula : NH3
  • Molecular weight : 17

Preparation of ammonia:
Principle Normally ammonia gas is prepared in the laboratory by heating a mixture of ammonium chloride (1 part) and slaked lime (3 parts). Instead of using slaked lime quick lime also be used.
Chemicals required: Ammonium chloride (NH4Cl) and quick lime (CaO) or dry slaked lime [Ca(OH)2]

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Chemical reactions:
2NH4Cl + Ca (OH)2 = 2NH3↑ + CaCl2 + 2H2O
2NH4Cl + CaO = 2NH3↑ + CaCl2 + H2O

Collection: Ammonia is lighter than air, it may be collected by the down­ward displacement of air. Ammonia is not collected through the downward displacement of water because, it is highly soluble in water. Precautions: The ingradients, the test-tube, the delivery pipes and the gas- jar should be absolutely dry. All the connections in the apparatus should be leak-proof.

Drying of ammonia : As ammonia is a basic substance, it cannot be direct by acidic drying agents like cone. H2SO4 or P2O5.
2NH3 + H2SO4 = (NH4)2SO4; 6NH3 + P2O5+ 3H2O = 2 (NH4)3PO

It forms an additive compound with calcium chloride. So fused CaCl2 cannot also be used to dry ammonia.
CaCl2 + 8NH3 = CaCl2-8HO (Additive compound)
It is best dried with the basic drying agent, quicklime (CaO).

2. Other methods of preparation of Ammonia

(a) By hydrolysis of a metallic nitride by heating with an alkali solution.
Mg3N2 + 6H2O= 3Mg (OH)2 + 2NH3
A1N + NaOH + H2O = NaAlO2 + NH3

(b) By heating a solution of a nitrate or nitrite with zinc and strong caustic soda solution.
3NaNO3 + 8Al + 5NaOH + 2H2O = 8NaAlO2 + 3NH3
NaNO2 + 3Zn + 5NaOH = 3N2ZnO2 + H2O + NH3

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

(c) By heating ammonium salts at high temperature.
(NH4)2SO4 = NH3↑ + NH4HSO4
2(NH4)3PO4 = 6NH3↑ + P2O5 + 3H2O

(d) By the reduction of the oxides of nitrogen (except nitrous oxide) e.g. by passing a mixture of nitric oxide and hydrogen over heated spongy platinum.
2NO + 5H2 = 2NH2+ 2H2O

(e) Calcium cyanamide is hydrolysed by highly heated steam to form ammonia.
CaCN2 + 3H2O = CaCO3 + 2NH3

(f) Harber’s process : In Haber’s synthetic process the mixture of one volume of nitrogen and three volume of hydrogen is heated at 550°C temperature and 200 atm pressure under the influence of iron as catalyst and MO as promoter to form ammonia.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 1

Physical properties of ammonia :

  • Smell Ammonia is a colourless gas with very pungent smell that affects eyes, nose and throat.
  • Density Ammonia has a density (vapour density = 8 5) less than air (vapour density of air = 144). The gas is readily liquefied by pressure alone (6 atm at 0°C).
  • It is highly soluble in water. 1 volume of water at 0°C dissolves 1150 volumes of ammonia and 739 at 20°C.

Liquor ammonia : A saturated aqueous solution of ammonia (sp. gr. = 0.88) contains only 35% NH3 by weight is known as liquor ammonia. Liquor ammonia is dangerous for eyes.
NH3 + H2O = NH4OH

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

The solution of ammonia is alkaline and turns red litmus to blue. Its basic character is due to the formation of OH.
NH3+ H2O \(\rightleftharpoons\) NH4++ OH
A bottle of liquor ammonia should be carefully opened after cooling in ice, as there is always a high pressure inside.

Chemical properties of NH3

(i) Reaction with oxygen:

(a) It does not support combustion, nor does it burn in air, but in oxygen it burns with a greenish-yellow flame, forming nitrogen and steam. 4NH3 + 3O2 = 2N2 + 6H2O

(b) In presence of heated platinum gauze catalyst at 500°C – 700°C, ammonia is oxidised to nitric oxide by air or oxygen. 4NH3 + 5O2 = 4NO + 6H2O

(ii) Reaction with acid : As ammonia is a base, it readily reacts with an acid to form salt.

NH3 + HCl = NH,Cl
NH3 + HNO3 = NH4NO3
2NH3 + H2SO4 = (NH4)2 SO4

The reaction of NH3 (gas) with HCl (gas) produces NH4Cl as a white solid.
This reaction shows the formation of a solid substance from the reaction of two gaseous substances.
NH3(gas) + HCl (gas) = NH4Cl (solid)

(iii) Reaction with alkali metals: Ammonia reacts with alkali metals at red-heat (360°C), forming amides which are violently decomposed by water, yielding ammonia.
2Na + 2NH3 = 2NaNH2 + H2
2K + 2NH3 = 2KNH2 + H2
NaNH2 + H2O = NaOH + NH3

(iv) Reaction with chlorine (non-metal): Ammonia reacts with chlorine in two ways.

(a) In excess ammonia Excess ammonia is oxidised by chlorine forming nitro­gen and is reduced itself to hydrochloric acid. Hydrochloric acid thus formed combines with ammonia producing ammonium chloride. This reaction also proves that ammonia contains nitrogen.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 15

(b) In excess chlorine : When excess chlorine reacts ammonia forming nasent nitrogen which again combines with chlorine producing nitrogen trichloride, an oily yellow explosive compound.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 16

(v) Reducing property: Ammonia reduces some metal oxides at high tempera­ture lo the F corresponding metals and itself gets oxidised to nitrogen gas. Ammonia gas is passed over heated black cupric oxide which is reduced to red metallic copper and ammonia is oxidised to nitrogen.

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 17

(vi) Formation of additive compounds :
Anhydrous CaCl2,  ZnCl2 etc. absorb ammonia to form additive compounds.
CaCl2 + 8NH3 = CaCl2 .8NH3
ZnCl2 + 8NH3 = ZnCl2 .8NH3

(vii) Reaction with carbon dioxide :
(a) Urea is formed with the reaction of ammonia and carbon dioxide at 200°C and 150 atm pressure.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 19

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

(b) Ammonia is converted into ammonium sulphate with the direct use of sul­phuric acid.
2NH3 + CO2 + H2O + CaSO4 = (NH4)2SO4 + CaCO3

(vii) Reactions with salts: Salts like ferric chloride, aluminium chloride react with ammonium hydroxide forming brown precipitate of ferric hydroxide and white gelatinus precipitate of aluminium hydroxide respectively.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 20
(ix) Formation of complex salts: In some cases the precipitated hydroxide dissolves in excess of ammonia forming a cationic complex e.g. (a) copper sulphate gives a pale blue precipitate of basic copper sulphate with ammo­nia which dissolves in excess of the precipitant forming a deep blue solution containing the complex salt, tetra-tetra-ammine cupric sulphate [Cu(NH3)4] SO4.WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 21

(b) Silver nitrate solution gives a white precipitate which quickly passess into brown oxide, soluble in excess of ammonia.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 22

(c) Ammonia produces brown colouration or precipitate in Nessler’s reagent (an alkaline solution of potassium mercuric iodide, k2Hgl4
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 23

Precaution to be taken to combat the effect of NH3 leaked from industires and ammonia tanks :
Ammonia has a strong pungent smell and is highly soluble in water. It is harmful for eyes. It gives troubles in eyes when exposed in small quantities but in excess it could damage eyes permanently.

Possible measure :
(a) As NH3 is highly soluble in cold water, it is neces­sary to spray cold water so that dispersed ammonia gas would be dissolved and minimizes its bad effect.

(b) If individual is affected then eyes should be washed readily with cold water time and again. If ammonia is leaked from ammonia tanks then the portion from which it is leaked should be kept immersed in cold water.

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Identification  of Ammonia : Ammonia can be identified by the following tests:

  • Ammonia is a colourless gas with a strong pungent smell.
  • White fumes are obtained when ammonia gas brought in contact with a glass rod moistened, with hydrochloric acid.
  • Ammonia (or its aqueous solution) forms a brown colour precipitate with Nessler’s reagent solution (an alkaline solution of K2Hgl2.
  • Ammonia turns moist red litmus paper into blue.
  • A strip of filter paper, soaked in mercurous salt solution, when exposed to ammonia gas, turns black.

Using Ammonia:
(a) Ammonia is used for the industrial preparation of nitric acid (by Ostwald process), sodium carbonate (by Solvay process).

(b) Large quantities of ammonia are used in the manufacture of fertilizers, such as-urea [CO(NH2)2], ammonium sulphate [(NH4)2SO] etc.

(c) It is also used in pharmaceutical industries and also in the preparation of smelling salt.

(d) Ammonia is used as solvent.

(e) It is used as laboratory reagent.

(f) Liquid ammonia is used as a refrigerant in ice making.

(g) Ammonia is used as cleaning agent for removing grease.

Structure of  Ammonia molecule :
Ammonia molecule has a pyramidal sharp vith nitrogen atom at the apex. The nitrogen atom of ammonia is sp3 hybridized and < H-N-H = 106°45C
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 24

(b) Hydrogen Sulphide (H2S)

Discovery: In 1774, Shele discovered the gas sulphurated hydrogen.

Occurrence : Hydrogen sulphide is found in volcanic gases and in many hot spring waters. Its presence in the atmosphere is in very small amount. Hydrogen sulphide is also present in rotten egg as also in leather.

Molecular formula : H2S

Molecular weight : 34

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Preparation of hydrogen sulphide :

1. Laboratory preparation:

Principle: At ordinary temperature ferrous sulphide taken in a woulfe’s bottle reacts with dilute sulphuric acid (1 volume acid and 6 volumes water), hydrogen sulphide is prepared.

Chemical required . Ferrous sulphide (FeS) and dilute sulphuric acid (H2SO4).
Chemical reaction : FeS + H2SO4 = FeSO4 + H2ST↑
Collection : Because the gas is heavier than air, it is collected over the upward displacement of air. Hydrogen sulphide is dissolved in water, so it is not colleced over downward displacement of water. It can also be collected in downward displacement of hot water because the gas is insoluble in hot water.

Purification : The hydrogen sulphide gas can be purified by absorbing it into a suspension of magnesium oxide in water and then regenerating it by heating magnesium bisulphide [Mg(HS)2] thus formed at 60°C.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 25
Pure H2S gas is obtained by heating antimony sulphide, Sb2S3 with pure cone hydrochloric acid.
Sb2S3 + 6HCl = 2SbCl3 + 3H2S

Drying of hydrogen sulphide :
Hydrogen sulphide may be dried by passing it through phosphorus pen oxide (P2O5)

(i) Anhydrous CaCl2 is not used for drying H2S gas.
CaCl2 + H2S = CaS + 2HCl

(ii) Conc. H2SO4 is not used for drying H2S gas.
H2S + H2SO4 = S↓ + SO2↑ + H2O

(iii) Quick lime also cannot be used for drying H2
CaO + H2S = CaS + H2O

(iv) Hg is not used for drying H2 But pure H2S does not react with Hg.

N.B. Dilute sulphuric acid is used in the preparation of H2S because it is non-volatile and dilute H2SO4 does not behave as an oxidising agent but cone. H2SO4 oxidises H2S when sulphur is precipited.
H2S + H2SO4 = S’ + SO2↑ + 2H2O

Nitric acid cannot be used for preparing H2S from a sulphide for H2S thus produced is oxidised by HNO3 into sulphur. Dilute HCl is not used for the preparation of H2S because HCl is a vol­atile acid.

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

It is mixed with hydrogen sulphide gas in such a way, separation becomes difficult.

Procautions should be taken not to inhale the gas and not to allow its prolonged contact with skin, during handling H2S.

2. Other methods of preparation of H2S
From metallic sulphide : Different metallic sulphides react with dilute HCl or H2SO4 to form H2S.
Na2S + 2HCl = 2NaCl + H2S↑
CaS + H2SO4 = CaSO4 + H2S↑
Reaction of insoluble sulphide with dilute acid reacts with insoluble .sulphide As2S3 to form H2S.
As2S3 + 12H (Zn + dil H2SO4) = 12 AsH3 + 3H2S↑

Synthetic process : Hydrogen gas and sulphur vapour are mixed with each other at 450°C in the presence of Ni-dust to produce hydrogen sulphide.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 26

Physical properties of hydrogen sulphide :

(i) Smell : Hydrogen sulphide is a colourless gas with bad smell like rotten eggs.
(ii) Density : The density of H2S (vapour density = 17) is higher than that of air (vapour density of air = 14 4)
(iii) Solubility in water: Hydrogen sulphide is appreciably soluble in cold water, (4-37 volumes at 0°C, 3-40 volumes at 10°C and 2-6 volumes at 20°C) but practically insoluble in hot water. The solution of H2S in water is acidic and it turns blue litmus red.
(iv) It is poisonous when inhaled in small amounts, it causes headache and is fatal in large amounts.
(v) It liquefies at 212 2K and freezes at -190 K to a transparent solid.

Chemical properties of H2S
(i) Combustibility : It does not support combustion, but it burns a blue flame in excess of air or oxygen, giving water and sulphur dioxide; but sulphur is deposited, if the supply of oxygen is limited.
2H2S + 3O2 = 2SO2 + 2H2O
2H2S + O2 = 2Sl + 2H2O

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

(ii) Acidic properties: Hydrogen sulphide is a weak dibasic acid. Its aqueous solution is acidic to litmus. With sodium hydroxide (NaOH), it forms sodium sulphide (NagS) as the normal salt and sodium hydrogen sulplide (NaHS) as the acid salt.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 27

Reducing property :

(a) Reaction with Cl2 : When hydrogen sulphide is passed over chlorine water, chlorine is reduced to HC1, on the other hand H2S is oxidised to sulphur.
\(\stackrel{0}{\mathrm{C}} l_2+\mathrm{H}_2 \stackrel{-2}{\mathrm{~S}}=2 \mathrm{H} \stackrel{-1}{\mathrm{C}} 1+\stackrel{0}{\mathrm{~S}} \downarrow\)

(b) Reaction with SO2 : Hydrogen sulphide reduces sulphur dioxide to sulphur and itself oxidised to sulphur.
\(\stackrel{+4}{\mathrm{~S}} \mathrm{O}_2+2 \mathrm{H}_2 \stackrel{-2}{\mathrm{~S}}=3 \stackrel{0}{\mathrm{~S}}+2 \mathrm{H}_2 \mathrm{O}\)

(c) Reaction with HNO3 : H2S reduces cone. HNOs to brown nitrogen dioxide (NO2) and is itself oxidised to sulphur.
\(2 \stackrel{+5}{\mathrm{H}^{\mathrm{N}}} \mathrm{O}_3+\mathrm{H}_2 \stackrel{-2}{\mathrm{~S}}=2 \stackrel{+4}{\mathrm{NO}_2}+\stackrel{0}{\mathrm{~S}}+2 \mathrm{H}_2 \mathrm{O}\)

(d) Reaction with cone. H2SO4 H2S reduces cone. H2SO4 to SO2 and is itself oxidised to sulphur.
\(\mathrm{H}_2 \stackrel{+6}{\mathrm{~S}} \mathrm{O}_4+\mathrm{H}_2 \stackrel{-2}{\mathrm{~S}}=\stackrel{+4}{\mathrm{~S}} \mathrm{O}_2+\stackrel{0}{\mathrm{~S}}+2 \mathrm{H}_2 \mathrm{O}\)

(e) When H2S is passed through a yellow solution of ferric chloride (FeCl3), the salt is reduced to colourless ferrous chloride (FeCl2) and H2S itself on oxidation gives a precipitate of sulphur.
\(2 \stackrel{+3}{\mathrm{~F}} \mathrm{eCl}_3+\mathrm{H}_2 \stackrel{-2}{\mathrm{~S}}=2 \stackrel{+2}{\mathrm{FeCl}_2}+2 \cdot \mathrm{\textrm {HCl }}+\stackrel{0}{\mathrm{~S}} \downarrow\)

(f) H2S redues pink solution of potassium permanganate acidified with dilute sulphuric acid to a colourless solution.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 28

(g) H2S also reduces potassium dichromate solution acidified with dilute sulphuric acid-the orange-red colour of the solution turns green.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 29

(h) H2S reduces H2O2 to water and H2S is oxidised to sulphur.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 31

(iv) Reactions with metallic salts: Hydrogen sulphide reacts with different metallic salts giving rise different coloured metallic sulphides. The basic redicals in the salts are separated from different in colours and solubilities.

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

(a) The following precipitates ate obtained in acid medium
CuSO4 + H2S = H2SO4 + CuSl ↓ (Black)
Pb(NO3)2 + H2S = 2HNO3 + PbSl ↓(Black)
CdCl2 + H2S = 2HCl + CdS ↓ (Yellow)
SnCl2 + H2S = 2HCl + SnS ↓ (Brown)

(b) Certain sulphides are precipitated only in alkaline solution. In ammonical solution zinc salts give a white sulphide, iron salts give black sulphide.
ZnSO4 + (NH4)2S = (NH4)2SO4 + ZnS ↓ (White)
FeSO2 + (NHS = (NH4)2SO4 + FeS ↓ (Black)

Identification of Hydrogen sulphide:

Hydrogen sulphide can be identified  by the following tests.

(i) It is a colourless gas with foul smell like rotten egg.
(ii) A bright silver coin, when held in the gas, turns black.
2Ag+ H2S = Ag2S4- (black) + H2
(iii) Lead acetate paper turns black when it is held in H2S gas.
Pb(CH3COO)2 + H2S = PbS↓ (black) + 2CH3COOH
(iv) The gas is passed through a solution of sodium hydroxide solution and then sodium nitroprusside solution is added to it. The solution turns purple colour.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 32

Absorbent: An acidic gas H2S is absorbed by the caustic alkalis, NaOH and KOH; lead nitrate solution also absorbs the gas.
Pb (NO3)2 + H2S = 2HNO3 + PbSl (black)

Uses of H2S :

  • H2S is used as a reagent in the separation of metal ions in group analysis.
  • It is sometimes used as a reducing agent.
  • It is used for preparing metallic sulphides. Some sulphides are used as pigments.

Structure of hydrogen  Sulphate : It has bent structure like water and H-S-H bond angle is 92-2°. In H2S molecule S atom is sp3 hybridized.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 33

C. Nitrogen (N2)

Nitrogen was discovered by Danil Rutherford in 1772 Laviosier showed that nitrogen is an elementary gas present in air and it is not a supporter of combustion and respiration.

Atomic Number : 7; Symbol: N ; Molecular formula : H2; Atomic weight : 14008

Electronic configuration : \(\text { Is } 2 s^2 2 p_x^1 2 p_y^1 2 p_z^{11}\)
Postion in priodic Table : Period 2, Group VA ;
Oxidation number : -3 to + 5.

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Comparation of Nitrogen :

A concentrated aqueous solution of ammonium nitrite is heated to produce nitrogen gas. But this reaction is very fast so there is a chance of explosion. For this reason, the concentrated aqueous solution of sodium nitrite and ammonium are mixed molar ratio then if this mixture is heated nitrogen gas is produced.

Chemical  Requird : (i) Sodium nitrite (NaNO2)
(ii) Ammonium Chloride (NH4Cl)

Chemical Reaction  :
(i) NaNO2 + NH4Cl = NH4NH2 + NaCl

Collection : Though nitrogen gas is slightly soluble in water, it is collected by the download displacement of water.

Drying : The gas is dried by passing through a U-tube containing cone. H2SO4.

2. Other methods of preparation of nitrogen :
(a) From ammonium dichromate : Ammonium dichromate on gentle heating decomposes violently, evolving nitrogen.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 34

(b) From ammonia : N2 is also formed by slowly passing chlorine into concen­trated ammonia, when ammonium chloride (NH4Cl) and nitrogen (N2) are formed.
8NH3 + 3Cl2 = 6NH4Cl + N2

(c) From urea Nitrogen is obtained by reaction with alkaline hypobromite solution.
CO (NH2)2 + 3NaOBr = CO2↑+ 3NaBr + 2H2O + N2
From azide compounds: Spectroscopically pure nitrogen is obtained by heating barium azide at 300°C. The metal remains.

(d) From nitric acid : Moderately dilute nitric acid reacts with copper evolving nitric oxide, which when passed over the heated metal yields nitrogen.
Ba(N3)2 = Ba + 3N2

(e) From nitric acid : Moderately dilute nitric acid reacts with copper evolving nitric oxide, which when passed over the heated metal yields nitrogen.
3Cu + 8HNO3 = 3Cu (NO3)2 + 2NO + 4H2O
2NO + 2Cu = 2CuO + N2

(f) From air : By passing over red-hot copper filings which fix the oxygen as oxide of copper : 2Cu + O2 = 2CuO, and nitrogen passes out. It contains about 1 percent argon.

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Physical properties of Nitrogen :

  • Nitrogen is a colourless gas without any smell or taste.
  • Vapour density of nitrogen (14) is slightly less than that of air (14 4).
  • It is slightly soluble in water (23 5 ml of N2 is dissolved in 1 lit water at STP).

Chemical properties : It is an inert element at ordinary temperature because of very large of dissociation of the molecules, but it enters into combination with many elements at higher temperature.

(i) Reactions with non-metals:

(a) Reaction with hydrogen : At 550°C and 200 atmopsheric pressure in presence of iron catalyst reacts with hydrogen to produce ammonia. This is an industrial process of NH3 (Haber’s process).
N2 + 3H2 \(\rightleftharpoons\) 2NH3 + 22.4 Kcal

(b) Reaction with oxygen : Nitrogen combines with oxygen forming nitric oxide under the influence of electric arc at a temperature of 3000°C.
N2 + O2 \(\rightleftharpoons\) 2NO – 43.3 Kcal

(c) Covalent nitrides are obtained by reacting nitrogen with Boron and Silican at high temperature.
2B + N2 = 2BN ; 6Si +4N2= 2Si3N4

Nitrogen is absorbed by heated metals like Ca, Mg and Al to form nitrides which by Hydrolysis gives ammonia.
3Ca + N2 = Ca3N2 ; Ca3N2 + 6H2O= 3Ca (OH)2 + 2NH3
3Mg + N2 = Mg3N2 ; Mg3N2 + 6H2O = 3Mg(OH)2 + 2NH3
2Al + N2= 2AlN ; AlN + 3H2O = Al (OH)3 + NH3

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

(iii) Reaction with compounds : calcium carbide is heated in a current of nitrogen at temperature of 1100° C, calcium cyanamide and carbon are formed which is commercially known as nitrolim.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 35
Significance of the presence of nitrogen in air : About 78% by volume of nitrogen is present in the air but neither plant nor animal tissues can directly absorb nitrogen from air except a few leguminous plants, such as pea, bean clover etc. There are two ways of fixation of nitrogen as

(i) By electric discharge
(ii) Bio-chemical reaction through bacteria.

(i) Electric discharge due to thundering : During electric discharge in the atmosphere nitrogen and oxygen present in air combine to form nitric oxide. This nitric oxide is then oxidised by atmospheric oxygen to produce nitrogen dioxide. Later this oxide upon mixing with water vapour or rain water forms nitric acid which falls upon our earth. Nitric acid then reacts with the bases present in the soil forming nitrate salts.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 36

(ii) Fixation of nitrogen due to bacteria : Some micro organism and blue green algae convert nitrogen present in air to ammonia and nitric salts by chemical process.

Uses of nitrogen

  • Atmospheric nitrogen is fixed in large quantities as ammonia, nitric acid and nitrolim.
  • Liquid nitrogen is a refirgerant.
  • Nitrogen gives an inert atmosphere in certain metallurgical operations.
  • In making gas thermometers and for filling electric bulbs.
  • Nitrogen is used in the preparation of different explosives.

d. Hydrogen Chloride (Hydrochloric Acid, HCl), Nitric Acid (HNOs), Sulphuric Acid (H2SO4)

Hydrogen Chloride œ Hyrochioric Acid (HCl)

  • Molecular weight; 365; Formula : HCl (Murlatic Acid)
  • Prepared from sea-salt first; Prestly (1772)
  • Devy established that it is a compound of hydrogen and chlorine.

Laboratory method of preparation of hydrogen chloride : Hydrogen chloride is obtained by heating a mixture of common salt or sodium chloride and concentrated sulphuric acid.

Reaction occurs in two steps :
(a) When the mixture is heated at 150°C – 200°C then sodium bisulphate and hydrogen chloride gas are obtained.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 37
(b) When it is heated at 500°C then sodium sulphate and hydrogen chloride gas are produced.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 38

  • In the laboratory the reaction is performed at lower temperature because :
  • At high temperature the flask used may be cracked.
  • Sodium sulphate at higher temperature forms a hand crust and sticks to the glass. Its removal is very difficult.
  • Collection : As dry hydrogen chloride gas is heavier than air, it is collected by downward displacement of air.
  • Hydrogen chloride is highly soluble in water, so it is not collected over displacement of water.

Drying agent To remove water vapour, hydrogen chloride is passed over concentrated sulphuric acid.
P2O5 is not used for drying HCl gas because

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

(ii) CaO, NaOH, KOH are not used for drying HCl gas because :
CaO + 2HCl = CaCl2 + H2O
NaOH + HCl = NaCl + H2O
KOH + HCl = KCl + H2O

Nitric acid la not used to prepare HCl gas :

(a) Nitric acid is not used because during the preparation of HCl gas, nitric acid oxidises the produced HCl gas into Cl2 gas.
HNO3 + 3HCl = HOCl + 2Cl + 2H2O

(v) Reaction with silver nitrate : An aqueous solution of hydrochloric acid (or any soluble metallic chloride) gives a curdy white precipitate of silver chloride with silver nitrate solution—the precipitate is soluble in ammonia but insoluble in nitric acid.
HCl + AgNO3 = AgCl + HNO3
AgCl + 2NH3 = [Ag(NH3)2]Cl

(vi) Reaction with Nitric acid : The mixture of three volume of cone. HCl acid and one volume of cone. HNO3 acid is called aqua-regia. Noble metals like gold, platinum etc. are dissolved in aquaregia.
3HCl + HNO3 = NOCl + 2 [Cl] + 2H2O
Au + 3[Cl] = AuCl3 ; AuCl3 + HCl = HAuCl4 (soluble)
Pt + 4[Cl] = PtCl4 ; PtCl4 + 2HCl = H2PtCl6 (soluble)

Identification of hydrogen chloride and hydrochloric acid :

  • HCl gas is identified by its strong choking smell.
  • It fumes in moist air. Dense white fumes are produced when a glass-rod, moistened with strong ammonia solution, is held in the gas.
  • HCl gas turns a moist blue litmus paper red. Hydrochloric acid also turns blue litmus paper red.
  • It forms a white curdy precipitate of AgCl with colourless. AgNO3 solution. The curdy precipitate is soluble in ammonium hydroxide solution.

Uses of hydrochloric acid :

  • It is used in dyeing and calico-printing.
  • It is used in the manufacture of glucose, glue and many useful metal-chlorides. HCl is used as reagents in chemical laboratories.
  • It is used in preparing aqua-regia.
  • It is used for washing (pickling) iron sheets before galvanization and tinning. It is used for the preparation of chlorine and chlorides.

Nitric Acid (HNO3)

Molecular weight: 63 ; Formula : HNO3 (aqua forties)
Prepared by distilling KNO3 (nitre) with concentrated sulphuric acid : Glauber (1650)

Preparation :
Laboratory method of preparation : Nitric acid is prepared in the lab­oratory by heating a mixture of sodium nitrate (or potassium nitrate) and concentrated sulphuric acid (in 3 : 2 mole proportion)

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Reaction occurs in two steps :
When the reaction is kept at 200°C -300°C temperature, then sodium bisul­phate or potassium bisulphate and nitric acid are produced.
NaNO3 + H2SO4 = NaHSO4 + HNO3 KNO3 + H2SO4 = KHSO4 + HNO3

At 800°C, sodium sulphate and nitric acid are formed in the reaction between sodium nitrate or potassium nitrate and concentrated sulphuric acid.
2NaNO3 + H2SO4 = Na2SO4 + 2HNO3
2KNO3 + H2SO4 = K2SO4 + 2HNO3

In the laboratory the reaction is performed at lower temperature because :
(a) At high temperature of about 800°C, nitric acid is decomposed to nitrogen dioxide (NO2) and oxygen (O2).
4HNO3 = 4NO2 + O2 + 2H2O

(b) Both HNO3 and HCl are volatile in nature. So, they will be collected jointly in a receiver where HNO3 may oxidise HCl into Cl2. This difficulty is removed by using cone. H2SO4 (b.p. 338°C) non-volatile acid.

Pure HCl preparation: Pure hydrogen chloride is prepared by the action of water upon silicon tetrachloride:
SiCl4 + 4H2O = Si(OH)4 + 4HCl↑

Hydrochloric acid is not prepared by dissolving hydrogen chloride gas in water directly because : HCl gas is highly soluble in water and the rate of dissolving of HCl gas in water is much higher than the rate of formation of HCl gas as a result there is a vacuum in the flask. To fill up the vacuum water in the beaker enters into the flask and creates explosion when it contacts with H2SO4.

Properties :

Physical properties :

  • Colour ; HCl gas and hydrochloric acid are both colourless.
  • Odour : HCl gas has a strong choking odour and the odour of hydrochloric acid is less choking.
  • Solubility : It is highly soluble in water. At 0°C, 450 c.c. of HCl are dis­solved in 1c. of water.
  • Boiling Point : HCl gas is easily converted to colourless liquid by applying pressure at low temperature. The liquid HCl is of boiling point -84’5°C. Liquid hydrochloride is transformed crystals at -111-4°C.
  • Density : The density of cone, hydrochloric acid is 119 g/ml. The vapour density of HCI gas is 18.25.

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Chemical properties :

(i) Reaction with alkali Hydrochloric acid reacts with alkalis forming salt and water.
NaOH + HCl = NaCl + H2O
Ca(OH)2 + 2HCl = CaCl2 + 2H2O

(ii) Reactions with metals : Metals lying above hydrogen in the electrochemical series react with dilute HCI forming hydrogen gas and chlorides of metals.
Mg + 2HCl = MgCl2 + H2
Fe + 2HCl = MgCl2 + H2

Noble metals, such as — gold, platinum, etc. are not reacts with the acid. Copper slowly dissolves in hot and concentrated acid and silver is slowly reacts in the presence of air only.
2Cu + 4HCl + O2 = 2CuCl2 + 2H2O
4Ag + 4HCl + O2 = 4AgCl + 2H2O

(iii) The aqueous acid dissolves metallic oxides, hydroxides and carbonates. Hy­drogen chloride reacts with ammonia in presence of trace of moisture, forming dense fumes of ammonium chloride.
CuO + 2HCl = CuCl2 + H2O
NaOH + HCl = NaCl + H2O
NH3(g) + HCl(g) = NH4Cl(s)
CaCO3 + 2HCl = CaCl2 + H2O + CO2

(iv) Reaction with MnO2 : HCl is readily oxidised to chlorine by manganese dioxide or potassium permanganate.

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 40

  • At high temperature glass retort may be cracked.
  • At high temperature sodium sulphate (Na2O4) or potassium sulphate (KgSO) produced forms a hard emst and sticks to tine glass. It cannot be removed easily.

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Concentrated HCl is not used in the preparation of HNOa because:

  • It is because of the fact that HCl is more volatile than HNO3.
  • During heating with HCl will be collected in the receiver, as a result HNO3 will not be produced.
    3HCl + HNO3 = NOCl 4- 2[Cl] + 2H2O

Fumming nitric acid : When NO2 is dissolved in concentrated HNO3, it is then called fumming nitric acid. NO2 is evolved as brown fumes from the acid. Hence, it is called fumming nitric acid. It is a strong oxidising agent. Fumming nitric acid is prepared when cone, nitric acid is distilled with As2O3 (arsenious oxide) or starch.

Large scale production : (Ostwald process, 1914) :
The following steps are followed :WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 41

Properties :

Physical properties:

  • Colour : Pure nitric acid is a colourless liquid.
  • Odour : Nitric acid has a choking smell.
  • Solubility : It is highly soluble in water.
  • Density : Cone. HNO3 has a density of 142 g/ml
  • Boiling point : Pure nitric acid boils at 86°C and freezes to a white solid at -42°C. The specific gravity of pure acid is T52.
  • It fumes in air if it is kept opened.
  •  Concentrated nitric acid is corrosive to skin.

Chemical properties:

(i) Acidic character : Aqueous solution of nitric acid is ionised to great extent, hence it is strong acid. It is a monobasic acid.
\(\mathrm{HNO}_3 \rightleftharpoons \mathrm{H}^{+}+\mathrm{NO}_3^{-}\)
HNO3 turns blue litmus to red.

(ii) Reactions with alkali : It reacts with alkalis forming salt and water.
NaOH + HNO3 = NaNOs + H2O
Ca(OH)2 + 2HNO3 = Ca(NO3)2 + 2H2O

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

(iii) Reaction with metals : Nitric acid reacts all metals with the exception of gold and platinum forming different products. The actual product formed depends upon the following factors :

  • nature of the metal.
  • concentration of the acid.
  • temperature

(a) Reaction with magnesium : Strong nitric acid on reaction with magnesium forms magnesium nitrate and nitric oxide.
3Mg + 8HNO3 = 3Mg(NO3)2 + 2NO + 4H2O
Very dilute nitric acid on treatment with magnesium and manganese liberates hydrogen.
Mg + 2HNO3 = Mg (NO3)2 + H2

(b) Reaction with iron : Hot and cone. HNO3 makes metallic iron passive. Passive iron does not exhibit its normal chemical properties.

(c) Reaction with copper : Hot concentrated nitric acid reacts with copper metal forming copper nitrate and nitrogen dioxide.
Cu + 4HNO3 = Cu(NO3)2 + 2NO2 + 2H2O

(iv) Reaction with AgNOs solution and BaCl2 solution : Nitric acid does not react with AgNO3 solution and BaCl2

(v) Decomposition : When nitric acid is strongly heated, it forms nitrogen dioxide and oxygen
4HNO3 = 2H2O + 4NO2 + O2

(vi) Oxidising nature: Since nitric acid has a strong tendency to give nascent oxygen it therefore, acts as a poweful oxidising agent both in the concentrated and dilute solutions. Concentrated nitric acid is generally related to nitrogen dioxide while the dilute acid is reduced to nitric oxide.
2HNO3 (cone.) = 2NO2 + H2O + [O]
2HNO3 (dilute) = 2NO + H2O + 3 [O]

(a) Concentrated nitric acid oxidises copper turnings to copper nitrate and itself is reduced to brown coloured nitrogen dioxide gas.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 47

(b) Concentrated nitric acid oxidises charcoal i.e. carbon to carbondioxide and itself reduced to nitrogen dioxide.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 43

(vii) Aquaregia : A mixture to concentrated nitric acid (1 volume) and hydrochloric acid (3 volume) is called aquaregia; it dissolves gold and platinum.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 44
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 45
(viii) A mixture of concentrated nitric acid and sulphuric acid is used in the nitration of aromatic compounds.

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 48

Identification of Nitric acid :

  • Nitric acid is a colourless liquid with a choking smell.
  • On heating with copper turnings, nitric acid produces broun fumes to NO2 and copper turnings dissolve to form a blue solution of copper nitrate.
  • Ring test (or a nitrate): Equal volumes of freshly prepared ferrous sulphate solution and dilute HNO3 (dilute solution of a nitrate) are mixed together in a test tube and cooled. Cone. H2SO4 is now added carefully into the inner side of the test tube so as to form a heavy bottom layer; a brown ring is formed at the junction of the two liquids.

6FeSO4 + 2HNO3 + 3H2SO4 = 3Fe2(SO4)3 + 4H2O +NO
FeSO4 + NO = FeSO4.NO
(nitroso ferrous sulphate, brown in colour)

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Uses of nitric acid :

  • In the manufacture of explosives like T.N.T. (trinitrotoluene), nitroglycerine, picric acid etc.
  • As a laboratory reagent.
  • It is used in the manufacture of fertilizers, such as — calcium ammonium nitrate.
  • In the manufacture of sulphuric acid by chamber process.
  • In the manufacture of dyes, artifical silk and perfumes.
  • It is used in the purification of gold and silver.
  • It is used in making celluloid, rayon and other nitro cellulose products.

Pollution of air, water from Goldsmith’s workshop : Pure gold is of 24 carats. But in ornaments it is 22 carats or less. Gold ornaments are prepared by mixing requisite amount of copper with gold. When these gold ornaments are redesigned to any other form it requires breaking. Goldsmiths use nitric acid to purify gold ornaments.’ Fumes of nitric acid and brown NO2 are formed in the process. These gases pollute air in the locality.
Cu + 4HNO3 = Cu(NO3)2 + 2NO2 + 2H2O

Air pollution : Excess amount of NO2 is coming from Goldsmith’s workshop is inhaled. There is an oozing of blood from lungs. Photochemical oxidant produced from NO and NO2 have harmful effect on plant and animals.

Water pollution : Copper nitrate produced in Goldsmith’s workshop is poisonous. This compound is mixed in ponds through drains. As a result of this, fishes and plants like algae in water of this ponds may be destroyed to certain extent.

Remedy from the pollution : The works in a Goldsmith factory should Be carried out in a fume cupboard with effective device for the gaseous and particulate pollutants to escape high in the atmosphere.

Acid rain: The acidic gaseous oxides, such as – SO2, NO2 and CO2 in the atmosphere in the formation of acid rain :
S + O2 = SO2 ; 2SO2 + O2 = 2SO3; SO3 + H2O= H2SO4
NO + O3 = NO2 + O2 ; NO2 + O3 = NO3 + O2
NO3 + NO2 = N2O5 ; N2O5 + H2O= 2HNO3

Acid rain affect

  • Plantation and agriculture by washing out the soil nutrients.
  • It also damages the building materials of houses, historic monuments and sculptures and may even destroy aquatic lives like fish.

Sulphuric Acid (H2SO4)

  • Moleculars weight : 98; Formula : H2O4 (oil of vitriol)
  • It is called : king of chemicals

Preparation:
Contact process :
Chemicals required : Sulphur or iron pyrities or spent sulphide, excess air, platinised asbestos or V2O5 (vanadium pentoxide) and water.

Principle :
(i) Sulphur or iron pyrites are burnt in aim to form SO2.
S + O2 = SO2
4FeS2 + 11O2 = 2Fe2O3 + 8SO2

(ii) The process is named as contact process because conversion of SO2 to SO3 carried out in presence of porous catalyst having large contact surface.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 49

According to Le-chatlier’s principle, the conditions for the maximum yield of SO3 are :

  • Excess of O2 (SO2 : O2 ratio is 2 : 3)
  • High pressure (2 atm to avoid corrosion of plant)
  • Low temperature (450°C)
  • Presence of catalyst (finely divided platinum or V2Os)

(iii) SO3 formed is absorbed into concentrated H2SO4 when oleum or fuming sulphuric acid gets formed.
SO3 + H2SO4 = H2S2O7 (oleum)

(iv) Oleum formed is diluted with water to get sulphuric acid or any desired concentration.
H2S2O7 + H2O = 2H2SO4

Sulphuric acid is not prepared directly by adding water to SO3 because:
The gas forms a dense sulphuric fog or mist with water on account of the following highly exothermic reaction. H2O + SO3 = H2SO4 + 89.2 KJ

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Fuming sulphuric acid or oleum : Fuming sulphuric acid or oleum is obtained when sulphur trioxide is passed over 98% sulphuric acid.
H2SO4 + SO3 = H2S2O7 (Pyro sulphuric acid)
Sulphuric acid can further be obtained if requisite amount of water is added to oleum.

Properties :

  • Physical properties :
  • Colour : It is a colourless oily liquid.
  • Odour : It is an odourless liquid.
  • Solubility : It is soluble in water in any proportions.
  • Density : Cone. H2SO4 is a heavy liquid with a density 1-84 g/ml.
  • Boiling point : Cone. H2SO4 boils at 338°C under atmospheric pressure.
  • It produces severe burns on the skin.

Chemical properties :

(i) Dissociation : It is quite stable but on strong heating, it dissociates into SO3 and H2
\(\mathrm{H}_2 \mathrm{SO}_4 \rightleftharpoons \mathrm{H}_2 \mathrm{O}+\mathrm{SO}_3\)

(ii) Acidic character : In aqueous solutions, sulphuric acid turns blue litmus red indicating its acidic character. It inoises in two steps as :
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 50
Sulphuric acid is dibasic acid and reacts with metals, metaloxides and car­bonates etc. which are the characteristic reactions of an acid.

Reactions with alkalis :
NaOH + H2SO4 = NaHSO4 (acid salt) + H2O
2NaOH + H2SO4 = Na2SO4 (normal salt) + 2H2O

Reaction with metals : Metals lying above hydrogen in the electrochemical series react with dilute H2SO4 yielding hydrogen gas.
Mg + H2SO4 = MgSO4 + H ↑
Fe + H2SO4 = FeSO4 + H2
Hot concentrated sulphuric acid reacts with metal copper forming sulphur dioxide.
Cu + 2H2SO4 = CuSO4 + SO2 + 2H2O

(iii) Reactions with solutions of AgNOs and BaCl2 : It does not react with AgNO3 sulphuric acid gives a heavy white precipitation with BaCl2 solution. This precipitation is in soluble in any mineral acids.
H2SO4 + BaCl2 = BaSO4 (white precipitate) ↓ + 2HCl

(iv) Oxidising action Hot concentrated sulphuric acid is an oxidising agent since it decomposes to give atomic oxygen. H2SO4 = H2O + SO2 + [O]

(a) Concentrated sulphuric acid oxidises copper metal to copper sulphate and itself reduced to sulphur dioxide.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 51

(b) Hot and concentrated sulphuric acid oxidises carbon to carbon dioxide and itself reduced to sulphur dioxide.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 52

(v) Dehydration action of cone. H2SO4 H2SO4 is a strong dehydrating agent and desiccating agent due to its great affinity for water.

(a) It dehydrates the white crystals of cane sugar to black mass of carbon.
C12H22O11(sugar) + [H2SO4] = 12C + [11H2O + H2SO4]

(b) Concentrated sulphuric acid absorbs water molecule from formic acid yielding carbon monoxide.
HCOOH + [H2SO4] = CO + [H2O + H2SO4]

(c) Oxalic acid decomposes to produce carbon dioxide and carbon monoxide.
WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 53

(d) The blue crystals of CuSO4, 5H2O are dehydrated to white, anhydrous CuSO4.

Indentification of H2SO4

  • H2SO4 is a heavy oilly liquid. It chars sugar and paper.
  • When barium chloride solution is added to dilute solution of H2SO4, a heavy white precipitate is formed. The precipitate is insoluble in HCl, HNO3 and NH4OH solutions.
    BaCl2 + H2SO4 = BaSO4 ↓ + 2HCl

Uses of sulphuric acid :

  • It is used to prepare HCl3, HNO3, H3PO4 etc
  • It is used to prepare the fertilisers like ammonium sulphate, superphosphate of lime etc.
  • In dyes, drugs and explosive industries for the manufacture of paints, pig­ments, dyes, drugs, picric acid and explosives like TNT.
  • For refining petroleum.
  • It is used for a laboratory reagent and for drying of gases.
  • It is used for cleaning the surface of metals before carrying out electroplating,
  • Sulphuric acid is used for the manufacture of rayon, photographic films, rubber and synthetic detergents.

Pollution due to SO2 : SO2, obtained from different sources, such as — ex­haust gas from motor vehicle, petroleum refining plant, extraction of metals etc. mixes in our atmosphere and creates environmental pollution.

Effect:

  • SO2 produces problems in eyes and also in lungs.
  • Possibility of cancer.
  • Asthama, broncrities, allergy in our body.

Stone cancer : Due to the corrosive action of SO2, SO3, and H2SO4 on marble stone i.e. on calcium carbonate are the main cause of damaging effect of historical monuments like the Tajmahal.
CaCO3 (Marble stone) + H2SO4 = CaSO4 + CO2 + H2O

An insoluble layer of CaSO4 is formed over monuments. As a result once layer is formed sulphuric acid does not come in contact with marble stone. The reaction is stopped and there is no further decay of the monuments. But if this layer is removed by any means then decay continuous. It is called stone cancer.

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry

Possible remedy :

  • Banning of acid (sulphuric) factories by an ordinance.
  • Excessive uses of motor vehicles should be restricted.
  • Petroleum refining plant should be kept 90 km apart.

Distinction of HCl, HNO3 and H2SO4 :

WBBSE Class 10 Physical Science Notes Chapter 8.4 Inorganic Chemistry in the Laboratory and in Industry 54

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

Well structured WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat can serve as a valuable review tool before exams.

Heat Class 9 WBBSE MCQ Questions

Multiple Choice Questions :

Question 1.
What is the unit of heat in SI system ?
(i) calorie
(ii) Joule
(iii) erg
(iv) Newton
Answer:
Joule

Question 2.
Who designed absolute scale of temperature ?
(i) Newton
(ii) Lord Kelvin
(iii) Celsius
(iv) Boyle
Answer:
Lord Kelvin

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

Question 3.
Which substance has the highest specific heat ?
(i) water
(ii) iron
(iii) gold
(iv) ice
Answer:
Water

Question 4.
The fundamental interval of celsius scale is divided into –
(i) 100
(ii) 180
(iii) 120
(iv) 102 division
Answer:
100

Question 5.
The fundamental interval of Fahrenheit scale is divided into
(i) 180
(ii) 100
(iii) 32
(iv) 212 division
Answer:
180

Question 6.
If C and F represent a particular temperature in Celsius and Fahrenheit scales respectively, then their relation is :
(i) \(\frac{C}{4}\) = \(\frac{F-32}{9}\)
(ii) \(\frac{C}{5}\) = \(\frac{F-32}{9}\)
(iii) \(\frac{C}{3}\) = \(\frac{F-32}{5}\)
(iv) \(\frac{C}{9}\) = \(\frac{F-32}{5}\)
Answer:
\(\frac{C}{5}\) = \(\frac{F-32}{9}\)

Question 7.
The temperature of a body is measured by the instrument :
(i) hydrometer
(ii) thermometer
(iii) voltameter
(iv) ammeter
Answer:
thermometer

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

Question 8.
How much calorie of heat would be produced by converting 42 joule of work completely into heat ?
(i) 100 calorie
(ii) 10 calorie
(iii) 1000 calorie
(iv) 90 calorie
Answer:
10 calorie

Question 9.
Normal body temperature of a man is –
(i) 98.4° C
(ii) 98.4° F
(iii) 98.4 K
(iv) 100° F
Answer:
98.4° F

Question 10.
Which substance has the highest specific heat?
(i) ethanol
(ii) water
(iii) acetone
(iv) ether
Answer:
water

Question 11.
If the specific heat of copper be 0.09, what will be the water equivalent of a block of copper of mass 50 g ?
(i) 45 g
(ii) 4.5 g
(iii) 450 g
(iv) 44 g
Answer:
4.5 g

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

Question 12.
What is the melting point of ice in Fahrenheit scale ?
(i) 0° F
(ii) 32° F
(iii) 180° F
(iv) 121° F
Answer:
32° F

Question 13.
What is the specific heat of water in SI system ?
(i) 4 2 joule
(ii) 4200 joule/kg
(iii) 4200 joule/kg/K
(iv) 42 erg
Answer:
4200 joule / kg / K

Question 14.
The dimension of heat is –
(i) M L T
(ii) M L2 T-2
(iii) M L-1 T-2
(iv) M L-1 T-1
Answer:
ML2 T-2

Question 15.
Generally body temperature of a man is measured in :
(i) Celsius scale
(ii) Fahrenheit Scale
(iii) Kelvin scale
(iv) none of these
Answer:
Fahrenheit

Question 16.
Heat reaches the earth from the sun by the process of –
(i) conduction
(ii) convection
(iii) radiation
(iv) expansion
Answer:
Radiation

Question 17.
The value of mechanical equivalent of heat in CGS system is :
(i) 4.18 × 107 erg / calorie
(ii) 4.8 erg / calorie
(iii) 5.18 × 107 erg / calorie
(iv) 5.18 erg / calorie
Answer:
4.18 × 107 erg / calorie

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

Question 18.
How much work is to be done to produce 50 calorie of heat ?
(i) 210 joule
(ii) 200 joule
(iii) 201 joule
(iv) 310 joule
Answer:
210 joule

Question 19.
The water equivalent of a body is equal to :
(i) mass of the body x specific heat
(ii) mass of the body / specific heat
(iii) mass of the body + specific heat
(iv) mass of the body – specific heat
Answer:
Mass of the body × specific heat

Question 20.
Mercury remains in liquid state for long range of temperature :
(i) -39° C to 357° C
(ii) -57° C to 457° C
(iii) -10° C to 257° C
(iv) -12° C to 122° C
Answer:
-39° C to 357° C

Question 21.
-40° C is equal to :
(i) -50° F
(ii) -40° F
(iii) -60° F
(iv) -80° F
Answer:
-40° F

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

Question 22.
1 Calorie is equal to –
(i) 4.18 Joule
(ii) 5.18 Joule
(iii) 3.18 Joule
(iv) 4.18 erg
Answer:
4-18 Joule.

Question 23.
If a temperature in Kelvin scale is TK and the temperature in Celsius scale is t° C, the relation is :
(i) TK = (273 + t°C)
(ii) TK = (273 – t°C)
(iii) TK = (273 / t°C)
(iv) none of these
Answer:
T K = (273 + t°C)

Question 24.
Calorimetry relates to the measurement of –
(i) heat
(ii) temperature
(iii) mechanical energy
(iv) none of these
Answer:
heat

Question 25.
Quantity of heat of a body depends on its –
(i) temperature
(ii) mass
(iii) material of the body
(iv) all of them
Answer:
all of them

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

Question 26.
Quantity of heat is given by –
(i) \(\frac{m s}{t}\)
(ii) mst
(iii) \(\frac{m t}{s}\)
(iv) \(\frac{m s t}{w}\)
Answer:
mst

Question 27.
Work done W and heat produced H are related to each other as(where J is the mechanical equivalent of heat)
(i) WH = J
(ii) H = JW
(iii) W = J H
(iv) W J H = 1
Answer:
W = JH

Question 28.
The value of mechanical equivalent of heat (in erg/calorie) is –
(i) 4.2 × 10-7
(ii) 4.2 × 107
(iii) \(\frac{1}{4.2}\)
Answer:
4.2 × 107

Question 29.
The latent heat of fusion of ice is –
(i) 80 cal g-1
(ii) 3.36 × 105 cal g-1
(iii) 80 joule kg-1
(iv) 3.36 cal g-1
Answer:
80 cal g-1

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

Question 30.
Saturated vapours obeys –
(i) Boyle’s law
(ii) Charles’ law
(iii) pressure law
(iv) none of these
Answer:
none of these

Question 31.
Dews are formed at a temperature –
(i) greater than dew point
(ii) less than dew point
(iii) equal to dew point
(iv) at critical temperature.
Answer:
equal to dew point

Question 32.
The volume of water is minimum at a temperature –
(i) -4°C
(ii) 4°C
(iii) 0°C
(iv) 1°C
Answer:
4°C

Question 33.
Thermal capacity of a body of mass m and specific heat s is –
(i) \(\frac{\mathrm{m}}{\mathrm{s}}\)
(ii) \(\frac{\mathrm{m}}{\mathrm{s}}\)
(iii) \(\frac{\mathrm{m}}{\mathrm{s}}\)
(iv) ms
Answer:
ms

Fill in the blanks :

1. Heat is the cause and temperature is the _____.
Answer:
effect

2. Heat is a form of _____ a body possesses.
Answer:
energy

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

3. The range of temperature between the upper and lower fixed point is known as _____ interval.
Answer:
fundamental

4. Kelvin scale of temperature starts from the temperature corresponding to _____, taken as zero Kelvin.
Answer:
-273° C

5. Mean calorie is _____ of the heat necessary to raise the temperature of 1 gram pure water from 0° C} to 100° C.
Answer:
(\(\frac{1}{100}\)) th

6. 1 calorie = 4.18 or approximately 4.2 _____.
Answer:
joule

7. Specific heat of _____ in SI system is 4200 J/kg/K.
Answer:
water

8. Thermal capacity or heat capacity of a body is defined as the ratio of the heat supplied to a body to its consequent rise in _____.
Answer:
temperature.

9. In _____ process heat is transmitted in wave form.
Answer:
radiation

10. In radiation process _____ is not required.
Answer:
medium

11. Heat reaches the earth from the sun by _____ process.
Answer:
radiation

12. Normal body temperature of a man is _____ degree celsius.
Answer:
36.9

13. 94° F = _____ °C.
Answer:
34.4

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

14. The water equivalent of a body is = mass x _____.
Answer:
specific heat

15. A body becomes hot by _____ heat and becomes cold on giving away heat.
Answer:
taking

16. Temperature is the condition of a body.
Answer:
thermal

17. Melting point of _____ under normal atmospheric pressure is taken as the lower fixed point of a scale of temperature.
Answer:
ice

18. The quantity of heat required to raise the _____ of unit mass of a substance through one degree is called the specific heat of the substance.
Answer:
temperature

19. When work is completely converted into heat or heat is compietely converted into work, one is _____ to the other.
Answer:
equivalent

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

20. Work done to produce unit heat is known as _____ equivalent of heat.
Answer:
mechanical

21. Heat is a _____ quantity.
Answer:
scalar

22. In CGS system the unit of specific heat is _____
Answer:
Calorie/g/°C

23. The unit of specific heat in SI system is _____
Answer:
J/kg/K

24. -40°C = _____ °F
Answer:
-40

25. What is the dimension of heat ?
Answer:
The dimension of heat is ML2 T-2.

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

26. 1°C = ______ °F
Answer:
\(\frac{9}{5}\)

27. in CGS system heat is measured by the unit called _____.
Answer:
Calorie

28. Temperature of a body depends on the _____ contained in it.
Answer:
heat

29. In SI system the unit of thermal capacity is _____.
Answer:
JK-1

30. 1 kcal = _____ cal
Answer:
1000

31. In case of change of state, heat is _____ or liberated.
Answer:
absorbed

32. Latent heat of fusion of ice is _____ Jkg-1
Answer:
3.36 × 105

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

33. Specific heat of _____ is highest than any other common substance.
Answer:
Water

34. The temperature at which dews are formed called _____point.
Answer:
dew

35. The density of water at 4°C is _____.
Answer:
maximum

36. In CGS system, latent heat of vapourization of water is _____.
Answer:
540 cal/g/°C