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Balbharti Maharashtra State Board 11th Biology Important Questions Chapter 11 Study of Animal Type: Cockroach Important Questions and Answers.

Maharashtra State Board 11th Biology Important Questions Chapter 11 Study of Animal Type: Cockroach

Question 1.
Why are cockroaches said to be omnipresent?
Answer:
Cockroaches are said to be omnipresent as they are present everywhere, all over the world. They are usually seen in damp and moist places, crevices.

Question 2.
Cockroaches are nocturnal and cursorial. Give reason.
Answer:
1. Cockroaches are active at night hence, they are termed as nocturnal.
2. They are cursorial insects as they show terrestrial adaptations for running.

Question 3.
Name the common species of cockroach found in India.
Answer:
Periplaneta americana, Blatta orientalis and Blatta germanica.

Question 4.
Describe in detail the external morphology of cockroach.
Answer:
External morphology of cockroach:
1. Shape and size: Cockroach has elongated, bilaterally symmetrical, dorso-ventrally flattened and truly segmented body.They are triploblastic and eucoelomate. The body cavity called haemocoel is filled with the fluid haemolymph.
2. Coloration: Their colour is glistening brown or reddish brown.
3. Exoskeleton: Tough, waxy, non-living chitinous exoskeleton protects the body of the cockroach. It is made up of nitrogenous polysaccharide – chitin that provides strength, elasticity and surface area for attachment of muscles. Each body segment of cockroach is covered by four chitinous plates called sclerites namely, dorsal tergum, ventral sternum and two lateral pleurons.
4. Body division: The body is divided into three regions viz. head, thorax and abdomen.
5. Head: It is formed by the fusion of six segments. It is triangular or ovate in shape. The head is highly mobile due to flexible neck. It bears a pair of long antennae, a pair of compound eyes and mouthparts adapted for biting and chewing of food.

Head of cockroach bears following structures:
(a) Antennae: They are also called as feelers.These are long, filamentous, segmented structures that can move in all directions.They are lodged in membranous pits known as antennal sockets.They are tactile (touch) as well as olfactory (smell) organs.
Function: They are useful in locating the food material in the vicinity.
(b) Fenestrae: Fenestrae also called as ocellar spots.
They are situated at the base of each antenna and appear as white spots.
(c) Compound eyes: Compound eyes are paired, dark, kidney – shaped structures placed on the dorsolateral sides of the head. They are made up of large number of hexagonal ommatidia i.e. around 2000 ommatidia (sing, ommatidium).
These ommatidia are the structural and functional unit of compound eye, each forming an image of very small part of visual field. Collectively, the compound eye produces a mosaic image. Even though the compound eye gives a mosaic or hazy vision yet the animal can detect the slightest movement of the object. Compound eyes provide low resolution and more sensitive vision.

d. Mouthparts: Cockroach has a pre-oral cavity in front of the mouth in which food is received. It is bounded by mouthparts which are of chewing and biting type. This includes: labrum, labium, a pair of mandibles, a pair of maxillae and tongue like hypopharynx (present at the centre of mouth). Salivary duct opens at the base of hypopharynx and the mouth opens into foregut.

6. Thorax: Thorax is made up of three distinct segments – prothorax (anterior segment), mesothorax and metathorax (posterior segment). Ventrally, the thorax bears three pairs of walking legs, one at each segment. Dorsally, the thorax bears two pairs of wings attached to mesothoracic and metathoracic segment of the body.

(a) Legs: Three pairs of walking legs are present on the ventral side. Each leg is formed of five podomeres namely coxa, trochanter, femur, tibia and tarsus.Tarsus is the last segment and it is made up of five movable segments or tarsomeres. The last tarsomere bears a pair of claws and cushion-like arolium helpful in clinging.

(b) Wings: Forewings and hindwings form the two pair of wings present on the dorsal side. Forewings are the first pair of dark, opaque, thick, leathery wings. Hindwings are thin, broad, membranous delicate and transparent.They are attached to tergum of metathorax.
Functions: Forewings are protective in function and hindwings are helpful in flight, thus are called true wings.

7. Abdomen:
a. The abdomen is elongated and made up of ten segments. Each segment has a dorsal tergum and ventral sternum.
b. Tergum is jointed to the sternum laterally by a soft cuticle called pleura.
c. The posterior segments are telescoped in. Due to this, the eighth and ninth terga get overlapped by the seventh. The tenth tergum projects backward and is deeply notched. It bears a pair of small, many jointed anal cerci.
d. The abdomen is narrow and tapering in males as compared to females.The ninth sternum of males also bears a pair of short, unjointed anal style.

Question 5.
Write about the compound eyes of cockroach.
Answer:
Compound eyes: Compound eyes are paired, dark, kidney — shaped structures placed on the dorsolateral sides of the head. They are made up of large number of hexagonal ommatidia i.e. around 2000 ommatidia (sing, ommatidium). These ommatidia are the structural and functional unit of compound eye, each forming an image of very small part of visual field. Collectively, the compound eye produces a mosaic image. Even though the compound eye gives a mosaic or hazy vision yet the animal can detect the slightest movement of the object. Compound eyes provide low resolution and more sensitive vision.

Question 6.
Explain the structure of legs in cockroach.
Answer:
Legs: Three pairs of walking legs are present on the ventral side. Each leg is formed of five podomeres namely coxa, trochanter, femur, tibia and tarsus.Tarsus is the last segment and it is made up of five movable segments or tarsomeres. The last tarsomere bears a pair of claws and cushion-like arolium helpful in clinging.

Question 7.
Sketch a neat and labelled diagram of dorsal and ventral view of cockroach.
Answer:

Question 8.
Write a short on body cavity of cockroach.
Answer:
1. Cockroach has a body cavity or true coelom present around the viscera.
2. The body cavity of cockroach is known as haemocoel as it is filled with haemolymph (blood). Cockroaches have open type of circulation thus; the body cavity is filled with haemolymph.
3. The body cavity contains fat bodies. These are in the form of loose, whitish mass of tissue. They are made up of large, polygonal cells which contain fat globules, proteins and sometimes glycogen.

Question 9.
What is the role of hypopharynx?
Answer:
Hypopharynx: Hypopharynx is also known as lingua. It is a somewhat cylindrical single structure, located in front of the labium and between first maxillae. The salivary duct opens at the base of hypopharynx. Hypopharynx bears comb-like plates called super-lingua on either side. Hypopharynx is present at the centre of the mouth.
Function: It is useful in the process of feeding and mixing saliva with food,

Question 10.
Sketch a neat and labelled diagram of gizzard of cockroach.
Answer:

Question 11.
Describe the heart in cockroach.
Answer:
Heart: It is about 2.5 cm long, narrow, muscular tube that is open anteriorly and closed posteriorly. It starts from 9^th abdominal segment and extends anteriorly upto 1^st thoracic segment. Heart of cockroach is 13 chambered, out of which 10 chambers are in abdominal region and 3 chambers are in thoracic region. Each chamber has a pair of vertical slit-like incurrent aperture or opening called ostium (plural: ostia). Ostia are present along lateral side in the posterior region of first 12 chambers. Each ostium has lip-like valves that allow the flow of blood from sinus to heart only.

Question 12.
What are alary muscles?
Answer:
Dorsal diaphragm: It has 12 pairs (10 abdominal and 2 thoracic) of fan-like alary muscles. Alary muscles are triangular with pointed end attached to terga at lateral side and broad end lies between the heart and dorsal diaphragm.

Question 13.
Explain the mechanism of blood circulation in cockroach.
Answer:
Mechanism of blood circulation:

1. Blood (haemolymph) circulates between sinuses and heart due to contraction and relaxation of heart and alary muscles.
2. The heart contracts (systole) and relaxes (diastole) alternatively. After diastole, there is a third phase in the heart cycle known as diastasis. During diastasis, heart remains in expanded state.
3. During diastole, heart expands and the alary muscles contract, making the dorsal diaphragm flat. As a result, blood passes from perivisceral sinus to pericardial sinus through fenestrae and finally to the heart through ostia.
4. During systole, contraction starts at the posterior end and the wave of contraction passes anteriorly. Due to this, blood is pushed towards the dorsal aorta.
5. Ostia remain closed with the help of valves, during systole. As a result, blood flushes into head region from where it goes to perivisceral and perineural sinuses.
6. During systole, alary muscles are relaxed and due to this, the dorsal diaphragm becomes convex.
7. The volume of pericardial sinus is now reduced. This makes the blood to move from pericardial sinus to perivisceral sinus through fenestrae.

Question 14.
Which muscles are involved in renewal of air in tracheal system?
Answer:
The rhythmic movements of thoracic and abdominal muscles are involved in the renewal of air in tracheal system.

Question 15.
Give the role of chitin in trachea of cockroach.
Answer:
The inner lining of chitin in trachea prevents the trachea from collapsing.

Question 16.
Write in short about the spiracles in cockroach.
Answer:
Spiracles: They are paired respiratory openings. Spiracles are present on the ventro-lateral side of the body, in pleural membrane. Cockroaches have two pairs of thoracic and eight pairs of abdominal spiracles.The spiracles open into a series of air sacs from which arise the tubes called The spiracles let the air into and out of trachea.

Question 17.
Describe the excretory system in cockroach.
Answer:

1. Malpighian tubules are the main excretory organs of cockroach.
2. They are thin, yellow coloured, ectodermal thread-like structures that lie in the haemocoel.
3. These tubules are 150 in number. Malpighian tubules are attached to the alimentary canal between the midgut and hindgut.
4. Each Malpighian tubule is lined with a single layer of glandular epithelial cells having microvilli. The distal portion of Malpighian tubule is secretory and the proximal part is absorptive in function.
5. They extract water and nitrogenous wastes from the haemocoel and convert them into uric acid and pass them into ileum. As the cockroach excretes uric acid, it is said to be uricotelic.
6. Also, fat bodies, nephrocytes and uricose glands (only in males) help in excretion.
7. In cockroach, nephrocytes (urate cells) associated with fat bodies and cuticle are also believed to be excretory in function. The nephrocytes are cells present along with the fat bodies or present along the heart and store nitrogenous wastes.
8. The excretory products later are removed in the haemocoel. Some nitrogenous wastes are deposited on the cuticle and eliminated during moulting.

Question 18.
Sketch and label the central nervous system of cockroach.
Answer:

Question 19.
Sense organs: Collect the information and complete the chart.
Answer:

Sense organ	Location	Function
1. Antenna	Head	Detect touch, smell and locate the food in vicinity
2. Eyes	Head	Provides mosaic vision, detect slightest movement of object, provide more sensitive vision but less resolution
3. Maxillary palp	Mouth	Detects smell and taste.
4. Labial palp	Mouth	Detects smell and taste.
5. Anal cerci	Abdomen	Detect touch and sound (respond to air or vibrations)

Question 20.
Describe the male reproductive system of cockroach.
Answer:

1. Male reproductive system includes primary and secondary reproductive organs.
2. Primary sex organs (male gonads) are called testes. They are paired and located in the 4th and 6th abdominal segments. Sperms produced in testis are carried by vasa deferentia.
3. Vasa deferentia is a pair of thin tubular structure that arise from the testes and open into ejaculatory duct through seminal vesicle. They carry sperms to ejaculatory duct.
4. Ejaculatory duct opens into male gonopore situated ventral to anus.
5. Sperms produced by testis are stored in seminal vesicles in the form of bundles called spermatophores. These spermatophores are deposited in female reproductive tract during copulation.
6. Mushroom shaped gland or utricular gland is an accessory reproductive gland. It is present in the 6th – 7th abdominal segments.
7. Male gonapophyses or phallomere form the external genitalia of male. These are three asymmetrical chitinous structures surrounding the male gonophore.

Question 21.
Name the following:

Question 1.
The other term for gizzard in cockroach.
Answer:
Proventriculus

Question 2.
After sufficient growth, nymph undergoes moulting and enters into this stage:
Answer:
Instar

Question 3.
Paired accessory sex glands present in female cockroach that open in genital chamber.
Answer:
Collaterial glands.

Question 4.
The structural and functional unit of eye of cockroach.
Answer:
Ommatidia

Question 5.
Body segment of cockroach is covered by these four chitinous plates.
Answer:
Dorsal tergum, ventral sternum and two lateral pleurons.

Question 22.
Explain the process of fertilization in cockroach.
Answer:

1. The process of fertilization in cockroach is internal.
2. Male and female cockroaches come together by their posterior phallomeres.
3. The spermatophores are transferred to the genital chamber of female cockroach.
4. Sperms released from the spermatophore reach the spermatheca.
5. The eggs are discharged from both the ovaries alternately into the common oviduct and pass into the genital chamber.
6. Sperms coming from the spermatheca fertilize the eggs in the genital chamber.

Question 23.
Name the gland whose secretions form ootheca or egg case.
Answer:
The secretion of collaterial glands forms a capsule around them is called as ootheca or egg case.

Question 24.
Describe the stages of development in cockroach.
Answer:

1. The development in cockroach (Periplaneta americana) is paurametabolous as the development occurs through nymphal stage.
   Fertilized egg → Nymph → Adult
2. The nymph looks like adult but it is smaller and sexually immature.
3. After sufficient growth, nymph undergoes moulting and enters into a stage between two successive moults known as instar.
4. Cockroaches may undergo moulting for around 13 times before reaching the adult stage.
5. The nymphal stages have wing pads but only adult cockroaches have wings.
6. The embryonic period in cockroach varies as per temperature and humidity. At 24°C, the duration is about 58 days and at 30° C, the duration is about 32 days.

Question 25.
Fill in the blanks.

1. The head of cockroach is formed by fusion of ________ segments.
2. Hindwings are attached to the tergum of ________.
3. _________ pairs of walking legs are present on ventral side of cockroach.
4. Laterally, the tergum in cockroach is jointed to sternum by soft cuticle called ________.
5. Tongue-like single structure present in front of the labium between first maxillae is called as ______.

Answer:

1. 6
2. Metathorax
3. 3
4. Pleura
5. Hypopharynx /lingua

Question 26.
Why are cockroaches considered as pests?
Answer:
Cockroaches are considered pests due to following reasons:
1. Cockroaches damage household materials like clothes, shoes, paper, etc.
2. Cockroaches eat and destroy the foodstuff. They contaminate food, which gives a typical smell to food and make it unpalatable.
3. They cany pathogens of diseases like cholera, diarrhoea, tuberculosis, typhoid, etc.

Question 27.
Cockroaches are considered as a part of food chain. Justify.
Answer:
Many amphibians, birds, lizards and rodents prey upon cockroaches and this makes them a part of food chain. They are also eaten by certain groups of people in South America, China and Myanmar.

Question 28.
Lata was surprised to see cockroaches used as specimen in her college laboratory. She asked her teacher the reason for cockroach being used as experimental animal in laboratory. What would be the probable reason given by her teacher?
Answer:
Cockroach are used in laboratories as experimental animal for biological research as they can be obtained easily without causing damage to ecological balance. Cockroaches are commonly used as experimental animal in laboratory because of their large size, omnivorous food habit, hardiness (chitinous exoskeleton), rapid growth and reproduction. Also they are available almost any season, at any locality.

Question 29.
Give the measures to control the population of cockroach.
Answer:
Cockroaches are economically harmful organism. They must be controlled in an efficient way.
Following are the measures to control the population of cockroach:
1. Maintain good sanitation: Dark and humid places of kitchen, cupboards, trolleys must be cleaned regularly. Cracks and crevices and other such areas must be filled. Accumulation of garbage at home should be avoided.
2. Keeping water in drainage trap: If the drain trap is dry, cockroaches frequently enter home by migrating up from sewer connections. So, we should always keep the drain trap filled with water.
3. Chemical control: Organophosphates, carbamates, pyrethroids and boric acid are efficient poisons of cockroaches. Various types of their formulations are available in market under various brand names.

Question 30.
Depending upon nature of food and feeding habits, different insects have different types of mouthparts. Collect images of different mouthparts and paste in appropriate boxes.
Answer:

Question 31.
Apply Your Knowledge

Question 1.
The cockroach was dissected so as to expose the digestive system. Student found tubules at two places in the digestive system:
1. Around the anterior part of stomach
2. At the junction of mid-gut and hind-gut Identify the tubules and give their functions.
Answer:
1. Hepatic caeca
2. Malpighian tubules
For functions: Hepatic caeca: These are thin, transparent, short, blind (closed) and hollow tubules.
Function: They secrete digestive enzymes.
For functions:

1. Malpighian tubules are the main excretory organs of cockroach.
2. They are thin, yellow coloured, ectodermal thread-like structures that lie in the
3. These tubules are 150 in number. Malpighian tubules are attached to the alimentary canal between the midgut and hindgut.
4. Each Malpighian tubule is lined with a single layer of glandular epithelial cells having microvilli. The distal portion of Malpighian tubule is secretory and the proximal part is absorptive in function.
5. They extract water and nitrogenous wastes from the haemocoel and convert them into uric acid and pass them into ileum. As the cockroach excretes uric acid, it is said to be uricotelic.
6. Also, fat bodies, nephrocytes and uricose glands (only in males) help in excretion.
7. In cockroach, nephrocytes (urate cells) associated with fat bodies and cuticle are also believed to be excretory in function. The nephrocytes are cells present along with the fat bodies or present along the heart and store nitrogenous wastes.
8. The excretory products later are removed in the haemocoel. Some nitrogenous wastes are deposited on the cuticle and eliminated during moulting.

Question 2.
Rita’s mother while cleaning the house spotted a darkish reddish to blackish brown coloured capsule glued on the crack. Her mother showed it to Rita and asked her about it. Rita recollected that it resembles to a picture shown by her teacher in classroom while teaching about cockroach. What it must be?
Answer:

1. The darkish reddish to blackish brown coloured capsule may be ootheca.
2. The secretion of collaterial glands forms a capsule around them is called as ootheca or egg case.
3. It is about 8 mm long and ranges from dark reddish to blackish brown.
4. Ootheca contains 14 to 16 fertilized eggs in two rows.
5. They are dropped or glued to a suitable surface, like a crack or crevice with good humidity near a food source.
6. A female cockroach on an average, produces 9 to 10 oothecae during its lifespan.

Question 32.
Quick Review

Question 33.
Exercise

Question 1.
Which species of cockroach are found in India?
Answer:
Periplaneta americana, Blatta orientalis and Blatta germanica.

Question 2.
Cockroaches are said to be omnipresent. Justify
Answer:
Cockroaches are said to be omnipresent as they are present everywhere, all over the world. They are usually seen in damp and moist places, crevices.

Question 3.
Classify cockroach giving reasons for its systematic position.
Answer:

Classification		Reasons
Kingdom	Animalia	Cell wall absent, heterotrophic nutrition.
Phylum	Arthropoda	They have jointed appendages. Body is chitinous and segmented.
Class	Insecta	They possess two pairs of wings and three pairs of walking legs.
Genus	Periplaneta	Straight wings and nocturnal.
Species	americana	Originated in the continent of America.

Question 4.
Describe in detail the body division of cockroach.
Answer:
Body division: The body is divided into three regions viz. head, thorax and abdomen.
Abdomen:

1. The abdomen is elongated and made up of ten segments. Each segment has a dorsal tergum and ventral sternum.
2. Tergum is jointed to the sternum laterally by a soft cuticle called pleura.
3. The posterior segments are telescoped in. Due to this, the eighth and ninth terga get overlapped by the seventh. The tenth tergum projects backward and is deeply notched. It bears a pair of small, many jointed anal cerci.
4. The abdomen is narrow and tapering in males as compared to females.The ninth sternum of males also bears a pair of short, unjointed anal style.

Question 5.
Name the last segment in the leg of cockroach.
Answer:
Legs: Three pairs of walking legs are present on the ventral side. Each leg is formed of five podomeres namely coxa, trochanter, femur, tibia and tarsus.Tarsus is the last segment and it is made up of five movable segments or tarsomeres. The last tarsomere bears a pair of claws and cushion-like arolium helpful in clinging.

Question 6.
What is arolium in cockroach?
Answer:
Legs: Three pairs of walking legs are present on the ventral side. Each leg is formed of five podomeres namely coxa, trochanter, femur, tibia and tarsus.Tarsus is the last segment and it is made up of five movable segments or tarsomeres. The last tarsomere bears a pair of claws and cushion-like arolium helpful in clinging.

Question 7.
What are ommatidia?
Answer:
Compound eyes: Compound eyes are paired, dark, kidney — shaped structures placed on the dorsolateral sides of the head. They are made up of large number of hexagonal ommatidia i.e. around 2000 ommatidia (sing, ommatidium). These ommatidia are the structural and functional unit of compound eye, each forming an image of very small part of visual field. Collectively, the compound eye produces a mosaic image. Even though the compound eye gives a mosaic or hazy vision yet the animal can detect the slightest movement of the object. Compound eyes provide low resolution and more sensitive vision.

Question 8.
What is the function of labium?
Answer:
Each gland has a salivary duct. Both the ducts unite to form a common salivary duct.

Question 9.
What are the antennae of cockroach also known as?
Answer:
Antennae: They are also called as feelers.These are long, filamentous, segmented structures that can move in all directions.They are lodged in membranous pits known as antennal sockets.They are tactile (touch) as well as olfactory (smell) organs.
Function: They are useful in locating the food material in the vicinity.

Question 10.
Write a short note on exoskeleton of cockroach.
Answer:
Exoskeleton: Tough, waxy, non-living chitinous exoskeleton protects the body of the cockroach. It is made up of nitrogenous polysaccharide – chitin that provides strength, elasticity and surface area for attachment of muscles. Each body segment of cockroach is covered by four chitinous plates called sclerites namely, dorsal tergum, ventral sternum and two lateral pleurons.

Question 11.
Describe the mouthparts of cockroach briefly.
Answer:

1. Hepatic caeca: These are thin, transparent, short, blind (closed) and hollow tubules.
2. Function: They secrete digestive enzymes.
3. Hindgut or proctodaeum: It consists of ileum, colon and rectum.
4. Ileum: It is short and narrow part of hindgut. Malpighian tubules open in the anterior lumen of ileum, near the junction of midgut and hindgut. Posterior region of ileum contains sphincter. Ileum directs the nitrogenous wastes and undigested food towards colon.
5. Colon: It is a longer and wider part of the hindgut. It directs waste material towards the rectum. It reabsorbs water from wastes as per the need.
6. Rectum: It is oval or spindle-shaped, terminal part of the hindgut. It contains six rectal pads along the internal surface for absorption of water. Rectum opens into anus. Anus is present on the ventral side of the 10^th segment. It is the last or posterior opening of the digestive system. The undigested food is released out of the body through anus.

Question 12.
Give another name for upper lip and lower lip of cockroach.
Answer:

1. Cockroach has a pair of salivary glands which secrete saliva.
2. Each gland has a salivary duct.
3. Both the ducts unite to form a common salivary duct.

Question 13.
Mention the three segments of thorax.
Answer:
Thorax: Thorax is made up of three distinct segments – prothorax (anterior segment), mesothorax and metathorax (posterior segment). Ventrally, the thorax bears three pairs of walking legs, one at each segment. Dorsally, the thorax bears two pairs of wings attached to mesothoracic and metathoracic segment of the body, a. Legs: Three pairs of walking legs are present on the ventral side. Each leg is formed of five podomeres namely coxa, trochanter, femur, tibia and tarsus.Tarsus is the last segment and it is made up of five movable segments or tarsomeres. The last tarsomere bears a pair of claws and cushion-like arolium helpful in clinging.

Question 14.
Which wings are known as true wings? Why?
Answer:
Wings: Forewings and hindwings form the two pair of wings present on the dorsal side. Forewings are the first pair of dark, opaque, thick, leathery wings. Hindwings are thin, broad, membranous delicate and transparent.They are attached to tergum of metathorax.
Functions: Forewings are protective in function and hindwings are helpful in flight, thus are called true wings.

Question 15.
Write a short note on haemocoel.
Answer:

1. Cockroach has a body cavity or true coelom present around the viscera.
2. The body cavity of cockroach is known as haemocoel as it is filled with haemolymph (blood). Cockroaches have open type of circulation thus; the body cavity is filled with haemolymph.
3. The body cavity contains fat bodies. These are in the form of loose, whitish mass of tissue. They are made up of large, polygonal cells which contain fat globules, proteins and sometimes glycogen.

Question 17.
Which region of the alimentary canal is also known as stomodaeum?
Answer:

1. Foregut or stomodaeum: It consists of pharynx, oesophagus, crop and gizzard.
2. Pharynx: It is very short, narrow but muscular tube that opens into oesophagus.
3. Function: Conduction of food into the oesophagus.
4. Oesophagus: It is slightly long and narrow tube which opens into crop.
5. Crop: Crop is a large, pear shaped and sac- like organ.
6. Function: It temporarily stores the food and then sends it to gizzard.
7. Gizzard: Gizzard or proventricuius is a small spherical organ. It is provided internally with a circlet of six chitinous teeth and backwardly directed bristles.The foregut ends with gizzard.

Function: The chitinous teeth present in gizzard are responsible for crushing the food and the bristles help to filter the food.

Question 18.
Explain in detail the hindgut of cockroach.
Answer:
Hindgut or proctodaeum: It consists of ileum, colon and rectum.

Question 19.
Write a short note on mesentron.
Answer:

1. Midgut or mesenteron: It consists of stomach and hepatic caeca.
2. Ventriculus or stomach: It is straight, short and narrow. Stomach is lined by glandular epithelium which secretes digestive enzymes.
   
3. Function: It is mainly responsible for digestion and absorption.

Question 20.
Explain the three parts of alimentary canal.
Answer:
1. Digestive system of cockroach consists of mouthparts, alimentary canal and salivary glands.
2. Mouthparts: Pre-oral cavity present in front of the mouth receives food. It is bounded by chewing and biting type of mouth parts.
3. These are movable, segmented appendages that help in ingestion of food. The mouthparts of cockroach comprises of:
a. Labrum: It forms the upper lip. It is a single flap-like movable part which covers the mouth from upper side. It forms an anterior wall of pre¬oral cavity.
Function: It is useful in holding the food during feeding.
b. Mandibles: These are two dark, hard, chitinous structures with serrated median margins.They are true jaws present on either side, behind the labrum.
Function: They perform co-ordinated side-wise movements with the help of adductor and abductor muscles to cut and crush the food.
c. Maxillae: These are the accesssory jaws. They are also called as first pair of maxillae. These are situated on the either side of mouth behind the mandibles. Each maxilla consists of sclerites like cardo, stipes, galea, lacinia and maxillary palps.
Functions: Maxillae hold food, help mandibles for mastication. They are also used for cleaning the antennae and front legs. Maxillary palps act as tactile organs.
d. Labium: It forms the lower lip. Labium is also known as second maxilla which covers the pre-oral cavity from the ventral side. It is firmly attached to the posterior part of head. It has three jointed labial palps which are sensory in function.
Function: It is useful in pushing the chewed food in the pre-oral cavity. It prevents the loss of food falling from the mandibles, while chewing.
e. Hypopharynx: Hypopharynx is also known as lingua. It is a somewhat cylindrical single structure, located in front of the labium and between first maxillae. The salivary duct opens at the base of hypopharynx. Hypopharynx bears comb-like plates called super-lingua on either side. Hypopharynx is present at the centre of the mouth.
Function: It is useful in the process of feeding and mixing saliva with food, iii. Alimentary canal: It is long a (6 – 7cm) tube of different diameters with two openings.

Question 21.
Explain the three sinuses in the coelom cockroach.
Answer:
Sinuses: The coelom of cockroach is divided into three sinuses – pericardial sinus, perivisceral sinus and perineural sinus.

1. Pericardial sinus: It is dorsal, very small and contains dorsal vessel.
2. Perivisceral sinus: It is middle and largest sinus. It contains fat bodies and almost all major visceral organs of alimentary canal and reproductive system.
3. Perineural sinus: It is ventral, small and contains ventral nerve cord. It is continuous into legs. All the three sinuses communicate with each other through the pores present between two successive points of attachments of diaphragms.

Question 22.
Define: alary muscles.
Answer:
Dorsal diaphragm: It has 12 pairs (10 abdominal and 2 thoracic) of fan-like alary muscles. Alary muscles are triangular with pointed end attached to terga at lateral side and broad end lies between the heart and dorsal diaphragm.

Question 24.
Write a short note on haemolymph.
Answer:
Haemolymph: Haemolymph is colourless as it is without any pigment. It consists of plasma and seven types of blood cells/haemocytes. Plasma consists of water with some dissolved organic and inorganic solutes. It is rich in nutrients and nitrogenous wastes like uric acid.

Question 26.
What are spiracles?
Answer:
Spiracles: They are paired respiratory openings. Spiracles are present on the ventro-lateral side of the body, in pleural membrane. Cockroaches have two pairs of thoracic and eight pairs of abdominal spiracles.The spiracles open into a series of air sacs from which arise the tubes called trachea. The spiracles let the air into and out of trachea.

Question 27.
With the help of neat and labelled diagram, explain the tracheal system of cockroach.
Answer:
1. Cockroach has an internal respiratory system of air tubes called tracheal system by which the air is brought into the body and is in contact with every part of the body. It allows the exchange of gases directly between the air and tissues without the need of blood.
These air tubes of internal respiratory system begin at the opening on body surface called spiracles.

2. Spiracles: They are paired respiratory openings. Spiracles are present on the ventro-lateral side of the body, in pleural membrane. Cockroaches have two pairs of thoracic and eight pairs of abdominal spiracles.The spiracles open into a series of air sacs from which arise the tubes called trachea. The spiracles let the air into and out of trachea.

3. Trachea: The trachea form a definite pattern of branching tubes arranged transversely as well as longitudinally. They are about 1mm thick and have spiral or annular thickening of chitin. The inner lining of chitin prevents the trachea from collapsing. Each trachea further branches into smaller tubes called tracheoles.

4. Tracheoles: These are fine intracellular tubes that penetrate deep into tissues. They are thin and not lined by chitin. They end blindly in the cells. Each tracheole at the blind end is filled with a watery fluid through which exchange of gases takes place. The content of this fluid keeps changing. At high muscular activity, part of fluid part is drawn into the tissues to enable more and rapid oxygen intake.

Question 28.
Write a short note on Malpighian tubules.
Answer:

1. Malpighian tubules are the main excretory organs of cockroach.
2. They are thin, yellow coloured, ectodermal thread-like structures that lie in the
3. These tubules are 150 in number. Malpighian tubules are attached to the alimentary canal between the midgut and hindgut.
4. Each Malpighian tubule is lined with a single layer of glandular epithelial cells having microvilli. The distal portion of Malpighian tubule is secretory and the proximal part is absorptive in function.
5. They extract water and nitrogenous wastes from the haemocoel and convert them into uric acid and pass them into ileum. As the cockroach excretes uric acid, it is said to be uricotelic.
6. Also, fat bodies, nephrocytes and uricose glands (only in males) help in excretion.
7. In cockroach, nephrocytes (urate cells) associated with fat bodies and cuticle are also believed to be excretory in function. The nephrocytes are cells present along with the fat bodies or present along the heart and store nitrogenous wastes.
8. The excretory products later are removed in the haemocoel. Some nitrogenous wastes are deposited on the cuticle and eliminated during moulting.

Question 29.
Write a short note on peripheral nervous system of cockroach.
Answer:
Peripheral nervous system (PNS):

1. The peripheral nervous system comprises of nerves that arise from various ganglia of CNS.
2. Six pairs of nerves arise from the supra-oesophageal ganglia.They supply to the eyes, antenna and labrum.
3. Nerves arising from the sub-oesophageal ganglion supply to the mandibles, maxillae and labium.
4. Nerves arising from the thoracic ganglia supply to the wings, legs and internal thoracic organs.
5. Nerves from abdominal ganglia go to the abdominal organs of respective abdominal segments.

Question 30.
Explain in detail the central nervous system of cockroach.
Answer:
Central nervous system (CNS): Central nervous system consists of nerve ring and ventral nerve cord.
Nerve ring consists of:

1. a pair of supra-oesophageal ganglia
2. a pair of circum-oesophageal connectives
3. a pair of sub-oesophageal ganglia
4. Supra-oesophageal ganglia or cerebral ganglia: A pair of supra-oesophageal ganglia is collectively known as the brain. Brain is present in head, above the oesophagus and between antennal bases. Each supra-oesophageal ganglion is formed by the fusion of three small ganglia – protocerebram, deutocerebrum and tritocerebrum.
5. Circum-oesophageal connectives: Supra-oesophageal ganglia are connected to sub-oesophageal ganglion by a pair of lateral nerves called as circum-oesophageal connectives. Connectives arise from supra-oesophagial ganglia.
6. Sub-oesophageal ganglia: It is a bilobed and present below the oesophagus, in head. It is also formed by the fusion of three pairs of ganglia.
7. Ventral nerve cord:
8. It arises from the sub-oesophageal ganglion. It is present along mid-ventral position, in perineural sinus.
9. It is double ventral nerve cord and consists of nine segmental, paired ganglia.
10. First three pairs of segmental ganglia are large and known as thoracic ganglia. The other six pairs of segmental ganglia are in abdomen (abdominal ganglia).
11. 6^th abdominal ganglion is the largest and it is present in 7^th abdominal segment.

Question 31.
Enlist the four ganglia of autonomous nervous system of cockroach.
Answer:
Autonomic nervous system (ANS): It consists of four ganglia and a retrocerebral complex.
The ganglia are as follows:

1. Frontal ganglion: It is present above the pharynx and in front of brain.
2. Hypocerebral ganglion: It is present on the anterior region of oesophagus.
3. Ingluvial ganglion: It is present on crop. It is also called as visceral ganglion.
4. Ventricular ganglion: It is present on gizzard.

Question 32.
Explain the structures and functions of different parts involved in female reproductive system of cockroach.
Answer:

1. Female reproductive system of cockroach consists a pair of ovaries, a pair of oviducts, vagina, spermatheca and accessory glands.
2. Ovaries are primary reproductive organs. They are paired and lie lateral in position in 2nd – 6lh abdominal segments. Each ovary is formed of a group of 8 ovarian tubules or ovarioles, containing a chain of developing ova. All ovarioles of an ovary open in lateral oviduct of respective side.
3. The lateral oviducts unite to form a common oviduct or vagina. Common oviduct or vagina opens into the Bursa copulatrix (genital chamber), the female organ of copulation.
4. Spermatheca, is a sperm storing structure present in the 6th segment opens into genital chamber. It receives the sperms during copulation and store them for fertilization.
5. Collaterial glands are accessory paired glands that open in genital chamber.
6. Female gonapophyses consists of six chitinous plates surrounding the genital pore. In males, genital pouch or genital chamber lies at the hind end of abdomen which is bounded dorsally by 9th and 10th terga and ventrally b; male genital pore and gonapophysis.

Question 33.
Explain in detail male reproductive system of cockroach.
Answer:

1. Male reproductive system includes primary and secondary reproductive organs.
2. Primary sex organs (male gonads) are called testes. They are paired and located in the 4th and 6th abdominal segments. Sperms produced in testis are carried by vasa deferentia.
3. Vasa deferentia is a pair of thin tubular structure that arise from the testes and open into ejaculatory duct through seminal vesicle. They carry sperms to ejaculatory duct.
4. Ejaculatory duct opens into male gonopore situated ventral to anus.
5. Sperms produced by testis are stored in seminal vesicles in the form of bundles called spermatophores. These spermatophores are deposited in female reproductive tract during copulation.
6. Mushroom shaped gland or utricular gland is an accessory reproductive gland. It is present in the 6th – 7th abdominal segments.
7. Male gonapophyses or phallomere form the external genitalia of male. These are three asymmetrical chitinous structures surrounding the male gonophore.

Question 34.
Multiple Choice Questions

Question 1.
Cockroach shows ________ adaptations.
(A) Cursorial
(B) Arboreal
(C) Fossorial
(D) Aquatic
Answer:
(A) Cursorial

Question 2.
Cockroach is a/an animal.
(A) omnivorous
(B) nocturnal
(C) cursorial
(D) all of these
Answer:
(D) all of these

Question 3.
Ocellar spots situated at the base of each antenna of cockroach is called as __________
(A) ommatidia
(B) lingua
(C) fenestrae
(D) proventrieulus
Answer:
(C) fenestrae

Question 4.
Foregut of cockroach is also known as ________ .
(A) stomodaeum
(B) mesenteron
(C) proctodaeum
(D) tergum
Answer:
(A) stomodaeum

Question 5.
Circlet of six chitinous teeth and backwardly directed bristles are present in _____ .
(A) fenestrae
(B) mesenteron
(C) rectum
(D) gizzard
Answer:
(D) gizzard

Question 6.
Blood filled cavity in cockroach is called ________
(A) haemocoel
(B) paracoel
(C) spongocoel
(D) metacoel
Answer:
(A) haemocoel

Question 7.
In cockroach, ventral nerve cord is present in the
(A) pericardial sinus
(B) perineural sinus
(C) head sinus
(D) perivisceral sinus
Answer:
(B) perineural sinus

Question 8.
In cockroach, malpighian tubules are present at junction of
(A) foregut and midgut
(B) midgut and hindgut
(C) hindgut and foregut
(D) foregut and hindgut
Answer:
(B) midgut and hindgut

Question 9.
The main excretory organ of cockroach is
(A) gizzard
(B) malpighian tubules
(C) utricular gland
(D) mushroom shaped gland
Answer:
(B) malpighian tubules

Question 10.
Central nervous system consists of
(A) nerve ring
(B) ingluvial ganglion
(C) ventral nerve cord
(D) both (A) and (C)
Answer:
(D) both (A) and (C)

Question 11.
Common oviduct opens into
(A) phallomere
(B) bursa copulatrix
(C) utricular gland
(D) vagina
Answer:
(B) bursa copulatrix

Question 12.
________ are external genitalia of male cockroach.
(A) Phallomeres
(B) Utricular gland
(C) Seminal vesicles
(D) Spermatheca
Answer:
(A) Phallomeres

Question 13.
The total number of ovarian tubules in a female cockroach is
(A) 8
(B) 2
(C) 16
(D) 26
Answer:
(C) 16

Question 14.
The secretion of collaterial glands forms a capsule around them is called
(A) spermatophore
(B) nymph
(C) ootheca
(D) gonophore
Answer:
(C) ootheca

Question 15.
Select the INCORRECT statement from the following.
(A) Head of cockroach is formed by the fusion of six segments.
(B) In cockroach, anus is present on ventral side of 10th segment.
(C) In cockroach, heart has 22 chambers.
(D) Cockroach has open type of circulatory system.
Answer:
(C) In cockroach, heart has 22 chambers.

Question 35.
Competitive Corner

Question 1.
Which of the following statements is INCORRECT?
(A) Female cockroach possesses sixteen ovarioles in the ovaries.
(B) Cockroaches exhibit mosaic vision with less sensitivity and more resolution.
(C) A mushroom-shaped gland is present in the 6th – 7th abdominal segments of male cockroach.
(D) A pair of spermatheca is present in the 6th segment of female cockroach.
Hint: Cockroaches exhibit mosaic vision with more sensitivity but less resolution.
Answer:
(B) Cockroaches exhibit mosaic vision with less sensitivity and more resolution.

Question 2.
Select the CORRECT sequence of organs in the alimentary canal of cockroach starting from mouth:
(A) Pharynx → Oesophagus → Gizzard → Ileum → Crop → Colon → Rectum
(B) Pharynx → Oesophagus → Ileum → Crop → Gizzard → Colon → Rectum
(C) Pharynx → Oesophagus → Crop Gizzard → Ileum → Colon → Rectum
(D) Pharynx → Oesophagus → Gizzard → Crop → Ileum → Colon → Rectum
Answer:
(C) Pharynx → Oesophagus → Crop Gizzard → Ileum → Colon → Rectum

Question 3.
Which of the following features is used to identify a male cockroach from a female cockroach?
(A) Forewings with darker tegmina
(B) Presence of caudal styles
(C) Presence of a boat shaped sternum on the 9th abdominal segment
(D) Presence of anal cerci
Answer:
(B) Presence of caudal styles

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 5 Chemical Bonding Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 5 Chemical Bonding

Question 1.
Explain electronic theory of valence.
Answer:
Electronic theory of valence:

• Electronic theory of valence was proposed by Kossel and Lewis in 1916.
• They gave a logical explanation of valence which was based on the inertness of noble gases (that is, octet rule developed by Lewis).
• According to Lewis, the atom can be pictured in terms of a positively charged ‘kernel’ (the nucleus plus inner electrons) and outer shell that can accommodate a maximum of eight electrons. This octet of electrons represents a stable electronic arrangement.
• Thus, according to this theory, during the formation of a chemical bond, each atom loses, gains or shares outer electrons so that it achieves stable octet.
• The formation of NaCl involves transfer of one electron from sodium (Na) to chlorine (Cl). Na⁺ and Cl^– ions are formed which are held together by chemical bond. The formation of H₂, F₂, Cl₂, HCl, etc., involves sharing of a pair of electrons between the atoms. In both the cases, each atom attains a stable outer octet of electrons.

Question 2.
Give the significance of octet rule. Explain why this rule is not valid for H and Li atoms.
Answer:
i. Significance: Octet rule is found to be very useful:

• in explaining the normal valence of elements
• in the study of the chemical combination of atoms leading to the formation of molecule.

ii. Octet rule is not valid for H and Li atoms. According to octet rule, during the formation of a chemical bond, each atom loses, gains or shares electrons so that it achieves stable octet (eight electrons in the valence shell). However, H and Li atoms tend to have only two electrons in their valence shell similar to that of Helium (1s²), which called duplet. Hence, octet rule is not valid for H and Li atoms.

Question 3.
Define ionic bond.
Answer:
The bond formed by complete transfer of one or more electrons from an electropositive atom to an electronegative atom, leading to formation of ions which are held together by electrostatic attraction is called ionic bond or electrovalent bond.

Question 4.
Explain the formation of ionic bond in sodium chloride (NaCl).
Answer:
Formation of sodium chloride (NaCl):
i. The electronic configurations of sodium and chlorine are:
Na (Z = 11): 1s² 2s² 2p⁶ 3s¹ or (2, 8, 1)
Cl (Z = 17): 1s² 2s² 2p⁶ 3s² 3p⁵ or (2, 8, 7)
ii. Sodium has one electron in its valence shell. It has tendency to lose one electron to acquire the electronic configuration of the nearest inert gas, neon (2, 8).
iii. Chlorine has seven electrons in its valence shell. It has tendency to gain one electron and thereby acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8).
iv. During the combination of sodium and chlorine atoms, the sodium atom transfers its valence electron to the chlorine atom.
v. Sodium atom changes into Na+ ion while the chlorine atom changes into Cl^– ion. The two ions are held together by strong electrostatic force of attraction.
vi. The formation of ionic bond between Na and Cl can be shown as follows:

Question 5.
Explain the formation of ionic bonds in calcium chloride (CaCl₂).
Answer:
Formation of calcium chloride (CaCl₂):
i. The electronic configurations of calcium and chlorine are:
Na (Z = 11): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² or (2, 8, 8, 2)
Cl (Z = 17): 1s² 2s² 2p⁶ 3s² 3p⁵ or (2, 8, 7)
ii. Calcium has two electrons in its valence shell. It has tendency to lose two electrons to acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8).
iii. Chlorine has seven electrons in its valence shell. It has tendency to gain one electron and thereby acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8).
iv. During the combination of calcium and chlorine atoms, the calcium atom transfers its valence electrons to two chlorine atoms.
v. Calcium atom changes into Ca²⁺ ion while the two chlorine atoms change into two Cl^– ions. These ions are held together by strong electrostatic force of attraction.
vi. The formation of ionic bond(s) between Ca and Cl can be shown as follows:

Question 6.
What are ionic solids?
Answer:
Ionic solids are solids which contain cations and anions held together by ionic bonds.
e.g. Sodium chloride (NaCl), Calcium chloride (CaCl₂)

Question 7.
Define lattice enthalpy.
Answer:
Lattice enthalpy of an ionic solid is defined as the energy required to completely separate one mole of solid ionic compound into the gaseous components.
Note: Lattice enthalpy values of some ionic compounds:

Compound	Lattice enthalpy kJ mol–1
LiCl	853
NaCl	788
BeF2	3020
CaCl2	2258
AlCl3	5492

Question 8.
Arrange NaCl, CaCl₂ and AlCl₃ in increasing order of lattice enthalpy (positive value). Justify your answer.
Answer:
Compounds having cations with higher charge have large lattice enthalpy (higher positive value) than compounds having cations with lower charge.
Hence, the correct order is NaCl < CaCl₂ < AlCl₃.

Question 9.
Lattice enthalpy of LiF is more than that of NaF. Explain.
Answer:
As the size of the cation decrease, lattice enthalpy increases. Li⁺ ion is smaller than Na⁺ ion. Hence, lattice enthalpy of LiF is more than that of NaF.

Question 10.
Define covalent bond.
Answer:
The attractive force which exists due to the mutual sharing of electrons between the two atoms of similar electronegativity or having small difference in electronegativities is called a covalent bond.

Question 11.
Explain the formation of covalent bond in H₂ molecule.
Answer:
Formation of H₂ molecule:
i. The electronic configuration of H atom is 1s¹.
ii. It needs one more electron to complete its valence shell.
iii. When two hydrogen atoms approach each other at a certain internuclear distance, they share their valence electrons.
iv. The shared pair of electrons belongs equally to both the hydrogen atoms. The two atoms are said to be linked by a single covalent bond and a H₂ molecule is formed.

Question 12.
Explain the formation of covalent bond in Cl₂ molecule.
Answer:
Formation of Cl₂ molecule:
i. The electronic configuration of Cl atom is [Ne] 3s² 3p⁵.
ii. It needs one more electron to complete its valence shell.
iii. When two chlorine atoms approach each other at a certain internuclear distance, they share their valence electrons. In the process, both the atoms attain the valence shell of octet of nearest noble gas, argon.
iv. The shared pair of electrons belongs equally to both the chlorine atoms. The two atoms are said to be linked by a single covalent bond and a Cl₂ molecule is formed.

Question 13.
What are the important features of covalent bond?
Answer:

• Each covalent bond is formed as a result of sharing of electron pair between the two atoms.
• When a covalent bond is formed, each combining atom contributes one electron to the shared pair.
• The combining atoms attain the outer shell noble gas configuration as a result of the sharing of electrons.

Question 14.
Explain the types of covalent bond with suitable examples.
Answer:
The three types of covalent bonds are as follows.
i. Single bond: When two combining atoms share one electron pair, the covalent bond between them is called single bond.
Single bond is observed in number of molecules.
e.g. H₂, Cl₂, water molecule, etc.

ii. Double bond: When two combining atoms share two pairs of electrons, the covalent bond between them is called a double bond, e.g. Double bond is present in C₂H₄ molecule

iii. Triple bond: When two combining atoms share three pairs of electrons, the covalent bond between them is called a triple bond, e.g. Triple bond is present in N₂ molecule.

Note: Formation of covalent bonds in CO₂, CCl₄ and C₂H₂:

Question 15.
Distinguish between ionic bond and covalent bond.
Answer:
Ionic bond:

1. It is formed by the transfer of electrons from one atom to another.
2. It is formed by the transfer of electrons from one atom to another.
3. In this, oppositely charged ions are formed.
4. There are no multiple ionic bonds.
5. This bond usually exists between metal and non-metal atoms.
6. e.g. NaCl, CaCl₂

Covalent bond:

1. It is formed by sharing of electrons.
2. Atoms are held together due to shared pair of electrons.
3. In this, oppositely charged ions are not formed.
4. Covalent bonds may be single or double or triple bonds.
5. This bond usually exists between non-metal atoms.
6. e.g. H₂, Cl₂

Question 16.
What are the steps to write Lewis dot structure?
Answer:
Steps to write Lewis dot structures:

• Add the total number of valence electrons of combining atoms in the molecule.
• In anions, add one electron for each negative charge.
• In cations, subtract one electron from valence electrons for each positive charge.
• Write skeletal structure of the molecule to show the atoms and number of valence electrons forming the single bond between the atoms.
• Add remaining electron pairs to complete the octet of each atom.
• If octet is not complete form multiple bonds between the atoms such that octet of each atom is complete.
• In polyatomic atoms and ions, the least electronegative atom is the central atom.
  e.g. In \(\mathrm{SO}_{4}^{2-}\) ion, ‘S’ is the central atom and in \(\mathrm{NO}_{3}^{-}\), ‘N’ is the central atom.
• After writing the number of electrons as shared pairs forming single bonds, the remaining electron pairs are used either for multiple bonds or they remain as lone pairs.

Question 17.
Write the Lewis structure of nitrite ion, \(\mathrm{NO}_{2}^{-}\).
Answer:
Step I: Count the total number of valence electrons of nitrogen atom, oxygen atom and one electron of additional negative charge.
Valence shell configuration of nitrogen and oxygen are:
N ⇒ (2s² 2p³), O ⇒ (2s² 2p⁴)
The total electrons available are:
5 + (2 × 6) + 1 = 18 electrons
Step II: The skeletal structure of \(\mathrm{NO}_{2}^{-}\) is written as O N O
Step III: Draw a single bond i.e., one shared electron pair between the nitrogen and each oxygen atoms. Then distribute the remaining electrons to achieve noble gas configuration for each atom. This does not complete the octet of nitrogen.

Hence, there is a multiple bond between nitrogen and one of the oxygen atoms (a double bond). The remaining two electrons constitute a lone pair on nitrogen.
Following are Lewis dot structures of \(\mathrm{NO}_{2}^{-}\).

Question 18.
Write the Lewis structure of CO molecule.
Answer:
Step I: Count number of electrons of carbon and oxygen atoms. The valence shell configuration of carbon and oxygen atoms are: 2s² 2p² and 2s² 2p⁴, respectively. The valence electrons available are:
4 + 6=10
Step II: The skeletal structure of CO is written as
C O
Step III: Draw a single bond (One shared electron pair) between C and O and complete the octet on O. The remaining two electrons is a lone pair on C.

The octet on carbon is not complete. Hence, there is a multiple bond between C and O (a triple bond between C and O atom). This satisfies the octet rule for carbon and oxygen atoms.
The Lewis structure of CO molecule is:

Question 19.
Explain the term formal charge.
Answer:
i. Formal charge is the charge assigned to an atom in a molecule, assuming that all electrons are shared equally between atoms, regardless of their relative electronegativities.
ii. While determining the best Lewis structure per molecule, the structure is chosen such that the formal charge is as close to zero as possible.
iii. The structure having the lowest formal charge has the lowest energy.
iv. Formal charge is assigned to an atom based on electron dot structures of the molecule/ion.
v. Formal charge on an atom in a Lewis structure of a polyatomic species can be determined using the following formula:

Question 20.
Explain the calculation of the formal charge on oxygen atoms in case of O₃ (ozone) molecule.
Answer:
i. Lewis dot structure of O₃ (ozone) molecule is:

Three oxygen atoms are present in the O₃ molecule and are labelled as 1, 2 and 3.
ii. Formal charges on oxygen atoms labelled as 1, 2, 3 are calculated as shown below:

iii. On the basis of the formal charge values, O₃ is shown as

[Note: Indicating the charges on the atoms in the Lewis structure helps in keeping track of the valence electrons in the molecule. Formal charges help in the selection of the lowest energy structure from a number of possible Lewis structures for a given species.]

Question 21.
CO₂ can be represented by following three structures:

Calculate the formal charge on each atom in all the three structures of CO₂ molecule. Identify the structure with lowest energy.
Answer: Formal charges on atoms labelled as 1, 2, 3 are calculated as shown below:
Structure (I):

Structure (II):

Structure (III):

While determining the best Lewis structure per molecule, the structure is chosen such that the formal charge is as close to zero as possible. The structure having the lowest formal charge has the lowest energy.

In structure (I), the formal charge on each atom is 0 while in structures (II) and (III) formal charge on carbon is 0 while oxygens have formal charge -1 or +1. Hence, the possible structure with the lowest energy will be structure (I). Thus, formal charges help in the selection of the lowest energy structure from a number of possible Lewis structures for a given species.

Question 22.
Find out the formal charges on S, C and N.
(S = C = N)^– ; (S – C ≡ N)^– ; (S ≡ C – N)^–
Answer:
Step I:
Write Lewis dot diagrams for the structures:

Step II:
Assign formal charges for all the atoms:
F.C. = V.E. – N.E. – 1/2 (B.E.)
Structure A:
Formal charge on S = 6 – 4 – 1/2(4) = 0
Formal charge on C = 4 – 0 – 1/2 (8) = 0
Formal charge on N = 5 – 4 – 1/2 (4) = -1
Structure B:
Formal charge on S = 6 – 6 – 1/2(2) = -1
Formal charge on C = 4 – 0 – 1/2 (8) = 0
Formal charge on N = 5 – 2 – 1/2 (6) = 0
Structure C:
Formal charge on S = 6 – 2 – 1/2(6) = +1
Formal charge on C = 4 – 0 – 1/2(8) = 0
Formal charge on N = 5 – 6 – 1/2(2) = -2

Question 23.
Give the limitations of octet rule:
Answer:
Limitations of octet rule:
i. Octet rule does not explain stability of some molecules.
The octet rule is based on the inert behaviour of noble gases, which have their octet complete i.e., have eight electrons in their valence shell. It is very useful to explain the structures and stability of organic molecules. However, there are many molecules whose existence cannot be explained by the octet theory. The central atoms in these molecules does not have eight electrons in their valence shell, and yet they are stable.
Such molecules can be categorized as having:
a. Incomplete octet
b. Expanded octet
c. Odd electrons

a. Molecules with incomplete octet: e.g. BF₃, BeCl₂, LiCl
In these covalent molecules, the atoms B, Be and Li have less than eight electrons in their valence shell but these molecules are stable.
Li in LiCl has only two electrons, Be in BeCl₂ has four electrons while B in BF₃ has six electrons in the valence shell.

b. Molecules with expanded octet: Some molecules like SF₆, PCl₅, H₂SO₄ have more than eight electrons around the central atom.

c. Odd electron molecules:
Some molecules like NO (nitric oxide) and NO₂ (nitrogen dioxide) do not obey the octet rule. These molecules, have odd number of valence electrons.

ii. The observed shape and geometry of a molecule, cannot be explained, by the octet rule.
iii. Octet rule fails to explain the difference in energies of molecules, though all the covalent bonds are formed in an identical manner, that is, by sharing a pair of electrons. The rule fails to explain the differences in reactivities of different molecules.
Note: Sulphur also forms many compounds in which octet rule is obeyed. For example, in sulphur dichloride, the sulphur atom has eight electrons around it.

Question 24.
State the basic idea on which VSEPR theory was proposed by Sidgwick and Powell.
Answer:
Valence Shell Electron Pair Repulsion (VSEPR) theory is based on the basic idea that the electron pairs on the atoms shown in the Lewis diagram repel each other. In the real molecule, they arrange themselves in such a way that there is minimum repulsion between them.

Question 25.
Give the rules of VSEPR Theory.
Answer:
Rules of VSEPR Theory:
i. Electron pairs arrange themselves in such a way that repulsion between them is minimum.
ii. The molecule acquires minimum energy and maximum stability.
iii. Lone pair of electrons also contribute in determining the shape of the molecule.
iv. Repulsion of other electron pairs by the lone pair (L.P.) stronger than that of bonding pair (B.P.).
Trend for repulsion between electron pair is as follows:
L.P. – L.P. > L.P. – B.P. > B.P. – B.P.
Lone pair-Lone pair repulsion is maximum because this electron pair is under the influence of only one nucleus while the bonded pair is shared between two nuclei.
Thus, the number of lone pair and bonded pair of electrons decide the shape of the molecules. Molecules having no lone pair of electrons have a regular geometry.

Note:
i. Electron pair geometry: The arrangement of electrons around the central atom is called as electron pair geometry. These electron pairs may be shared in a covalent bond or they may be lone pairs.
ii. Geometry of some molecules (having no lone pair of electrons):


iii. Geometry of some molecules (having one or more lone pairs of electrons):

Question 26.
Match the following:

	Molecule		Shape
i.	SbF5	a.	Trigonal bipyramidal
ii.	SO2	b.	Bent
iii.	SF4	c.	Square pyramidal
iv.	IF5	d.	See-saw

Answer:
i – a,
ii – b,
iii – d,
iv – c

Question 27.
Complete the following table:

Answer:

Question 28.
Explain geometry of NH₃ molecule according to VSEPR theory.
Answer:

• In NH₃ molecule, the central atom nitrogen has five electrons in its valence shell. On bond formation with three hydrogen atoms, there are 8 electrons in the valence shell of nitrogen. Out of these, three pairs are bond pairs (N – H covalent bonds) and one forms lone pair. The expected geometry is tetrahedral and bond angle is 109° 28′.
• There are two types of repulsions between the electron pairs: Lone pair – bond pair and bond pair – bond pair
• The lone pair – bond pair repulsions are stronger and the bonded pairs are pushed inwards. Thus, reducing the bond angle to 107°18′ and shape of the molecule becomes trigonal pyramidal.

Diagram:

Question 29.
Explain geometry of H₂O molecule according to VSEPR theory.
Answer:

• In H₂O molecule, the central atom oxygen has six electrons in its valence shell. On bond formation with two hydrogen atoms, there are 8 electrons in the valence shell of oxygen. Out of these, two pairs are bond pairs and two are lone pairs.
• Due to lone pair – lone pair repulsion, the lone pairs are pushed towards the bond pairs and bond pair – lone pair repulsions become stronger thereby reducing the H-O-H bond angle from 109° 28′ to 104° 35′ and the geometry of the molecule becomes angular (bent).

Diagram:

Question 30.
Write the postulates of Valence Bond Theory.
Answer:
Postulates of Valence Bond Theory:

• A covalent bond is formed when the half-filled valence orbital of one atom overlaps with the half-filled valence orbital of another atom.
• The electrons in the half-filled valence orbitals must have opposite spins.
• During bond formation, the half-filled orbitals overlap and the opposite spins of the electrons get neutralized. The increased electron density decreases the nuclear repulsion and energy is released during overlapping of the orbitals.
• Greater the extent of overlap, stronger is the bond formed. However, complete overlap of orbitals does not take place due to intemuclear repulsions.
• If an atom possesses more than one unpaired-electrons, then it can form more than one bond. So, number of bonds formed will be equal to the number of half-filled orbitals in the valence shell i.e., number of unpaired electrons.
• The distance at which the attractive and repulsive forces balance each other is the equilibrium distance between the nuclei of the bonded atoms. At this distance, the total energy of the two bonded atoms is minimum and stability of the molecule is maximum.
• Electrons which are paired in the valence shell cannot participate in bond formation. However, in an atom if there is one or more vacant orbital present then these electrons can unpair and participate in bond formation provided the energies of the filled and vacant orbitals differ slightly from each other.
• During bond formation, the ‘s’ orbital which is spherical can overlap in any direction. The ‘p’ orbitals can overlap only in the x, y or z directions. Similarly, ‘d’ and ‘f orbitals are oriented in certain directions in space and overlap only in these directions. Thus, the covalent bond is directional in nature.

Note: In order to explain the covalent bonding, Heitler and London developed the valence bond theory on the basis of wave mechanics. This theory was further extended by Pauling and Slater.

Question 31.
Explain the formation of hydrogen molecule with the help of potential energy curve.
OR
Explain the formation of H₂ on the basis of VBT.
Answer:
Formation of H₂ on the basis of VBT:

• Hydrogen atom has electronic configuration 1s¹. It contains one unpaired electron in its valence shell.
• When the two hydrogen atoms containing unpaired electrons with opposite spins are separated by a large distance, they can neither attract nor repel each other (there are no interactions between them). The energy of the system is the sum of the potential energies of the two atoms which is arbitrarily taken as zero.
• When the two atoms approach each other, attractive and repulsive forces begin to operate on them. Experimentally, it has been found that during formation of hydrogen molecule, the magnitude of the newly developed attractive forces contributes more than the newly developed repulsive forces. As a result, the potential energy of the system begins to decrease.
• As the atoms come closer to one another the energy of the system decreases. The overlap of the atomic orbitals increases only up to a certain distance between the two nuclei, where the attractive and repulsive forces balance each other and the system attains minimum energy. At this stage, a stable bond is formed between the two atoms.
• If the distance between the two atoms is decreased further, the repulsive forces exceed the attractive forces and the energy of the system increases and stability decreases.
• When two hydrogen atoms with electrons having parallel spin approach each other, the potential energy of the system increases and bond formation does not take place.

Question 32.
Define:
i. Sigma overlap
ii. pi overlap
Answer:
i. Sigma overlap (σ bond): When two half-filled orbitals of two atoms overlap along the internuclear axis, it is called as sigma overlap or sigma bond.
ii. pi overlap (π bond): When two half-filled orbitals of two atoms overlap side-ways (laterally), it is called as π overlap or π bond.

Question 33.
Explain with example:
i. s-s σ overlap
ii. p-p σ overlap
iii. s-p a overlap
Answer:
i. s-s σ overlap:
a. The overlap between two half-filled s orbitals of two different atoms containing unpaired electrons with opposite spins is called s-s overlap.
e.g. Formation of H₂ molecule by s-s overlap:
Hydrogen atom (Z = 1) has electronic configuration: 1s¹. The 1s¹ orbitals of two hydrogen atoms overlap along the internuclear axis to form a σ bond between the atoms in H₂ molecule.
b. Diagram:

ii. p-p σ overlap:
a. This type of overlap takes place when two p orbitals from different atoms overlap along the internuclear axis.
e.g. Formation of F₂ molecule by p-p overlap:
Fluorine atom (Z = 9) has electronic configuration 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{2}\) \(2 \mathrm{p}_{\mathrm{y}}^{2}\) \(2 \mathrm{p}_{\mathrm{z}}^{1}\).
During the formation of F₂ molecule, half-filled 2p_z orbital of one F atom overlaps with similar half-filled 2p_z orbital containing electron with opposite spin of another F atom axially and a p-p σ bond is formed.
b. Diagram:

iii. s-p σ overlap:
a. In this type of overlap one half filled s orbital of one atom and one half filled p orbital of another orbital overlap along the internuclear axis.
e.g. Formation of HF molecule by s-p overlap:
Hydrogen atom (Z = 1) has electronic configuration: 1s¹ and fluorine atom (Z = 9) has electronic configuration 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{2}\) \(2 \mathrm{p}_{\mathrm{y}}^{2}\) \(2 \mathrm{p}_{\mathrm{z}}^{1}\). During the formation of HF molecule, half-filled 1s orbital of hydrogen atom overlaps coaxially with half-filled 2p_z orbital of fluorine atom with opposite electron spin and an s-p σ bond is formed.
b. Diagram:

Question 34.
Explain the formation of π bond with diagram.
OR
Explain π overlap with diagram.
Answer:

• When two half-filled p orbitals of two atoms overlap side-ways (laterally), it is called π overlap and the bond formed is called π bond.
• π bond is perpendicular to the intemuclear axis.

Diagram:

Question 35.
Identify the type of bond formed:


Answer:
i. σ bond
ii. π bond
iii. σ bond
iv. σ bond

Question 36.
Explain divalency of beryllium, though number of unpaired electrons in a beryllium atom is zero.
Answer:
Beryllium (Z = 4) has electronic configuration 1s² 2s² in its ground state.
When one of the 2s electrons of Be is promoted to the vacant 2p orbital, the electronic configuration of Be in its excited state becomes 1s² 2s¹ \(2 \mathrm{p}_{\mathrm{x}}^{1}\). This is called formation of an excited state and it has two unpaired electrons. Hence, though the number of unpaired electrons in the ground state of Be atom is zero, beryllium shows divalency.

Question 37.
Define the term hybridization.
Answer:
Hybridization is defined as the process of mixing of valence orbitals of same atom and recasting them into equal number of new equivalent orbitals (hybrid orbitals).

Question 38.
Explain in detail the steps involved in hybridization.
Answer:
Steps involved in hybridization:
i. Formation of the excited state:
a. The paired electrons in the ground state are uncoupled and one electron is promoted to the vacant orbital having slightly higher energy.
b. Now, total number of half-filled orbitals is equal to the valency of the element in the stable compound, e.g. In BeF₂, valency of Be is two. In the excited state, one electron from 2s orbital is uncoupled and promoted to 2p orbital.

ii. Mixing and recasting:

• In this step, the two ‘s’ and ‘p’ orbitals having slightly different energies mix with each other.
• Redistribution of electron density and energy takes place and two new orbitals having exactly same shape and energy are formed.
• These new orbitals arrange themselves in space in such a way that there is minimum repulsion and maximum separation between them. e.g. During formation of sp hybrid orbitals as in Be, the two sp hybrid orbitals form an angle of 180° with each other.

Question 39.
List the important conditions required for hybridization.
Answer:
Conditions for hybridization:

• Orbitals belonging to the same atom can participate in hybridization.
• Orbitals having nearly same energy can undergo hybridization.

[Note: 2s and 2p orbitals of the same atom undergo hybridization but 3s and 2p orbitals of the same atom do not.]

Question 40.
Enlist the characteristic features of hybrid orbitals.
Answer:
Characteristic features of hybrid orbitals:

• Number of hybrid orbitals formed is exactly the same as the participating atomic orbitals.
• They have same energy and shape.
• Hybrid orbitals are oriented in space in such a way that there is minimum repulsion and thus are directional in nature.
• The hybrid orbitals are different in shape from the participating atomic orbitals, but they bear the characteristics of the atomic orbitals from which they are derived.
• Each hybrid orbitals can hold two electrons with opposite spins.
• A hybrid orbital has two lobes on the two sides of the nucleus. One lobe is large and the other small.
• Covalent bonds formed by hybrid orbitals are stronger than those formed by pure orbitals, because the hybrid orbital has electron density concentrated on the side with a larger lobe and the other is small allowing greater overlap of the orbitals.

Question 41.
Explain with diagrams:
i. sp³ hybridization
ii. sp² hybridization
iii. sp hybridization
Answer:
i. sp³ hybridization:
In this type, one s and three p orbitals having comparable energy mix and recast to form four sp³ hybrid orbitals, ‘s’ orbital is spherically symmetrical while the p_x, p_y, p_z, orbitals have two lobes and are directed along x, y and z axes, respectively.

The four sp³ hybrid orbitals formed are equivalent in energy and shape. They have one large lobe and one small lobe. They are at an angle of 109° 28′ with each other in space and point towards the comers of a tetrahedron. CH₄, NH₃, H₂O are examples where the orbitals on central atom undergo sp³ hybridization.

Diagram:

ii. sp² hybridization:
This hybridization involves the mixing of one s and two p orbitals to give three sp² hybrid orbitals of same energy and shape. The three orbitals are maximum apart and oriented at an angle of 120° and are in one plane. The third p orbital does not participate in hybridization and remains at right angles to the plane of the sp² hybrid orbitals. BF₃, C₂H₄ molecules are examples of sp² hybridization.
Diagram:

iii. sp hybridization:
In this type, one s and one p orbital undergo mixing and recasting to form two sp hybrid orbitals of same energy and shape. The two hybrid orbitals are placed at an angle of 180°. Other two p orbitals do not participate in hybridization and are at right angles to the hybrid orbitals. For example, BeCl₂ and acetylene molecule (HC ≡ CH).
Diagram:

Question 42.
Explain the formation of an ammonia molecule on the basis of hybridization.
Answer:
Formation of an ammonia (NH₃) molecule on the basis of sp³ hybridization:
i. Ammonia molecule (NH₃) has one nitrogen atom and three hydrogen atoms.
ii. The ground state electronic configuration of nitrogen (Z = 7) is 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\) \(2 \mathrm{p}_{\mathrm{y}}^{1}\) \(2 \mathrm{p}_{\mathrm{z}}^{1}\)
Electronic configuration of nitrogen:

iii. The ground state electronic configuration explains the observed valency of nitrogen in NH₃ which is three.
iv. The 2s, 2p_x, 2p_y and 2p_z orbitals of nitrogen atom mix and recast to form four sp³ hybrid orbitals of equivalent energy. These orbitals are tetrahedrally oriented in space. One of the sp³ hybrid orbital contains a lone pair of electrons.
v. Three half-filled sp³ hybrid orbitals of N atom overlap axially with half-filled 1s orbital of three different hydrogen atoms to form three N-H (sp³-s) sigma covalent bonds.
vi. Since, there is one lone pair of electrons in one of the sp³ hybrid orbitals of nitrogen, there is repulsion between lone pair and bonding pair of electrons. As a result, the H-N-H bond angle is reduced from regular tetrahedral angle 109° 28′ to 107° 18′. Geometry of NH₃ molecule is pyramidal or distorted tetrahedral.

Question 43.
Explain the formation of water (H₂O) molecule on the basis of hybridization.
Answer:
Formation of water (H₂O) molecule on the basis of sp³ hybridization:
i. Water molecule (H₂O) has one oxygen atom and two hydrogen atoms.
ii. The ground state electronic configuration of oxygen (Z = 8) is 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{2}\) \(2 \mathrm{p}_{\mathrm{y}}^{1}\) \(2 \mathrm{p}_{\mathrm{z}}^{1}\).
Electronic configuration of oxygen:

iii. The ground state electronic configuration explains the observed valency of oxygen in H₂O molecule which is 2.
iv. The 2s, 2p_x, 2p_y and 2p_z orbitals of oxygen atom mix and recast to form four sp³ hybrid orbitals of equivalent energy. These orbitals are tetrahedrally oriented in space. Two of the sp hybrid orbitals contain lone pair of electrons.
v. Two half-filled sp³ hybrid orbitals of O atom overlap axially with half-filled 1s orbitals of two different hydrogen atoms to fonn two O-H (sp³-s) sigma covalent bonds.
vi. Since, there are two lone pairs of electrons in two of the sp³ hybrid orbitals of oxygen, there is repulsion between lone pair and bonding pair of electrons. As a result, the H-O-H bond angle is reduced from regular tetrahedral angle 109°28′ to 104°35′. The geometry of H₂O molecule is angular or V shaped.

Diagram:

Question 44.
Explain the formation of an ethene molecule on the basis of hybridization.
Answer:
Formation of an ethene (ethylene) molecule on the basis of sp² hybridization:
i. Ethene molecule (C₂H₄) has two carbon atoms and four hydrogen atoms.
ii. The ground state electronic configuration of C (Z = 6) is 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\) \(2 \mathrm{p}_{\mathrm{y}}^{1}\) \(2 \mathrm{p}_{\mathrm{z}}^{0}\).
Electronic configuration of carbon:

iii. One electron from 2s orbital of each carbon atom is excited to the 2p_z orbital. Then each carbon atom undergoes sp2 hybridization.
iv. One ’s’ orbital and two ‘p’ orbitals on carbon hybridize to form three sp² hybrid orbitals of equal energy and symmetry.
v. Two sp² hybrid orbitals overlap axially two ‘s’ orbitals of hydrogen to form sp²-s σ bond. The unhybridized ‘p’ orbitals on the two carbon atoms overlap laterally to form a π bond. Thus, the C₂H₄ molecule has four sp²-s σ bonds, one sp²-sp² σ bond and one p-p π bond.
vi. Each H-C-H and H-C-C bond angle in ethene molecule is 120°. All the six atoms in ethene (ethylene) molecule are in one plane. Geometry of the molecule at each carbon atom is trigonal planar.

Question 45.
Explain the formation of boron trifluoride on the basis of hybridization.
Answer:
Formation of boron trifluoride on the basis of sp² hybridization:
i. Boron trifluoride (BF₃) has one boron atom and three fluorine atoms.
ii. Observed valency of boron in BF₃ molecule is three and its geometry is trigonal planar. This can be explained on the basis of sp² hybridization.
iii. The ground state electronic configuration of B (Z = 5) is 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\) \(2 \mathrm{p}_{\mathrm{y}}^{0}\) \(2 \mathrm{p}_{\mathrm{z}}^{0}\).
Electronic configuration of boron:

iv. One electron from 2s orbital of boron atom is uncoupled and promoted to vacant 2p_y orbital.
v. The three orbitals i.e. 2s, 2p_x of and 2p_y of boron undergoes sp² hybridization to form three sp² hybrid orbitals of equivalent energy, which are oriented along the three comers of an equilateral triangle making an angle of 120°.
vi. Each sp² hybrid orbital of boron atom having unpaired electron overlaps axially with half-filled 2p_z orbital of fluorine atom containing electron with opposite spin to form three B-F sigma bonds by sp²-p type of overlap.
vii. Each F-B-F bond angle in BF₃ molecule is 120°. The geometry of BF₃ molecule is trigonal planar.

Diagram:

Question 46.
Explain the formation of an acetylene molecule on the basis of hybridization.
Answer:
Formation of acetylene (ethyne) molecule on the basis of sp hybridization:
i. Acetylene molecule (C₂H₂) has two carbon atoms and two hydrogen atoms.
ii. The ground state electronic configuration of C (Z = 6) is 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\) \(2 \mathrm{p}_{\mathrm{y}}^{1}\) \(2 \mathrm{p}_{\mathrm{z}}^{0}\).
Electronic configuration of carbon:

iii. Each carbon atom undergoes sp hybridization. One s and one p orbitals mix and recast to give two sp hybrid orbitals arranged at 180° to each other.
iv. Out of the two sp hybrid orbitals of carbon atom, one overlaps axially with s orbital of hydrogen while the other sp hybrid orbital overlaps with sp hybrid orbital of other carbon atom to form the sp-sp σ bond. The C H σ bond is formed by sp-s overlap.
v. The remaining two unhybridized p orbitals overlap laterally to form two p-p π bonds. So, there are three bonds between the two carbon atoms: one C-C σ bond (sp-sp) overlap, two C-C π bonds (p-p) overlap. There are two sp-s σ bonds in acetylene (one between each C and H).
vi. Each H-C-C bond angle in ethyne molecule is 180°. All the four atoms in ethyne molecule are in a straight line. The geometry of acetylene molecule is linear.

Diagram:

Question 47.
Explain the formation of BeCl₂.
Answer:
Formation of BeCl₂:
i. BeCl₂ molecule has one Be atom and two chlorine atoms.
ii. Electronic configuration of Be is 1s² 2s² \(2 \mathrm{p}_{\mathrm{z}}^{0}\).
Electronic configuration of beryllium:

iii. The 2s and 2p_z orbitals undergo sp hybridization to form two sp hybrid orbitals oriented at 180° with each other. 2p_z orbitals of two chlorine atoms overlap with the sp hybrid orbitals to form two sp-p σ bonds.
Cl – Be – Cl bond angle is 180°. The geometry of the molecule is linear.

Diagram:

Question 48.
Match the following:

	Molecule		Hybridization and bond angle
i.	Water	a.	Sp2, 120°
ii.	Boron trifluoride	b.	Sp3, 104.5°
iii.	Beryllium fluoride	c.	Sp3, 109.5°
iv.	Methane	d.	Sp, 180°

Answer:
i – b
ii – a
iii – d
iv – c

Question 49.
Give the importance of valence bond theory.
Answer:
Valence Bond theory introduced five new concepts in chemical bonding:

• Delocalization of electron over the two nuclei
• Shielding effect of electrons
• Covalent character of bond
• Partial ionic character of a covalent bond
• The concept of resonance and connection between resonance energy and molecular stability

Question 50.
What are the limitations of valence bond theory?
Answer:
Limitations of valence bond theory:

• Valence Bond theory explains only the formation of covalent bond in which a shared pair of electrons comes from two bonding atoms. However, it offers no explanation for the formation of a coordinate covalent bond in which both the electrons are contributed by one of the bonded atoms.
• Oxygen molecule is expected to be diamagnetic according to this theory. The two atoms in oxygen molecule should have completely filled electronic shells which give no unpaired electrons to the molecule making it diamagnetic. However, experimentally the molecule is found to be paramagnetic having two unpaired electrons. Thus, this theory fails to explain paramagnetism of oxygen molecule.
• Valence bond theory does not explain the bonding in electron deficient molecules like B₂H₆ in which the central atom possesses a smaller number of electrons than required for an octet of electron.

Question 51.
What are the two ways in which two atomic orbitals combine to form molecular orbitals (MOs)?
Answer:
Two atomic orbitals can combine in two ways to form molecular orbitals:
i. By addition of their wave functions.
ii. By subtraction of their wave functions.
Addition of the atomic orbtials wave functions results in formation of a molecular orbital which is lower in energy than atomic orbitals and is termed as Bonding Molecular Orbital (BMO). Subtraction of the atomic orbitals results in the formation of a molecular orbital which is higher in energy than the atomic orbitals and is termed as Antibonding Molecular Orbital (AMO).

Question 52.
State True or False. Correct the false statement.
i. According to MO theory, the formation of molecular orbitals from atomic orbitals is expressed in terms of Linear Combination of Atomic Orbitals (LCAO).
ii. An MO contains maximum two electrons with opposite spins.
iii. Interference of electron waves of combining atoms can only be constructive.
iv. In bonding molecular orbital, the large electron density is observed between the nuclei of the bonding atoms than the individual atomic orbitals.
v. In the antibonding molecular orbital, the electron density is nearly zero between the nuclei.
Answer:
i. True
ii. True
iii. False
Interference of electron waves of combining atoms can be constructive or destructive.
iv. True
v. True

Question 53.
What are the conditions required for linear combination of atomic orbitals to form molecular orbitals?
Answer:
The following conditions are required for the linear combination of atomic orbitals (LCAO) to form molecular orbitals:
i. The combining atomic orbitals must have comparable energies.
So, Is orbitals of one atom can combine with 1 s orbital of another atom but not with 2s orbital, because energy of 2s orbital is much higher than that of 1 s orbital.

ii. The combining atomic orbitals must have the same symmetry along the molecular axis. Conventionally, z axis is taken as the internuclear axis. So even if atomic orbitals have same energy but their symmetry is not same they cannot combine. For example, 2s orbital of an atom can combine only with 2p_z orbital of another atom, and not with 2p_x or 2p_y orbital of that atom because the symmetries are not same. p_z is symmetrical along z axis while px is symmetrical along x axis.

iii. The combining atomic orbitals must overlap to the maximum extent. Greater the overlap, greater is the electron density between the nuclei and so stronger is the bond formed.

Question 54.
Explain and draw an energy level diagram obtained by the linear combination of two 1s atomic orbitals.
Answer:
The s-orbitals are spherically symmetrical along x, y and z axis. Two Is atomic orbitals combine to form σ 1s (bonding molecular orbital) and σ*1s (antibonding molecular orbital). Both the a bonding and σ* antibonding orbitals are symmetrical along the bond axis.
Diagram:

[Note: If we consider ‘z’ to be internuclear axis then linear combination of p_z orbitals from two atoms can form σ 2p_z bonding σ*(2p_z) molecular orbitals.]

Question 55.
Explain the formation of π and π* molecular orbitals with the help of a diagram.
Answer:
When the atomic orbitals overlap laterally, a pi (π) molecular orbital is formed.
The px and py orbitals are not symmetrical along the bond axis. They have a positive lobe above the axis and negative lobe below the axis. Hence, linear combination of such orbitals leads to the formation of molecular orbitals with positive and negative lobes above and below the bond axis. These are designed as π bonding and π antibonding orbitals. The electron density in such π orbitals is concentrated above and below the bond axis. The π molecular orbitals has a node between the nuclei.
Diagram:

Question 56.
Write the increasing order of energies of molecular orbitals in various diatomic molecules of second row elements.
Answer:
The increasing order of energies of molecular orbitals for molecules (except O₂ and F₂) is:
σ1s < σ*1s < σ2s < σ*2s < (π2p_x = π2p_y) < σ2p_z < (π*2p_x = π*2p_y) < σ*2p_z
The increasing order of energy of molecular orbitals for diatomic molecules like O₂ and F₂ is:
σ1s < σ*1s < σ2s < σ*2s < σ2p_z < (π2p_x = π2p_y) < (π*2p_x = π*2p_y) < σ*2p_z

Question 57.
Explain briefly the information provided by the electronic configuration of molecules.
Answer:
The electronic configuration of molecules provides the following information:

• Stability of molecules: If the number of electrons in bonding MOs is greater than the number in antibonding MOs the molecule is stable.
• Magnetic nature of molecules: If all MOs in a molecule are completely filled with two electrons each, the molecule is diamagnetic (i.e., repelled) by magnetic field. However, if at least one MO is half-filled with one electron, the molecule is paramagnetic (i. e., attracted by magnetic field).
• Bond order of molecule: The bond order of the molecule can be calculated from the number of electrons in bonding MOs (N_b) and antibonding MOs (N_a).

Question 58.
What are the key ideas of MO theory?
OR
What are the salient features of MO theory?
Answer:
Key ideas of MO Theory:

• MOs in molecules are similar to AOs of atoms. Molecular orbital describes region of space in the molecule representing the probability of an electron.
• MOs are formed by combining AOs of different atoms. The number of MOs formed is equal to the number of AOs combined.
• Atomic orbitals of comparable energies and proper symmetry combine to form molecular orbitals.
• MOs those are lower in energy than the starting AOs are bonding MOs and those higher in energy are antibonding MOs.
• The electrons are filled in MOs beginning with the lowest energy.
• Only two electrons occupy each molecular orbital and they have opposite spins, that is, their spins are paired.
• The bond order of the molecule can be calculated from the number of bonding and antibonding electrons.

Question 59.
Explain the formation of the following molecules on the basis of MOT. Also find the bond order.
i. H₂
ii. Li₂
iii. N₂
iv. O₂
v. F₂
Answer:
i. Hydrogen molecule (H₂):

a. Hydrogen atom (Z = 1) has electronic configuration as 1s¹.
b. Hydrogen atom contains one electron, hence hydrogen molecule which is diatomic contains two electrons.
c. Linear combination of two 1s atomic orbitals gives rise to two molecular orbitals σ1s and σ*1s.
d. The two electrons from the hydrogen atoms occupy the σ1s molecular orbital and σ*1s remains vacant.
e. Thus, electronic configuration of H₂ molecule is σ1s².
f. Since, no unpaired electron is present in hydrogen molecule, it is diamagnetic.
g. There are no electrons in the antibonding molecular orbital (σ*1s).
h. The bond order of H₂ molecule is
Bond order = \(\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}=\frac{2-0}{2}\) = 1
Thus, a single covalent bond is present between two hydrogen atoms.
[Note: The bond length is 74 pm and the bond dissociation energy is 438 kJ mol^-1.]

ii. Lithium molecule (Li₂):

a. Lithium atom (Z = 3) has electronic configuration as 1s² 2s¹.
b. Lithium atom has 3 electrons, hence Li₂ molecule has 6 electrons.
c. Linear combination of four atomic orbitals gives rise to four molecular orbitals namely σ1s, σ*1s, σ2S and σ*2s.
d. The electronic configuration of Li₂ molecule is (σ1s)² (σ*1s)² (σ2s)².
e. Since no unpaired electron is present in lithium molecule, it is diamagnetic.
f. Bond order of Li₂ molecule
\(=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}=\frac{4-2}{2}=1\)
Thus, a single covalent bond is present between two Li atoms. Hence, Li₂ is a stable molecule.

iii. Nitrogen molecule (N₂):

a. Nitrogen atom (Z = 7) has electronic configuration as 1s² 2s² 2p³.
b. Nitrogen atom contains 7 electrons, hence nitrogen molecule contains 14 electrons.
c. Linear combination of atomic orbitals gives rise to different molecular orbitals.
d. The electronic configuration of N₂ molecule is
N₂: (σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (π2p_x)² (π2p_y)² (σ2p_z)²
e. Since N₂ molecule does not have unpaired electron, it is diamagnetic.
f. Bond order of N₂ molecule
\(=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}=\frac{10-4}{2}=3\)
Thus, there are three bonds in N₂ molecule.

Question 60.
Study the following tables showing bond enthalpies of single and multiple bonds.

Bond	ΔaH/kJ mol-1
C-H	400-415
N-H	390
O-H	460-464
C-C	345
C-N	290-315
C-O	355-380
C-Cl	330

Bond	ΔaH/kJ mol-1
C-Br	275
O-O	175-184
C=C	610-630
C≡C	835
C=O	724-757
C≡N	854

i. Among single bonds, which bond is the strongest?
ii. How is bond enthalpy related to bond strength?
Answer:
i. Among single bonds, O-H bond is the strongest.
ii. Larger the bond enthalpy, stronger is the bond.

Question 61.
Write a short note on bond length.
Answer:
Bond length:

• Bond length is defined as the equilibrium distance between the nuclei of two covalently bonded atoms in a molecule.
• Each atom of the bonded pair contributes to the bond length.
• Bond length depends upon the size of atoms and multiplicity of bonds. It increases with increase in size of atom and decreases with increase in multiplicity of bond.
  e.g. C – C single bond is longer than C ≡ C triple bond.
• Bond lengths are measured by X-ray and electron diffraction techniques.

Question 62.
Cl-Cl covalent bond length is smaller than Br-Br covalent bond length. Explain.
Answer:
Bond length increases with increase in size of atom. Cl atom is smaller than Br atom. Hence, Cl-Cl covalent bond length is smaller than Br-Br covalent bond length.

Question 63.
Arrange the following bonds in decreasing order of bond strength: C-N, C=N, C≡N
Answer:
C≡N > C=N > C-N
Note: Average bond lengths for some single, double and triple bonds:

Type of bond	Covalent bond length (pm)
O-H	96
C-H	107
N-O	136
C-O	143
C-N	143
C-C	154
C=C	121
N=O	122
C=C	133
C=N	138
C≡N	116
C≡C	120

Type of bond	Covalent bond length (pm)
H2(H-H)	74
F2(F-F)	144
Cl2(Cl-Cl)	199
Br2(Br-Br)	228
I2(I-I)	267
N2(N≡N)	109
O2(O-O)	121
HF (H-F)	92
HCl (H-Cl)	127
HBr (H-Br)	141
HI (H-I)	160

Question 64.
Write a short note on bond order.
Answer:
i. According to the Lewis theory, bond order is given by the number of bonds between the two atoms in a molecule.
e.g. a. In hydrogen molecule, bond order between hydrogen atoms is one as one electron pair is shared.
b. In oxygen molecule, bond order between oxygen atoms is two as two electron pairs are shared.
c. In acetylene molecule, bond order between two carbon atoms is three as three electron pairs are shared.

ii. The isoelectronic molecules and ions have identical bond orders.
e.g. a. The bond order of F₂ and \(\mathrm{O}_{2}{ }^{2-}\) is one.
b. The bond order of N₂, CO and NO⁺ is 3.

iii. As the bond order increases, the bond enthalpy increases and bond length decreases.
iv. With the help of bond order, the stability of a molecule can be predicted.
[Note: N₂ molecule has bond enthalpy of 946 kJ mol^-1. It is one of the highest for diatomic molecules.]

Question 65.
Explain how polarity (ionic character) is developed in a covalent bond.
Answer:
i. Covalent bonds are formed between two atoms of the same or different elements.
ii. When a covalent bond is formed between atoms of same element such as H-H, F-F, Cl-Cl, etc., the shared pair of electrons is attracted equally to both atoms and is situated midway between two atoms. Such covalent bond is termed as nonpolar covalent bond.
iii. When a covalent bond is formed between two atoms of different elements that have different electronegativities, the shared electron pair does not remain at the centre. The electron pair is pulled towards the more electronegative atom resulting in the separation of charges. This give rise to dipole. The more electronegative atom acquires a partial -ve charge and the other atom gets a partial +ve charge. Such a bond is called as polar covalent bond. The examples of polar molecules include HF, HC1, etc.

Fluorine is more electronegative than hydrogen, therefore, the shared electron pair is more attracted towards fluorine and the atoms acquire partial +ve and -ve charges, respectively.
iv. Polarity of the covalent bond increases as the difference in the electronegativity between the bonded atoms increases. When the difference in electronegativities of combining atom is about 1.7, ionic percentage in the covalent bond is 50%.

Question 66.
Define and explain the term dipole moment.
Answer:
i. Dipole moment (μ) is the product of the magnitude of charge and distance between the centres of +ve and -ve charges.
ii. It is given by, µ = Q × r
where, Q = charge, r = distance of separation.
iii. Unit of dipole moment is Debye (D).
iv. Dipole moment being a vector quantity is represented by a small arrow with the tail on the positive centre and head pointing towards the negative centre.

Note: 1 D = 3.33564 × 10^-30 C m
where C is coulomb and m is meter.

Question 67.
Dipole moment in case of BeF2 is zero. Explain.
Answer:

• Dipole moment is a vector quantity. Therefore, the resultant dipole moment of a molecule is the vector sum of dipole moments of various bonds in the molecule.
• In BeF₂ molecule, Be-F bond is polar and has a bond dipole moment.
• BeF₂ is a linear molecule with two Be-F bonds oriented at 180° (opposite to each other).
• The two bond dipoles are equal in magnitude and act in opposite direction to cancel each other. Therefore, the net dipole moment in case of BeF₂ is zero.

Question 68.
Dipole moment in case of BF₃ is zero. Explain.
Answer:
i. Dipole moment is a vector quantity. Therefore, the resultant dipole moment of a molecule is the vector sum of dipole moments of various bonds in the molecule.
ii. In BF₃ molecule, B-F bond is polar and has a bond dipole moment.
iii. Also, in BF₃, the three B-F bonds are oriented at an angle of 120° to one another.
iv. The resultant of any two bond moments is equal in magnitude and opposite in direction to that of third. Hence, the net sum is zero and the dipole moment of tetra-atomic BF₃ molecule is zero.

Question 69.
Dipole moment of H₂O is higher than that of NH₃. Explain.
Answer:
In both NH₃ and H₂O, the central atom undergoes sp³ hybridization. In both the molecules, the orbital dipole due to the lone pair increases the effect of resultant dipole moment. However, in NH₃, nitrogen has only one lone pair while in H₂O, oxygen has two lone pairs. Hence, dipole moment of H₂O is higher than that of NH₃.

Question 70.
Dipole moment of NF₃ is less than that of NH₃, even though N-F bond is more polar than N-H bond. Explain.
Answer:

• Both NH₃ and NF₃ have pyramidal structure with a lone pair on the N atom. In NF₃, F is more electronegative than N while in NH₃, N is more electronegative.
• In NH₃, the orbital dipole due to the lone pair is in the same direction as the resultant dipole moment of N-H bonds, whereas in NF₃ the orbital dipole is in the direction opposite to the resultant dipole moment of three N-F bonds.
• The orbital dipole because of lone pair decreases the effect of the resultant N-F bond moments, which results in the low dipole moment of NF₃ (0.8 × 10^-30 C m) as compared to NH₃ (4.90 × 10^-30 C m).

Question 71.
CHCl₃ is polar. Explain.
Answer:
In CHCl₃, the dipoles are not equal and do not cancel each other. Hence, CHCl₃ is polar with a non-zero dipole moment.

Note: Dipole moments and geometry of some molecules are given in the following table:

Question 72.
Explain Fajan’s rule with suitable examples.
Answer:
Fajan’s rule:
i. Smaller the size of the cation and larger the size of the anion, greater is the covalent character of the ionic bond. For example, Li⁺ Cl^– is more covalent than Na⁺Cl. Similarly, Li⁺I^– is more covalent than Li⁺Cl^–.

ii. Greater the charge on cation, more is covalent character of the ionic bond. For example, covalent character of AlCl₃, MgCl₂ and NaCl decreases in the following order Al³⁺(Cl^–)₃ > Mg²⁺(Cl^–)₂ > Na⁺ Cl^–

iii. A cation with the outer electronic configuration of the s²p⁶d¹⁰ type possesses greater polarising power compared to the cation having the same size and same charge but having outer electronic configuration of s²p⁶ type.

This is because d electrons of the s²p⁶d¹⁰ shell screen nuclear charge less effectively compared to s and p electrons of s²p⁶ shell. Hence, the effective nuclear charge in a cation having s²p⁶d¹⁰ configuration is greater than that of the one having s²p⁶ configuration. For example: Cu⁺Cl^– is more covalent than Na⁺Cl^–. Here,
(Cu⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰; Na⁺ = 1s² 2s² 2p⁶)

Question 73.
Explain resonance with respect to \(\mathrm{CO}_{3}^{2-}\) ion.
Answer:
i. Three structures written for \(\mathrm{CO}_{3}^{2-}\) as follows:

ii. Each structure differs from the other only in the position of electrons without changing positions of the atoms. None of these individual structures is adequate to explain the properties of \(\mathrm{CO}_{3}^{2-}\).
iii. The actual structure of \(\mathrm{CO}_{3}^{2-}\) is a combination of three Lewis structures and is called as the resonance hybrid.
iv. Energy of the resonance hybrid structure is less than the energy of any single canonical form. Hence, resonance stabilizes certain polyatomic molecules or ions.
v. The average of all resonating structures contributes to overall bonding characteristic features of the molecule or ion.

Question 74.
Explain O₃ molecule is the resonance hybrid.
Answer:
Ozone is a resonance hybrid of structures I and II. The structures I and II are canonical forms while structure III is a resonance hybrid. The energy of structure III is less than that of I and II.

Question 75.
Define resonance energy.
Answer:
Resonance energy is defined as the difference in energy of the most stable contributing structure and the resonating forms.

Question 76.
Write the resonance structures of \(\mathrm{NO}_{3}^{-}\) ion.
Answer:
Resonance structures of \(\mathrm{NO}_{3}^{-}\) :

Question 77.
A student represents the Lewis dot structure of AlCl₃ molecule as shown below:

i. Is the representation correct? Justify your answer.
ii. If the chlorine atoms are replaced by bromine atoms, what will be the number of electrons present in the valence shell of aluminium?
Answer:
i. No, the representation is incorrect. There will be no lone pair of electrons on aluminium.
ii. The number of electrons present in the valence shell of aluminium will be six.

Question 78.
Below is an incomplete Lewis structure for glycine. Complete the following Lewis structure and answer the following questions. (Hint: Add lone pairs and multiple bonds to the structure below to give each atom a formal charge of zero.)

i. How many lone pairs of electrons are present on N-atom in the structure?
ii. How many pi bonds are present in the structure?
iii. How many sigma bonds are present in the structure?
Answer:
The correct Lewis structure is:

i. The number of lone pairs of electrons on N-atom is 1.
ii. The number of pi bonds in the structure is 1.
iii. The number of sigma bonds in the structure is 9.

Question 79.
Consider the following four species and answer the below given questions.
\(\mathrm{O}_{2}^{-}\), O₂ \(\mathrm{O}_{2}^{+}\), \(\mathrm{O}_{2}^{2-}\)
i. What is the bond order of \(\mathrm{O}_{2}^{+}\) ?
ii. Which species is least stable?
Answer:
i. Electronic configuration of \(\mathrm{O}_{2}^{+}\) can be given as:
(σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (σ2p_z)² (π2p_x)² (π2p_y)² (π*2p_x)¹ (π*2p_y)⁰
∴ Bond order = \(\frac {1}{2}\) (10 – 5) = 2.5

ii. Stability of the molecule or species ∝ Bond order
Bond order decreases as: \(\mathrm{O}_{2}^{+}\) > O₂ > \(\mathrm{O}_{2}^{-}\) > \(\mathrm{O}_{2}^{2-}\)
∴ Stability decreases as: \(\mathrm{O}_{2}^{+}\) > O₂ > \(\mathrm{O}_{2}^{-}\) > \(\mathrm{O}_{2}^{2-}\)
Hence, least stable species is \(\mathrm{O}_{2}^{2-}\).

Multiple Choice Questions

1. The CORRECT Lewis structure of CO molecule is:

Answer:

2. Which of the following molecule does NOT obey octet rule?
(A) BF₃
(B) CO₂
(C) H₂O
(D) N₂
Answer:
(A) BF₃

3. In BF₃, bond angle is .
(A) 90°
(B) 109°
(C) 120°
(D) 180°
Answer:
(C) 120°

4. Identify the geometry represented by the following diagram.

(A) Trigonal bipyramidal
(B) T-shape
(C) square planar
(D) square pyramidal
Answer:
(D) square pyramidal

5. The geometry of H₂S is ………….
(A) tetrahedral
(B) angular
(C) linear
(D) trigonal planar
Answer:
(B) angular

6. What will be the shape of molecule whose central atom is associated with 3 bonds and one lone pair?
(A) Trigonal pyramidal
(B) Tetrahedral
(C) Square planar
(D) Triangular planar
Answer:
(A) Trigonal pyramidal

7. Pair of molecules having identical geometry is …………..
(A) BF₃, NH₃
(B) BF₃, AlF₃
(C) BeF₂, H₂O
(D) BCl₃, PCl₃
Answer:
(B) BF₃, AlF₃

8. Which of the following molecule has bent shape?
(A) PCl₃
(B) OF₂
(C) BH₃
(D) BeBr2
Answer:
(B) OF₂

9. Which of the following is INCORRECT?
(A) The strength of the bond depends on the extent of overlap of the atomic orbitals.
(B) The extent of overlap depends on the shape and size of the atomic orbitals.
(C) The energy of the bonded atoms is more than that of the free atoms.
(D) During overlap of atomic orbitals, the electron density increases in between the two nuclei.
Answer:
(C) The energy of the bonded atoms is more than that of the free atoms.

10. In the potential energy curve for hydrogen molecule, the maximum stability is achieved when …………..
(A) potential energy of the system is maximum
(B) potential energy of the system is minimum
(C) force of repulsion become greater than force of attraction
(D) no bond formation takes place
Answer:
(B) potential energy of the system is minimum

11. In acetylene, C-C σ bond is formed by …………. overlap.
(A) sp²-sp²
(B) sp-sp
(C) sp-s
(D) p-p
Answer:
(B) sp-sp

12. The formation of O-H bonds in a water molecule involves …………. overlap.
(A) sp³-s
(B) sp¹-s
(C) sp-p
(D) sp³-p
Answer:
(A) sp³-s

13. The molecular orbital shown in the diagram can be described as ………….

(A) σ
(B) σ*
(C) π*
(D) π
Answer:
(C) π*

14. The bond order of lithium molecule is ………….
(A) one
(B) two
(C) three
(D) four
Answer:
(A) one

15. The bond order in N₂ molecule is …………
(A) 1
(B) 2
(C) 3
(D) 4
Answer:
(C) 3

16. The bond energies of F₂, Cl₂, Br₂ and I₂ are 37, 58, 46, and 36 kcal/mol respectively. The strongest bond is present in …………..
(A) Br₂
(B) I₂
(C) Cl₂
(D) F₂
Answer:
(C) Cl₂

17. The common features among the species CO and NO⁺ are: ……………
(A) isoelectronic species and bond order 3
(B) isoelectronic species and bond order 2
(C) odd electron species and unstable
(D) odd electron species and bond order 1
Answer:
(A) isoelectronic species and bond order 3

18. Which of the following is CORRECT for H₂O ?

	H=O bond	H2O molecule
(A)	polar	nonpolar
(B)	nonpolar	polar
(C)	polar	polar
(D)	nonpolar	nonpolar

Answer:
(C)

19. Each of the following molecules has a non-zero dipole moment EXCEPT:
(A) NF₃
(B) BF₃
(C) SO₂
(D) LiH
Answer:
(B) BF₃

20. Which of the following compounds is non-polar?
(A) HCl
(B) CH₂Cl₂
(C) CHCl₃
(D) CCl₄
Answer:
(D) CCl₄

21. The dipole moment of …………..
(A) NF₃ is higher than that of NH₂
(B) BF₃ is higher than that of NH₃
(C) H₂S is higher than that of H₂O
(D) HCl is higher than that of HBr
Answer:
(D) HCl is higher than that of HBr

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 4 Structure of Atom Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 4 Structure of Atom

Question 1.
Complete the information about the properties of subatomic particles in the following table:

Answer:

Question 2.
What are three important subatomic particles of an atom?
Answer:
Electron, proton and neutron are the three important subatomic particles of an atom.

Question 3.
Write a short note on discovery of electron.
Answer:

• In the year 1897, J J. Thomson studied the properties of cathode rays through a cathode ray tube experiment and found that the cathode rays are a stream of very small, negatively charged particles.
• These particles are 1837 times lighter than a hydrogen atom and are present in all atoms.
• These particles were later named as electrons.

Question 4.
Draw labelled diagram of Rutherford’s α-particle scattering experiment.
Answer:

Question 5.
Write a short note on discovery of proton.
Answer:

• After the discovery of nucleus in an atom, Rutherford found that the fast moving α-particles transmuted nitrogen into oxygen with simultaneous liberation of hydrogen.
  \({ }_{7}^{14} \mathrm{~N}+{ }_{2}^{4} \alpha \longrightarrow{ }_{8}^{17} \mathrm{O}+{ }_{1}^{1} \mathrm{H}\)
• He further showed that other elements could also be transmuted similarly and hydrogen was always emitted in the process.
• Based on these observations, he proposed that the hydrogen nucleus must be contained inside nuclei of all the elements. Hence, he renamed hydrogen nucleus as proton.

Question 6.
Write a short note on discovery of neutron.
Answer:

• In 1920, Ernest Rutherford proposed the existence of an electrically neutral and massive particle in the nucleus of an atom in order to account for the disparity in atomic number and atomic mass of an element.
• In 1932, James Chadwick measured the velocity of protons ejected from paraffin by an unidentified radiation from beryllium (Be).
• From that he determined the mass of the particles of this unidentified neutral radiation, which was found to be almost same as that of the mass of a proton.
• He named this neutral particle as ‘neutron’, which was earlier predicted by Rutherford.

Question 7.
State true or false. Correct the false statement.
i. An electron is 1837 times lighter than a proton.
ii. In Rutherford’s experiment of scattering of α-particles by thin gold foil, most of the α-particles
bounced back.
iii. Cathode rays are a stream of very small, positively charged particles.
Answer:
i. True
ii. False,
In Rutherford’s experiment of scattering of α-particles by thin gold foil, very few α-particles bounced back.
iii. False,
Cathode rays are a stream of very small, negatively charged particles.

Question 8.
Explain the term: Atomic number
Answer:

• Atomic number is defined as the number of protons present in the nucleus of an atom of a particular element.
• Atomic number is represented by Z.
• An atom is electrically neutral. Hence, the number of protons equals to the number of electrons. In other words, the atomic number of an atom is equal to the number of electrons.

∴ Atomic Number (Z) = Number of protons = Number of electrons

Question 9.
Give reason: The approximate atomic mass in Daltons is numerically equal to the number of nucleons in the atom.
Answer:

• The electrons possess negligible mass. They do not contribute much to the mass of an atom.
• Therefore, the entire mass of an atom is supposed to be present in the nucleus which consists of protons and neutrons, which are collectively called as nucleons.

Hence, approximate atomic mass in Daltons is numerically equal to the number of nucleons in the atom.

Question 10.
Explain: Atomic mass number.
Answer:

• The sum of the total number of protons and neutrons present in the nucleus of an atom is called the atomic mass number of that atom.
• Atomic mass number is represented by A
• Mass number (A) = Number of protons (Z) + Number of neutrons (N) = Total number of nucleons
  ∴ A = Z + N OR N = A – Z

Question 11.
How is an atom of an element ‘X’ having atomic number ‘Z’ and atomic mass number ‘A’ represented?
Answer:
An atom of an element ‘X’ having atomic number ‘Z’ and atomic mass number ‘A’ is represented as: \({ }_{Z}^{A} X\)

Question 12.
What is a nuclide?
Answer:
The atom or nucleus having a unique composition as specified by \({ }_{Z}^{A} X\) is called as a nuclide.

Question 13.
If an element ‘X’ has 6 protons and 8 neutrons, then write its representation.
Answer:
The representation of the given element is \({ }_{6}^{14} \mathrm{X}\).

Question 14.
Three elements Q, R and T have mass number 40. Their atoms contain 22, 21 and 20 neutrons respectively. Represent their atomic composition with appropriate symbol.
Answer:
Mass number (A) = Number of protons (Z) + Number of neutrons (N) .
A = Z + N
∴ Z = A – N
For the given three elements, A = 40. Values of their atomic numbers Z, are calculated from the given values of the number of neutrons, N, using the above formula.
For Q: Z = A – N = 40 – 22 = 18
For R: Z = A – N = 40 – 21 = 19
For T: Z = A – N = 40 – 20 = 20
Now, atomic composition of an element (X) is represented as \({ }_{Z}^{A} X\).
The atomic compositions of the three elements are written as follows:
\({ }_{18}^{40} \mathrm{Q}\), \({ }_{19}^{40} \mathrm{R}\), \({ }_{20}^{40} \mathrm{T}\)

Question 15.
Find out the number of protons, electrons and neutrons in the nuclide \({ }_{18}^{40} \mathrm{Ar}\).
Solution:
For the given nuclide,
Atomic number, Z = 18, Mass number, A = 40
Number of protons = Number of electrons = Z = 18
Number of neutrons (N) = A – Z = 40 – 18 = 22
Ans: Number of protons = 18, Number of electrons = 18, Number of neutrons = 22

Question 16.
How many protons, electrons and neutrons are there in the following nuclei?
i. \({ }_{8}^{17} \mathrm{O}\)
ii. \({ }_{12}^{25} \mathrm{Mg}\)
iii. \({ }_{35}^{80} \mathrm{Br}\)
Solution:
i. \({ }_{8}^{17} \mathrm{O}\)
Atomic number, Z = 8, Mass number, A = 17
Number of protons = Number of electrons = Z = 8
Number of neutrons (N) = A – Z = 17 – 8 = 9
Ans: Number of protons = 8, Number of electrons = 8, Number of neutrons = 9

ii. \({ }_{12}^{25} \mathrm{Mg}\)
Atomic number, Z = 12, Mass number, A = 25
Number of protons = Number of electrons = Z = 12
Number of neutrons (N) = A – Z = 25 – 12 = 13
Ans: Number of protons = 12, Number of electrons = 12, Number of neutrons = 13

iii. \({ }_{35}^{80} \mathrm{Br}\)
Atomic number, Z = 35, Mass number, A = 80
Number of protons = Number of electrons = Z = 35
Number of neutrons (N) = A – Z = 80 – 35 = 45
Ans: Number of protons = 35, Number of electrons = 35, Number of neutrons = 45

Question 17.
Define isotopes.
Answer:
Isotopes are defined as the atoms of an element having the same number ofprotons but different number of neutrons in their nuclei.
e.g. \({ }_{6}^{12} \mathrm{C}\), \({ }_{6}^{13} \mathrm{C}\) and \({ }_{6}^{14} \mathrm{Br}\) are isotopes.

Question 18.
Complete the information about the isotopes of carbon in the following table:

Answer:

Question 19.
Define isobars.
Answer:
Isobars are defined as the atoms of different elements having the same mass number but different atomic number.
e.g. \({ }_{6}^{14} \mathrm{C}\) and \({ }_{7}^{14} \mathrm{N}\) are isobars.

Question 20.
Complete the following table:

Answer:

Question 21.
The two natural isotopes of chlorine viz. \({ }_{17}^{35} \mathrm{Cl}\) and \({ }_{17}^{37} \mathrm{Cl}\) exist in relative abundance of 3 : 1. Find out the average atomic mass of chlorine.
Solution:
Given: Isotopes of chlorine \({ }_{17}^{35} \mathrm{Cl}\) and \({ }_{17}^{37} \mathrm{Cl}\).
Ratio of relative abundance of these isotopes is 3 : 1.
To find: Average atomic mass of chlorine
Calculation: From the relative abundance 3 : 1, it is understood that out of 4 chlorine atoms, 3 atoms have mass 35 and 1 atom has mass 37.
Therefore, the average atomic mass of chlorine = \(\frac{3 \times 35+1 \times 37}{4}\) = 35.5
∴ Average atomic mass of chlorine = 35.5 u
Ans: The average atomic mass of chlorine is 35.5 u.

Question 22.
Find out the average atomic mass of lithium (Li) from the following data:

Isotope	Atomic mass (u)	Abundance
6Li	6.015	7.59%
7Li	7.016	92.41%

Solution:
Given: Three isotopes of lithium along with respective atomic mass and % abundance.
To find: Average atomic mass of lithium
Calculation:

Ans: Average atomic mass of lithium is 6.940 u.

Question 23.
Certain results were obtained when scientists studied the interactions of radiation with matter. What were the two results, utilized by Neils Bohr to overcome the drawbacks of Rutherford model?
Answer:
The two results utilized by Neils Bohr to overcome the drawbacks of Rutherford model were:

• Wave particle duality of electromagnetic radiation
• Line emission spectra of hydrogen

Question 24.
Explain in short the wave particle duality of light (electromagnetic radiation).
Answer:
Wave particle duality of light (electromagnetic radiation):

• Light has both particle and wave like nature.
  Phenomena such as diffraction and interference of light could be explained by treating light as electromagnetic wave.
• However, the black-body radiation or photoelectric effect could not be explained by wave nature of light. This could be accounted for by considering particle nature of light. Thus, both phenomena could be explained only by accepting that light has dual behaviour.
• When light interacts with matter it behaves as a stream of particles (called photons) and when light propagates, it behaves as an electromagnetic wave.

Question 25.
Observe the following figure of an electromagnetic wave and answer the questions given below:

i. What does ‘x’ represent?
ii. What does ‘y’ represent?
Answer:
i. ‘x’ represents amplitude of the wave.
ii. ‘y’ represents wavelength of the wave.

Question 26.
Define and explain the following terms:
i. Wavelength (λ).
ii. Frequency (ν)
iii. Wavenumber (\(\bar{v}\))
iv. Amplitude (A)
v. Velocity (c)
Answer:
i. Wavelength (λ):

• The distance between two consecutive crests or two consecutive troughs in a wave is called wavelength.
• It is represented by Greek letter λ (lambda).
• The SI unit for wavelength is metre (m).

Note: The other units include Angstrom, nanometre, picometer (1 pm = 10^-12 m) and micron (1µ = 10^-6 m).
1Å = 10^-8 cm = 10^-10 m
1nm = 10^-9 m = 10Å

ii. Frequency (ν):

• The number of waves that pass a given point in one second is called frequency.
• It is represented by Greek letter ‘ν’ (nu).
• The SI unit of frequency is Hertz (Hz) or s^-1.

Note: 1 Hz = 1 cycle per second (1 cps)
The units, kilo Hertz (kHz) and mega Hertz (mHz) are commonly used.
1 kHz = 10³ Hz = 10³ cps
1 mHz = 10⁶ Hz = 10⁶ cps

iii. Wavenumber (\(\bar{v}\)):

• The number of wavelengths per unit length is called the wavenumber.
• It is represented by \(\bar{v}\) (nu bar).
• The commonly used unit for wavenumber is cm^-1 while its SI unit is m^-1.
• Wavenumber of a wave is related to the wavelength as follows:
  \(\bar{v}\) = \(\frac{1}{\lambda}\)

iv. Amplitude (A):

• The height of a crest or the depth of a trough from the line of propagation of the wave is called
  amplitude.
• It is represented by letter ‘A’.
• The square of the amplitude represents the intensity (brightness) of the radiation.

v. Velocity (c):

• The distance travelled by a wave in one second is called the velocity of the wave.
• It is denoted by letter c.
• It is the product of the frequency and wavelength. Hence, c = νλ
• The velocity of all types of electromagnetic radiations (in space or in vacuum) is the same and it is equal to the velocity of light (3 × 10¹⁰ cm s^-1 or 3 × 10⁸ m s^-1. However, they may have different wavelengths and frequencies.

Question 27.
Write a short note on quantum theory of radiation.
Answer:
i. Max Planck put forward a theory known as Planck’s quantum theory to explain black-body radiation.
ii. According to this theory, the energy of electromagnetic radiation depends upon the frequency and not the amplitude.
iii. The smallest quantity of energy that can be emitted or absorbed in the form of electromagnetic radiation is called as ‘quantum’.
iv. The energy (E) of each quantum of radiation is directly proportional to its frequency (ν).
i.e., E ∝ ν ; E = hν
where, h = Planck’s constant = 6.626 × 10^-34 J s.

Question 28.
Parameters of blue and red light are 400 nm and 750 nm respectively. Which of the two is of higher energy?
Answer:
400 nm and 750 nm are the wavelengths of blue and red light respectively. Energy of radiation is given by the expression E = hν and the frequency (ν), of radiation is related to the wavelength by the expression.
ν = \(\frac{c}{\lambda}\)
∴ E = \(\frac{\mathrm{hc}}{\lambda}\)
Therefore, shorter the wavelength, λ, larger the frequency, ν, and higher the energy, E. Thus, blue light which has shorter λ (400 nm) than red light (750 nm) has higher energy.

Question 29.
What is an emission spectrum?
Answer:

• When a substance is irradiated with electromagnetic radiation, it absorbs energy. Atoms, molecules or ions, which have absorbed radiation are said to be ‘excited’. Heating can also result in an excited state.
• The excited species emits the absorbed energy in the form of radiation. This process is called emission of radiation and the recorded spectrum of this emitted radiation is called ‘emission spectrum’.

Question 30.
Give some examples of commonly used light sources that work on atomic emission.
Answer:
Examples are fluorescent tube, sodium vapor lamp, neon sign and halogen lamp.

Question 31.
Write a short note on emission spectrum of hydrogen. Also, list all the five series of lines in the hydrogen spectrum.
Answer:
i. When electric discharge is passed through gaseous hydrogen, it emits radiation. The recorded spectrum of this emitted radiation is called hydrogen emission spectrum.

ii. This spectrum falls in different regions of electromagnetic radiation and it is comprised of a series of lines corresponding to different frequencies. That is, the spectrum was discontinuous.

iii. In the year, 1885, Balmer expressed the wave numbers of the emission lines in the visible region of electromagnetic spectrum by the formula:
\(\overline{\mathrm{v}}=109677\left[\frac{1}{2^{2}}-\frac{1}{\mathrm{n}^{2}}\right] \mathrm{cm}^{-1}\)
where, n = 3, 4, 5,…
These lines are known as Balmer series.

iv. Rydberg found that other series of lines could be described by the following formula:
\(\bar{v}=109677\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right] \mathrm{cm}^{-1}\)
where, 109677 cm^-1 is called Rydberg constant for hydrogen (R_H).

v. Different series of emission spectral lines for hydrogen are as follows:

Series	n1	n2	Region
Lyman	1	2, 3, 4, ….	Ultraviolet
Balmer	2	3, 4, 5, ….	Visible
Paschen	3	4, 5, 6, ….	Infrared
Bracket	4	5, 6, 7, ….	Infrared
Pfund	5	6, 7, 8,….	Infrared

Question 32.
Observe the emission spectrum of hydrogen and answer the following questions.

i. Is the spectrum continuous?
ii. In which region of electromagnetic radiation does the Paschen series belong?
iii. Which series falls in the visible region of electromagnetic radiation?
Answer:
i. The spectra is not continuous and comprises of a series of lines corresponding to different frequencies.
ii. Paschen series falls in the infrared region of electromagnetic radiation.
iii. Balmer series falls partly in the visible region of electromagnetic radiation.

Question 33.
Give the expression to calculate wavenumber of the emission lines in the Balmer series.
Answer:
\(\bar{v}=109677\left[\frac{1}{2^{2}}-\frac{1}{\mathrm{n}^{2}}\right] \mathrm{cm}^{-1}\)
where, n = 3, 4, 5, ….

Question 34.
Visible light has wavelengths ranging from 400 nm (violet) to 750 nm (red). Express these wavelengths in terms of frequency (Hz). (1 nm = 10^-9 m)
Solution:
Given: Wavelengths: λ₁ = 400 nm (for violet light), λ₂ = 750 nm (for red light)
To find: Frequencies: ν₁, ν₂
Formula: ν = \(\frac{c}{\lambda}\)
Calculation: i. Wavelength of violet light, λ₁ = 400 nm = 400 × 10^-9 m
Frequency, ν = \(\frac{c}{\lambda}\) where c is speed of light = 3.0 × 10⁸ ms^-1
∴ ν₁ = \(\frac{3.0 \times 10^{8} \mathrm{~ms}^{-1}}{400 \times 10^{-9} \mathrm{~m}}\) = 7.50 × 10¹⁴ Hz
ii. Wavelength of red light, λ₂ = 750 nm = 750 × 10^-9 m
Frequency, ν = \(\frac{c}{\lambda}\)
∴ ν₂ = \(\frac{3.0 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}}{750 \times 10^{-9} \mathrm{~m}}\) = 4.00 × 10¹⁴ Hz
Ans: The frequency of violet light is 7.50 × 10¹⁴ Hz and that of red light is 4.00 × 10¹⁴ Hz.

Question 35.
Yellow light emitted from a lamp has a wavelength of 580 nm. Find the frequency and wavenumber of this light.
Solution:
Given: Wavelength (λ) = 580 nm
To find: Frequency (ν), Wave number \(\bar{v}\)
Formulae : \(v=\frac{c}{\lambda}, \bar{v}=\frac{1}{\lambda}\)
Calculation: Wavelength of yellow light (λ) = 580 nm = 580 × 10^-9 m [1 nm = 10^-9 m]
We know that frequency (ν) is related to wavelength as: ν = \(\frac{c}{\lambda}\)
where, c, velocity of light = 3.0 × 10^-8 m s^-1
∴ ν = \(\frac{3.0 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}}{580 \times 10^{-9} \mathrm{~m}}=5.17 \times 10^{14} \mathrm{~s}^{-1}\)
Again, Wave number, \(\bar{v}=\frac{1}{\lambda}=\frac{1}{580 \times 10^{-9} \mathrm{~m}}=1.72 \times 10^{6} \mathrm{~m}^{-1}\)
Ans: Frequency = 5.17 × 10¹⁴ s^-1 and wave number = 1.72 × 10⁶ m^-1

Question 36.
Calculate the energy of a photon of radiation having wavelength 300 nm. [h = 6.63 × 10^-34 J s]
Solution:
Given: Wavelength (λ) = 300 nm
To find: Energy of a photon (E)
Formulae: E = \(\frac{\mathrm{hc}}{\lambda}\)
Calculation: From formula,
E = \(\frac{6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s} \times 3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}}{300 \times 10^{-9} \mathrm{~m}}=6.63 \times 10^{-19} \mathrm{~J}\)
Ans: Energy of a photon is 6.63 × 10^-19 J.

Question 37.
Explain briefly the results of Bohr’s theory for hydrogen atom.
Answer:
i. The stationary states for electrons are numbered n = 1, 2, 3……. These integers are known as principal quantum numbers.
ii. The radii of the stationary states are r_n = n² a₀, where a₀ = 52.9 pm (picometer). Thus, the radius of the first stationary state, called the Bohr radius is 52.9 pm.
iii. The most important property associated with the electron is the energy of its stationary state. It is given by the formula:
E_n = -R_H (1/n²), where n = 1, 2, 3, …..
R_H is the Rydberg constant for hydrogen and its value in joules is 2.18 × 10^-18 J.
The lowest energy state is called the ground state. Energy of the ground state (n = 1) is:
E₁ = -2.18 × 10^-18 × 1/1² = -2.18 × 10^-18 J
Energy of the stationary state corresponding to n = 2 is
E₂ = -2.18 × 10^-18 × (1/(2)²) = -0.545 × 10^-18 J.
iv. Bohr theory can be applied to hydrogen like species. For example, He⁺, Li²⁺, Be³⁺ and so on. Energies and radii of the stationary states associated with these species are given by:

where, Z is the atomic number. From the above expressions, it can be seen that the energy decreases (becomes more negative) and radius becomes smaller as the value of Z increases.
v. Velocities of electrons can also be calculated from the Bohr theory. Qualitatively, it is found that the magnitude of velocity of an electron increases with increase of Z and decreases with increase in the principal quantum number (n).

Question 38.
How many electrons are present in \({ }_{1}^{2} \mathrm{H}\), ₂He and He⁺ ? Which of these are hydrogen-like species?
Answer:
Hydrogen-like species contain only one electron.
Consider \({ }_{1}^{2} \mathrm{H}\):
Number of protons = Number of electrons = 1
Consider ₂He:
Number of protons = Number of electrons = 2
Consider He⁺:
Number of electrons = (Number of electrons in He – 1) = 2 – 1 = 1
Thus, \({ }_{1}^{2} \mathrm{H}\) and He⁺ are hydrogen-like species.
[Note: Bohr’s theory is applicable to hydrogen atom and hydrogen-like species, which contain only one electron.]

Question 39.
Describe how the line spectrum of hydrogen is explained by Bohr theory.
Answer:
i. According to second postulate of Bohr theory, radiation is emitted when an electron moves from an outer orbit of higher principal quantum number (n_i) to an inner orbit of lower principal quantum number (n_f). The energy difference (ΔE) between the initial and final orbit of the electronic transition corresponds to the energy of the emitted radiation.
ii. From the third postulate of Bohr theory, ΔE can be expressed as
ΔE = E_i – E_f …….(1)
iii. According to the results derived from Bohr theory, the energy (E_n) of an orbit is related to its principal quantum number ‘n’ by the equation:
E = \(-\mathrm{R}_{\mathrm{H}}\left(\frac{1}{\mathrm{n}^{2}}\right)\) ……(2)
iv. On combining these two equations, we get:

v. Substituting the value of R_H in joules, we get

vi. This expression can be rewritten in the terms of wavenumber of the emitted radiation in the following steps:
(ΔE) J = (h) J s × (ν) Hz
and

This equation appears like the Rydberg equation, where, n_f = n₁ and n_i = n₂.
In other words, Bohr theory successfully accounts for the empirical Rydberg equation for the line emission spectrum of hydrogen.

Question 40.
Observe the following diagram showing electronic transition in the hydrogen spectrum.

i. Electron jumps from higher energy level to n = 1. Which series does it correspond to?
ii. Electron jumps from higher energy level to n = 4. Which series does it correspond to?
iii. Which transition will give second line of Balmer series?
Answer:
i. When electron jumps from higher energy level to n = 1, it corresponds to Lyman series.
ii. When electron jumps from higher energy level to n = 1, it corresponds to Bracket series.
iii. When electron jumps from n = 4 to n = 1, the second line of Balmer series is observed.

Question 41.
Explain de Broglie’s equation.
Answer:
de Broglie’s equation:

• de Broglie proposed (in 1924) that matter should exhibit a dual behaviour. That is, every object which possesses a mass and velocity behaves both as a particle and as a wave. An electron has mass and velocity. This means that an electron should have momentum (p), a property of particle as well as wavelength (λ), a property of wave.
• According to de Broglie, the wavelength λ of a particle of mass m moving with a velocity v is
  λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) where, h is Planck’s constant.
• The quantity mv gives the momentum of the particles.
  λ = \(\frac{\mathrm{h}}{\mathrm{p}}\) where, p represents the momentum of the particle.
• de Broglie’s prediction was confirmed by diffraction experiments (a wave property).

[Note: According to de Broglie’s equation, the wavelength of a moving particle is inversely proportional to its mass. Therefore, heavier particles have much smaller wavelength than lighter particles like electrons.]

Question 42.
Write a note on Heisenberg’s uncertainty principle.
Answer:
i. Uncertainty principle was proposed by Wemer Heisenberg in 1927. It can be stated as “It is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron ”.
ii. If Δx is the uncertainty in the determination of the position of a very small moving particle and Δpx is the uncertainty in the determination of its momentum, then
Δx Δp_x ≥ \(\frac{\mathrm{h}}{4 \pi}\) …….(1) where h is Planck’s constant
iii. The above equation can alternatively be stated as,
Δx × m × Δv_x ≥ \(\frac{\mathrm{h}}{4 \pi}\), because Δp_x = m × Δv_x ……..(2)
where Δv_x is the uncertainty in the determination of velocity and m is the mass of the particle.

Question 43.
Calculate the radius and energy associated with the first orbit of He.
Solution:
Given: n = 1
To find: Radius and energy associated with the first orbit of He⁺
Formulae:

Calculation: He⁺ is a hydrogen-like species having Z = 2.
Using formula (i),
Radius of the first orbit of He⁺ = r₁ = \(\frac{52.9 \times(1)^{2}}{2} \mathrm{pm}\)
= 26.45 pm
Using formula (ii),
Energy of the first orbit of He⁺ = E₁ = -2.18 × 10^-18 \(\left(\frac{2^{2}}{1^{2}}\right) \mathrm{J}\)
= -8.72 × 10^-18 J
Ans: Radius of the first orbit of He⁺ is 26.45 pm and energy of the first orbit of He⁺ is -8.72 × 10^-18 J.

Question 44.
What is the wavelength of the photon emitted during the transition from the orbit of n = 5 to that of n = 2 in hydrogen atom?
Solution:
Given: n_i = 5, n_f = 2
To find: Wavelength of the photon emitted

Ans: Wavelength of the photon emitted is 434 nm.

Question 45.
Calculate the mass of a hypothetical particle having wavelength 5894 A and velocity 1.0 × 10⁸ ms^-1.
Solution:
Given: Wavelength (λ) = 5894 Å, Velocity (ν) = 1.0 × 10⁸ ms^-1
To find: Mass of a particle
Formula: λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) or m = \(\frac{\mathrm{h}}{\lambda \mathrm{v}}\) (according to de-Broglie equation)
Calculation: m = \(\frac{\mathrm{h}}{\lambda \mathrm{v}}\)
∴ m = \(\frac{6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-1}}{\left(5894 \times 10^{-10} \mathrm{~m}\right) \times\left(1.0 \times 10^{8} \mathrm{~ms}^{-1}\right)}\)
= 1.124 × 10^-35 kg
Ans: Mass of a particle is 1.124 × 10^-35 kg.

[Calculation using log table:
\(\frac{6.626 \times 10^{-34}}{5894 \times 10^{-10} \times 1.0 \times 10^{8}}=\frac{6.626}{5894} \times 10^{-32}\)
= Antilog₁₀ [log₁₀ 6.626 – log₁₀ 5894] × 10^-32
= Antilog₁₀ [0.8213 – 3.7704] × 10^-32
= Antilog₁₀ [latex]\overline{3} .0509[/latex] × 10^-32 = 1.124 × 10^-35]

Question 46.
Write a short note on Schrodinger equation.
Answer:
Schrodinger equation or wave equation:
i. Schrodinger developed the fundamental equation of quantum mechanics which incorporates wave particle duality of matter. The Schrodinger equation or wave equation is written as:
It \(\hat{\mathbf{H}}\)ψ = Eψ
Here \(\hat{\mathbf{H}}\) is a mathematical operator called Hamiltonian, ψ (psi) is the wave function and E is the total energy of the system.

ii. When Schrodinger equation is solved for an electron in hydrogen atom, the possible values of energy states (E) that the electron may have along with the corresponding wave function (ψ) are obtained. As a consequence of solving this equation, a set of three quantum numbers characteristic of the quantized energy levels and the corresponding wave functions are obtained. These are: Principal quantum number (n), azimuthal quantum number (l) and magnetic quantum number (m_l).

iii. The solution of Schrodinger wave equation led to three quantum numbers and successfully predicted features of hydrogen atom emission spectrum.

iv. Splitting of spectral lines in multi-electron atomic emission spectra could not be explained through such model. These were explained by George Uhlenbeck and Samuel Goudsmit (1925) who proposed the presence of the fourth quantum number called electron spin quantum number, m_s.

Question 47.
Write a short note on magnetic orbital quantum number (m_l).
Answer:
Magnetic orbital quantum number (m_l):

• Magnetic orbital quantum number describes the relative spatial orientation of the orbitals in a given subshell.
• It is denoted by m; and it has values from -l to +l through zero, giving total values or total orientations equal to (2l + 1).
• For s-subshell, 1 = 0, hence, m_l = 0. Thus, s-subshell contains only one orbital.
• For p-subshell, l = 1, hence, m_l = +1, 0, -1. Thus, p-subshell contains three orbitals having distinct orientations.

Question 48.
If n = 2, what are the values of quantum number l and m_l ?
Answer:
For a given n, l = 0 to (n – 1) and for given l, m_l = -l…., 0…. + l
Therefore, the possible values of l and ml for n = 2 are:

Value of n	Value of l	Value of ml
2	0	ml = 0
	1	ml = -1 ml = 0 ml = +1

Question 49.
What are the values of m_l for f-subshell?
Answer:
For f-subshell, l = 3. Therefore, m_l has seven values: + 3, + 2, + 1, 0, -1, -2, -3.

Question 50.
How many orbitals make the N-shell? What is the subshell wise distribution of orbitals in the N-shell?
Answer:
For N-shell principal quantum number n = 4
∴ Total number of orbitals in N-shell = n² = 4² = 16. The total number of subshells in N-shell = n = 4.
The four subshells with their azimuthal quantum numbers and the constituent number of orbitals are as shown below:

Azimuthal quantum number (l)	Symbol of subshell	Number of orbitals (2l + 1)
l = 0	s	(2 × 0) + 1 = 1
l = 1	P	(2 × 1) + 1 = 3
l = 2	d	(2 × 2) + 1 = 5
l = 3	f	(2 × 3) + 1 = 7

Question 51.
Complete the following flow chart:

Answer:

Question 52.
Write a short note on electron spin quantum number.
Answer:
Electron spin quantum number (m_s):

• Electron spin quantum number describes the spin state of the electron in an orbital. It is designated as ms.
• An electron spins around its axis and this imparts spin angular momentum to it.
• The two orientations which the spin angular momentum of an electron can take up give rise to the spin states which can be distinguished from each other by the spin quantum number, m_s, which can be either +1/2 or -1/2.
• The two spin states are represented by two arrows, ↑ (pointing up) and ↓ (pointing down) and thus have opposite spins.

Question 53.
An atom has two electrons in its 4s orbital. Write the values of the four quantum numbers for each of them.
Answer:
For the 4s orbital, 4 stands for the principal quantum number n; s stands for the subshell s having the value of azimuthal quantum number, l = 0. In the ‘s’ subshell, there is only one orbital and has magnetic quantum number, ml = 0. The two electrons in this orbital have opposite spins. Thus, the four quantum numbers of two electrons in 4s orbital are:

Question 54.
Write a short note on probability density of electron.
Answer:
i. The probability of finding an electron at a given point in an atom is proportional to the square of the wave function at that point (ψ²).
ii. According to Max Born, the square of wave function at a point in an atom is the probability density of the electron at that point.
The following figure shows the probability density diagrams of Is and 2s atomic orbitals. These diagrams appear like a cloud.
The electron cloud of 2s orbital shows one node, which is a region with nearly zero probability density and displays the change of sign for its corresponding wavefunction.
Diagram:

Question 55.
What is meant by the term ‘boundary surface diagram’?
Answer:
A boundary surface is drawn in space for an orbital such that the value of probability density (ψ²) is constant and encloses a region where the probability of finding electron is typically more than 90%. Such a boundary surface diagram is a good representation of shape of an orbital.
e.g. Boundary surface diagram of Is and 2s orbitals are spherical in shape.

Question 56.
Describe the shape of s orbital.
Answer:
Shape of s orbital:

• For each value of principal quantum number ‘n’, there is only one s-orbital.
• For s-orbital, l = 0 and m_l = 0, hence s-orbital has only one orientation i.e., the probability of finding the electrons is same in all directions. Thus, s-orbital is spherically symmetrical around the nucleus.
• The value of n determines the size of an orbital. With increase in the value of n, the size of the s-orbital increases.

Diagram:

Question 57.
Describe the shape of 2p orbitals.
Answer:
Shape of 2p orbitals:
i. For p orbital, l = 1. For l = 1, m_l = +1, 0, -1. Thus, p orbitals have three orientations.
ii. Each orbital has two lobes on the two sides of a nodal plane passing through the nucleus.
iii. Shape of 2p orbitals resembles a dumbbell.
iv. The size and energy of the three 2p orbitals are the same. However, their orientations in space are different. The lobes of the three 2p orbitals are along the x, y and z axes. Accordingly, the corresponding orbitals are designated as 2p_x, 2p_y and 2p_z. The size and energy of the orbitals in p subshell increase with the increase of principal quantum number.
Diagram:

Question 58.
Describe the shape of 3d orbitals.
Answer:
Shape of 3d orbitals:

• For d orbital, l = 2. For l = 2, m_l = +2, +1, 0, -1, -2. Thus, d orbitals have five orientations.
• They are designated as d_xy, d_yz, d_zx, d_x²-y² and d_z.
• Shape of 3d orbitals are shown in the following figure. The first three have double dumb-bell shape. They lie in xy, yz and xz plane, respectively. The d_x²-y² is also dumb-bell shaped and lies along the x and y axes. d_z² is dumb-bell shaped along z axis with a dough-nut shaped ring of high electron density around the nucleus in xy plane.
• In spite of difference in their shapes, the five d orbitals are equivalent in energy. The shapes of 4d, 5d, 6d…….. orbitals are similar to those of 3d orbitals, but their respective size and energies are large or they are said to be more diffused.

Diagram:

Question 59.
Explain (n + l) rule with respect to energies of orbitals.
Answer:
The lower the sum (n + l) for an orbital, the lower is its energy. If two orbitals have the same (n + l) values, then the orbital with the lower value of n is of lower energy. This is called the (n + l) rule.

Question 60.
What are the two methods of representing electronic configuration of an atom?
Answer:
The two methods of representing electronic configuration of an atom are:
i. Orbital notation: ns^a np^b nd^c …..
In the orbital notation method, a shell is represented by the principal quantum number (n) followed by respective symbol of the subshell. The number of electrons occupying that subshell being written as superscript on right side of the symbol.

ii. Orbital diagram:
In the orbital diagram method, each orbital in a subshell is represented by a box and the electrons represented by an arrow (↑ for up spin and ↓ for down spin) are placed in the respective boxes. In this method, all the four quantum numbers of electron are accounted for.

Note: Consider two electrons in 3s orbital:

Question 61.
Complete the following table:

Answer:

Question 62.
Explain condensed orbital notation of electronic configuration of an atom.
Answer:

• The orbital notation of electronic configuration of an element with high atomic number comprises a long train of symbols of orbitals with an increasing order of energy.
• It can be condensed by dividing it into two parts: Inner or core part of electronic configuration and outer electronic configuration.
• Electronic configuration of the preceding inert gas is a part of the electronic configuration of any element. In the condensed orbital notation, it is implied by writing symbol of that inert gas in a square bracket. It is core part of the electronic configuration of that element. The outer electronic configuration is specific to a particular element and written immediately after the bracket.
• For example, the orbital notation of potassium ‘K (Z = 19) is Is² 2s² 2p⁶ 3s² 3p⁶ 4s¹’. Its core part is the electronic configuration of the preceding inert gas argon ‘Ar: 1s² 2s² 2p⁶ 3s² 3p⁶, while ‘4s¹’ is an outer part. Therefore, the condensed orbital notation of electronic configuration of potassium is ‘K: [Ar] 4s².’

Note: Electronic configuration of the elements with atomic numbers 1 to 30 is as follows:

Question 63.
Write electronic configuration of ₁₈Ar and ₁₉K using orbital notation and orbital diagram method.
Answer:
From the atomic numbers, it is understood that 18 electrons are to be filled in Ar atom and 19 electrons are to be filled in K atom. These are to be filled in the orbitals according to the Aufbau principle. The electronic configuration of these atoms can be represented as:

Question 64.
Write condensed orbital notation of electronic configuration of the following elements:
i. Fluorine (Z = 9)
ii. Scandium (Z = 21)
iii. Cobalt (Z = 27)
iv. Zinc (Z = 30)
Answer:

No.	Element	Condensed orbital notation
i.	Fluorine (Z = 9)	[He] 2s2 2p5
ii.	Scandium (Z = 21)	[Ar] 4s2 3d1
iii.	Cobalt (Z = 27)	[Ar] 4s2 3d7
iv.	Zinc (Z = 30)	[Ar] 4s2 3d10

Question 65.
Find out one dinegative anion and one unipositive cation which are isoelectronic with Ne atom. Write their electronic configuration using orbital notations and orbital diagram method.
Answer:
Atomic number (Z) of Ne is 10. Therefore, Ne and its isoelectronic species contain 10 electrons each. The dinegative anionic species, isoelectronic with Ne is obtained by adding two electrons to the atom with Z = 8. This is O^2- ion.
The unipositive cationic species, isoelectronic with Ne is obtained by removing one electron from an atom with Z = 11. It is Na⁺ ion.
These species and their electronic configurations are shown below:

Question 66.
A student pictorially represented the electronic configuration of cobalt (Z = 27) in ground state as shown in the following figure.

i. Is this the correct representation?
ii. Identify the rules of electron filling that are violated (if any) in the above answer and give the correct representation.
Answer:
i. No, the electronic configuration of cobalt in ground state is incorrectly represented.
ii. The mles of electron filling that are violated in the above diagram are Pauli’s exclusion principle and Hund’s rule. The correct electronic configuration is:

Question 67.
With reference to the representative model of the gold foil experiment, answer the following questions:

i. What evidence regarding an atom do lines A, B and C provide?
ii. How does Rutherford’s model contradict Thomson’s plum-pudding model?
iii. What results would you expect from the experiment if Thomson’s plum-pudding model was correct?
Answer:
i. Line A – Most of the space inside the atom is empty because most of the α-particles passed through the gold foil without getting deflected.
Line B – The nucleus is positively charged since some α-particles were deflected at small angles.
Line C – The nucleus contains most of the atoms mass since few α-particles were deflected backward, i.e. toward the radioactive source.

ii. According to plum-pudding model, an atom was considered a positively charged sphere with negatively charged electrons embedded in it. However, Rutherford’s gold foil experiment proved that an atom consists of large empty space with positive charge concentrated only at the centre (nucleus) and negatively charged electrons revolve around the nucleus in various orbits.

iii. If Thomson’s plum-pudding model was correct, then in the gold foil experiment we would not expect to see any significant deflection of the α-particles, i.e., Most α-particles would pass through the foil with very small or no deflections.

Question 68.
Complete the following table:

Answer:

Multiple Choice Questions

1. Which of the following statements about the electron is INCORRECT?
(A) It is a negatively charged particle.
(B) The mass of electron is equal to the mass of neutron.
(C) It is a basic constituent of all atoms.
(D) It is a constituent of cathode rays.
Answer:
(B) The mass of electron is equal to the mass of neutron.

2. The isotopes of an element differ in
(A) the number of neutrons in the nucleus
(B) the charge on the nucleus
(C) the number of extra-nuclear electrons
(D) both the nuclear charge and the number of extra-nuclear electrons
Answer:
(A) the number of neutrons in the nucleus

3. The difference between U²³⁵ and U²³⁸ atoms is that U²³⁸ contains ………….
(A) 3 more protons
(B) 3 more protons and 3 more electrons
(C) 3 more neutrons and 3 more electrons
(D) 3 more neutrons
Answer:
(D) 3 more neutrons

4. The number of electrons, protons and neutrons in ³¹P^3- ion is respectively ……………
(A) 15, 15, 16
(B) 15, 16, 15
(C) 18, 15, 16
(D) 15, 16, 18
Answer:
(C) 18, 15, 16

5. In vacuum, the speed of all types of electromagnetic radiation is equal to ………….
(A) 3.0 × 10⁶ m s^-1
(B) 3.0 × 10⁸ m s^-1
(C) 3.0 × 10¹⁰ m s^-1
(D) 3.0 × 10¹² m s^-1
Answer:
(B) 3.0 × 10⁸ m s^-1

6. In the electromagnetic spectrum, the ultraviolet region is around ………….. Hz.
(A) 10⁶
(B) 10¹⁰
(C) 10¹⁶
(D) 10²⁶
Answer:
(C) 10¹⁶

7. In hydrogen spectrum, the series of lines appearing in ultraviolet region of electromagnetic spectrum are called ………….
(A) Lyman series
(B) Balmer series
(C) Pfund series
(D) Brackett series
Answer:
(A) Lyman series

8. The energy of electron in the n^th Bohr orbit of H-atom is …………

Answer:
(C) \(-\frac{2.18 \times 10^{-18}}{\mathrm{n}^{2}} \mathrm{~J}\)

9. Which of the following is a hydrogen-like species?
(A) He²⁺
(B) Be³⁺
(C) Li⁺
(D) H⁺
Answer:
(B) Be³⁺

10. The de Broglie wavelength associated with a particle of mass 10^-6 kg moving with a velocity of 10 m s^-1 is ………..
(A) 6.63 × 10^-22 m
(B) 6.63 × 10^-29 m
(C) 6.63 × 10^-31 m
(D) 6.63 × 10^-34 m
Answer:
(B) 6.63 × 10^-29 m

11. An orbital is designated by …………. quantum numbers while an electron in an atom is designated by …………. quantum numbers.
(A) two, three
(B) three, two
(C) four, two
(D) three, four
Answer:
(D) three, four

12. In a multi-electron atom, the energy of the orbital depends on two quantum numbers: ……….
(A) n and ms
(B) n and ml
(C) ml and ms
(D) n and l
Answer:
(D) n and l

13. The number of subshells in a shell is equal to ………..
(A) n
(B) n²
(C) n – 1
(D) l + 1
Answer:
(A) n

14. The maximum number of electrons in a subshell for which l = 3 is …………..
(A) 14
(B) 10
(C) 8
(D) 4
Answer:
(A) 14

15. Which of the following is INCORRECT?
(A) A nodal plane has ψ² very close to zero.
(B) The value of ψ² at any finite distance from the nucleus is always zero.
(C) A boundary surface diagram enclosing 100 % probability density cannot be drawn.
(D) A boundary surface diagram is a good representation of shape of an orbtial.
Answer:
(B) The value of ψ² at any finite distance from the nucleus is always zero.

16. p_z-Orbital has ……….. nodal plane/planes.
(A) zero
(B) one
(C) two
(D) three
Answer:
(B) one

17. Which of the following pairs of d-orbitals will have electron density along the axis?
(A) d_xy, d_x²-y²
(B) d_z², d_xz
(C) d_xz, d_yz
(D) d_z², d_x²-y²
Answer:
(D) d_z², d_x²-y²

18. The two electrons have the following set of quantum numbers:
P = 3, 2, -2, +\(\frac {1}{2}\)
Q = 3, 0, 0, +\(\frac {1}{2}\)
Which of the following statement is TRUE?
(A) P and Q have same energy
(B) P has greater energy than Q
(C) P has lesser energy than Q
(D) P and Q represent same electron
Answer:
(B) P has greater energy than Q

19. For 3d orbital, the values of n and l are ………… respectively.
(A) 0, 3
(B) 3, 2
(C) 3, 0
(D) 3, 3
Answer:
(B) 3, 2

20. For the electron present in 1 s orbital of helium atom, the correct set of values of quantum numbers is ………..
(A) 1, 0, 0, +1/2
(B) 1, 1, 0, +1/2
(C) 1, 1, 1, +1/2
(D) 2, 0, 0, +1/2
Answer:
(A) 1, 0, 0, +1/2

21. The ground state electronic configuration for chromium atom (Z = 24) is …………
(A) [Ar] 3d⁵ 4s¹
(B) [Ar] 3d⁴ 4s²
(C) [Ar] 3d⁸
(D) [Ar] 4s¹ 4p⁵
Answer:
(A) [Ar] 3d⁵ 4s¹

22. The electronic configuration of Ni²⁺ is ……….. (Atomic number of Ni = 28)
(A) [Ar] 4s² 3d⁶
(B) [Ar] 4s¹ 3d⁸
(C) [Ar] 3d⁸
(D) [Ar] 4s² 3d⁸
Answer:
(C) [Ar] 3d⁸

Balbharti Maharashtra State Board 11th Commerce Book Keeping & Accountancy Important Questions Chapter 9 Final Accounts of a Proprietary Concern Important Questions and Answers.

Maharashtra State Board 11th Commerce BK Important Questions Chapter 9 Final Accounts of a Proprietary Concern

1. Answer in One Sentence.

Question 1.
What do you mean by pre-received income?
Answer:
Income that is received before it is due for receipt is called pre-received income.

Question 2.
What do you mean by bad debts?
Answer:
The debts which are not recoverable in spite of repeated efforts to collect the same are called bad debts.

Question 3.
What is Capital?
Answer:
Amount invested by the proprietor from time to time in business is known as capital.

Question 4.
What are the adjustments?
Answer:
Adjustments are additional information given below the trial balance and are to be considered for arriving at the correct profit or loss.

Question 5.
State the meaning of Current assets.
Answer:
Assets that are purchased with the intention of converting them into cash during the operating year are called current assets.
E.g. stock of goods.

2. Give a word, term, or phrase which can substitute each of the following statements:

Question 1.
Debit balance of Trading Account.
Answer:
Gross Loss

Question 2.
The credit balance of the Trading Account.
Answer:
Gross Profit

Question 3.
Debit balance of Profit and Loss Account.
Answer:
Net Loss

Question 4.
The credit balance of Profit and Loss Account.
Answer:
Net Profit

Question 5.
A debt that cannot be recovered.
Answer:
Bad debts

Question 6.
Reduction in the value of fixed assets due to its continuous use.
Answer:
Depreciation

Question 7.
Carriage paid on the purchase of goods.
Answer:
Carriage Inwards

Question 8.
Statement of balances of various ledger accounts.
Answer:
Trial Balance

Question 9.
An amount withdraws by a proprietor from a business in cash or kind.
Answer:
Drawings

Question 10.
Account prepared to find out gross profit or gross loss.
Answer:
Trading A/c

Question 11.
Income received before it is due.
Answer:
Pre-received Income

Question 12.
Unpaid expenses of a business.
Answer:
Outstanding Expenses

Question 13.
Account prepared on the basis of direct expenses and direct income of the business.
Answer:
Trading A/c

Question 14.
Account prepared on the basis of indirect expenses and indirect incomes of the business.
Answer:
Profit and Loss A/c

Question 15.
Group of accounts which gives the result of business activities.
Answer:
Final Accounts

3. Select the most appropriate alternatives given below and rewrite the sentence:

Question 1.
A list of balances of all the accounts in ledger is called _______________
(a) Balance Sheet
(b) Profit and Loss A/c
(c) Trading A/c
(d) Trial Balance
Answer:
(d) Trial Balance

Question 2.
Opening stock is entered in a Trading Account on the _______________ side.
(a) credit
(b) debit
(c) asset
(d) liabilities
Answer:
(b) debit

Question 3.
Drawing account is closed by transferring the balance to the _______________ account.
(a) Drawing
(b) Liabilities
(c) Assets
(d) Capital
Answer:
(d) Capital

Question 4.
Outstanding expenses is a _______________ account.
(a) Real
(b) Personal
(c) Nominal
(d) None of them
Answer:
(b) Personal

Question 5.
Depreciation is always charged on _______________ assets.
(a) Current
(b) Fixed
(c) Fictitious
(d) Intangible
Answer:
(b) Fixed

Question 6.
Pre-received income is shown on _______________ side of Balance sheet.
(a) Assets
(b) Liabilities
(c) Credit
(d) Debit
Answer:
(b) Liabilities

Question 7.
Royalty on production is a _______________ expenses.
(a) direct
(b) indirect
(c) capital
(d) none of them
Answer:
(a) direct

Question 8.
Interest on investment is _______________ of business concern.
(a) a profit
(b) a loss
(c) an expense
(d) an income
Answer:
(d) an income

Question 9.
All items of indirect income are shown on the credit side of the _______________ Account.
(a) Balance Sheet
(b) Profit and Loss
(c) Manufacturing
(d) None of them
Answer:
(b) Profit and Loss

Question 10.
Reserve for discount on debtor has a _______________ balance.
(a) credit
(b) debit
(c) nil
(d) positive
Answer:
(a) credit

4. State True or False with reasons:

Question 1.
Closing Stock is valued at cost or market price whichever is more.
Answer:
This statement is False.
Closing stock is valued at cost or market price whichever is less. It is based on the theory of anticipated profit is not brought with the account before actual realization or the Principle of conservatism.

Question 2.
Income received in advance is a liability.
Answer:
This statement is True.
Income received in advance is the liability to the business because it has not yet earned the money and the business has an obligation to deliver the goods or services to the customer.

Question 3.
Prepaid expenses are a liability.
Answer:
This statement is False.
Prepaid means paid in advance for the future period. It is the amount paid but has not yet been used up or has not yet expired. So prepaid expenses are an asset.

Question 4.
A balance Sheet is a real account.
Answer:
This statement is False.
The balance sheet is a statement prepare at the end of the financial or accounting year. It is not an Account. It gives an idea of Total Assets and liabilities on a particular day.

Question 5.
Depreciation need not be provided if the asset is not in use.
Answer:
This statement is False.
The working life of fixed assets decreases with passes of time. The value of these assets decreases every year as new technology introduced in the market old becomes outdated so it is necessary to depreciate an asset even it is not in use.

5. Fill in the blanks:

Question 1.
A Copy Right is _______________ Asset.
Answer:
An intangible

Question 2.
Wages paid for installation of Machinery should be debited to _______________ A/c.
Answer:
Machinery

Question 3.
A provision made for debts irrecoverable from debtors is called _______________
Answer:
Reserve for doubtful debts

Question 4.
In the absence of information interest on Drawings is charged for _______________ months.
Answer:
six

Question 5.
Return outward are deducted from _______________
Answer:
purchases

Question 6.
Net profit is transferred to _______________
Answer:
Balance Sheet Capital A/c

Question 7.
If cash/goods withdrawn by proprietor for domestic use, it is called _______________
Answer:
Drawings

Question 8.
Payment made in advance are shown on _______________ side of Balance Sheet.
Answer:
Asset

Question 9.
Royalty on production is debited to _______________ A/c of final A/c.
Answer:
Trading

Question 10.
General Expenses are recolored to the debit side of _______________ A/c when Office expenses, Sundry expenses, or General expenses are given in the trial balance.
Answer:
Trading

6. Find the odd one:

Question 1.
Salary, Sundry expenses, General Expenses.
Answer:
General Expenses

Question 2.
Creditors, Bank Loan, Investment.
Answer:
Investment

Question 3.
Royalty on purchases, Octroi, Discount Allowed.
Answer:
Discount Allowed

Question 4.
Rent, Factory Lighting, Freight.
Answer:
Rent

Question 5.
Outstanding Salary, Accrued Interest, Outstanding Rent.
Answer:
Accrued Interest

7. Do you agree or disagree with the following statement:

Question 1.
Software expenses paid for the installation of the computer should be debited to Software A/c.
Answer:
Disagree

Question 2.
In absence of information interest on the drawing is charged for twelve months.
Answer:
Disagree

Question 3.
Only carriage means carriage on sales.
Answer:
Disagree

Question 4.
Final accounts are prepared at the end of the month.
Answer:
Disagree

Question 5.
Return Inwards means purchase return.
Answer:
Disagree

8. Correct and Rewrite the following statements:

Question 1.
Bank overdraft is an Asset of the business concern.
Answer:
Bank overdraft is a liability of business concern.

Question 2.
Discount allowed is an income for business
Answer:
Discount allowed is an expense for the business.

Question 3.
To Goods withdrawn by proprietor A/c
Capital A/c……………Dr
(Being goods withdrawn by the proprietor for personal use)
Answer:
To Goods withdrawn by proprietor A/c
To Capital A/c………….Dr
(Being goods withdrawn by the proprietor for use)

Question 4.
Sundry Debtors A/c……………Dr
To Bad debts A/c
(Being Bad debts written off)
Answer:
Bad Debts A/c……………Dr
To Sundry Debtors A/c
(Being Bad debts written off)

Question 5.
Opening stock A/c…………….Dr
Direct expenses A/c…………….Dr
Purchase A/c………………Dr
To Trading A/c
(Being Opening Stock, Direct expenses, and purchase transferred to Trading A/c)
Answer:
Trading A/c……….Dr
To Opening stock A/c
To Direct expenses A/c
To Purchase A/c
(Being opening stock, Direct expenses and purchase transferred to Trading A/c)

Balbharti Maharashtra State Board 11th Commerce Book Keeping & Accountancy Important Questions Chapter 10 Single Entry System Important Questions and Answers.

Maharashtra State Board 11th Commerce BK Important Questions Chapter 10 Single Entry System

1. Answer in One sentence only.

Question 1.
In which method statement of affairs is prepared?
Answer:
Under the Net worth method of a single entry, a statement of affairs is prepared.

Question 2.
How is closing capital calculated under a single entry system?
Answer:
Under single entry system closing capital is calculated by deducting the total closing value of liabilities from the total closing value of assets.

Question 3.
Which statement is prepared under the single entry system to ascertain profit?
Answer:
A statement of profit or loss is prepared to ascertain profit under a single entry system.

Question 4.
What is a Statement of Profit or Loss?
Answer:
A statement that is prepared under a single entry system to calculate profit or loss is called a statement of profit or loss.

2. Write a word, term, or phrase which can substitute each of the following statements.

Question 1.
A system of book-keeping in which both the aspects of transactions are recorded.
Answer:
Double Entry System

Question 2.
Name the method of accounting in which only cash and personal transactions are recorded.
Answer:
Single Entry System

Question 3.
A statement is similar to the Balance sheet was prepared to ascertain the amounts of closing capital.
Answer:
Closing Statement of Affairs

Question 4.
The system of accounting is most scientific and reliable.
Answer:
Double Entry System

Question 5.
Name the statement prepared to find out profit or loss under a single entry system.
Answer:
Statement of Profit or Loss

Question 6.
Excess of opening capital over closing capital of proprietor under single entry system.
Answer:
Loss

Question 7.
Method of accounting in which real accounts and nominal accounts are not maintained.
Answer:
Single Entry System

Question 8.
A statement that shows profit or loss of business under a single entry system.
Answer:
Statement of Profit or Loss

Question 9.
An accounting system where rules of debit and credit are not followed.
Answer:
Single Entry System

Question 10.
The incomplete method of the accounting system.
Answer:
Single Entry System

Question 11.
A system of book-keeping which records only one aspect of business transactions and ignores other aspects.
Answer:
Single Entry System

Question 12.
A statement that shows the balances of various assets and liabilities at their approximate or estimated values as on a particular date.
Answer:
Statement of Affairs

3. Select the most appropriate answer from the alternatives given below and rewrite the sentence.

Question 1.
The difference between the capital at the end of the year and capital at the beginning of the year is called _____________
(a) Profit
(b) Income
(c) Drawings
(d) Expenses
Answer:
(a) Profit

Question 2.
A statement of _____________ is to be prepared in under to find out profit or loss under a single entry system.
(a) Income
(b) Affairs
(c) Revenue
(d) Profit or Loss
Answer:
(d) Profit or Loss

Question 3.
A statement of affairs is a summarised statement of an estimated _____________
(a) Financial Position
(b) Profit
(c) Income
(d) Loss
Answer:
(a) Financial Position

Question 4.
If closing capital is ₹ 30,000 and profit is ₹ 5,000 opening capital was _____________
(a) ₹ 35,000
(b) ₹ 30,000
(c) ₹ 25,000
(d) ₹ 15,000
Answer:
(c) ₹ 25,000

Question 5.
Under single Entry system, Profit = Closing Capital less _____________
(a) Opening Capital
(b) Opening Assets
(c) Opening Liabilities
(d) Drawings
Answer:
(a) Opening Capital

Question 6.
The capital at the end of the accounting year is ascertained by preparing _____________
(a) Cash Account
(b) Closing Statement of Affairs
(c) Total Debtors Account
(d) Opening Statement of Affairs
Answer:
(b) Closing Statement of Affairs

Question 7.
The capital at the beginning of the accounting year is ascertained by preparing _____________
(a) Receipt and Payment Account
(b) Cash Account
(c) Opening Statement of Affairs
(d) Closing Statement of Affairs
Answer:
(c) Opening Statement of Affairs

Question 8.
Under Single Entry System only _____________ are opened.
(a) Cash and Personal Accounts
(b) Real Accounts
(c) Nominal Accounts
(d) Real and Nominal Accounts
Answer:
(a) Cash and Personal Accounts

Question 9.
Statement of Affairs is just like _____________
(a) Profit and Loss A/c
(b) Real A/c
(c) Trading A/c
(d) Balance Sheet
Answer:
(d) Balance Sheet

Question 10.
Under the Net worth method, the basis for ascertaining profit or loss is the difference between _____________
(a) Capital on two dates
(b) Gross assets on two dates
(c) Liabilities on two dates
(d) Net assets on two dates
Answer:
(a) Capital on two dates

4. State True or False with reasons:

Question 1.
Statement of profit is just like Profit and Loss Account.
Answer:
This statement is False.
The profit and loss account has the debit and credit side which shows all expenses on the debit side and all incomes on the credit side and the differences are profit and loss for the year. Whereas statement of profit just adds and less income and expenses to find profit or loss.

Question 2.
The single Entry System is based on certain rules and principles.
Answer:
This statement is False.
A single Entry System is an ancient method of recording business transactions. It is a simple method of book-keeping. It is not a scientific and accurate system of Accounting. This system has no proper set of rules to be followed.

Question 3.
All transactions are recorded in the Single Entry System.
Answer:
This statement is False.
All transactions are not recorded in a single entry system. Only cash books and personal accounts of debtors and creditors are maintained. All transactions are recorded in the Double Entry System of Book-Keeping and Accountancy.

Question 4.
Arithmetical accuracy cannot be checked in Single Entry.
Answer:
This statement is True.
All transactions and accounts are not recorded in the Single Entry System. So it is impossible to prepare a Trial balance under this system without which Arithmetical accuracy cannot be checked.

Question 5.
Drawings made during the year decrease the profit under the Single Entry System.
Answer:
This statement is False.
Drawings made during the year are added to the closing capital in the statement of profit, it increases the profit under the Single Entry System.

5. Do you agree with the following statements?

Question 1.
The single Entry System of Book-keeping is a scientific method of books of accounts.
Answer:
Disagree

Question 2.
The single Entry System is useful only for large organizations.
Answer:
Disagree

Question 3.
Statement of Affairs is just like a profit and loss account.
Answer:
Disagree

Question 4.
The difference between Assets and Liabilities is called net profit.
Answer:
Disagree

Question 5.
The single Entry System follows the golden rules of accounts.
Answer:
Disagree

6. Fill in the Blanks.

Question 1.
In _____________ Book-keeping system, only Cash/Bank A/c and Personal accounts of Debtors and Creditors are opened.
Answer:
Single Entry

Question 2.
Capital is the difference between _____________ and _____________
Answer:
Assets, Liabilities

Question 3.
Single Entry System of Book-keeping is _____________ system of books of accounts.
Answer:
Conventional Accounting

Question 4.
_____________ accuracy is not guaranteed under Single Entry System.
Answer:
Arithmetical

Question 5.
In statement of profit or loss, profit on sale of assets are _____________ to closing capital.
Answer:
Added

Question 6.
Bad debts are _____________ from closing capital in statement of profit or loss.
Answer:
Deducted

Question 7.
_____________ unscientific system of Book-keeping.
Answer:
Single Entry System

Question 8.
Under the Single Entry System, profit or loss is calculated by deducting the opening capital balance from _____________ at the end of the year.
Answer:
the closing capital balance

7. Find the odd one.

Question 1.
Stock in trade, Bank overdraft, Bills receivable.
Answer:
Bank overdraft

Question 2.
Interest on Loan, Interest on Investment, Income receivable.
Answer:
Interest on Loan

Question 3.
Bad debts, Reserve for Bad debts, Reserve for a discount on creditors.
Answer:
Reserve for a discount on creditors

Question 4.
Income received in advance, Prepaid Expenses, Outstanding Expenses.
Answer:
Prepaid Expenses

8. Complete the following table:

Question 1.

Answer:
₹ 13,000

Question 2.

Answer:
₹ 45,000

Question 3.

Answer:
₹ 85,000

Question 4.

Answer:
₹ 40,000, ₹ 18,000

Question 5.

Answer:
₹ 6,000

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues

Multiple choice questions

Question 1.
Diversity in living beings is due to ……………………
(a) mutation
(b) long term evolutionary change
(c) gradual change
(d) short term evolutionary change
Answer:
(b) long term evolutionary change

Question 2.
Diversification of plant life appeared ……………………
(a) due to long periods of evolutionary changes
(b) due to abrupt mutations
(c) suddenly on the earth
(d) by seed dispersal
Answer:
(a) due to long periods of evolutionary changes

Question 3.
Rauwolfia vomitoria shows …………………. in terms of the potency and concentration of reserpine that it produces.
(a) genetic diversity
(b) species diversity
(c) ecological diversity
(d) biodiversity
Answer:
(a) genetic diversity

Question 4.
Which of the following country has the greatest ecosystem diversity?
(a) Norway
(b) India
(c) Sweden
(d) Finland
Answer:
(b) India

Question 5.
India has deserts, rain forests, mangroves, coral reefs, wetlands, estuaries and alpine meadows. What kind of biodiversity is depicted in this statement?
(a) Geographic diversity
(b) Species diversity
(c) Ecological diversity
(d) Genetic diversity
Answer:
(c) Ecological diversity

Question 6.
Biodiversity and its conservation are vital environmental issues of international concern because ……………….
(a) it fetches more economic progress and development
(b) it can attract more international tourists
(c) biodiversity and its conservation is essential for our survival and well-being of earth
(d) all the animals and plants would be extinct if not taken care of
Answer:
(c) biodiversity and its conservation is essential for our survival and well-being of earth

Question 7.
Biodiversity of geographical region represents …………………….
(a) endangered species found in the region
(b) the diversity in the organisms living in the region
(c) genetic diversity present in the dominant species of the region
(d) species endemic to the region
Answer:
(b) the diversity in the organisms living in the region

Question 8.
The latitudinal gradient in the pattern of biodiversity shows that …………………
(a) species diversity decreases as one moves away from the equator towards the poles
(b) species diversity increases as we move away from the equator towards the poles
(c) species diversity decreases as we move away from the poles towards the equator
(d) species diversity remains constant as we move away from the poles towards the equator
Answer:
(a) species diversity decreases as one moves away from the equator towards the poles

Question 9.
Which of the following is the possible cause for greater biodiversity in the tropics?
(a) Higher productivity due to more solar energy.
(b) Lesser technological development.
(c) Traditional and religious practices for conservation of nature.
(d) Lesser natural calamities.
Answer:
(a) Higher productivity due to more solar energy.

Question 10.
Log S = log C = Z log A is the equation that depicts relation between ………………….
(a) population density and time
(b) population growth and time
(c) species richness and area
(d) area and species migrations
Answer:
(c) species richness and area

Question 11.
Regression coefficient is shown by ……………………….. in the Humboldt’s equation [Log S = log C + Z log A] of species richness.
(a) S
(b) Z
(c) A
(d) C
Answer:
(b) Z

Question 12.
Choose an incorrect statement:
(a) The relation between species richness and area for a wide variety of taxa turns out to be a rectangular hyperbola.
(b) The relation between species richness and area on a logarithmic scale, the relationship is a straight line.
(c) For the species-area relationships among very large areas like the entire continents, the slope of the line appears to be much steeper.
(d) Value of Z always keep on changing for every taxonomic group or the region.
Answer:
(d) Value of Z always keep on changing for every taxonomic group or the region.

Question 13.
Who observed that within a region, species richness increased with increasing explored area, to a certain limit ?
(a) Robert May
(b) John Muir
(c) Alexander von Humboldt
(d) David Tilman
Answer:
(c) Alexander von Humboldt

Question 14.
Who was Alexander von Humboldt ?
(a) American Population biologist
(b) German naturalist and geographer
(c) Dutch Botanist
(d) French Zoologist
Answer:
(b) German naturalist and geographer

Question 15.
Which is the most well-known pattern of biodiversity ?
(a) Species-Area relationship
(b) Latitudinal gradient
(c) Longitudinal gradient
(d) Altitudinal gradient
Answer:
(b) Latitudinal gradient

Question 16.
Name the scientist who studied ecosystem by using analogy of ‘The rivet popper hypothesis’.
(a) John Muir
(b) Alexander von Humboldt
(c) Robert May
(d) Paul Ehrlich
Answer:
(d) Paul Ehrlich

Question 17.
When is the serious threat developed to an ecosystem ?
(a) When any one or two species become extinct.
(b) When key species that drive ecosystem become extinct.
(c) When native species are replaced by exotic species.
(d) When human beings take conservation measures.
Answer:
(b) When key species that drive ecosystem become extinct.

Question 18.
Which animal group is more vulnerable to the process of extinction ?
(a) Amphibian
(b) Reptilia
(c) Aves
(d) Mammalia
Answer:
(a) Amphibian

Question 19.
Deforestation does not lead to
(a) quick nutrient cycling
(b) soil erosion
(c) alteration of local weather condition
(d) destruction of natural habitat of wild animals
Answer:
(a) quick nutrient cycling

Question 20.
What are the ‘The Evil Quartet’ for the loss of biodiversity ?
(a) Habitat loss and fragmentation, Over exploitation, Alien species invasion, Co-extinctions
(b) Pollution, Global warming, Increasing population, Reclamation
(c) Greenhouse effect, Sea level rise, Air pollution, Deforestation
(d) Agriculture, Industrialization, Urbanization, Constructing transport facilities
Answer:
(a) Habitat loss and fragmentation, Over exploitation, Alien species invasion, Co-extinctions

Question 21.
Introduction of which aquaculture fish has created threat to the indigenous catfishes in Indian rivers ?
(a) Clarias gariepinus
(b) Arius sps.
(c) Heteropneustus Jossilis
(d) Pangasius pangasius
Answer:
(a) Clarlas gariepinus

Question 22.
The phenomena of co-extinction is observed when ………………….
(a) host fish gets extinct, its parasites also meet the same fate
(b) parasites are killed, the host too suffers
(c) host-parasite relationship is terminated
(d) parasites overpower the host
Answer:
(a) host fish gets extinct, its parasites also meet the same fate

Question 23.
Which of the following suffers due to co-extinction ?
(a) Selection of mates for reproduction
(b) Plant-pollinator mutualism
(c) Feeding preferences
(d) Prey-predator relationships
Answer:
(b) Plant-pollinator mutualism

Question 24.
The region with very high levels of species richness is called
(a) Biodiversity hotspot
(b) National Park
(c) Sanctuary
(d) Biosphere
Answer:
(a) Biodiversity hotspot

Question 25.
Which of the following does not offer ex-situ conservation to the flora and fauna ?
(a) Zoological parks
(b) Botanical gardens
(c) Sanctuaries
(d) Gene banks
Answer:
(c) Sanctuaries

Question 26.
Gametes of threatened species can be preserved in viable and fertile condition for long periods using ………………… techniques.
(a) cryopreservation
(b) tissue culture
(c) formalin preservation
(d) DNA hybridization
Answer:
(a) cryopreservation

Question 27.
Find the odd one out
(a) Seed banks
(b) Gene banks
(c) In vitro fertilization
(d) Electrophoresis
Answer:
(d) Electrophoresis

Question 28.
Chipko andolan movement is to protect the ………………….
(a) flora
(b) fauna
(c) trees
(d) rivers
Answer:
(c) trees

Question 29.
Hotspots are the examples of …………………..
(a) in-situ conservation
(b) ex-situ conservation
(c) wildlife protection
(d) water conservation
Answer:
(a) in-situ conservation

Question 30.
Which of the following is not the outcome of preserving biodiversity ?
(a) To maintain the ecological processes.
(b) To build national economy.
(c) To study life in its natural habitats.
(d) To disturb ecological balance.
Answer:
(d) To disturb ecological balance.

Question 31.
Which of the following is not an example of ex-situ conservation of biodiversity ?
(a) Botanical gardens
(b) Culture collections
(c) Zoological parks
(d) Colleges teaching courses on biodiversity
Answer:
(d) Colleges teaching courses on biodiversity

Question 32.
Cryopreservation of gametes of threatened species in viable and fertile condition can be referred to as ………………..
(a) In-situ cryo-conservation of biodiversity
(b) In-situ conservation of biodiversity
(c) Advanced ex-situ conservation of biodiversity
(d) In-situ conservation by sacred groves
Answer:
(c) Advanced ex-situ conservation of biodiversity

Question 33.
In which of the following, both pairs have correct combination ?
(a) In-situ conservation : Tissue culture Ex-situ conservation : Sacred groves
(b) In-situ conservation : National Park Ex-situ conservation : Botanical Garden
(c) In-situ conservation : Cryopreservation Ex-situ conservation : Wildlife Sanctuary
(d) In-situ conservation : Seed Bank Ex-situ conservation : National Park
Answer:
(b) In-situ conservation : National Park, Ex-situ conservation : Botanical Garden

Question 34.
The organization which publishes the Red List of species is …………………
(a) UNEP
(b) WWF
(c) ICFRE
(d) IUCN
Answer:
(d) IUCN

Question 35.
The World Biodiversity Day is observed on ………………..
(a) 22nd April
(b) 5th June
(c) 3rd March
(d) 22nd May
Answer:
(d) 22nd May

Question 36.
Which of the following expanded form of the given acronyms is correct?
(a) IPCC = International Panel for Climate Change.
(b) UNEP = United Nations Environment Policy.
(c) EPA = Environmental Pollution Agency.
(d) IUCN = International Union for Conservation of Nature and Natural Resources.
Answer:
(d) IUCN = International Union for Conservation of Nature and Natural Resources

Question 37.
The Air Prevention and Control of Pollution Act came into force in …………………
(a) 1975
(b) 1981
(c) 1985
(d) 1990
Answer:
(b) 1981

Question 38.
Which is the most dangerous and common kind of environmental pollution ?
(a) Air
(b) Water
(c) Noise
(d) Radioactive
Answer:
(a) Air

Question 39.
How does carbon monoxide, a poisonous gas emitted by automobiles, prevent transport of oxygen into the body tissues ?
(a) By destroying haemoglobin.
(b) By forming a stable compound with haemoglobin.
(c) By obstructing the reaction of oxygen with haemoglobin.
(d) By changing oxygen into carbon dioxide.
Answer:
(b) By forming a stable compound with haemoglobin

Question 40.
A scrubber in the exhaust of a chemical industrial plant removes
(a) gases like ozone and methane
(b) particulate matter of the size 2.5 micrometer or less
(c) gases like sulphur dioxide
(d) particulate matter of the size 5 micrometer or above
Answer:
(c) gases like sulphur dioxide

Question 41.
Which of the following can be considered as the most hazardous effect of air pollution ?
(a) Reduction in the growth and yield of the crop.
(b) Premature death of the plants.
(c) Effect on the monuments.
(d) Deleterious effects on the respiratory system of all animals.
Answer:
(d) Deleterious effects on the respiratory system of all animals.

Question 42.
Harmful effects of air pollution does not depend on the
(a) concentration of pollutants
(b) duration of exposure
(c) the type of organism
(d) time of the day
Answer:
(d) time of the day

Question 43.
Which equipment is most widely used for filtering out particulate matter ?
(a) Scrubber
(b) Electrostatic precipitator
(c) Filters
(d) Centrifuges
Answer:
(b) Electrostatic precipitator

Question 44.
What is the percentage of particulate matter removed from the thermal power exhaust with the help of electrostatic precipitator ?
(a) 25%
(b) 50%
(c) 80%
(d) 99%
Answer:
(d) 99%

Question 45.
Scrubber removes gases like
(a) Ozone
(b) Carbon dioxide
(c) Sulphur dioxide
(d) Methane
Answer:
(c) Sulphur dioxide

Question 46.
Which part of the electrostatic precipitator is maintained at several thousand volts ?
(a) Collection plates
(b) Electrode wires
(c) Corona
(d) Water line spray
Answer:
(b) Electrode wires

Question 47.
Which particles are responsible for causing the greatest harm to human health ?
(a) Particulates with size 1.00 micrometres in diameter.
(b) Particulates with size of 10 mm.
(c) Particulates with size 2.5 micrometres or less in diameter.
(d) Particulates with size of 100 micrometres in diameter
Answer:
(c) Particulates with size 2.5 micrometres or less in diameter.

Question 48.
According to Central Pollution Control Board (CPCB), which particulate size in diameter (in micrometers) of the air pollutants is responsible for greatest harm to human health ?
(a) 2.5 or less
(b) 1.5 or less
(c) 1.0 or less
(d) Between 2.5-5.3
Answer:
(a) 2.5 or less

Question 49.
When the exhaust passes through the catalytic converters, what happens to the unburnt hydrocarbons ?
(a) They are converted to oxygen and water.
(b) They are converted to energy to run the car.
(c) They are converted to carbon dioxide and water.
(d) They are converted to carbonates.
Answer:
(c) They are converted to carbon dioxide and water.

Question 50.
When the exhaust passes through the catalytic converters, what happens to the carbon monoxide and nitric oxide ?
(a) They are changed to carbon dioxide and nitrogen gas, respectively.
(b) They are changed to oxygen and carbon monoxide respectively.
(c) They are converted into hydrocarbons.
(d) They remain unchanged.
Answer:
(a) They are changed to carbon dioxide and nitrogen gas, respectively.

Question 51.
Motor vehicles equipped with catalytic converter should use unleaded petrol because
(a) lead causes pollution
(b) lead in the petrol inactivates the catalyst
(c) lead makes automobile machinery inefficient
(d) lead causes more consumption of petrol
Answer:
(b) lead in the petrol inactivates the catalyst

Question 52.
Which expensive metals are fitted into catalytic converters of the automobiles for reducing emission of poisonous gases ?
(a) Platinum-palladium and rhodium
(b) Silver, Gold
(c) Platinum and Gold
(d) Rhodium and Silver
Answer:
(a) Platinum-palladium and rhodium

Question 53.
Which part of the electrostatic precipitator attract the charged dust particles ?
(a) Collection plates
(b) Electrode wires
(c) Corona
(d) Dust particles
Answer:
(a) Collection plates

Question 54.
Two thirds of sulphur dioxides are produced by
(a) heating plants
(b) industrial processes
(c) automobile traffic
(d) electric power plants
Answer:
(d) electric power plants

Question 55.
Which is the worst polluted city among the world with respect to air pollution ?
(a) New York
(b) Tokyo
(c) New Delhi
(d) Dubai
Answer:
(c) New Delhi

Question 56.
Which are the other equivalent norms to Euro-II norms ?
(a) Bharat stage II
(b) Euro-III
(c) Euro-IV
(d) Air (Prevention and Control of Pollution) Act
Answer:
(a) Bharat stage II

Question 57.
Which of the following statements is inaccurate ?
(a) All automobiles should have Euro-III emission norm compliant automobiles and fuels by 2010.
(b) All automobiles and fuel-petrol and diesel – were to have met the Euro-III emission specifications in major 11 cities from April 1, 2005.
(c) All automobiles should have to meet the Euro-IV norms by April 1, 2010.
(d) Quality of Delhi air has significantly deteriorated due to all the above norms.
Answer:
(d) Quality of Delhi air has significantly deteriorated due to all the above norms.

Question 58.
What was observed within a period of 1997 and 2005 in Delhi as regards to air quality ?
(a) There was net increase in all the types of air pollutants.
(b) Delhi became totally pollution-free during this period only.
(c) There was substantial fall in concentrations of CO₂ and SO₂.
(d) There was decrease in the concentration of H₂S and CO.
Answer:
(c) There was substantial fall in concentrations of CO₂ and SO₂.

Question 59.
When was Air (Prevention and Control of Pollution) Act amended to include noise as an air Pollutant ?
(a) 1981
(b) 1984
(c) 1986
(d) 1987
Answer:
(d) 1987

Question 60.
Which of the following is not the measure to reduce the noise pollution ?
(a) Delimitation of horn-free zones around hospitals and schools.
(b) Permissible sound-levels of crackers and of loudspeakers.
(c) Time limits after which loudspeakers cannot be played.
(d) Complete ban on processions playing loud percussion instruments.
Answer:
(d) Complete ban on processions playing loud percussion instruments.

Question 61.
How does CO affect plant respiration ?
(a) By yellowing leaves
(b) By closing stomatal openings
(c) By reacting with cytochrome oxidase enzyme system
(d) By causing defoliation and leaf lesions
Answer:
(c) By reacting with cytochrome oxidase enzyme system

Question 62.
Acid rains are produced by
(a) excess emissions of NO₂ and SO₂ from burning fossil fuels
(b) excess production of NH₃ by industry and coal gas
(c) excess release of carbon monoxide by incomplete combustion
(d) excess formation of CO₂ by combustion and animal respiration
Answer:
(a) excess emissions of NO₂ and SO₂ from burning fossil fuels

Question 63.
How much decibel sound is produced by the jet plane or rocket ?
(a) 50 dB
(b) 80 dB
(c) 150 dB
(d) 200 dB
Answer:
(c) 150 dB

Question 64.
To what decibel level noise rises during festive seasons due to crackers ?
(a) 20 dB
(b) 50 dB
(c) 100 dB
(d) 150 dB
Answer:
(c) 100 dB

Question 65.
dB is the standard abbreviation used for the quantitative expression of
(a) the density of bacteria in a medium
(b) a particular pollutant
(c) the dominant Bacillus in a culture
(d) a certain pesticide
Answer:
(b) a particular pollutant

Question 66.
Sound becomes hazardous noise pollution at level
(a) above 30 dB
(b) above 80 dB
(c) above 100 dB
(d) above 120 dB
Answer:
(b) above 80 dB

Question 67.
When was Water (Prevention and Control of Pollution) Act passed by the Government of India ?
(a) 1974
(b) 1984
(c) 1986
(d) 1992
Answer:
(a) 1974

Question 68.
A mere ………………… impurities make water contaminated with domestic sewage unfit for human use.
(a) 0.8%
(b) 0.7%
(c) 0.5%
(d) 0.1%
Answer:
(d) 0.1%

Question 69.
What is the outcome of algal bloom ?
(a) Lots of algae available for fodder.
(b) Deterioration of water quality and fish mortality.
(c) Cleaning up of the ambient water.
(d) Decrease in BOD amount.
Answer:
(b) Deterioration of water quality and fish mortality

Question 70.
When there are excessive microorganisms in the water that cause biodegradation there is
(a) sharp rise in the dissolved oxygen content
(b) sharp decline of dissolved oxygen content
(c) refreshing odour to the water
(d) loss of algal population
Answer:
(b) sharp decline of dissolved oxygen content

Question 71.
Which is world’s most problematic aquatic weed ?
(a) Hydrilla
(b) Pistia
(c) Eichhornia crassipes
(d) Duckweed
Answer:
(c) Eichhornia crassipes

Question 72.
Which two substances are well-known for biomagnification ?
(a) Mercury and DDT
(b) Cadmium and Lead
(c) Petroleum hydrocarbons and sewage
(d) Paper manufacturing effluents and copper
Answer:
(a) Mercury and DDT

Question 73.
Choose the correct statement
(a) Concentration of DDT in the water declined with the passing time.
(b) Concentration of DDT in the water remains the same over many years without any effect.
(c) If concentration of DDT starts at 0.003 ppb in water, it can ultimately reach 25 ppm in fish-eating birds.
(d) If concentration of DDT is 25 ppm in water, in fish-eating bird population it reduces to 0.003 ppb later.
Answer:
(c) If concentration of DDT starts at 0.003 ppb in water, it can ultimately reach 25 ppm in fish-eating birds

Question 74.
Which major pollutant is released from electricity generating units ?
(a) heated water
(b) arsenic
(c) cadmium
(d) ammonia
Answer:
(a) heated water

Question 75.
Which of the statement is incorrect with reference to thermal waste water ?
(a) Thermal wastewater eliminates or reduces the number of organisms sensitive to high temperature.
(b) Thermal wastewater may enhance the growth of plants and fish in extremely cold areas.
(c) Thermal wastewater causes damage to the indigenous flora and fauna.
(d) Thermal waste water is not an important category of pollutants.
Answer:
(d) Thermal waste water is not an important category of pollutants.

Question 76.
Choose the incorrect statement from the following statements
(a) Ecological sanitation is a sustainable system for handling human excreta.
(b) One can save lot of water if flush is not used for sanitation.
(c) Ecological sanitation is a practical, hygienic, efficient and cost-effective solution to human waste disposed.
(d) Human excreta cannot be recycled into any resource or natural and safe fertilizer.
Answer:
(d) Human excreta cannot be recycled into any resource or natural and safe fertilizer.

Question 77.
Eutrophication is caused by
(a) acid rain
(b) nitrates and phosphates
(c) sulphates and carbonates
(d) CO₂ and CO
Answer:
(b) nitrates and phosphates

Question 78.
Measuring Biochemical Oxygen Demand (BOD) is a method used for
(a) estimating the amount of organic matter in sewage water.
(b) working out the efficiency of oil driven automobile engines.
(c) measuring the activity of Saccharomyces cerevisae in producing curd on a commercial scale.
(d) working out the efficiency of RBCs about their capacity to carry oxygen.
Answer:
(a) estimating the amount of organic matter in sewage water.

Question 79.
High value of BOD (Biochemical Oxygen Demand) indicates that
(a) consumption of organic matter in the water is higher by the microbes
(b) water is pure
(c) water is highly polluted
(d) water is less polluted
Answer:
(c) water is highly polluted

Question 80.
When huge amount of sewage is dumped into a river, its BOD will
(a) increase
(b) decrease
(c) sharply decrease
(d) remain unchanged
Answer:
(a) increase

Question 81.
Which river of India is considered as an unending sewer ?
(a) Mula in Pune
(b) Panchaganga in Kolhapur
(c) Ganga from Haridwar to Kolkata
(d) Patalganga in Panvel
Answer:
(c) Ganga from Haridwar to Kolkata

Question 82.
Sewage drained into water bodies kill fishes because
(a) excessive carbon dioxide is added to water
(b) it gives off a bad smell
(c) it removes the food eaten by the fish
(d) it increases competition fishes for dissolved oxygen
Answer:
(d) it increases competition fishes for dissolved oxygen

Question 83.
Which method was commonly practised for managing solid waste generated by municipal bodies ?
(a) Open dumps
(b) Open burning dumps
(c) Accumulation in trenches
(d) Accumulation in water bodies
Answer:
(b) Open burning dumps

Question 84.
Biochemical Oxygen Demand (BOD) may not be a good index for pollution of water bodies receiving effluents from
(a) domestic sewage
(b) dairy industry
(c) petroleum industry
(d) sugar industry
Answer:
(c) petroleum industry

Question 85.
What is the appropriate scientific method for waste disposed ?
(a) Land fill
(b) Open burning dump
(c) Sanitary landfill
(d) Open dumps (Junk yards)
Answer:
(c) Sanitary landfill

Question 86.
Which statement correctly describes the process of waste disposal in sanitary landfill ?
(a) Solid wastes are dumped in a trench or depression.
(b) Solid wastes are dumped in a depression and compacted.
(c) Solid wastes are dumped in a trench, compacted and covered by soil.
(d) Solid wastes are dumped in a trench and burnt to reduce volume.
Answer:
(c) Solid wastes are dumped in a trench, compacted and covered by soil.

Question 87.
Which is adverse effect of sanitary land fill noticed occasionally?
(a) Breeding place for rats and flies.
(b) Seepage of chemicals polluting ground water.
(c) Burning of hazardous waste.
(d) Accumulation of biodegradable materials.
Answer:
(b) Seepage of chemicals polluting ground water.

Question 88.
From the following, which is not a category of sorting of waste ?
(a) Recyclable
(b) Biodegradable
(c) Non-biodegradable
(d) Explosive
Answer:
(d) Explosive

Question 89.
From the following, which is a recyclable waste ?
(a) Food waste
(b) Newspaper
(c) Leather
(d) Rubber
Answer:
(b) Newspaper

Question 90.
Which is appropriate method for disposal of hospital wastes ?
(a) Sanitary landfills
(b) Open dumps
(c) Use of incinerators
(d) Composting
Answer:
(c) Use of incinerators

Question 91.
Which is valid suggestion for responsible citizen to deal with managing non- biodegradable waste ?
(a) Mixing of biodegradable and recyclable waste material.
(b) Mixing of biodegradable and non- biodegradable waste material.
(c) Use of more and more disposable material.
(d) Use of cloth bags and avoid plastic carry bags.
Answer:
(d) Use of cloth bags and avoid plastic carry bags.

Question 92.
For treatment of e-waste, which is the most suitable solution ?
(a) Recycling and recovery
(b) Buried in landfills
(c) Incineration
(d) Disposal and storage in open
Answer:
(a) Recycling and recovery

Question 93.
Which metals are recovered from recycling of E-waste ?
(a) Copper, iron, silicon, nickel and gold.
(b) Platinum, aluminium, silicon, silver and rhodium.
(c) Sodium, iron, silicon, uranium and potassium.
(d) Copper, radium, zinc, cobalt and titanium.
Answer:
(a) Copper, iron, silicon, nickel and gold.

Question 94.
Greenhouse effect is warming due to
(a) infra-red rays reaching earth
(b) moisture layer in the atmosphere
(c) increase in temperature due to increase in carbon dioxide concentration of the atmosphere
(d) ozone layer in the atmosphere
Answer:
(c) increase in temperature due to increase in carbon dioxide concentration of the atmosphere

Question 95.
Which one of the following is the correct percentage of the two (out of the total of 4) greenhouse gases that contribute to the total global warming ?
(a) CFCs 14%, Methane 20%
(b) CO₂ 40%, CFCs 30%
(c) N₂O 6%, CO₂ 86%
(d) Methane 20%, N₂O 18%
Answer:
(a) CFCs 14%, Methane 20%

Question 96.
The two gases making highest relative contribution to the greenhouse gases are
(a) CO₂ and CH₄
(b) CH₄ and N₂O
(c) CFCs and N₂O
(d) CO₂ and N₂
Answer:
(a) CO₂ and CH₄

Question 97.
Which is the chief reason of global warming ?
(a) Greenhouse effect
(b) Absorption of UV radiations by ozone
(c) Effect of visible light
(d) Trapping of radio waves
Answer:
(a) Greenhouse effect

Question 98.
Why naturally occurring greenhouse effect is important?
(a) It maintains the temperature of earth at average 15 °C.
(b) It maintains the temperature of earth at 18 °C.
(c) It maintains lot of greenery on the surface of the earth.
(d) It maintains level of ozone in atmosphere.
Answer:
(a) It maintains the temperature of earth at average 15 °C.

Question 99.
Clouds and gases reflect about ………………… of the incoming solar radiation.
(a) one half
(b) one-fourth
(c) one tenth
(d) three-fourth
Answer:
(b) one-fourth

Question 100.
Montreal Protocol aims at ……………………
(a) Biodiversity conservation
(b) Control of water pollution
(c) Control of CO₂ emission
(d) Reduction of ozone depleting substances
Answer:
(d) Reduction of ozone depleting substances

Question 101.
Which are the four most affected aspects due to climate change caused due to global warming?
(a) Energy, Agricultural research, Sewage disposed, Entertainment.
(b) Food supply, Water, Health, Infrastructure.
(c) Political views, Equality, Natural Resources, Safety of women.
(d) Education, Empowerment of people, Shelter, Processed food.
Answer:
(b) Food supply, Water, Health, Infrastructure.

Question 102.
What is exactly measured in Dobson units or DU?
(a) The thickness of the ozone in a column of air from the ground to the top of the Atmosphere.
(b) The noise level in the circumscribed area.
(c) The amount of chlorofluorocarbons.
(d) The hole in the ozone umbrella.
Answer:
(a) The thickness of the ozone in a column of air from the ground to the top of the Atmosphere.

Question 103.
‘Good ozone’ is found in the while the bad ozone is in ………………….
(a) Mesosphere, Ionosphere
(b) Mesosphere, Troposphere
(c) Stratosphere, Troposphere
(d) Stratosphere, Ionosphere
Answer:
(c) Stratosphere, Troposphere

Question 104.
UV-B does not cause ……………………
(a) aging of skin
(b) damage to skin cells
(c) various types of skin cancers
(d) albinism or lightning of the skin
Answer:
(d) albinism or lightning of the skin

Question 105.
What is snow-blindness cataract ?
(a) Cataract noticed in people living in snow clad areas.
(b) Cataract that shows symptom of white dense patch.
(c) Cataract resulted due to inflammation of cornea.
(d) Cataract that causes total blindness.
Answer:
(c) Cataract resulted due to inflammation of cornea.

Question 106.
Which policy is introduced by Government of India to conserve forests effectively with local people ?
(a) Wildlife protection
(b) Chipko movement
(c) Joint forest movement
(d) Joint tree plantation
Answer:
(c) Joint forest movement

Question 107.
Name the award declared by Government of India to motivate people for protecting wildlife.
(a) Amrita Devi-Bishnoi Tree Protection Award.
(b) Amrita Devi – Bishnoi Wildlife Protection Award.
(c) Amrita Devi-Chipko Movement Award.
(d) Bahuguna – Chipko Movement Award.
Answer:
(b) Amrita Devi – Bishnoi Wildlife Protection Award.

Match the columns

Question 1.

Column A	Column B
(1) Walter Rosen	(a) Popularisation of term biodiversity
(2) David Tillman	(b) Rivet Popper Hypothesis
(3) Paul Ehrlich	(c) Productivity Stability Hypothesis
(4) Edward Wilson	(d) Coined

Answer:

Column A	Column B
(1) Walter Rosen	(d) Coined
(2) David Tillman	(c) Productivity Stability Hypothesis
(3) Paul Ehrlich	(b) Rivet Popper Hypothesis
(4) Edward Wilson	(a) Popularisation of term biodiversity

Question 2.

Column I (Phenomena)	Column II (Effect)
(1) Eutrophication	(a) Soil erosion
(2) Biomagnification	(b) Prevention of extinction
(3) Conservation	(c) Accumulation of non-biodegradable substance
(4) Deforestation	(d) Death of aquatic ecosystem

Answer:

Column I (Phenomena)	Column II (Effect)
(1) Eutrophication	(d) Death of aquatic ecosystem
(2) Biomagnification	(c) Accumulation of non-biodegradable substance
(3) Conservation	(b) Prevention of extinction
(4) Deforestation	(a) Soil erosion

Question 3.

Indian region shows	Number
(1) Biosphere reserves	(a) 448
(2) National parks	(b) 14
(3) Wildlife sanctuaries	(c) 90

Answer:

Indian region shows	Number
(1) Biosphere reserves	(b) 14
(2) National parks	(c) 90
(3) Wildlife sanctuaries	(a) 448

Classify the following to form Column B as per category given in Column A.

Question 1.
Karnataka, Chanda, Khasi, Rajasthan, Sarguja, Jaintia, Maharashtra, Bastar.

Region	Sacred groves seen at
(1) Meghalaya	—————
(2) Western ghat regions	—————
(3) Aravali hills	—————
(4) Madhya Pradesh	————–

Answer:

Region	Sacred groves seen at
(1) Meghalaya	Khasi, Jaintia
(2) Western ghat regions	Karnataka,
(3) Aravali hills	Maharashtra
(4) Madhya Pradesh	Rajasthan, Bastar

Question 2.
Dust, Carbon monoxide, Lead, Smog, Methane, Mercury, DDT, Cadmium.

Column A	Column B (Examples)
(1) Particulate pollutant	—————
(2) Gaseous pollutant	—————
(3) Biomagnification	—————
(4) Heavy metals	————–

Answer:

Column A	Column B (Examples)
(1) Particulate pollutant	Dust, Smog
(2) Gaseous pollutant	Carbon monoxide, Methane
(3) Biomagnification	Mercury, DDT
(4) Heavy metals	Lead, Cadmium

Very short answer questions

Question 1.
Which are the biodiversity hotspots in India?
Answer:
India has three of world’s biodiversity viz. Western Ghats, Indo-Burma and Eastern- Himalayas.

Question 2.
How many national parks and sanctuaries are present in Maharashtra?
Answer:
In Maharashtra, there are 5 national parks and 11 sanctuaries.

Question 3.
What is included under biodiversity?
Answer:
Biodiversity includes a vast array of species of microorganisms – viruses, algae, fungi, plants and animals occurring in all the different habitats on Earth and forming the ecological complexes.

Question 4.
Who coined the biodiversity?
Answer:
The term biodiversity was coined by Walter Rosen in 1982.

Question 5.
Who popularised the term biodiversity?
Answer:
Edward Wilson popularized the term biodiversity to describe combined diversity at all the levels of biological organisation.

Question 6.
How long it has taken for Biodiversity to form on the earth?
Answer:
Biodiversity which is currently present took over 3.5 billion of years of evolutionary history to form on earth.

Question 7.
Where does India stand as far as species diversity is concerned?
Answer:
Answer:
India is one among 15 nations that are rich in species diversity.

Question 8.
Which habitats do not show latitudinal and altitudinal gradients?
Answer:
Arid and Semiarid habitats and aquatic habitat do not show latitudinal and altitudinal gradient.

Question 9.
What is the relationship between species richness and latitudinal gradient?
Answer:
Species richness is high at lower latitudes and there is a steady decline towards the poles, i.e. species richness for plants and animals decreases as we move away from equator to the poles.

Question 10.
What is the relationship between species diversity and altitude?
Answer:
Species diversity is more at lower altitudes than at the higher altitude.

Question 11.
At which latitude there is maximum diversity?
Answer:
Biodiversity is maximum in tropical rain forests at equator.

Question 12.
Enumerate the biodiversity in Amazon rain forest
Answer:
The world’s largest tropical rainforest of Amazon, there are around 40,000 plant species, nearly 1,300 bird species, 3,000 types of fish, 427 species of mammals and more than 1,25,000 invertebrates.

Question 13.
How many species have been documented as per IUCN data of 2004?
Answer:
Over 1.5 million species have been documented as per IUCN data (2004).

Question 14.
How much of global biodiversity wealth has been recorded as per May’s estimate?
Answer:
As per May’s estimate of global biodiversity, there are 22% of our natural wealth recorded.

Question 15.
Which are the activities of humans that could cause loss of biodiversity even before they are recorded?
Answer:
Human activities like reclamation and deforestation can cause loss of varieties even before they are identified and recorded.

Question 16.
Which one is considered the sixth extinction? Why?
Answer:
The current loss of biodiversity is considered to be the sixth extinction. It is considered so because the loss of biodiversity is progressing at an alarming rate of 100 to 1000 times faster than pre-human times.

Question 17.
Enlist the causes of Biodiversity losses.
Answer:
There are four major causes of biodiversity losses which are popularly known as, ‘The Evil Quartet’ which are habitat loss and fragmentation, over exploitation, alien species invasion and co-extinction.

Question 18.
Why alien species of animals become invasive and harmful to local species?
Answer:
There is lack of local predator for the invasive species, therefore, the invasive species cannot be controlled.

Question 19.
What is the work of The International Union for Conservation of Nature and Natural Resources (IUCN)?
Answer:
The International Union for Conservation of Nature and Natural Resources or IUCN maintains a Red Data Book or Red list which includes records of conservation status of plant and animal species.

Question 20.
What was the main reason of loss of natural resources in last ten decades?
Answer:
Human population has grown exponentially along with industrial development, both of these have resulted in the rampant loss of natural resources in last ten decades.

Question 21.
Which Act was passed to protect and improve quality of environment?
Answer:
In order to protect and improve the quality of our environment, the Government of India has passed the Environment Protection Act 1986.

Question 22.
Which is the Indian law that includes noise as an air pollutant?
Answer:
In India, the Air (Prevention and Control of Pollution) Act 1981, Amendment 1987, includes noise as an air pollutant.

Question 23.
What are the sources of noise pollution?
Answer:
The common sources of noise pollution are machines, transportation, construction sites, industries, etc.

Question 24.
How do we identify polluted water?
Answer:
Polluted water is usually turbid, foul smelling, coloured and contains number of pathogens, heavy metals, oils, etc.

Question 25.
What is the most common source of water pollution?
Answer:
Domestic sewage is one of the most common sources of water pollution.

Question 26.
What is algal bloom?
Answer:
Algal bloom is excessive growth of planktonic freely floating blue-green algae caused due to presence of large amount of nutrients in water.

Question 27.
Why is algal bloom considered to be bad?
Answer:
Algal bloom makes the water coloured and unpotable, releasing toxins in water which can kill the fish.

Question 28.
Why water pollution act was passed in India? When was it passed?
Answer:
When Government realised the importance of maintaining the cleanliness of the water bodies, the Water (Prevention and Control of Pollution) Act was passed in year 1974 to safeguard our water resources.

Question 29.
Which substances can cause biomagnification ?
Answer:
Those substance like some pesticides which are non-biodegradable and those which get accumulated in the tissues of living organisms without metabolism or excretion, cause biomagnification.

Question 30.
Which method of recycling of sewage is used in Tirumala hills?
Answer:
At Tirumala hills there are reverse osmosis units set up to recycle sewage water, which helps in solving the huge water demand.

Question 31.
What is the advantage of recycling of sewage water by reverse osmosis (RO)?
Answer:
Recycling sewage water by RO System helps to solve the problem of scarcity of water and also disposal of sewage water.

Question 32.
Which systems are now made mandatory by Municipal Corporation for new constructions ?
Answer:
Rainwater harvesting is made mandatory for new constructions by Municipal Corporation.

Question 33.
Which ban was imposed by Maharashtra Government on 23rd June 2018?
Answer:
Maharashtra government sent a notification to ban use, sale, distribution and storage of plastic material to fight pollution caused due to extensive use of plastic.

Question 34.
What are biomedical wastes?
Answer:
The harmful wastes generated by the hospitals that contains disinfectants, harmful chemicals, discarded body parts, blood and also pathogenic microorganisms are called biomedical wastes.

Question 35.
Why recycling of e-wastes is considered dangerous?
Answer:
In developing countries, due to lack of facilities for recycling of e-waste, it is done by manual methods, which is harmful as the workers are exposed to toxic substances from e-waste.

Question 36.
Which are commonly called greenhouse gases?
Answer:
CO₂ and methane are commonly called greenhouse gases.

Question 37.
What are Dobson units?
Answer:
Thickness of the ozone in a column of air from the ground to the top of the atmosphere is measured as Dobson units (DU).

Question 38.
If ozone layer is intact, which UV radiations are absorbed by earth’s atmosphere?
Answer:
UV radiations of wavelength shorter than UV-B i.e. 100-280 nm are almost completely absorbed by earth’s atmosphere, given that the ozone layer is intact.

Question 39.
What is the aim of National Forest Policy?
Answer:
National Forest Policy (NFP) aims at maintaining 33% forest cover in the country.

Question 40.
How much forest cover is being lost due to deforestation?
Answer:
Almost 40% tropical forests and 1% temperate forests are lost due to deforestation.

Give definition of the following

Question 1.
Biodiversity
Answer:
Biodiversity is part of nature which includes the differences in the genes among the individuals of a species; the variety and richness of all plants and animal species at different scales in a space – local regions, country and the world; and the types of ecosystem, both terrestrial and aquatic, within a defined area.

Question 2.
Extinct species
Answer:
The species which gets totally eliminated from the earth is called extinct species.

Question 3.
Endangered species
Answer:
When the number of members of a species starts dwindling, it is said to be endangered species.

Question 4.
Invasive species
Answer:
The species that does not belong to the region or locality but is introduced accidentally or intentionally which causes harmful effects to the already existing local species, are called invasive species.

Question 5.
Bioprospecting
Answer:
Bioprospecting is systematic search for development of new sources of chemical compounds, genes, microorganisms, macroorganism and other valuable products from nature.

Question 6.
Biochemical Oxygen Demand (BOD)
Answer:
BOD is the amount of dissolved oxygen required by microorganisms for decomposing the organic matter present in water which is expressed in milligram of oxygen per litre (mg/L) of water.

Question 7.
Natural Eutrophication
Answer:
Natural eutrophication is the process of aging of a lake due to nutrient enrichment of water.

Question 8.
Cultural or Accelerated eutrophication
Answer:
The process of aging of the water body due to pollutants from human activities such as effluents from agricultural lands, industries and homes (household).

Question 9.
Biological Magnification (Biomagnification) :
Answer:
Biological magnification is the phenomenon through which certain pollutants get accumulated in tissues in increasing concentration along the food chains in successive trophic levels.

Question 10.
Deforestation
Answer:
Deforestation is conversion of forest area into non-forest area.

Question 11.
Reforestation
Answer:
Reforestation is the natural process of restoring a forest that once existed but was destroyed or removed at some time in past.

Question 12.
Chipko Movement
Answer:
Chipko Movement is people’s participation for the protection of trees in which people hug the trees and save it from the axe of tree-cutters. Initially it happened in 1974 in Garhwal region of Himalayas and now it is spread world-wide.

Question 13.
Joint Forest Management (JFM)
Answer:
Joint Forest Management is an attempt to conserve forests in a sustainable matter which has been introduced by the Government of India in 1980s for working with the local communities for protection and management of the forests.

Name the following/Give examples

Question 1.
Levels of biodiversity
Answer:

1. Genetic diversity
2. Species diversity or community diversity
3. Ecosystem diversity or ecological diversity

Question 2.
Two patterns of biodiversity
Answer:

1. Latitudinal and Altitudinal gradient
2. Species-area relationship.

Question 3.
Three types of extinction
Answer:

1. Natural extinction
2. Mass extinction
3. Manmade (anthropogenic) extinction

Question 4.
Three examples of animals that are now extinct due to over-exploitation
Answer:

1. Dodo
2. Stellar sea cow
3. Passenger pigeon

Question 5.
Types of air pollutants
Answer:

1. Particulate pollutants
2. Gaseous pollutants.

Question 6.
Name the different types of pollution.
Answer:

1. Water pollution
2. Air pollution
3. Radioactive pollution
4. Noise pollution
5. Soil pollution.

Question 7.
Main types of wastes
Answer:

1. Biodegradable
2. Recyclable
3. Non-biodegradable.

Distinguish between the following

Question 1.
Genetic diversity and Species diversity.
Answer:

Genetic diversity	Species diversity
(1) Genetic diversity is intraspecific diversity.	(1) Species diversity is interspecific diversity.
(2) Genetic diversity is due to number and types of genes and chromosomes present in different species.	(2) Species diversity is due to number of species of plants and animals that are present in a region.
(3) Genetic diversity includes variations in genes, alleles and chromosomes	(3) Species diversity includes species richness and species evenness.

Question 2.
In-situ and ex-situ conservation.
Answer:

In-situ conservation	Ex-situ conservation
(1) In-situ conservation is a onsite conservation.	(1) Ex-situ conservation is done outside the habitat of plants and animals.
(2) Plant and animal species are conserved in their natural habitat for protecting endangered species.	(2) Plant and animal species are conserved in artificial or manmade place.
(3) It is done in natural environment.	(3) It is done in manmade environment.
(4) National parks, Sanctuaries, biosphere reserve, etc. are set up for in-situ conservation.	(4) Zoo, aquarium, seed banks are the examples of ex-situ conservation
(5) It is a dynamic process. Cheap and convenient to conduct.	(5) It is static process. Its expensive and commercial process.
(6) Captive breeding is not successful in all cases of in-situ conservation method.	(6) Captive breeding is successful and can help in increasing the number of endangered organisms.

Give reasons

Question 1.
There is decrease in species diversity at higher altitude.
Answer:

1. At higher altitudes, there are different climatic conditions such as drastic season variations and lesser ambient temperature.
2. Survival of organisms thus becomes difficult. Therefore, at such altitudes, species diversity decreases.

Question 2.
Loss of biodiversity leads to the overall imbalance in the ecosystem.
Answer:

1. Loss of biodiversity in any area leads to the decline in plant production.
2. There is lower resilience to environmental disturbance like flood.
3. It may also lead to alteration in environmental processes like disease cycles, plant productivity, etc.
4. The food chains and food webs are disturbed.
5. The productivity of the ecosystem is reduced and this results into overall imbalance in the ecosystem.

Question 3.
Motor vehicles equipped with catalytic converter should use unleaded petrol.
Answer:

1. There is lead in the petrol which can inactivate the catalyst present in catalytic converter.
2. Due to this inactivation catalytic converter will not work properly. Therefore motor vehicles equipped with catalytic converter should use unleaded petrol.

Question 4.
Using CNG as a fuel is more eco-friendly.
Answer:

1. Diesel or petrol cause air pollution.
2. On the other hand, CNG is advantageous because it is cheaper and thus people find it economical to use CNG.
3. CNG burns efficiently and causes lesser pollution.
4. CNG cannot be adulterated. Thus, it is, the only adulteration-proof fuel which is eco-friendly.

Question 5.
High BOD indicates intense level of microbial pollution.
Answer:

1. When BOD of a water sample is high, it denotes that the amount of dissolved oxygen in the water which is required by the microorganisms to decompose the organic matter in that water is high.
2. This shows that there are many microbes in the water body. Thus, high BOD indicates intense level of microbial pollution.

Question 6.
If microorganisms are more in a water body, other aquatic creatures find it difficult to survive.
Answer:

1. Microorganisms involved in biodegradation of organic matter in water body consume lot of dissolved oxygen.
2. Due to this consumption there is sharp decline in oxygen level of water which leads to mortality of fish and other aquatic creatures. Therefore, other aquatic creatures find it difficult to survive.

Question 7.
Lake can literally get choked to death due to eutrophication.
Answer:

1. Eutrophication means enrichment due to nutrients.
2. Pollution due to human activities releases effluents through agricultural lands, industries and houses.
3. Many of these contain phosphates and sulphates which cause excessive eutrophication.
4. This leads to non-availability of oxygen for other aquatic organisms, mainly causing death of fish.
5. The processes of decomposition of these dead fish adds to further depletion of oxygen and hence lake can literally get choked to death due to eutrophication.

Question 8.
Water hyacinth is called ‘Terror of Bengal’.
Answer:

1. Water hyacinth (Eichhornia crassipes) is native plant of amazon basin which was introduced in India for its beautiful, purple flowers.
2. It has become an invasive species.
3. This plant is a nuisance as it grows excessively and covers entire water body in which it is present.
4. It grows faster than our ability to remove it, so it is commonly called ‘Terror of Bengal’.

Question 9.
Most of the water pollution is manmade.
Answer:

1. There Eire many activities of human beings which dump the wastes into water bodies.
2. The industrial processes also cause dumping of hazardous waste into surrounding water bodies.
3. Human activities is the only factor that causes water pollution and there are no natural causes for water pollution.

Question 10.
There is steady concentration of ozone in the stratosphere.
Answer:

1. Ozone is a form of oxygen which is photo-dissociated and is generated by absorption of short wavelength UV radiations.
2. Both generation and dissociation of ozone is in equilibrium.
3. This is the equation which shows these conversions.
   O₃ → O₂ + [O]
   O₂ + [O] → UV RAYS → O₃
4. Therefore, there is steady concentration of ozone in the stratosphere.

Question 11.
The CO₂ has crucial role in global warming.
Answer:
(1) Carbon dioxide is produced by many activities of human beings such as destruction of forests, combustion of fossil fuels, cement plants and other industries, burning and respiration by all living organisms.

(2) This CO₂ forms a layer in the upper atmosphere.

(3) When solar energy reaches the earth surface, the infrared radiations from this are trapped by the layer of CO₂.

(4) Along with CO₂ other gases such as methane, CFCs and nitrogen oxides also form a blanket in the atmosphere which traps the reflected infrared rays.

(5) This results in greenhouse effect and global warming. CO₂ alone can increase the temperature by about 50%. During the last 50 years the average temperature of the earth has increased due to steadily increasing CO₂ concentration. Thus, CO₂ plays a crucial role in global warming.

Question 12.
Global warming is caused by ‘greenhouse effect’.
Answer:

1. Carbon dioxide along with methane, nitrogen oxides and CFCs can absorb infrared radiations reflected from the earth’s surface.
2. The blanket formed by these gases in the atmosphere traps the reflected infrared rays and produces heat on the earth’s surface which results in greenhouse effect.
3. The greenhouse effect in turn causes global warming.

Question 13.
Ozone present in the stratosphere is called good ozone.
Answer:

1. The ozone present in the upper atmospheric region, i.e. in the stratosphere, absorbs the ultraviolet radiations present in the sunlight.
2. These radiations are harmful for living organisms.
3. Since ozone protects the living organisms from such dangerous UV radiations, therefore, the ozone present in the stratosphere is called good ozone.

Question 14.
The UV radiations are injurious.
Answer:

1. When ultraviolet (UV) radiation falls on the cells of living organisms it is absorbed in the DNA and proteins present in the nucleus.
2. The high energy of UV radiations break the chemical bonds within these molecules.
3. This results in damage to the skin cells and cause skin cancers of various types.
4. High doses of UV-B radiations also cause inflammation of cornea called snow blindness, cataract, etc. The UV radiations therefore are injurious.

Question 15.
Plastic ban in Maharashtra is an essential step.
Answer:
1. Plastic is non-biodegradable and man-made substance which cannot be decomposed naturally.

2. If burnt it causes toxic fumes. If buried it will contaminate the soil. If thrown anyhow, it can damage other animals like cattle. If thrown in water bodies, it kills the aquatic organisms. It clogs the water outlets and can cause flooding of cities during rainy season. Therefore, disposal of plastic has become a major issue.

3. On the contrary, the users of plastic have increased tremendously due to ease in using the plastic items.

4. Because plastic causes damage to ecosystem, it was very essential to curtail its use. Banning its use reduces is helpful in managing the excessive and unnecessary use, therefore, plastic ban in Maharashtra is an essential step.

Write short notes

Question 1.
Ecological (Ecosystem) diversity.
Answer:

1. Different types of ecosystems or habitats within a given geographical area forms ecosystem diversity.
   On Earth there are a large variety of ecosystems.
2. Each ecosystem has its own complement of distinctive interlinked species, based on the differences in the habitat. It is also specific for particular geographical region.
3. In one region, generally, there may be one or many different types of ecosystems.
4. In India, there are varieties of ecosystems such as deserts, rain forests, deciduous forests, estuaries, wetlands, grasslands, etc.
5. In India, the Western Ghats show great ecosystem diversity. However, regions like Ladakh and Rann of Kutch have less ecosystem diversity.

Question 2.
Habitat loss and fragmentation.
Answer:

1. Habitat loss and fragmentation is the prime cause of destruction of biodiversity.
2. Due to degradation and pollution there is reduction in vast natural habitats. This creates crisis situation for resident living organisms.
3. This is largely due to human activities.
4. There is also a threat to migratory birds and for animals which need larger territories.
5. Reduction in tropical rain forests has been reduced from 14% to 6% over the years, which has surely destroyed many species.

Question 3.
Over-exploitation.
Answer:

1. Humans have exploited natural resources beyond their needs.
2. The excessive consumption and accumulation have resulted into problem of over¬exploitation.
3. Overexploitation of resources has caused threats to various organisms.
4. Dodo bird, stellar sea cow and passenger pigeon are extinct due to overexploitation.
5. Over exploitation of fish from sea has also resulted into dearth of fish.

Question 4.
Alien species invasion.
Answer:

1. When invasive species are accidentally or intentionally introduced into a particular region which causes extinction of local and already existing species, it is called alien species invasion.
2. Examples of such invasive plant species are
   (a) the carrot grass (Parthenium)
   (b) Lantana(c) Water hyacinth (Eichhornia).
3. Invasive animal species is African catfish Clarias gariepinus, introduced for aquaculture purpose. This has caused harm to endemic catfish varieties.
4. Invasion by such species is one of the major reasons for extinction of local species.

Question 5.
Co-extinctions.
Answer:

1. When one organism is associated with other one in an obligatory way, then, if one is extinct, the other also gets extinct.
2. Extinction of one variety leads to loss of associate variety from the ecosystem. Such phenomena is called co-extinction.
3. Extinction of host fish causes extinction of unique parasites.
4. Coevolved plant-pollinator also will have such a threat.

Question 6.
Write a note on BD Act 2002.
Answer:

1. The law broadly defines biodiversity.
2. It includes plants, animals and microorganisms and their parts, their genetic materials and by-products.
3. It excludes value added products and human genetic material.
4. Regulation of access to Indian biological resources as well as scientific cataloguing of traditional knowledge about ethnobiological materials were the main objectives for proposing this Act.
5. There is three-tier system, viz. National Biodiversity Authority (NBA) at the national level, the State Biodiversity Boards (SBBs) at the state level and Biodiversity Management Committees (BMCs) at the local level that gives approval of utilization of any biological resource for commercial or research purpose.
6. It is mandatory for foreigners, NRIs as well as Indian citizens and institutions to seek permission from NBA before exploiting local resource.
7. NBA has powers of civil court. Not seeking approval of NBA, can incur jail and fine up to 10 lakh rupees.

Question 7.
Particulate air pollutants.
Answer:

1. Particulate air pollutants are either solids or liquids.
2. Particles having larger diameter of 10 pm settle in the soil but finer particles with 1 pm or less remain suspended in the air.
3. Central Pollution Control Board (CPCB) has declared that, particulate matter of size 2.5 pm or less in diameter (PM 2.5) are responsible for causing the greatest harm to humans.
4. These fine particulates can be inhaled deep into the lungs and are responsible for irritation, inflammation and damage to lungs.
5. In addition to this, it causes breathing and respiratory disorders and premature deaths.
6. Examples of particulate pollutants are : Smoke, smog, pesticides, heavy metals, dust and radioactive elements.

Question 8.
Gaseous pollutants.
Answer:
(1) Gaseous pollutants are gases which cause air pollution.

(2) Some common gaseous pollutants are CO₂, CO, SO₂, NO, NO₂, etc.

(3) Carbon dioxide (CO₂) : It is a greenhouse gas which is continuously produced due to human activities such as burning of fossil fuels and rampant deforestation. Photosynthesis carried out by plants can balance CO₂ : O₂ ratio in the air, provided there is good green cover. CO₂ is also removed from the air by weathering of silicate rocks forming limestone. Aeroplane traffic especially release lot of CO₂. CO₂ is the main cause for global warming and climate change.

(4) Carbon monoxide (CO) : CO is a toxic gas produced by incomplete combustion of different fuels. Therefore, vehicular exhausts are the largest source of CO.

(5) Nitrogen dioxide (NO₂) and nitrogen monoxide (NO) : Nitrogen oxides are released through automobiles and chemical industries as waste gases. NOa can form nitric acid after reacting with water vapour, this causes irritation to eyes and lungs. Injury to lungs, liver and kidneys is caused due to these gases.

Question 9.
Thermal pollution.
Answer:

1. Thermal pollution of water is caused when heated water is added to the water body which results into rise in temperature of water.
2. Thermal and nuclear power plants cause thermal pollution of adjoining water bodies.
3. The power plants use water as coolant and release back this hot water.
4. Many resident organisms which are sensitive to temperature die due to sudden rise in temperature.
5. This leads to loss of flora and fauna of the water body.

Question 10.
Ecosan.
Answer:

1. Ecological sanitation (Ecosan) is a sanitation provision that safely reuses excreta in agriculture as a manure.
2. By Ecosan the need for chemical fertilizers is reduced. Ecosan toilet is a closed system without water and it is an alternative to leach pit toilets.
3. They are useful in places of water scarcity or places with risk of ground water contamination.
4. The principle of recovery and recycling of nutrients from excreta to create a valuable resource for agriculture is used here in Ecosan.
5. The pit of an ecosan toilet fills up after some time, then it is closed and sealed for about 8-9 months.
6. In this time the faeces gets completely composted to organic manure.
7. It is a practical, efficient and cost-effective solution for human waste disposal.
8. There are working Ecosan toilets in many areas of Gujarat, Kerala, Tamil Nadu and Sri Lanka.

Question 11.
Sanitary landfills.
Answer:

1. Sanitary landfills are the places where wastes are dumped in depression or trench. Wastes are dumped here on everyday basis.
2. In large metro cities, landfills are over-filled rapidly. Landfills are unhealthy and they may emit foul odour.
3. Also there is a danger of chemicals percolating and reaching down to ground water and contaminate this water source.

Question 12.
Greenhouse effect.
Answer:
1. Greenhouse effect is caused due to heating up of earth’s surface and atmosphere. This heating is due to trapped infrared rays that are reflected from the earth’s surface by atmospheric gases.

2. The solar energy reaches the earth in the form of ultraviolet radiations, visible light and infrared and radio waves. Clouds and gases reflect about Vith radiations and absorb some of it.

3. Harmful UV radiations are absorbed by the ozone layer of the stratosphere and thus do not reach the earth’s surface.

4. Infrared radiation has heating effect thus it warms up the earth’s atmosphere and various objects. A part of the infrared radiations falling on the earth surface which have longer wavelength is reflected back into the outer space.

5. But there is a layer of carbon dioxide in the lower region of the atmosphere along with other atmospheric gases such as methane, nitrogen oxide, etc.

6. Due to these gases radiations cannot escape out. Carbon dioxide, along with methane, nitrogen oxides and CFCs can absorb infrared radiations reflected from the earth’s surface. They are therefore called greenhouse gases.

7. The blanket of these greenhouse gases in the atmosphere traps the reflected infrared rays and produces heat on the earth’s surface. This results in greenhouse effect which in turn causes global warming.

Question 13.
Mission Harit Maharashtra.
Answer:

1. Government of Maharashtra in 2016, undertook an ambitious project of planting 50 crore trees in four years.
2. Yearly targets were given to each district for plantations.
3. The plantations are under the guidelines of National Forest Policy (NFP).
4. For information about plantation, protection and mass awareness, a 24-hour toll free helpline number 1926 called ‘Hello Forest’ has been set up.
5. There is also a mobile application called ‘My Plants’ which is set up by Forest Department. It records details of the plantation such as numbers, species and location.
6. Following number of saplings were planted. 2.87 crore saplings in 2016, 5.17 crore saplings in 2017, 15.17 crore saplings in 2018, 33 crore saplings in 2019. Authorities are taking care of these plantations.
7. Also Japanese Miyawaki method of plantation is adapted by State Forest Department and Social Forestry Department. Such plantations are in districts of Beed, Hingoli, Pune, Jalgaon, Aurangabad, etc.

Short answer questions

Question 1.
Why and how did biodiversity evolve on the earth?
Answer:

1. There is great diversity with respect to size from microscopic to macroscopic, shape, colour, form, mode of nutrition, type of habitat, reproduction, motility, duration of life cycle span, etc.
2. This diversity evolved in living beings for surviving and perpetuating to accommodate with different environmental conditions such as climatic, edaphic, topographic, geographic, etc. and different situations.
3. For achieving this, living organisms adapted to different conditions and various habitats.
4. This lead in formation of different features which lead to diversity in them.
5. These adaptations in different environments serve as basis for diversity.

Question 2.
Explain species diversity with suitable examples.
Answer:

1. Species diversity is interspecific diversity due to number of species of plants and animals that are present in a region.
2. It can be expressed by variety of species i.e. species richness as well as number of individuals of different species i.e. species evenness.
3. E.g. Species diversity is more in Western Ghats than in Eastern Ghats.
4. Natural undisturbed tropical forests have much greater species richness than monoculture plantation of timber plant, developed by forest plantation.

Question 3.
What are latitudes and longitudes ? Which of these imaginary lines are more significant with reference to diversification of living beings?
Answer:

1. Latitude is a geographic coordinate that specifies the north-south position of a point on the Earth’s surface. It is an angle which ranges from 0° at the Equator to 90° (North or South) at the poles.
2. Longitude is a geographic coordinate that specifies the east-west position of a point on the Earth’s surface, or the surface of a celestial body. It is an angular measurement, usually expressed in degrees.
3. Latitude is the imaginary line that is more significant to diversification of living beings.
4. Species richness shows latitudinal gradient for many plants and animal species. Species richness is high at lower latitudes and there is a steady decline towards the poles.

Question 4.
Explain Productivity Stability Hypothesis.
Answer:

1. Productivity Stability Hypothesis emphasises the importance of species diversity to the ecosystem. This hypothesis was given by David Tillman.
2. It states that rich diversity leads to lesser variation in biomass production over a period of time and species richness is not needed for maintaining the stability of an ecological community.
3. If average biomass production remains fairly constant over a period of time, then that community remains stable.
4. The stable community remains strong to withstand disturbances and also recover quickly. Such community is resistant to invasive species.

Question 5.
What are the shortcomings of the graphic representation diagrams that give data of existing organisms?
Answer:

1. In the diagrams of graphic representation of known animal and plant groups there is no data about prokaryotes.
2. Several moneran species which are not cultivable under laboratory conditions are also not included.
3. Conventional taxonomic methods are not suitable for identification of prokaryotic species. Therefore, such species are not included in these diagrams.

Question 6.
What is India’s wealth in biodiversity?
Answer:

1. India has a share of 8.1% of total biodiversity wealth of the earth.
2. India is one of the 12 megadiversity countries of the globe.
3. India has 2.4% of total land area of the world but there are around 45000 identified plant species and nearly double the number of animal varieties in India.

Question 7.
What are the reasons for loss of biodiversity?
Answer:
Loss of biodiversity occurs due to two main causes:

1. Natural reasons : E.g. Forest fires, earthquakes, volcanic eruptions, etc.
2. Manmade reasons : E.g. Habitat destruction, hunting, settlement, overexploitation and reclamation.

Question 8.
When did major mass extinction events occurred ?
Answer:
In the following geological time scale, plants as well as animal groups underwent major mass extinctions.

1. Between Cretaceous and Coenozoic period.
2. Between Triassic and Jurassic period.
3. Between Permian and Triassic period.
4. Between Devonian and Carboniferous period.
5. Between Ordovician and Silurian period.

Question 9.
What do you understand by invasive species? How does it affect local population?
Answer:

1. New species are introduced into any ecosystem either accidentally or intentionally.
2. Such introduction proves harmful for existing species.
3. Sometimes even local species get extinct.
4. If such extinction happens, then this new species is called an invasive species. E.g. Parthenium or carrot grass, Lantana and water hyacinth (Eichhornia) are such invasive plant species.
5. Nile perch which is a predator fish in Lake Victoria cause harm to 200 local species of Cichlid fish.
6. Clarias gariepinus (African catfish) was brought to India for aquaculture purpose. This catfish species has proved harmful to endemic catfish varieties.
7. Since there is lack of local predator, this alien species survives and cause harmful effect on local species.

Question 10.
Explain how loss of species diversity can harm ecosystem?
Answer:

1. When species diversity is lost, the ecosystem enters into imbalance.
2. The biodiversity loss results into lesser plant production.
3. This causes disruption of further food chains and food webs.
4. The environmental processes such as disease cycles, plant productivity, etc. are also adversely affected.
5. The productivity of the ecosystem is reduced and this results into overall imbalance in the ecosystem.

Question 11.
What is the basic cause of pollution?
Answer:

1. Exponential growth of human population coupled with industrial development is the main cause of imbalance in all the ecosystems.
2. There is also excessive utilization and production of synthetic materials and construction activities.
3. These together have caused several undesired substances in ecosphere.
4. This dumping of substances has resulted in severe pollution.

Question 12.
What are the effects of air pollution?
Answer:

1. Air pollutants affect the surfaces of the respiratory system of all living beings. Therefore, any type of air pollution affects the process of respiration and respiratory system.
2. The concentration of pollutants decide the severity of damage caused to body.
3. Duration of exposure and the type of the organism also decide the effects of air pollution.
4. In plants, air pollution results in poor yield of crops and premature death of plants.
5. Particulate pollutants are very harmful for human beings. Fine particulates which enter the depths of lungs are responsible for irritation, inflammation and damage to lungs.
6. This causes breathing and respiratory disorders and premature deaths.

Question 13.
Which is the major cause of atmospheric air pollution and how can it be avoided?
Answer:

1. Automobiles are the major cause for atmospheric (air) pollution.
2. Regular maintenance of vehicles and use of lead-free petrol or diesel can reduce air pollution as lesser pollutants are released from the exhausts.
3. Using public transport and carpooling can be done to reduce number of automobiles on the road.

Question 14.
Does particulate matter help to reduce atmospheric temperature?
Answer:
Particulate matter plays a complicated role when it comes to influencing the temperature of the earth. The particles are light-absorbing and consequently contribute to the rise in global temperatures, but they also reflect a portion of the sunlight and so play a role in increasing the albedo, which moderates the temperature increase. This is what is known as negative radiative forcing.

Question 15.
Give any norms for reducing sulphur and aromatic contents of petrol and diesel.
Answer:
Euro I, II, III and IV norms were suggested in world to reduce sulphur and aromatic contents of petrol and diesel. In India, the aim is to reduce sulphur emission to 50 ppm in petrol and diesel along with aromatic hydrocarbons to 35%. Government has directly adapted BS VI norm.

Question 16.
What are the ill effects of noise pollution on human health?
Answer:

1. Noise causes psychological and physiological changes in human beings. It causes sleeplessness, increased heartbeat, altered breathing pattern and psychological stress.
2. Extremely high sound level of more than 150 decibels or more generated during a take-off of a jet plane or rocket, may damage ear drums, resulting into permanent hearing loss.
3. Noise negatively interferes with child’s learning and behaviour pattern.

Question 17.
What are the ways to reduce noise pollution?
Answer:

1. Using sound absorbent materials or by muffling the noise, especially in the industrial areas.
2. Laws to reduce noise pollution should be strictly implemented.
3. Blowing of horns should be discouraged in the areas of schools and hospitals.
4. Firecrackers and loudspeakers should be completely banned. Government rules regarding this should be strictly followed.
5. Supreme Court of India has banned loudspeakers at public gatherings after 10 pm.
6. Awareness about noise pollution caused during festivals and processions should be spread among masses.

Question 18.
Explain BOD and its effects on aquatic ecosystem.
Answer:

1. BOD is the amount of dissolved oxygen required by microorganisms for decomposing the organic matter present in water.
2. This can be measured by chemical testing and is expressed in milligram of oxygen per litre (nigT) of water.
3. If BOD of water is high, it denotes that the water is highly polluted with decomposing organic matter.
4. High BOD shows that water is not potable but is polluted with microorganisms and organic debris.
   Waters with higher BOD will also cause death of resident aquatic organisms.
5. Lower BOD on the other hand can vouch for cleaner water.
6. BOD is an indicator of polluted ecosystem.

Question 19.
Why do you think the amount of DDT is maximum in birds?
Answer:

1. Birds like hawk or kingfisher occupy the higher trophic levels.
2. They feed on smaller prey like small birds, frogs, fishes, snakes, etc. These smaller animals feed on insects which already may have accumulated some amount of DDT due to their feeding on plants having DDT content.
3. Since non-biodegradable DDT remains in tissues of organisms, larger birds will receive the maximum dose of DDT.
4. As the phenomenon of biomagnification occurs in case of non-biodegradable DDT, the amount of DDT will go on rising from plants → insects → smaller animals → then to larger birds. Birds of prey will thus have maximum DDT in their tissues.

Question 20.
Ecological sanitation is the need of the day. Justify.
Answer:

1. Large megacities have excessive human population.
2. The problem of water shortage is also severe in some cities.
3. Many slum-dwelling people do not have access to clean and safe toilet too.
4. Open-air defecation results into outbreak of many communicable diseases.
5. It also results into unclean and unhygienic atmosphere. In such case, ecological sanitation is a much better option.

Question 21.
What is solid waste? How is it disposed?
Answer:

1. Solid waste means all the trash which is not needed by a person who throws it. Wastes from home, offices, stores, schools, hospitals, etc. form the wastes which is collected and disposed by municipality.
2. Municipal solid waste is composed of paper, food, plastic, glass, metals, rubber, leather, textile, etc.
3. Burning reduces volume of the waste. But it produces toxic air pollution.
4. The incomplete burning also causes release of CO.
5. Open dumps of wastes are breeding ground for rats and flies, so sanitary landfills are created.

Question 22.
Why is solid waste management facing endless problems?
Answer:

1. Disposal of solid wastes is huge problem and personnel looking after this are overburdened.
2. Open burning causes obnoxious gases which lead to air pollution.
3. Open dumps serve as breeding grounds for rats and flies causing contamination of the environment.
4. Epidemics of infectious diseases can spread if waste is kept unattended.
5. Method of landfills is also inadequate. Such sites in large metros are getting filled. From such landfills, dangerous chemicals also seep and pollute underwater ground resources.
6. Plastic and other synthetic items in the solid waste cause further problems in decomposition and hence they damage the ecosystem.

Question 23.
How pollution by domestic garbage can be controlled?
Answer:

1. Everybody should learn to segregate garbage at source. The wet i.e. biodegradable and dry i.e. non-biodegradable materials should be separated, instead of throwing them callously in trash.
2. Non-biodegradable material can be recycled or reused. It should be disposed off in proper way.
3. Biodegradable material should be composted at home by simple methods of vermicomposting.
4. Every household should practise these methods which will reduce the volume of garbage that is sent out to land filling or burning.
5. The excessive load on Municipalities can also be reduced if each one takes the responsibility of his or her own garbage.
6. Larger establishments like offices, schools or industries should provide space for solid waste and other garbage management.

Question 24.
What is e-waste? How is it disposed?
Answer:

1. Any useless part of electronic origin or irreparable computers and other electronic goods as well as electrical waste are known as e-wastes.
2. e-wastes are buried in landfills or are completely burnt.
3. e-waste is also exported to developing countries like China, India and Pakistan from developed countries for recycling.
4. Recycling is done for e-wastes in which metals like copper, iron, silicon, nickel and gold are recovered.

Question 25.
How citizens should sort out solid wastes?
Answer:

1. All the collected wastes should be categorized into three types, viz. biodegradable, non- biodegradable and recyclable.
2. Each person should sort out the waste. The biodegradable wastes should be put into deep pits and allowed to undergo slow degradation. Every building should have pit for biodegradable domestic wastes.
3. Recyclable material like newspaper, plastic bags, empty milk pouches should be sold off.
4. Every citizen should reduce his or her garbage generation and should also minimize the use of non-biodegradable products. Simple change such as refusing a plastic bag and using a reusable cloth bag can solve the problem of plastic garbage.

Question 26.
What are the preventive measures against Global warming?
Answer:

1. Reducing use of fossil fuel.
2. Efficient use of energy. Use of alternative energy sources like solar or wind energy.
3. Reducing deforestation.
4. Tree plantation and afforestation activities.
5. Reducing the rate of growth of human population.
6. International initiatives for reduction of greenhouse gases.

Question 27.
What is global warming? On what does it depend?
Answer:

1. Global warming is increased temperature of the earth. During past century, the temperature of the
2. Earth has increased by 0.6 °C, most of it during last three decades.
3. This is mainly caused by greenhouse effect.
4. Global warming depends upon the amount of CO₂ and other greenhouse gases present in the atmosphere.
5. An increase in CO₂ concentration increases earth’s temperature by retaining more heat.
6. Carbon dioxide increases temperature by about 50%, CFCs increase it by 20% and methane increases it by 15% whereas other pollutants increase it by 10%.
7. Atmospheric air pollution, industrialization and greenhouse effect cause global warming.

Question 28.
Give an account of possible effects of global warming.
Answer:

1. This rise in temperature leads to unfavourable changes in environment and resulting in odd climatic changes. (E.g. El Nino effect).
2. Global warming results in melting of polar ice caps and Himalayan snow caps which may be the cause for submerging of the coastal areas.
3. The frequency and severity of cyclones, hurricanes and unseasonal rains has increased tremendously causing series of natural disasters all over the world. This results into loss of infrastructure.
4. Outbreak of various vector borne diseases has increased in last two decades due to global warming.
5. The marine environment is worst affected due to excessive heat being absorbed into oceans.

Question 29.
Why greenhouse gases are increasing in their proportion?
Answer:

1. Greenhouse gases are increasing due to excessive burning of fossil fuels in industries, by automobiles and air transportation, by burning of agricultural wastes, etc.
2. All the combustion is causing levels of CO₂ to rise.
3. Biogas plants, paddy fields, cattle sheds add methane to atmosphere.
4. More amount of chlorofluorocarbons are emitted due to fire extinguishers and air conditioners.
5. Due to loss of forest cover, there are few trees left to absorb atmospheric CO₂.
6. Man is making development and progress due to which green house gases are increasing in proportion.

Question 30.
How is ozone hole formed?
OR
Give effect of CFC on ozone shield.
Answer:

1. The ozone molecules are attacked by chlorofluorocarbon molecules. This action causes disturbances in the ozone shield due to increased rate of ozone degradation by Chlorofluorocarbon (CFC).
2. CFCs can move upwards to reach stratosphere. Here UV rays act on them and release Cl^– atoms. The released Cl-degrades ozone, which subsequently releasing molecular oxygen.
3. Cl^– atoms act as catalyst. So, they remain in the stratosphere and continue the effect of ozone degradation.
4. This results in ozone depletion. This leads to the formation of ozone hole which is a large area of thinned ozone layer. One such ozone hole was observed over the Antarctic region.

Question 31.
What is the effect of UV-B radiation on body?
Answer:

1. UV-B radiations of wavelength 280-322nm cause following deleterious effects:
2. Damaging effect to DNA.
3. Formation of mutation.
4. Aging of skin and damage to skin cells including various types of skin cancers.
5. Cornea of human eye absorbs UV-B radiations. High dose of UV-B causes inflammation of cornea called snow blindness, cataract, etc. leading to permanent damage to cornea.

Question 32.
What is Montreal Protocol?
Answer:

1. Montreal Protocol was signed agreement of an international treaty among different nations who had recognised the harmful effects of ozone depletion.
2. It was signed at Montreal (Canada) in 1987 to control emission of ozone depleting substances.
3. After signing of Montreal protocol and its subsequent execution in 1989, the ozone depletion has been reduced worldwide.

Question 33.
‘There is a hole in the ozone layer.’ What do you understand by this ?
Answer:

1. The area in Antarctica region with a thin ozone layer is known as ozone hole.
2. CFCs (chlorofluorocarbons) which -are widely used as refrigerants disturb the balance between the production and degradation of ozone.
3. Ultraviolet (UV) rays and chlorofluorocarbons (CFCs) release chlorine (Cl^–) ions which act as catalyst in the degradation of ozone layer.
4. Chloride ions also degrade the ozone layer.
5. The degradation of ozone layer results in the depletion of ozone causing a hole in the ozone layer.

Question 34.
What is the scenario of deforestation in India?
Answer:

1. The scenario of deforestation is grim in India because we have cut down many trees from forests due to various developmental activities.
2. At the beginning of 20th century, 30% was the forest cover.
3. By the end of the 20th century, it became 19.4%.
4. The National Forest Policy 1988 of India has recommended 33% forest cover for the plains and 67% for the hills.

Question 35.
What are the causes of deforestation?
Answer:
Causes of deforestation are as follows:

1. Conversion of forest to agricultural land for growing food for ever-increasing human population.
2. Trees are cut for timber, firewood, for keeping cattle in farm and for other purposes.
3. For constructions of dams, road, railways, metros, residential complexes, etc.
4. For any kind of developmental activities due to Government Policies.

Question 36.
What is Jhum cultivation?
Answer:

1. Jhum cultivation is practised in north eastern India.
2. This is also called slash and burn agriculture in which farmers cut down trees of the forest and burn the plant remains.
3. This ash from burnt trees is used as fertilizer. The land is used for farming and cattle grazing.
4. When cultivation is harvested, the area is left for several years so as to allow its recovery.

Question 37.
How Jhum cultivation has lead to deforestation in recent years?
Answer:

1. Because in the Jhum cultivation, the forest trees are burnt to make space for farming and for obtaining ash as fertilizer, it leads to loss of precious forest cover.
2. Once the trees are destroyed, farmers use this land for farming. After the cultivation and harvest, the land is left barren.
3. It is used for cattle grazing.
4. Since long period is required for the recovery of land back into forest patch, the deforestation results.

Question 38.
What are the major effects of deforestation ?
Answer:
Major effects of deforestation:

1. Increased concentration of COa in the atmosphere.
2. Trees hold lot of carbon in their biomass which is lost with deforestation.
3. Loss of biodiversity due to habitat destruction.
4. Disturbances in hydrologic cycle.
5. Soil erosion and desertification in extreme cases.

Question 39.
Why Amrita Devi Bishnoi Wildlife Protection Award is started by Government of India? To whom is it given?
Answer:

1. In 1731, a Bishnoi woman Amrita Devi hugged the trees to save them but was killed by King of Jodhpur who wanted the wood for his palace. For protecting the forest Amrita Devi and her three daughters along with hundreds of other Bishnois lost their lives.
2. The Government of India in memory of this sacrifice has started Amrita Devi Bishnoi Wildlife Protection Award.
3. This award is given to individuals or community from rural areas who protect wildlife.

Question 40.
What is Joint Forest Management?
Answer:

1. Government of India has introduced the concept of Joint Forest Management (JFM) for working closely with local communities who protect and manage forests.
2. In return, for their services to the forest, the communities can get benefit of various forest products (Fruits, gum, rubber, medicine, etc.).
3. This is done with the idea to conserve the forest in a sustainable manner.
4. JFM has started from 1980.

Question 41.
Floods in Sangli and Kolhapur in August 2019, were responsible for many problems during and after the floods. Think and enlist different types of problems faced by flood affected areas.
Answer:

1. Floods of Western Maharashtra were mainly due to unseasonal and extremely heavy rain caused by climate change events.
2. The floods had hit the districts of Kolhapur and Sangli hard. Sangli got completely marooned: There was no electric supply for extended period.
3. Over two lakh people were living without electricity in affected areas.
4. Lack of food and drinking water was a main problem.
5. Several water-supply schemes became dysfunctional.
6. Fields were completely ruined as crops were damaged. In Kolhapur district alone crops over 67,000 hectares were damaged.
7. Large number of cattle were dead.
8. Houses were destroyed completely. Thousands of people were shifted to safer places.
9. About 223 villages in district of Kolhapur suffered a lot. 18 out of these have been completely marooned.
10. About 28,897 persons were affected out of which 8,923 people were shifted.
11. 813 houses were affected in the district, out of which 89 are completely damaged.
    There was also the scarcity of petrol and diesel.
12. The Mumbai-Bengaluru National Highway, which passes through Kolhapur was dysfunctional and hence transport was also affected.

Chart based/Table based questions

Question 1.
Complete the given table:

Vehicle	Norms	Cities of Implementation
—————-	Bharat Stage II	——————-
4 wheelers	Bharat Stage III	——————
4 wheelers	—————	13 mega cities (Delhi and NCR, Mumbai, Kolkata, Chennai, Bengaluru, Surat, Kanpur, Agra, Lucknow, Solapur) since April 2010.
2 wheelers	Bharat Stage III	——————–
—————	Bharat Stage III	Throughout the country since October 2010

Answer:

Vehicle	Norms	Cities of Implementation
4 wheelers	Bharat Stage II	All metro cities
4 wheelers	Bharat Stage III	Throughout the country since October 2010
4 wheelers	Bharat Stage IV	13 mega cities (Delhi and NCR, Mumbai, Kolkata, Chennai, Bengaluru, Surat, Kanpur, Agra, Lucknow, Solapur) since April 2010.
2 wheelers	Bharat Stage III	Throughout the country since October 2010
3 wheelers	Bharat Stage III	Throughout the country since October 2010

Question 2.
Complete the given table

Name of polluting gas	Source	Effect
——————	————–	Combining with haemoglobin and causing respiratory problems
NO2	Chemical industries	—————–
CO2	—————	——————
——————-	Biogas plants	Greenhouse effect

Answer:

Name of polluting gas	Source	Effect
CO	Vehicular exhaust	Combining with haemoglobin and causing respiratory problems
NO2	Chemical industries	Irritation to lungs and eyes
CO2	Combustion of any kind	Greenhouse effect
CH4	Biogas plants	Greenhouse effect

Diagram based questions

Question 1.
What can you say about species diversity A and B?

Answer:

1. Size of species A and B are same. The number of species shown in both A and B are also same as both A and B have 4 different species each.
2. Group A : In this species 4 different species are seen, but the density of the plants is not much. Group B also shows 4 different species. But two of these are in less number.
3. Group A is more diverse that species B, but species B is showing more population of one species.

Question 2.
Observe the figure and answer the following questions.
(a) Identify the given instrument.
(b) What is its significance?
(c) Explain its working in brief.

Answer:
(a) Electrostatic precipitator.

(b) This is an important equipment which is used to remove particulate matter like soot and dust present in industrial exhaust. It is capable of removing almost 99% particulate matter present in exhaust of a thermal power plant.

(c) (1) In Electrostatic precipitator, high voltage is applied which produces electric discharge.
(2) This discharge causes ionisation of air in the smokestack.
(3) As a result, the free electrons are formed.
(4) These electrons in the ionised air attach to the gaseous or dust particles moving up the stack.
(5) Negatively charged particles move towards the positive electrode and settle down there.
(6) They are then removed by vibrations of the electrodes and collected in the reservoir.

Question 3.
Observe the given diagram and answer the questions given below.

(a) What is the name of this equipment?
(b) What is the function of such equipment?
(c) Explain its working in brief.
Answer:
(a) Exhaust gas scrubber.
(b) Exhaust gas scrubbers are used to clean air by removing both dust and gases.
(c) The exhaust is passed through dry or wet packing material. When it is done, gases like SO₂ are removed. For this purpose, the exhaust is passed through a spray of water or lime.

Question 4.
Observe the given diagram and answer the questions based on it.

(a) Which apparatus is shown in the given diagram ?
(b) What is the function of this apparatus?
(c) What are the reactions that can take place in blocks 1, 2, and 3?
Answer:
(a) Catalytic converter.

(b) The harmful gases like CO and nitrogen oxides which are present in the automobile exhausts are removed by catalytic converters. Thereby, harmful effects of air pollution are reduced.

(c) In block 1 : Nitrogen oxides are present in the exhaust gases. They enter into reduction block of catalyst. The oxides of nitrogen react forming nitrogen and oxygen.
In block 2 : The exhaust gases enter the next block called oxidation block of the catalyst. Here, hydrocarbons and the newly formed oxygen react to form carbon dioxide.
In block 3 : The exhaust gases enter into last block from here the least harmful gases are released out.

Question 5.
Observe the graph and explain Alexander von Humboldt’s views about species richness and area relationship.

Answer:

1. Scientists have tried to establish relationship between species diversity and the size of the habitat. It is considered that number of species present is directly proportional to the area.
2. It is understood that larger areas may have more resources that can be distributed amongst the inhabitant species.
3. Alexander von Humboldt observed that species richness does increase with the increase in area but only till a certain limit.
4. For many species this curve is a rectangular hyperbola.
5. If we consider S to be species richness, A as area under study, C as the Y intercept and Z as the slope of the line, this relationship can be described by the equation, log S = log C + Z log A.
6. On logarithmic scale this relationship is a straight line, as observed in the figure above. For smaller areas, value of Z ranges between 0.1 to 0.2 regardless of species or region under study.
7. But for the larger areas like the entire continents, slopes are closer to vertical axis i.e. steeper.
8. This observation indicates that in very large areas, number of species found, increase faster than the area explored.

Question 6.
Explain the phenomenon of biomagnification by observing the diagram given below.
OR
Explain the phenomenon of biological magnification.

Answer:

1. Some non-biodegradable substances like pesiticides or heavy metals have the tendency to accumulate in the tissues of living organisms.
2. When such organisms are eaten by their predators, these pollutants enter the bodies of predators.
3. At lower trophic level the concentration of such pollutant may be low, but when they are fed upon by their predator the amount of pollutant goes on increasing.
4. As shown in the diagram there is only 0.000003 ppm DDT in the water. This DDT level is meagre but when zooplankton survive in this water, DDT concentration increases in their body and becomes 0.04 ppm.
5. When many of these zooplankton are eaten by small fish, it rises to 0.5 ppm.
6. In turn, the several smaller fish are eaten by a large fish and in it the concentration rises to 2 ppm.
7. When such larger fishes are consumed by a bird, it receives maximum amount of DDT which might kill this bird.
8. In this way, the DDT level shows biomagnification. Biomagnification is thus the phenomena of increase in the concentration of non-biodegradable substances according to the food chain or trophic relationships.

Long answer questions

Question 1.
What is biodiversity? Explain genetic diversity with suitable example.
Answer:
1. Biodiversity is the part of nature which includes the differences in the genes among the individuals of a species; the variety and richness of all plants and animal species at different scales in a space – local regions, country and the world; and the types of ecosystem, both terrestrial and aquatic, within a defined area.

2. Genetic diversity:
Genetic diversity is the intraspecific diversity in the number and types of genes and chromosomes present in different species.

• It also includes variation in the genes and their alleles in the same species. Variation within a population and diversity between populations that are associated with adaptation to local conditions.
• Genetic diversity or variability is essential for a healthy breeding population of a species.
• Genetic variations are changes in the allelic genes which lead to individual differences within species.
• Such variations help in the evolution. The chances of continuation of species in the changing environmental conditions are caused due to such variation and it allows the best organisms to get adapted to survive. Races and subspecies are formed due to genetic diversity.
• Examples of genetic diversity : (a) There are 1000 varieties of mangoes and 50.000 varieties of rice or wheat in India, (b) Rauwolfia vomitoria is a medicinal plant that secretes reserpine.
• This plant is inhabitant of different Himalayan ranges. There is variations in terms of potency and concentration of reserpine, from different locations.

Question 2.
Species richness goes on decreasing as we move from equator to pole. Explain.
Answer:

1. In tropical regions, there are lesser climatic changes throughout the year and availability of plenty of sunlight.
2. Moreover, in tropical areas there are lesser disturbances like periodic glaciations as compared to those seen in the polar regions.
3. In tropical regions, there is a stability over millions of years which favoured speciation and hence there is more species richness.
4. Also in tropical regions, there are lesser migrations which reduce gene flow between geographically isolated regions. This too favoured speciation.
5. There is more availability of intense sunlight, warmer temperatures and higher annual rainfall in tropics. These factors have brought higher species richness in tropics.
6. Constant climatic conditions and abundance of resources in tropical regions provide more food preferences for animals species.
7. E.g. fruits are available throughout the year in rain forests, therefore variety of frugivorous animals are seen here, as compared to the temperate regions.

In short, species richness or diversity for plants and animals decreases as we move away from equator to the poles.

Question 3.
Explain Rivet Popper Hypothesis.
Answer:

1. Rivet Popper hypothesis was given by Paul Ehrlich to emphasise significance of diversity.
2. For explaining the hypothesis, he gave an analogy between aeroplane and ecosystem.
3. As the rivets keep all parts of the aeroplane together, similarly, all species keep the diversity of an ecosystem in functional.
4. Just as if one species gets extinct, initially not much of a problem will take place in an ecosystem, just as in case of a single rivet mission cannot cause problem in flight. However, if the same damage is continued, the turbulence will be experienced.
5. When more rivets are popped out gradually, there will be a serious threat to the safety of the aeroplane.
6. Also the rivets in key positions can cause serious situation. With same analogy he explained that if loss of species occurs, initially the problem will not be obvious but later if similar damage continues, there will be a threat to the ecosystem.
7. Thus, there is a relationship between diversity and well-being of ecosystem which is not linear.
8. Loss of key species causes threat in very short span of time by affecting food chains, food web, energy flow, natural cycles, etc. This will disturb the balance of the ecosystem.

Question 4.
Give various categories of endangered species explained by IUCN.
Answer:
Following are the categories of endangered species as explained by IUCN.

1. Extinct (EX) : EX is added to species in which the last individual has died or is not recorded. So now on the earth not a single organism of this kind can be seen.
2. Extinct in the Wild (EW) : This category contains those species whose members survive only in captivity.
3. Critically Endangered (CR) : Critically endangered is a category containing those species that possess an extremely high risk of extinction with very few surviving members around 50 or so.
4. Endangered (EN) : EN is added to a species that possess a very high risk of extinction as a result of rapid population decline of 50 to more than 70% over past three generations or the previous 10 years.
5. Vulnerable (VU) : VU is a category containing those species that possess a very high risk of extinction as a result of rapid population decline of 30 to more them 50% over last three generations or the previous 10 years.
6. Near Threatened (NT) : NT are species that are close to becoming threatened or may meet the criteria for threatened status in the near future.
7. Least Concern (LC) : LC is a category containing species that are pervasive and abundant after careful assessment.
8. Data Deficient (DD) : DD is a condition applied to species in which the amount of available data related to its risk of extinction, is lacking in some way.
9. Not Evaluated (NE) In NE category any of the nearly 1.9 million species described by scientists are included, but not assessed by the IUCN.

Question 5.
What were the measures taken by Delhi Government to combat air pollution in Delhi?
Answer:
Following measures were taken to combat air pollution by Delhi Government:

1. All the city buses were converted to run on CNG (i.e. compressed natural gas) by year 2002. CNG causes less pollution and is less expensive fuel.
2. There was new fuel policy drafted by the Government.
3. The norms were set to reduce sulphur and aromatic content of petrol and diesel.
4. Engines of the automobiles were upgraded.
5. Bharat stage emission standards (BS) were set. These standards are equivalent to Euro norms and have evolved on similar lines as Bharat Stage II (BS II) to BS VI from 2001 to 2017.

Question 6.
Why is Delhi worst polluted city as far as air pollution is concerned?
Answer:

1. Delhi has colder weather during winters.
2. There are stagnant winds which trap smoke from various sources like automobile exhausts and firecrackers.
3. Many farms surrounding Delhi resort to burning crop stubbles.
4. Even in the city garbage is lit.
5. There is more dust on the roads.
6. Automobile traffic is heavy and public transport systems were not efficient, till the metro was started.

Effects caused by this air pollution are as follows:

1. Citizens suffered breathlessness and chest muscle contraction.
2. The irritation in eyes, asthma and allergy were frequently reported.

Question 7.
What were the control measures taken by Delhi Government for combating air pollution ?
Answer:
Following measures were taken by Delhi Government:

1. By 2002, all the city buses of Delhi were converted to CNG buses which now do not run on diesel.
2. The new fuel policy was introduced and the norms were set to reduce sulphur and aromatic content of petrol and diesel.
3. Upgradation of engines was done.
4. Bharat stage emission standards (BS) were set which were equivalent to Euro norms.
5. Bharat Stage II (BS II) to BS VI norms were given from 2001 to 2017.
6. Administration took certain measures like closing educational institutions, suspending of construction or demolition work, undertaking vacuum cleaning of roads, etc.
7. The polluting industries were penalized and Badarpur thermal power plant was temporarily closed down.

Question 8.
How is water polluted due to domestic sewage and Industrial Effluents?
Answer:

1. When water is impure, it cannot be used for human consumption.
2. Small amount about 0.1% impurities in water also makes the water polluted.
3. In domestic sewage, there are dissolved salts such as nitrates, phosphates, other nutrients and toxic metal ions as well as organic compounds.
4. Sewage also contains biodegradable organic matter and harmful bacterial and virus.
5. Organic matter can be decomposed by bacteria and other microorganisms. But such water is not potable.
6. Industrial effluents also contain harmful heavy metals and other solids.
7. Solids can be easily removed from water but the dissolved salts cannot be separated.
8. Biodegradable organic matter in sewage water is calculated by measuring Biochemical Oxygen Demand (BOD).

Question 9.
What is eutrophication? Describe the phenomena of eutrophication.
Answer:

1. Eutrophication is the phenomena caused when a body of water becomes overly enriched with minerals and nutrients which induce excessive growth of algae.
2. This process may result in oxygen depletion of the water body.
3. Planktonic algae and algal bloom are the result of such nutrient-enriched water.
4. Due to excessive growth of these plant species, the light in the lower layer of water is reduced.
5. These plants perform photosynthesis during daytime but at night they compete with animal species for oxygen.
6. There organic load of water body increases causing reduction in dissolved oxygen content.
7. This results in death of fishes and other aquatic organisms. Once these dead bodies start decomposing in the water, there is more oxygen depletion.
8. It then results in loss of species diversity.
9. The water body which was once eutrophic now turns into stinking and turbid water body with coloured water. This is death of an ecosystem.

Question 10.
What are the two types of eutrophication? What is the main difference in them?
Answer:

1. Natural Eutrophication and Cultural or Accelerated Eutrophication are two types of eutrophication.
2. Natural eutrophication is caused due to natural processes. It is also called aging of a lake due to nutrient enrichment of water.
3. It is a very slow and gradual process.
4. Natural aging of lakes may take thousands of years, depending on the size of the lake, climatic conditions and other factors.
5. Cultural or Accelerated eutrophication is caused due to human activities and chiefly due to pollution. Effluents from agricultural lands, industries and homes (household) cause such type of eutrophication.
6. This phenomenon is called Cultural or Accelerated eutrophication.
7. Due to excessive growth of algae there is lesser amount of dissolved oxygen for aquatic organisms, especially during night time. Due to this death of fish and other aquatic organisms take place.
8. The dead bodies of aquatic organisms decompose causing further depletion of the dissolved oxygen.
9. This results into complete collapse of the ecosystem of water body.

Question 11.
Comment on deforestation status of the world and its major effects.
Answer:
I. Deforestation status of the world:

1. Forest area has declined all across the world in the past three decades. But in last decade, it is the reverse trend as the rate of forest loss has declined due to the growth of sustainable management.
2. Global Forest Resources Assessment 2020 (FRA 2020) has given the figures of the rate of forest loss in 2015-2020. It is said to be declined to 10 million hectares (mha), reducing from 12 million hectares (mha) in 2010-2015.
3. The FRA 2020 was released by the United Nations Food and Agriculture Organization (FAO) on May 13, 2020. It has examined the status of, and trends in, more than 60 forest-related variables in 236 countries and territories in the period 1990-2020.
4. The world lost 178 mha of forest since 1990. However, the rate of net forest loss decreased substantially during 1990-2020 due to a reduction in deforestation in some countries, plus increases in forest area in others through afforestation and the natural expansion of forests.
5. The rate of net forest loss declined from 7.8 mha per year in the decade 1990-2000 to 5.2 mha per year in 2000-2010 and 4.7 mha per year in 2010-2020.
6. Among the world’s regions, Africa had the largest annual rate of net forest loss in 2010-2020, at 3.9 mha, followed by South America, at 2.6 mha.
7. On the other hand, Asia had the highest net gain of forest area in 2010-2020, followed by Oceania and Europe.
8. The world’s total forest area was 4.06 billion hectares (bha), which was 31% of the total land area. This area was equivalent to 0.52 ha per person, the report noted.
9. The largest proportion of the world’s forests were tropical (45%), followed by boreal, temperate and subtropical.
10. More than 54% of the world’s forests were in only five countries – the Russian Federation, Brazil, Canada, the United States of America and China. The area of naturally regenerating forests worldwide decreased since 1990, but the area of planted forests increased by 123 mha.

The rate of increase in the area of planted forest slowed in the last ten years.
(Source : https : //www.downtoearth.org.in/ news/forests/deforestation-rate-globally- declined-between-2015-and-2020-fao- report-71107)

II. The effects of deforestation:

1. Increased concentration of COa in the atmosphere.
2. Trees hold lot of carbon in their biomass which is lost with deforestation.
3. Loss of biodiversity due to habitat destruction.
4. Disturbances in hydrologic cycle.
5. Soil erosion and desertification in extreme cases.