Class 12

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 4 Reproductive Health Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 4 Reproductive Healths

PSEB 12th Class Biology Guide Reproductive Health Textbook Questions and Answers

Question 1.
What do you think is the significance of reproductive health in a society?
Answer:
Reproductive health is the total well being in all aspects of reproduction. It includes physical, emotional, behavioural, and social well being. Sexually transmitted diseases such as AIDS, gonorrhoea, etc. are transferred from one individual to another through sexual contact. It can also lead to unwanted pregnancies.

Hence, it is necessary to create awareness among people, especially the youth, regarding various reproduction related aspects as the young individuals are the future of the country and they are most susceptible of acquiring sexually transmitted diseases. Creating awareness about the available birth control methods, sexually transmitted diseases and their preventive measures, and gender equality will help in bringing up a socially conscious healthy family. Spreading. awareness regarding uncontrolled population growth and social evils among young individuals will help in building up a reproductively healthy society.

Question 2.
Suggest the aspects of reproductive health which need to be given special attention in the present scenario.
Answer:
Reproductive health is the total well being in all aspects of reproduction. The aspects which have to be given special attention in the present scenario are as follows:
1. Counselling and creating awareness among people, especially the youth, about various aspects of reproductive health, such as sexually transmitted diseases, available contraceptive methods, case of pregnant mothers, adolescence, etc.

2. Providing support and facilities such as medical assistance to people during pregnancy, STDs, abortions, contraceptives, infertility, etc. for building a reproductively healthy society.

Question 3.
Is sex education necessary in schools? Why?
Answer:
Yes, introduction of sex education in schools is necessary. It would provide right information to young individuals at the right time about various aspects of reproductive health such as reproductive organs, puberty, and adolescence related changes, safe sexual practices, sexually transmitted diseases, etc.

The young individual or adolescents are more susceptible in acquiring various sexually transmitted diseases. Hence, providing information to them at the right time would help them to lead a reproductively healthy life and also protect them from the myths and misconceptions about various sex related issues.

Question 4.
Do you think that reproductive health in our country has improved in the past 50 years? If yes, mention some such areas of improvement.
Answer:
Yes, the reproductive health has tremendously improved in India in the last 50 years. The areas of improvement are as follows:

1. Massive child immunisation programme, which has lead to a decrease in the infant mortality rate.
2. Maternal and infant mortality rate, which has been decreased drastically due to better post natal care.
3. Family planning, which has motivated people to have smaller families.
4. Use of contraceptive, which has resulted in a decrease in the rate of sexually transmitted diseases and unwanted pregnancies.

Question 5.
What are the suggested reasons for population explosion?
Answer:
The human population is increasing day by day, leading to population explosion. It is because of the following two major reasons:
(a) Decreased death rate
(b) Increased birth rate and longevity
The death rate has decreased in the past 50 years. The factor leading to decreased death rate and increased birth rate are control of diseases, awareness and spread of education, improvement in medical facilities, ensured food supply in emergency situation, etc. All this has resulted in an increase in the longevity of an individual.

Question 6.
Is the use of contraceptives justified? Give reasons.
Answer:
Yes, the use of contraceptives is absolutely justified. The human population is increasing tremendously. Therefore, to regulate the population growth by regulating reproduction has become a necessary demand in the present times. Various contraceptive devices have been devised to reduce unwanted pregnancies, which help in bringing down the increased birth rate and hence, in checking population explosion.

Question 7.
Removal of gonads cannot be considered as a contraceptive option. Why?
Answer:
Contraceptive devices are used to prevent unwanted pregnancy and to prevent the spreading of STDs. There are many methods, such as natural, barrier, oral, and surgical methods, that prevent unwanted pregnancy. However, the complete removal of gonads cannot be a contraceptive option because it will lead to infertility and unavailability of certain hormones that are required for normal functioning of accessory reproductive parts. Therefore, only those contraceptive methods can be used that prevent the chances of fertilisation rather than making the person infertile forever.

Question 8.
Amniocentesis for sex determination is banned in our country. Is this ban necessary? Comment.
Answer:
Amniocentesis is a pre-natal diagnostic technique that is used to determine the sex and metabolic disorders of the developing foetus in the mother’s uterus through the observation of the chromosomal patterns. This method was developed so as to determine any kind of genetic disorder present in the foetus. However, unfortunately, this technique is being misused to detect the sex of the child before birth and the female foetus is then aborted. Thus, to prevent the increasing female foeticides, it is necessary to ban the usage of amniocentesis technique for determining the sex of a child.

Question 9.
Suggest some methods to assist infertile couples to have children.
Answer:
Infertility is the inability of a couple to produce a baby even after unprotected intercourse. It might be due to abnormalities present in either male or female, or might be even both the partners. The techniques used to assist infertile couples to have children are as follows:
(a) Test Tube Babies: This involves in-vitro fertilisation where the sperms meet the egg outside the body of a female. The zygote, hence produced, is then transferred in the uterus or fallopian tube of a normal female. The babies produced from this method are known as test tube babies.

(b) Gamete Intra Fallopian Transfer (GIFT): It is a technique that involves the transfer of gamete (ovum) from a donor into the fallopian tube of the recipient female who is unable to produce eggs, but has the ability to conceive and can provide right conditions for the development of an embryo.

(c) Intra Cytoplasmic Sperm Injection (ICSI): It is a method of injecting sperm directly into the ovum to form an embryo in laboratory.

(d) Artificial Insemination: Artificial insemination is a method of transferring semen (sperm) from a healthy male donor into the vagina or uterus of the recipient female. It is employed when the male partner is not able to inseminate the female or has low sperm counts.

Question 10.
What are the measures one has to take to prevent from contracting STDs?
Answer:
Sexually transmitted diseases (STDs) get transferred from one individual to the other through sexual contact. Adolescents and young adults are at the greatest risk of acquiring these sexually transmitted diseases. Hence, creating awareness among the adolescents regarding its after-effects can prevent them from contracting STDs. The use of contraceptives, such as condoms, etc. while intercourse, can prevent the transfer of these diseases. Also, sex with unknown partners or multiple partners should be avoided as they may have such diseases. Specialists should be consulted immediately in case of doubt so as to assure early detection and cure of the disease.

Question 11.
State True/False with explanation.
(a) Abortions could happen spontaneously too. (True/False)
(b) Infertility is defined as the inability to produce a viable offspring and is always due to abnormalities/defects in the female partner. (True/False)
(c) Complete lactation could help as a natural method of contraception. (True/False)
(d) Creating awareness about sex related aspects is an effective method to improve reproductive health of the people. (True/False)
Answer:
(a) False
Abortion is term given for medical termination of pregnancy.

(b) False
Infertility is defined as the inability of the couple to produce baby even after unprotected coitus. It might occur due to abnormalities/defects in either male or female or both.

(c) False
Complete lactation or lactational amenorrhea is a natural method of contraception. Flowever, it is limited till lactation period, which continues till six months after parturition.

(d) True.

Question 12.
Correct the following statements:
(a) Surgical methods of contraception prevent gamete formation.
(b) All sexually transmitted diseases are completely curable.
(c) Oral pills are very popular contraceptives among the rural women.
(d) In E. T. techniques, embryos are always transferred into the uterus.
Answer:
(a) Surgical methods of contraception prevent the flow of gamete during intercourse.
(b) Some of the sexually transmitted diseases are curable if they are detected early and treated properly. AIDS is still an incurable disease.
(c) Oral pills are very popular contraceptives among urban women.
(d) In embryo transfer technique, 8 celled embryos are transferred into the fallopian tube while more than 8 celled embryos are transferred into the uterus.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 3 Human Reproduction Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 3 Human Reproduction

PSEB 12th Class Biology Guide Human Reproduction Textbook Questions and Answers

Question 1.
Fill in the blanks:
(a) Humans reproduce …………………. .(asexually/sexually)
Answer:
sexually.

(b) Humans are …………………… .(oviparous/viviparous/ovoviviparous)
Answer:
viviparous.

(c) Fertilisation is ………………………. in humans, (external/internal)
Answer:
internal

(d) Male and female gametes are …………………. .(diploid/haploid)
Answer:
haploid.

(e) Zygote is ………………….. .(diploid/haploid)
Answer:
diploid.

(f) The process of release of ovum from a mature follicle is called ……………………. .
Answer:
ovulation.

(g) Ovulation is induced by a hormone called …………………… .
Answer:
luteinising hormone.

(h) The fusion of male and female gametes is called ……………………… .
Answer:
fertilisation.

(i) Fertilisation takes place in ………………… .
Answer:
fallopian tube (ampullary-isthmic junction).

(j) Zygote divides to form ……………….., which is implanted in uterus.
Answer:
blastocyst

(k) The structure which provides vascular connection between foetus and uterus is called …………………… .
Answer:
placenta.

Question 2.
Draw a labelled diagram of male reproductive system.
Answer:

Question 3.
Draw a labelled diagram of female reproductive system.
Answer:

Question 4.
Write two major functions each of testis and ovary.
Answer:
Functions of the testis
(a) They produce male gametes called spermatozoa by the process of spermatogenesis.
(b) The leydig cells of the seminiferous tubules secrete the male sex hormone called testosterone. Testosterone aids the development of secondary sex characteristics in males.

Functions of the ovary
(a) They produce female gametes called ova by the process of oogenesis.
(b) The growing Graafian follicles secrete the female sex hormone called estrogen. Estrogen aids the development of secondary sex characteristics in females.

Question 5.
Describe the structure of a seminiferous tubule.
Answer:
The production of sperms in the testes takes place in a highly coiled structure called the seminiferous tubules. These tubules are located in the testicular lobules. Each seminiferous tubule is lined by germinal epithelium. It is lined on its inner side by two types of cells namely spermatogonia aid sertoli cells respectively. Spermatogonia are male germ cells which produce primary spermatocytes by meiotic divisions. Primary spermatocytes undergo further meiotic division to form secondary spermatocytes and finally, spermatids. Spermatids later metamorphoses into male gametes called spermatozoa. Sertoli cells are known as nurse cells of the testes as they provide nourishment to the germ cells. There are large polygonal cells known as interstitial cells or leydig cells just adjacent to seminiferous tubules. These cells secrete the male hormone called testosterone.

Question 6.
What is spermatogenesis? Briefly describe the process of spermatogenesis.
Answer:
Spermatogenesis is the process of the production of sperms from the immature germ cells in males. It takes place in seminiferous tubules present inside the testes. During spermatogenesis, a diploid spermatogonium (male germ cell) increases its size to form a diploid primary spermatocyte. This diploid primary spermatocyte undergoes first meiotic division (meiosis I), which is a reductional division to form two equal haploid secondary spermatocytes. Each secondary spermatocyte then undergoes second meiotic division (meiosis II) to form two equal haploid spermatids. Hence, a diploid spermatogonium produces four haploid spermatids. These spermatids are transformed into spermatozoa (sperm) by the process called spermiogenesis.

Question 7.
Name the hormones involved in regulation of spermatogenesis.
Answer:
Follicle-stimulating hormones (FSH) and luteinising hormones (LH) are secreted by gonadotropin releasing hormones from the hypothalamus. These hormones are involved in the regulation of the process of spermatogenesis. FSH acts on sertoli cells, whereas LH acts on leydig cells of the testis and stimulates the process of spermatogenesis.

Question 8.
Define spermiogenesis and spermiation.
Answer:
Spermiogenesis : It is the process of transforming spermatids into matured spermatozoa or sperms.
Spermiation : It is ‘the process when mature spermatozoa are released from the sertoli cells into the lumen of seminiferous tubules.

Question 9.
Draw a labelled diagram of sperm.
Answer:

Question 10.
What are the major components of seminal plasma?
Answer:
Semen (produced in males) is composed of sperms and seminal plasma. The major components of the seminal plasma in the male reproductive system are mucus, spermatozoa, and various secretions of accessory glands. The seminal plasma is rich in fructose, calcium, ascorbic acid, and certain enzymes. It provides nourishment and protection to sperms.

Question 11.
What are the major functions of male accessory ducts and glands?
Answer:
The male accessory ducts are vasa efferentia, epididymis, vas deferens, and rete testis. They play an important role in the transport and temporary storage of sperms. On the contrary, male accessory glands are seminal vesicles, prostate glands, and bulbourethral glands. These glands secrete fluids that lubricate the reproductive system and sperms. The sperms get dispersed in the fluid which makes their transportation into the female body easier. The fluid is rich in fructose, ascorbic acid, and certain enzymes. They also provide nutrients and activate the sperm.

Question 12.
What is oogenesis? Give a brief account of oogenesis.
Answer:
Oogenesis is the process of the formation of a mature ovum from the oogonia in females. It takes place in the ovaries. During oogenesis, a diploid oogonium or egg mother cell increases in size and gets transformed into a diploid primary oocyte. This diploid primary oocyte undergoes first meiotic division i.e., meiosis I or reductional division to form two unequal haploid cells. The smaller cell is known as the first polar body, while the larger cell is known as the secondary oocyte. This secondary oocyte undergoes second meiotic division i.e., meiosis II or equational division and gives rise to a second polar body and an ovum. Hence, in the process of oogenesis, a diploid oogonium produces a single haploid ovum while two or three polar bodies are produced.

Question 13.
Draw a labelled diagram of a section through ovary.
Answer:

Question 14.
Draw a labelled diagram of a Graafian follicle.
Answer:

Question 15.
Name the functions of the following:
(a) Corpus luteum
(b) Endometrium
(c) Acrosome
(d) Sperm tail
(e) Fimbriae
Answer:
(a) Corpus Luteum: It is formed from the ruptured Graafian follicle. It secretes progesterone hormone during the luteal phase of the menstrual cycle. A high level of progesterone inhibits the secretions of FSH and LH, thereby preventing ovulation. It also allows the endometrium of the uterus to proliferate and to prepare itself for implantation.

(b) Endometrium: It is the innermost lining of the uterus. It is rich in glands and undergoes cyclic changes during various phases of the menstrual cycle to prepare itself for the implantation of the embryo.

(c) Acrosome: It is a cap-like structure present in the anterior part of the head of the sperm. It contains hyaluronidase enzyme, which hydrolyses the outer membrane of the egg, thereby helping the sperm to penetrate the egg at the time of fertilisation.

(d) Sperm Tail: It is the longest region of the sperm that facilitates the movement of the sperm inside the female reproductive tract.

(e) Fimbriae: They are finger-like projections at the ovarian end of the fallopian tube. They help in the collection of the ovum (after ovulation), which is facilitated by the beating of the cilia.

Question 16.
Identify True/False statements. Correct each false statement to make it true.
(a) Androgens are produced by Sertoli cells. (True/False)
(b) Spermatozoa get nutrition from Sertoli cells. (True/False)
(c) Leydig cells are found in ovary. (True/False)
(d) Leydig cells synthesise androgens. (True/False)
(e) Oogenesis takes place in corpus luteum. (True/False)
(f) Menstrual cycle ceases during pregnancy. (True/False)
(g) Presence or absence of hymen is not a reliable indicator of virginity or sexual experience. (True/False)
Answer:
(a) Androgens are produced by Sertoli cells.
False Correct : Leydig cells.

(b) Spermatozoa get nutrition from Sertoli cells.
True

(c) Leydig cells are found in ovary.
False Correct : spermatogonia.

(d) Leydig cells synthesise androgens.
True

(e) Oogenesis takes place in corpus luteum.
False Correct : ovaries

(f) Menstrual cycle ceases during pregnancy.
True

(g) Presence or absence of hymen is not a reliable indicator of virginity or sexual experience.
True

Question 17.
What is menstrual cycle? Which hormones regulate menstrual cycle? ,
Answer:
The menstrual cycle is a series of cyclic physiologic changes that take place inside the female reproductive tract in primates. The whole cycle takes around 28 days to complete. The end of the cycle is accompanied by the breakdown of uterine endothelium, which gets released in the form of blood and mucus through the vagina. This is known as menses.

The follicle stimulating hormone (FSH), luteinising hormone (LH), L estrogen, and progesterone are the various hormones that regulate the menstrual cycle. The level of FSH and LH secreted from the anterior pituitary gland increases during the follicular phase. FSH secreted under the influence of RH (releasing hormone) from the hypothalamus , stimulates the conversion of a primary follicle into a graafian follicle.

The level of LH increases gradually leading to the growth of follicle and f secretion of estrogen. Estrogen inhibits the secretion of FSH and stimulates the secretion of luteinising hormone. It also causes the thickening of the uterine endometrium. The increased level of LH causes the rupturing of the graafian follicle and release the ovum into the fallopian tube. The ruptured graafian follicle changes to corpus luteum and starts secreting progesterone hormone during the luteal phase.

Progesterone hormone helps in the maintenance and preparation of endometrium for the implantation of the embryo. High levels of progesterone hormone in the blood decrease the secretion of LH and FSH, therefore inhibiting further ovulation.

Question 18.
What is parturition? Which hormones are involved in induction of parturition?
Answer:
Parturition is the process of giving birth to a baby as the development of the foetus gets completed in the mother’s womb. The hormones involved in this process are oxytocin and relaxin. Oxytocin leads to the contraction of smooth muscles of myometrium of the uterus, which directs the full term foetus towards the birth canal. On the other hand, relaxin hormone causes relaxation of the pelvic ligaments and prepares the uterus for child birth.

Question 19.
In our society the women are often blamed for giving birth to [ daughters. Can you explain why this is not correct?
Answer:
All human beings have 23 pairs of chromosomes. Human males have 22 pairs of autosomes and contain one or two types of sex chromosome. They are either X or Y. On the contrary, human females have 22 pairs of autosomes and contain only the X sex chromosome. The sex of an individual is determined by the type of the male gamete (X or Y), which fuses with the X chromosome of the female. If the fertilising sperm is X, then the baby will be a girl and if it is Y, then the baby will be a boy.
Hence, it is incorrect to blame a woman for the gender of the child.

Question 20.
How many eggs are released by a human ovary in a month? How many eggs do you think would have been released if the mother gave birth to identical twins? Would your answer change if the twins bom were fraternal?
Answer:
An ovary releases an egg every month. When two babies are produced in succession, they are called twins. Generally, twins are produced from a single egg by the separation of early blastomeres resulting from the first zygotic cleavage. As a result, the young ones formed will have the same genetic make-up and are thus, called identical twins.

If the twins born are fraternal, then they would have developed from two separate eggs. This happens when two eggs’ (one from each ovary) are released at the same time and get fertilised by two separate sperms. Hence, the young ones developed will have separate genes and are therefore, called non-identical or fraternal twins.

Question 21.
How many eggs do you think were released by the ovary of a female dog which gave birth to 6 puppies?
Answer:
Dogs and rodents are polyovulatory species. In these species, more than one ovum is released from the ovary at the time of ovulation. Hence, six eggs were released by the ovary of a female dog to produce six puppies.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 15 Biodiversity and Conservation Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 15 Biodiversity and Conservation

PSEB 12th Class Biology Guide Biodiversity and Conservation Textbook Questions and Answers

Question 1.
Name the three important components of biodiversity.
Answer:
Biodiversity is the variety of living forms present in various ecosystems. It includes variability among life forms from all sources including land, air, and water. Three important components of biodiversity are as follows :

• Genetic diversity
• Species diversity
• Ecological diversity

Question 2.
How do ecologists estimate the total number of species present in the world?
Answer:
The diversity of living organisms present on the Earth is very vast. According to an estimate by researchers, it is about seven million. the total number of species present in the world is calculated by statistical comparison between species richness of a well-studied group of insects of temperate and tropical regions. Then, these ratios are extrapolated with other groups of plants and animals to calculate the total species richness present on the Earth.

Question 3.
Give three hypotheses for explaining why tropics show greatest levels of species richness.
Answer:
There are three different hypotheses proposed by scientists for explaining species richness in the tropics.

1. Tropical latitudes receive more solar energy than temperate regions, which leads to high productivity and high species diversity.
2. Tropical regions have less seasonal variations and have a more or less constant environment. This promotes the niche specialization and thus, high species richness.
3. Temperate regions were subjected to glaciations during the ice age, while tropical regions remained undisturbed which led to an increase in the species diversity in this region.

Question 4.
What is the significance of the slope of regression in a species-area relationship?
Answer:
The slope of regression (z) has a great significance in order to find a species-area relationship. It has been found that in smaller areas (where the species-area relationship is analyzed), the value of slopes of regression is similar regardless of the taxonomic- group or the region. However, when a similar analysis is done in larger areas, then the slope of regression is much steeper.

Question 5.
What are the major causes of species losses in a geographical region?
Answer:
Biodiversity is the variety of living forms present in various ecosystems. It includes variability among life forms from all sources including land, air, and water. Biodiversity around the world is declining at a very fast pace. The following are the major causes for the loss of biodiversity around the world :

(i) Habitat Loss and Fragmentation: Habitats of various organisms are altered or destroyed by uncontrolled and unsustainable human activities such as deforestation, slash and burn agriculture, mining, and urbanisation. This results in the breaking up of the habitat into small pieces, which effects the movement of migratory animals and also, decreases the genetic exchange between populations leading to a declination of species.

(ii) Over-exploitation: Due to over-hunting and over-exploitation of various plants and animals by humans, many species have become endangered or extinct (such as the tiger and the passenger pigeon).

(iii) Alien Species Invasions: Accidental or intentional introduction of non-native species into a habitat has also led to the declination or extinction of indigenous species. For example, the Nile perch introduced in Lake Victoria in Kenya led to the extinction of more than two hundred species of native fish in the lake.

(iv) Co-extinction: In a native habitat, one species is connected to the other in an intricate network. The extinction of one species causes the extinction of other species, which is associated with it in an obligatory way. For example, the extinction of the host will cause the extinction of its parasites.

Question 6.
How is biodiversity important for ecosystem functioning?
Answer:
An ecosystem with high species diversity is much more stable than an ecosystem with low species diversity. Also, high biodiversity makes the ecosystem more stable in productivity and more resistant towards disturbances such as alien species invasions and floods.

If an ecosystem is rich in biodiversity, then the ecological balance would not get affected. As we all know, various trophic levels are connected through food chains. If anyone organism or all organisms of any one trophic level is killed, then it will disrupt the entire food chain. For example, in a food chain, if all plants are killed, then all deer’s will die due to the lack of food.

If all deer’s are dead, soon the tigers will also die. Therefore, it can be concluded that if an ecosystem is rich in species, then there will be other food alternatives at each trophic level which would not allow any organism to die due to the absence of their food resource. Hence, biodiversity plays an important role in maintaining the health and ecological balance of an ecosystem.

Question 7.
What are sacred groves? What is their role in conservation?
Answer:
Sacred groves are tracts of forest which are regenerated around places of worship. Sacred groves are found in Rajasthan, Western Ghats of Karnataka, and Maharashtra, Meghalaya, and Madhya Pradesh. Sacred grows help in the protection of many rare, threatened, and endemic species of plants and animals found in an area. The process of deforestation is strictly prohibited in this region by tribals. Hence, the sacred grove biodiversity is a rich area.

Question 8.
Among the ecosystem services are control of floods and soil erosion. How is this achieved by the biotic components of the ecosystem?
Answer:
The biotic components of an ecosystem include the living organisms such as plants and animals. Plants play a very important role in controlling floods and soil erosion. The roots of plants hold the soil particles together, thereby preventing the top layer of the soil to get eroded by wind or running water. The roots also make the soil porous, thereby allowing groundwater infiltration and preventing floods. Hence, plants are able to prevent soil erosion and natural calamities Fucii as floods and droughts. They also increase the fertility of soil and biodiversity.

Question 9.
The species diversity of plants (22 percent) is much less than that of animals (72 percent). What could be the explanations to how animals achieved greater diversification?
Answer:
72 percent of species recorded on the Earth are animals and only 22 percent species are plants. There is quite a large difference in their percentage This is because animals have adapted themselves to ensure their survival in changing environments in comparison to plants. For example, insects and other animals have developed a complex nervous system to control and coordinate their body structure. Also, repeated body/ segments with paired appendages and external cuticles have made insects versatile and have given them the ability to survive in vain JUS habitats as compared to other life forms.

Question 10.
Can you think of a situation where we deliberately want to make a species extinct? How would you justify it?
Answer:
Yes, there are various kinds of parasites and disease-causing microbes that we deliberately want to eradicate from the Earth. Since these micro-organisms are harmful to human beings, scientists are working hard to fight against them.

Scientists have been able to eliminate smallpox virus from the world through the use of vaccinations. This shows than humans deliberately want to make these species extinct. Several other eradication programs such as polio and Hepatitis B vaccinations are aimed to eliminate these disease-causing microbes.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 2 Sexual Reproduction in Flowering Plants Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants

PSEB 12th Class Biology Guide Sexual Reproduction in Flowering Plants Textbook Questions and Answers

Question 1.
Name the parts of an angiosperm flower in which development of male and female gametophyte take place.
Answer:
The male gametophyte or the pollen grain develops inside the pollen chamber of the anther, whereas the female gametophyte (also known as the embryo sac) develops inside the nucellus of the ovule from the functional megaspore.

Question 2.
Differentiate between microsporogenesis and megasporogenesis. Which type of cell division occurs during these events? Name the structures formed at the end of these two events.
Answer:

Microsporogenesis	Megasporogenesis
1. It is the process of the formation of microspore tetrads from a microspore mother cell through meiosis.	It is the process of the formation of the four megaspores from a megaspore mother cell in the region of the nucellus through meiosis.
2. It occurs inside the pollen sac of the anther.	It occurs inside the ovule.

(b) Both events (microsporogenesis and megasporogenesis) involve the process of meiosis or reduction division which results in the formation of haploid gametes from the microspore and megaspore mother cells.

(c) Microsporogenesis results in the formation of haploid microspores from a diploid microspore mother cell. On the other hand, megasporogenesis results in the formation of haploid megaspores from a diploid megaspore mother cell.

Question 3.
Arrange the following terms in the correct developmental sequence:
Pollen grain, sporogenous tissue, microspore tetrad, pollen mother cell, male gametes.
Answer:
The correct developmental sequence is as follows:
Sporogenous tissue, pollen mother cell, microspore tetrad, pollen grain, male gametes.
During the development of microsporangium, each cell of the sporogenous tissue acts as a pollen mother cell and gives rise to a microspore tetrad, containing four haploid microspores by the process of meiosis (microsporogenesis). As the anther matures, these microspores dissociate and develop into pollen grains. The pollen grains mature and give rise to male gametes.

Question 4.
With a neat, labelled diagram, describe the parts of a typical ’ angiosperm ovule.
Answer:
An ovule is a female megasporangium where the formation of megaspores takes place.

The various parts of a typical angiospermic ovule are as follows :
1. Funiculus: It is a stalk-like structure which represents the point of attachment of the ovule to the placenta of the ovary.

2. Hilum: It is the point where the body of the ovule is attached to the funiculus.

3. Integuments: They are the outer layers surrounding the ovule that provide protection to the developing embryo.

4. Micropyle: It is a narrow pore formed by the projection of integuments. It marks the point where the pollen tube enters the ovule at the time of fertilisation.

5. Nucellus: It is a mass of the parenchymatous tissue surrounded by the integuments from the outside. The nucellus provides nutrition to the developing embryo. The embryo sac is located inside the nucellus.

6. Chalaza: It is the based swollen part of the nucellus from where the integuments originate.

Question 5.
What is meant by monosporic development of female gametophyte?
Answer:
The female gametophyte or the embryo sac develops from a single functional megaspore. This is known as monosporic development of the female gametophyte. In most flowering plants, a single megaspore mother cell present at the micropylar pole of the nucellus region of the ovule undergoes meiosis to produce four haploid megaspores. Later, out of these four megaspores, only one functional megaspoxe develops into the female gametophyte, while the remaining three degenerate.

Question 6.
With a neat diagram explain the 7-celled, 8-nucleate nature of the female gametophyte.
Answer:
Structure of the mature embryo sac

The female gametophyte (embryo sac) develops from a single functional megaspore. This megaspore undergoes three successive mitotic divisions to form eight nucleate embryo sacs.
The first mitotic division in the megaspore forms two nuclei. One nucleus moves towards the micropylar end while the other nucleus moves towards the chalazal end. Then, these nuclei divide at their respective ends and re-divide to form eight nucleate stages. As a result, there are four nuclei each at both the ends i.e., at the micropylar and the chalazal end in the embryo sac. At the micropylar end, out of the four nuclei only three differentiate into two synergids and one egg cell. Together they are known as the egg apparatus.

Similarly, at the chalazal end, three out of four nuclei differentiates as antipodal cells. The remaining two cells (of the micropylar and the chalazal end) move towards the centre and are known as the polar nuclei, which are situated in a large central cell. Hence, at maturity, the female gametophyte appears as a 7-celled structure, though it has 8-nucleate.

Question 7.
What are chasmogamous flowers? Can cross-pollination occur in cleistogamous flowers? Give reasons for your answer.
Answer:
There are two types of flowers present in plants namely Oxalis and Viola – chasmogamous and cleistogamous flowers. Chasmogamous flowers have exposed anthers and stigmata similar to the flowers of other species.

Cross-pollination cannot occur in cleistogamous flowers. This is because cleistogamous flowers never open at all. Also, the anther and the stigma lie close to each other in these flowers. Hence, only self-pollination is possible in these flowers.

Question 8.
Mention two strategies evolved to prevent self-pollination in flowers.
Answer:
Self-pollination involves the transfer of pollen from the stamen to the pistil of the same flower. Two strategies that have evolved to prevent self-pollination in flowers are given on next page:
1. Self-incompatibility: In certain plants, the stigma of the flower has the capability to prevent the germination of pollen grains and hence, prevent the growth of the pollen tube. It is a genetic mechanism to prevent self-pollination called self-incompatibility. Incompatibility may be between individuals of the same species or between individuals of different species. Thus, incompatibility prevents breeding.

2. Protandry: In some plants, the gynoecium matures before the androecium or vice-versa. This phenomenon is known as protogyny or protandry respectively. This prevents the pollen from coming in contact with the stigma of the same flower.

Question 9.
What is self-incompatibility? Why does self-pollination not lead to seed formation in self-incompatible species?
Answer:
Self-incompatibility is a genetic mechanism in angiosperms that prevents self-pollination. It develops genetic incompatibility between ‘ individuals of the same species or between individuals of different
species.

The plants which exhibit this phenomenon have the ability to prevent germination of pollen grains and thus, prevent the growth of the pollen tube on the stigma of the flower. This prevents the fusion of the gametes along with the development of the embryo. As a result, no seed formation takes place.

Question 10.
What is bagging technique? How is it useful in a plant breeding programme?
Answer:
Various artificial hybridisation techniques (under various crop improvement programmes) involve the removal of the anther from bisexual flowers without affecting the female reproductive part (pistil) through the process of emasculation. Then, these emasculated flowers are wrapped in bags to prevent pollination by unwanted pollen grains. This process is called bagging.

This technique is an important part of the plant breeding programme as
it ensures that pollen grains of only desirable plants are used for fertilisation of the stigma to develop the desired plant variety.

Question 11.
What is triple fusion? Where and how does it take place? Name the nuclei involved in triple fusion.
Answer:
Triple fusion is the fusion of the male gamete with two polar nuclei inside the embryo sac of the angiosperm. This process of fusion takes place inside the embryo sac.

When pollen grains fall on the stigma, they germinate and give rise to the pollen tube that passes through the style and enters into the ovule. After this, the pollen tube enters one of synergids and releases two male gametes there.

Out of the two male gametes, one gamete fuses with the nucleus of the egg cell and forms the zygote (syngamy). The other male gamete fuses with the two polar nuclei present in the central cell to form a triploid primary endosperm nucleus. Since this process involves the fusion of three haploid nuclei, it is known as triple fusion. It results in the formation of the endosperm.
One male gamete nucleus and two polar nuclei are involved in this process.

Question 12.
Why do you think the zygote is dormant for sometime in a fertilised ovule?
Answer:
The zygote is formed by the fusion of the male gamete with the nucleus of the egg cell. The zygote remains dormant for some time and waits for the endosperm to form, which develops from the primary endosperm cell resulting from triple fusion. The endosperm provides food for the growing embryo and after the formation of the endosperm, further development of the embryo from the zygote starts.

Question 13.
Differentiate between:
(a) hypocotyl and epicotyl;
(b) coleoptile and coleorrhiza;
(c) integument and testa;
(d) perisperm and pericarp.
Answer:
(a)

Hypocotyl	Epicotyl
1. The portion of the embryonal axis which lies below the cotyledon in a dicot embryo is known as the hypocotyl.	The portion of the embryonal axis which lies above the cotyledon in a dicot embryo is known as the epicotyl.
2. It terminates with the radicle.	It terminates with the plumule.

(b)

Coleoptile	Coleorrhiza
It is a conical protective sheath that encloses the plumule in a monocot seed.	It is an undifferentiated sheath that encloses the radicle and the root cap in a monocot seed.

(c)

Integument	Testa
It is the outermost covering of an ovule. It provides protection to it.	It is the outermost covering of a seed. It provides protection to the young embryo.

(d) Perisperm

Perisperm	Pericarp
It is the residual nucellus which persists. It is present in some seeds such as beet and black pepper.	It is the ripened wall of a fruit, which develops from the wall of an ovary.

Question 14.
Why is apple called a false fruit? Which part(s) of the flower forms the fruit?
Answer:
Fruits derived from the ovary and other accessory floral parts are called false fruits. On the contrary, true fruits are those fruits which develop from the ovary, but do not consist of the thalamus or any other floral part. In an apple, the fleshy receptacle forms the main edible part. Hence, it is a false fruit.

Question 15.
What is meant by emasculation? When and why does a plant breeder employ this technique?
Answer:
Emasculation is the process of removing anthers from bisexual flowers without affecting the female reproductive part (pistil), which is used in various plant hybridisation techniques.

Emasculation is performed by plant breeders in bisexual flowers to obtain the desired variety of a plant by crossing a particular plant with the desired pollen grain. To remove the anthers, the flowers are f covered with a bag before they open. This ensures that the flower is pollinated by pollen grains obtained from desirable varieties only. Later, the mature, viable, and stored pollen grains are dusted on the
bagged stigma by breeders to allow artificial pollination to take place and obtain the desired plant variety.

Question 16.
If one can induce parthenocarpy through the application of growth substances, which fruits would you select to induce parthenocarpy and why?
Answer:
Parthenocarpy is the process of developing fruits without involving the process of fertilisation or seed formation. Therefore, the seedless varieties of economically important fruits such as orange, lemon, water melon etc. are produced using this technique. This technique involves inducing fruit formation by the application of plant growth hormones such as auxins.

Question 17.
Explain the role of tapetum in the formation of pollen-grain wall.
Answer:
Tapetum is the innermost layer of the microsporangium. It provides nourishment to the developing pollen grains. During microsporogenesis, the cells of tapetum produce various enzymes, hormones, amino acids, and other nutritious material required for the development of pollen grains. It also produces the exine layer of the pollen grains, which is composed of the sporopollenin.

Question 18.
What is apomixis and what is its importance?
Answer:
Apomixis is the mechanism of seed production without involving the process of meiosis and syngamy. It plays an important role in hybrid seed production. The method of producing hybrid seeds by cultivation is very expensive for farmers. Also, by sowing hybrid seeds, it is difficult to maintain hybrid characters as characters segregate during meiosis. Apomixis prevents the loss of specific characters in the hybrid. Also, it is a cost-effective method for producing seeds.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 16 Environmental Issues Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 16 Environmental Issues

PSEB 12th Class Biology Guide Environmental Issues Textbook Questions and Answers

Question 1.
What are the various constituents of domestic sewage? Discuss the effects of sewage discharge on a river.
Answer:
Domestic sewage is the waste originating from the kitchen, toilet, laundry, and other sources. It contains impurities such as suspended solid (sand, salt, clay), colloidal, material (faecal matter, bacteria, plastic and cloth fibre), dissolved materials (nitrate, phosphate, calcium, sodium, ammonia), and disease-causing microbes. When organic wastes from the sewage enter the water bodies, it serves as a food source for micro-organisms such as algae and bacteria. As a result, the population of these micro-organisms in the water body increases.

Here, they utilise most of the dissolved oxygen for their metabolism. This results in an increase in the levels of Biological oxygen demand (BOD) in river water and results in the death of aquatic organisms. Also, the nutrients in the water lead to the growth of planktonic algal, causing algal bloom. This causes deterioration of water quality and fish mortality.

Question 2.
List all the wastes that you generate, at home, school or during your trips to other places. Could you very easily reduce the generation of these wastes? Which would be difficult or rather impossible to reduce?
Answer:
Wastes generated at home include plastic bags, paper napkins, toiletries, kitchen wastes (such as peelings of vegetables and fruits, tea leaves), domestic sewage, glass, etc.

Wastes generated at schools include waste paper, plastics, vegetable and fruit peels, food wrappings, sewage etc.
Wastes generated at trips or picnics include plastic, paper, vegetable and fruit peels, disposable cups, plates, spoons etc.

Yes, wastes can be easily reduced by the judicious use of the above materials. Wastage of paper can be minimised by writing on both sides of the paper and by using recycled paper. Plastic and glass waste can also be reduced by recycling and re-using.

Also, substituting plastic bags with biodegradable jute bags can reduce wastes generated at home, school, or during trips. Domestic sewage can be reduced by optimising the use of water while bathing, cooking, and other household activities. Non-biodegradable wastes such as plastic, metal, broken glass, etc. are difficult to decompose because micro-organisms do riot have the ability to decompose them.

Question 3.
Discuss the causes and effects of global warming. What measures need to be taken to control global warming?
Answer:
Global warming is defined as an increase in the average temperature of the Earth’s surface. Causes of Global Warming: Global warming occurs as a result of the increased concentration of greenhouse gases in the atmosphere. Greenhouse gases include carbon dioxide, methane, and water vapour. These gases trap solar radiations released back by the Earth. This helps in keeping our planet warm and thus, helps in human survival. However, an increase in the amount of greenhouse gases can lead to an excessive increase in the Earth’s temperature, leading to global warming. Global warming is a result of industrialisation, burning of fossil fuels, and deforestation.

Effects of Global Warming: It has been observed that in the past three decades, the average temperature of the Earth has increased by 0.6°C. As a result, the natural water cycle has been disturbed resulting in changes in the pattern of rainfall. It also changes the amount of rainwater. Also, it results in the melting of Polar ice caps and mountain glaciers, which has caused a rise in the sea level, leading to the inundation of coastal regions.

Control Measures for Preventing Global Warming:

• Reducing the use of fossil fuels
• Use of bio-fuels
• Improving energy efficiency
• Use of renewable source of energy such as CNG etc.
• Reforestation.
• Recycling of materials

Question 4.
Match the items given in column A and B:

Column A	Column B
(a) Catalytic converter	(i) Particulate matter
(b) Electrostatic precipitator	(ii) Carbon monoxide and nitrogen oxides
(c) Earmuffs	(iii) High noise level
(d) Landfills	(iv) Solid wastes

Answer:

Column A	Column B
(a) Catalytic converter	(ii) Carbon monoxide and nitrogen oxides
(b) Electrostatic precipitator	(i) Particulate matter
(c) Earmuffs	(iii) High noise level
(d) Landfills	(iv) Solid wastes

Question 5.
Write critical notes on the following:
(a) Eutrophication
(b) Biological magnification
(c) Groundwater depletion and ways for its replenishment
Answer:
(a) Eutrophication:
It is the natural ageing process of a lake caused due to nutrient enrichment. It is brought down by the runoff of nutrients such as animal wastes, fertilisers, and sewage from land which leads to an increased fertility of the lake. As a result, it causes a tremendous increase in the primary productivity of the ecosystem. This leads to an increased growth of algae, resulting, into, algal blooms. Later, the decomposition of these algae depletes the supply of oxygen, leading to the death of other aquatic animal life.

(b) Biological Magnification: Unknowingly some harmful chemicals enter our bodies through the food chain. We use several pesticides and other chemicals to protect our crops from diseases and pests. These chemicals are either washed down into the soil or into the water bodies. From the soil, these are absorbed by the plants along with water and minerals, and from the water bodies, these are taken up by aquatic plants and animals. This is one of the ways in which they enter the food chain. As these chemicals are not degradable, these get accumulated progressively at each trophic level. As human beings occupy the topmost level in any food chain, the maximum concentration of these chemicals get accumulated in our bodies. This phenomenon is known as biological magnification.

(c) Ground Water Depletion and Ways for its Replenishment: The level of groundwater has decreased in the recent years. The source of water supply is rapidly diminishing each year because of an increase in the population and water pollution. To meet the demand of water, water is withdrawn from water bodies such as ponds, rivers etc. As a result, the source of groundwater is depleting.

This is because the amount of groundwater being drawn for human use is more than the amount replaced by rainfall. Lack of vegetation cover also results in very small amounts of water seeping through the ground. An increase in water pollution is another factor that has reduced the availability of groundwater.

Measures for Replenishing Ground Water:

• Preventing over-exploitation of groundwater
• Optimising water use and reducing water demand
• Rainwater harvesting
• Preventing deforestation and plantation of more trees.

Question 6.
Why does ozone hole form over Antarctica? How will enhanced ultraviolet radiation affect us?
Answer:
The ozone hole is more prominent over the region of Antarctica. It is formed due to an increased concentration of chlorine in the atmosphere. Chlorine is mainly released from chlorofluorocarbons (CFC’s) widely used as refrigerants. The CFC’s magnate from the troposphere to the stratosphere, where they release chlorine atoms by the action of UV rays on them.

The release of Chlorine atoms causes the conversion of ozone into molecular oxygen. One atom of chlorine can destroy around 10,000 molecules of ozone and causes ozone depletion. The formation of the ozone hole will result in an increased concentration of UV – B radiations on the Earth’s surface. UV -B damages DNA and activates the process of skin ageing. It also causes skin darkening and skin cancer. High levels of UV -B cause corneal cataracts in human beings.

Question 7.
Discuss the role of women and communities in protection and conservation of forests.
Answer:
Women and communities have played a major role in environmental conservation movements.
(i) Case Study of the Bishnoi Community: The Bishnoi community in Rajasthan strictly believes in the concept of living peacefully with nature. In 1731, the king of Jodhpur ordered his ministers to arrange wood for the construction of his new palace. For this purpose, the minister and 1116 workers went to Bishnoi village.

There, a Bishnoi woman called Amrita Devi along with her daughter and hundreds of other Bishnois showed the courage to step forward and stop them from cutting trees. They embraced the trees and lost their lives at the hands of soldiers of the king. This resistance by the people of the village forced the king to give up the idea of cutting trees.

(ii) Chipko Movement: The Chipko movement was started in 1974 in the Garhwal region of the Himalayas. In this movement, the women from the village stopped the contractors from cutting forest trees by embracing them.

Question 8.
What measures, as an individual, would you take to reduce environmental pollution?
Answer:
The following initiatives can be taken to prevent environmental pollution:
Measures for Preventing Air Pollution

• Planting more trees
• Use of clean and renewable energy sources such as CNG and bio-fuels
• Reducing the use of fossil fuels
• Use of catalytic converters in automobiles

Measures for Preventing Water Pollution:

• Optimising the use of water
• Using kitchen wastewater in gardening and other household purposes

Measures for Controlling Noise Pollution:

• Avoid burning crackers on Diwali
• Plantation of more trees

Measures for Decreasing Solid Waste Generation:

• Segregation of waste
• Recycling and reuse of plastic and paper
• Composting of biodegradable kitchen waste
• Reducing the use of plastics.

Question 9.
Discuss briefly the following:
(a) Radioactive wastes
(b) Defunct ships and e-wastes
(c) Municipal solid Wastes
Answer:
(a) Radioactive Wastes: Radioactive wastes are generated during the process of generating nuclear energy from radioactive materials. Nuclear waste is rich in radioactive materials that generate large quantities of ionising radiations such as gamma rays. These rays cause mutation in organisms, which often results in skin cancer. At high dosage, these rays can be lethal.

Safe disposal of radioactive wastes is a big challenge. It is recommended that nuclear wastes should be stored after pre-treatment in suitable shielded containers, which should then be buried in rocks.

(b) Defunct Ships and E-wastes: Defunct ships are dead ships that are no longer in use. Such ships are broken down for scrap metal in countries such as India and Pakistan. These ships are a source of various toxicants such as asbestos, lead, mercury etc. Thus, they contribute to solid* wastes that are hazardous to health.

E-waste or electronic wastes generally include electronic goods such as computers etc. Such wastes are rich in metals such as copper, iron, silicon, gold etc. These metals are highly toxic and pose serious health hazards. People of developing countries are involved in the recycling process of these metals and therefore, get exposed to toxic substances present in these wastes.

(c) Municipal Solid Wastes: Municipal solid wastes are generated from schools, offices, homes, and stores. It is generally rich in glass, metal, paper waste, food, rubber, leather, and textiles. The open dumps of municipal wastes serve as a breeding ground for flies, mosquitoes, and other disease-causing microbes. Hence, it is necessary to dispose of municipal solid waste properly to prevent the spreading of diseases. Sanitary landfills and incineration are the methods for the safe disposal of solid wastes.

Question 10.
What initiatives were taken for reducing vehicular air pollution in Delhi? Has air quality improved in Delhi?
Answer:
Delhi has been categorised as the fourth most polluted city of the world in a list of 41 cities. Burning of fossil fuels has added to the pollution of air in Delhi.
Various steps have been taken to improve the quality of air in Delhi.

(a) Introduction of CNG (Compressed Natural Gas): By the order of the supreme court of India, CNG-powered vehicles were introduced at the end of year 2006 to reduce the levels of pollution in Delhi. CNG is a clean fuel that produces very little unburnt particles.
(b) Phasing out of old vehicles
(c) Use of unleaded petrol
(d) Use of low-sulphur petrol and diesel
(e) Use of catalytic converters
(f) Application of stringent pollution-level norms for vehicles
(g) Implementation of Bharat stage I, which is equivalent to euro II norms in vehicles of major Indian cities.

The introduction of CNG-powered vehicles has improved Delhi’s air quality, which has lead to a substantial fall in the level of CO₂ and SO₂. However, the problem of suspended particulate matter (SPM) and respiratory suspended particulate matter,(RSPM) still persists.

Question 11.
Discuss briefly the following:
(a) Greenhouse gases
(b) Catalytic converter
(c) Ultraviolet B
Answer:
(a) Greenhouse Gases: The greenhouse effect refers to an overall increase in the average temperature of the Earth due to the presence of greenhouse gases. Greenhouse gases mainly consist of carbon dioxide, methane, and water vapour. When solar radiations reach the Earth, some of these radiations are absorbed.

These absorbed radiations are Released back into the atmosphere. These radiations are trapped by the greenhouse gases present in the atmosphere. This helps in keeping our planet warm and thus, helps in human survival.
However, an increase in the amount of greenhouse gases can lead to an excessive increase in the Earth’s temperature, thereby causing global warming.

(b) Catalytic Converter: Catalytic converters are devices fitted in automobiles to reduce automobile or vehicle pollution. These devices contain expensive metals such as platinum, palladium, and rhodium that act as catalysts.
As the vehicular discharge -passes through the catalytic converter, the unburnt hydrocarbons present in it get converted into carbon dioxide and water. Carbon monoxide and nitric oxide released by catalytic converters are converted into carbon dioxide and nitrogen gas respectively.

(c) Ultraviolet B: Ultraviolet-B is an electromagnetic radiation which has a shorter wavelength than visible light.
It is a harmful radiation that comes from sunlight and penetrates through the ozone hole onto the Earth’s surface.
It induces many health hazards in humans. UV -B damages DNA and activates the process of skin ageing. It also causes skin darkening and skin cancer. High levels of UV -B cause corneal cataracts in human beings.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 1 Reproduction in Organisms Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 1 Reproduction in Organisms

PSEB 12th Class Biology Guide Reproduction in Organisms Textbook Questions and Answers

Question 1.
Why is reproduction essential for organisms?
Answer:
Reproduction is a fundamental feature of all living organisms. It is a biological process through which living organisms produce offsprings (young ones) similar to them. Reproduction ensures the continuance of various species on the Earth. In the absence of reproduction, the species will not be able to exist for a long time and may soon get extinct.

Question 2.
Which is a better mode of reproduction sexual or asexual? Why?
Answer:
Sexual reproduction is a better mode of reproduction. It allows the formation of new variants by the combination of the DNA from two different individuals, typically one of each sex. It involves the fusion of the male and the female gamete to produce variants, which are not identical to their parents and to themselves. This variation allows the individual to adapt constantly changing and challenging environment. Also, it leads to the evolution of better suited organisms which ensures greater survival of a species. On the contrary, asexual reproduction allows very little or no variation at all. As a result, the individuals produced are exact copies of their parents and themselves.

Question 3.
Why is the offspring formed by asexual reproduction referred to as clone?
Answer:
A clone is a group of morphologically and genetically identical individuals. In the process of asexual reproduction, only one parent is involved and there is no fusion of the male and the female gamete. As a result, the offsprings so produced are morphologically and genetically similar to their parents and are thus, called clones.

Question 4.
Offspring formed due to sexual reproduction have better chances of survival. Why? Is this statement always true?
Answer:
Sexual reproduction involves the fusion of the male and the female gamete. This fusion allows the formation of new variants by the combination of the DNA from two (usually) different members of the species. The variations allow the individuals to adapt under varied environmental conditions for better chances of survival.

However, it is not always necessary that the offspring produced due to sexual reproduction has better chances of survival. Under some circumstances, asexual reproduction is more advantageous for certain organisms. For example, some individuals who do not move from one place to another and are well settled in their environment. Also, asexual reproduction is a fast and’ a quick mode of reproduction which does not consume much time and energy as compared to sexual reproduction.

Question 5.
How does the progeny formed from asexual reproduction differ from those formed by sexual reproduction?
Answer:

Progeny formed from asexual reproduction	Progeny formed from sexual reproduction
1. Asexual reproduction does not involve the fusion of the male and the female gamete. Organisms undergoing this kind ofreproduction produce offsprings that are morphologically and genetically identical to them.	Sexual reproduction involves the fusion of the male and the female gamete of two individuals, typically one of each sex. Organisms undergoing this kind of reproduction produce offsprings that are not identical to them.
2. Offsprings thus produced do not show variations and are called clones.	Offsprings thus produced show variations from each other and their parents.

Question 6.
Distinguish between asexual and sexual reproduction. Why is
vegetative reproduction also considered as a type of asexual reproduction?
Answer:

Sexual reproduction	Asexual reproduction
1. It involves the fusion of the male and the female gamete.	It does not involves the fusion of the male and the female gamete.
2. It requires two (usually) different individuals.	It requires only one individual.
3. The individuals produced are not identical to their parents and show variations from each other and also, from their parents.	The individuals produced are identical to the parent and are hence, called clones.
4. Most animals reproduce sexually. Both sexual and asexual modes of reproduction are found in plants.	Asexual modes of reproduction are common in organisms having simple organisations such as algae and fungi.
5. It is a slow process.	It is a fast process.

Vegetative reproduction is a process in which new plants are obtained without the production of seeds or spores. It involves the propagation of plants through certain vegetative parts such as the rhizome, sucker, tuber, bulb, etc. It does not involve the fusion of the male and the female gamete and requires only one parent. Hence, vegetative reproduction is considered as a type of asexual reproduction.

Question 7.
What is vegetative propagation? Give two suitable examples.
Answer:
Vegetative propagation is a mode of asexual reproduction in which new plants are obtained from the vegetative parts of plants. It does not involve the production of seeds or spores for the propagation of new plants. Vegetative parts of plants such as runners, rhizomes, suckers, tubers, etc. can be used as propagules for raising new plants.

Examples of vegetative reproduction are given below:
1. Eyes of potato: The surface of a potato has several buds called eyes. Each of these buds when buried in soil develops into a new plant, which is identical to the parent plant.

2. Leaf buds of Bryophyllum: The leaves of Bryophyllum plants bear several adventitious buds on their margins. These leaf buds have the ability to grow and develop into tiny plants when the leaves get detached from the plant and come in contact with moist soil.

Question 8.
Define:
(a) Juvenile phase,
(b) Reproductive phase,
(c) Senescent phase.
Answer:
(a) Juvenile Phase: All organisms have to reach a certain stage of growth and maturity in their life, before they can be reproduce sexually. This phase of growth is called the juvenile phase or vegetative phase in plants.

(b) Reproductive Phase: When the juvenile phase is over the organisms enter the period of reproductive phase or sexual maturity. It is indicated by showing various morphological and physiological changes such as development of secondary sexual characters in animals and by flowering in plants. This is the actual period of the life span of any organism when it is capable of producing offsprings. This phase is of variable duration in different organisms.

(c) Senescent Phase : This is the final and third stage of growth cycle. It can be considered as the end of reproductive phase. It is accompanied by reduction in functional capacity and increase in cellular break down and metabolic failures. It ultimately leads to death.

Question 9.
Higher organisms have resorted to sexual reproduction in spite of its complexity. Why?
Answer:
Although sexual reproduction involves more time and energy, higher organisms have resorted to sexual reproduction in spite of its complexity. This is because this mode of reproduction helps introduce new variations in progenies through the combination of the DNA from two (usually) different individuals. These variations allow the individual to cope with various environmental conditions and thus, make the organism better suited for the environment. Variations also lead to the evolution of better organisms and therefore, provide better chances of survival. On the other hand, asexual reproduction does not provide genetic differences in the individuals produced.

Question 10.
Explain why meiosis and gametogenesis are always interlinked?
Answer:
Meiosis is a process of reductional division in which the amount of genetic material is reduced. Gametogenesis is the process of the formation of gametes. Gametes produced by organisms are haploids (containing only one set of chromosomes), while the body of an organism is diploid. Therefore, for producing haploid gametes (gametogenesis), the germ cells of an organism undergo meiosis. During the process, the meiocytes of an organism undergo two successive nuclear and cell divisions with a single cycle of DNA replication to form the haploid gametes.

Question 11.
Identify each part in a flowering plant and write whether it is haploid (n) or diploid (2n).
(a) Ovary …………………………
(b) Anther ……………………..
(c) Egg ………………………….
(d) Pollen ……………………..
(e) Male gamete ……………….
(f) Zygote ………………..
Answer:
(a) diploid (2n)
(b) diploid (2n)
(c) haploid (n)
(d) haploid (n)
(e) haploid (n)
(f) diploid (2n)

Question 12.
Define external fertilisation. Mention its disadvantages.
Answer:
External fertilisation is the process in which the fusion of the male and the female gamete takes place outside the female body in an external medium, generally water. Fish, frog, starfish are some organisms that exhibit external fertilisation.

Disadvantages of external fertilisation
In external fertilisation, eggs have less chances of fertilisation. This can lead to the wastage of a large number of eggs produced during the process.
Further, there is an absence of proper parental care to the offspring, which results in a low rate of survival in the progenies.

Question 13.
Differentiate between a zoospore and a zygote.
Answer:

Zoospore	Zygote
1. A zoospore is a motile asexual spore that utilises the flagella for movement.	A zygote is a noh-motile diploid cell formed as a result of fertilisation.
2. It is an asexual reproductive structure.	It is formed as a result of sexual reproduction.
3. Zoospores are formed in simple plants like, algae or fungi.	Zygote is formed in complex organism.

Question 14.
Differentiate between gametogenesis from embryogenesis.
Answer:

Gametogenesis	Embryogenesis
1. It is the process of the formation of haploid male and female gametes from diploid meiocytes (gamete mother cell) through the process of meiosis.	It is the process of the development of the embryo from the repeated mitotic divisions of the diploid zygote.
2. Gametes may be either homogametes or heterogametes.	Animals may be either oviparous or viviparous.

Question 15.
Describe the post-fertilisation changes in a flower.
Answer:
As a result of double fertilisation in flowering plants, zygote (2n) and the primary endosperm nucleus (3n) is produced. The calyx, corolla, stamens and style wither away. The calyx may persist or even show growth in certain cases. The post fertilisation changes which take place are

• Endosperm formation
• Embryo formation
• Seed formation and
• Fruit formation.

The primary endosperm nucleus becomes active and forms a nutritive vegetative tissue. The endosperm at the expense of food present in the nucellus, Endosperm may be completely used up by the developing embryo (non-endospermic seeds e.g., pea) or may persist in the seed (endospermic seed e.g., castor). The zygote, waits for sometime till the formation of endosperm and then develops into embryo, by withdrawing nutrition from the endosperm. Ultimately the ovules are transformed into seeds and the ovary becomes a fruit. The formation of fruit helps in the nourishment and protection to the developing seeds and later helps in seed dispersal. Under favourable conditions the seeds germinate to form new plants.

Question 16.
What is a bisexual flower? Collect five bisexual flowers from your neighbourhood and with the help of your teacher find out their common and scientific names.
Answer:
A flower that contains both the male and female reproductive structure (stamen and pistil) is called a bisexual flower.
Examples of plants bearing bisexual flowers are:

1. Water lily {Nymphaea odorata)
2. Rose (Rosa multiflora)
3. Hibiscus (Hibiscus Rosa-sinensis)
4. Mustard (Brassica nigra)
5. Petunia (Petunia hybrida)

Question 17.
Examine a few flowers of any cucurbit plant and try to identify the staminate and pistillate flowers. Do you know any other plant that hears unisexual flowers?
Answer:
Cucurbit plant bears unisexual flowers as these flowers have either the stamen or the pistil. The staminate flowers bear bright, yellow coloured petals along with stamens that represent the male reproductive structure. On the other hand, the pistillate flowers bear only the pistil that represents the female reproductive structure.
Other examples of plants that bear unisexual flowers are corn, papaya, cucumber, etc.

Question 18.
Why are offspring of oviparous animals at a greater risk as compared to offspring of viviparous animals?
Answer:
Oviparous animals lay eggs outside their body. As a result, the eggs of these animals are under continuous threat from various environmental factors. On the other hand, in viviparous animals, the development of the egg takes place inside the body of the female. Hence, the offspring of an egg-laying or oviparous animal is at greater risk as compared to the offspring of a viviparous animal, which gives birth to its young ones.

Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 9 Coordination Compounds Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds

PSEB 12th Class Chemistry Guide Coordination Compounds InText Questions and Answers

Question 1.
Explain the bonding in coordination compounds in terms of Werner’s postulates.
Answer:
(i) The primary valencies are satisfied by negative ions and equal-to the oxidation state of the metal.

(ii) The secondary valencies can be satisfied by neutral or negative ions. It is equal to the coordination number of the central metal atom and is fixed.

(iii) The ions bound to the central metal ion to secondary linkages have definite spatial arrangements and give geometry to the complex. While primary valency is non-directional.

Question 2.
FeSO₄ solution mixed with (NH₄)₂SO₄ solution in 1 : 1 molar ratio gives the test of Fe²⁺ ion but CuSO₄ solution mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu²⁺ ion. Explain why?
Answer:
FeSO⁴ solution mixed with (NH⁴)²SO⁴ solution in 1 : 1 molar ratio forms double salt, FeSO⁴∙(NH⁴)²SO⁴∙6H²O which ionises in the solution to give Fe2+ ions. Hence, it gives the test of Fe²⁺ ions.

CuSO₄ solution mixed with aqueous ammonia in 1 : 4 molar ratio forms a complex, with the formula [Cu(NH₃)₄]SO₄. The complex ion, [CU(NH₃)]²⁺ does not ionise to give Cu²⁺ ions. Hence, it does not give the test of Cu²⁺ ion.

Question 3.
Explain with two examples each of the following: Coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.
Answer:
Coordination entity: A coordination entity constitutes usually a central metal atom or ion, to which a fixed number of other atoms or ions or groups are attached by coordinate bonds. A coordination entity may be neutral, positively or negatively charged. For examples : [Ni(CO)₄], [CoCl₃(NH₃)₃], [Co(NH₃)₆]^3+.

Ligand : A ligand is an ion or a small molecule having at least one lone pair of electrons and capable of forming a coordinate bond with central atom or ion in the coordination entity. For example: Cl^–, OH^–, CN^–, CO, NH₃, H₂O etc.

Coordination number : The coordination number of the central atom or ion is determined by the number of a bonds between the ligands and the central atom or ion. n bonds are not consider for the determination of coordination number. The a bonding electrons may be indicated by a pair of dots (:). For example, [Co(:NH₃)₆]³⁺ and [Fe(:CN)₆]^3-.

Coordination polyhedron : The spatial arrangement of the ligands which are directly attached to the central atom or ion called coordination polyhedron.
For example: [Co(NH₃)₆]³⁺ is octahedral, [Ni(CO)₄] is tetrahedral and [PtCl₄ ]^2- is square planar.

Homoleptic and heteroleptic : Complexes in which a metal is bound to only one type of donor groups are known as homoleptic.
For example : [Co(NH₃)₆]³⁺, [PtCl₆]^2- .
Complexes in which a metal is bound to more than one kind of donor groups are known as heteroleptic. ‘
For example : [Co(NH₃)₄Cl₂]⁺, [PdI₂(ONO)₂ (H₂O)₂],

Question 4.
What is meant by unidentate, didentate and ambidentate ligands? Give two examples for each.
Answer:
A molecule or an ion which has only one donor atom to form one coordinate bond with the central metal atoms is called unidentate ligand, e.g., Cl^– and NH₃.

A molecule or an ion which contains two donor atoms and hence forms two coordinate bonds with the central metal atoms is called a didentate ligand, e.g., NH₂—CH₂—CH₂—NH₂ and ^–OOC — COO^–.
A molecule or an ion which contains two donor atoms but only one of them forms a coordinate bond at a time with the central metal atom is called ambidentate ligand, e.g., CN^–or NC^– and \(\) or : ONO^–.

Question 5.
Specify the oxidation numbers of the metals in the following coordination entities:
(i) [Co(H₂O)(CN)(en)₂]²⁺
(ii) [CoBr₂(en)₂]⁺
(iii) [PtCl₂]^2-
(iv) K₃Fe(CN)₆]
(v) [Cr(NH₃)₂Cl₃]
Answer:
(i) x + (-1) + (0) + (0) = + 2 so x = +3 (III)
(ii) x + 2(-1) + 0 = +1 so x = +3 (III)
(iii) x + 4(-1) = -2 so x = +2(11)
(iv) x + 6(-1) = -3 so x = +3 (III)
(v) x + 3(-1) + 0 = 0 so x = +3 (III)

Question 6.
Using IUPAC norms write the formulas for the following:
(i) Tetrahydroxozincate(II)
(ii) Potassium tetrachloridopalladate(II)
(iii) Diamminedichloridoplatinum(II)
(iv) Potassium tetracyanonickelate(II)
(v) Pentaamminenitrito-O-cobalt(III)
(vi) Hexaamminecobalt(III) sulphate
(vii) Potassium tri(oxalato)chromate(III)
(viii) Hexaammineplatinum(IV)
(ix) Tetrabromidocuprate(Il)
(x) Pentaamminenitrito-N-cobalt(lll)
Answer:
(i) [Zn(OH)₄]^2-
(ii) K₂[PdCl₄]
(iii) pt(NH₃)₂Cl₂]
(iv) K₂[Ni(CN)₄]
(v) [Co(ONO) (NH₃)₅]²⁺
(vi) [CO(NH₃)₆]₂ (SO₄)₃
(vii) K₃[Cr(C₂O₄)₃]
(viii) [Pt(NH₃)₆]⁴⁺
(ix) [Cu(Br)₄]^2-
(x) [Co (NO₂) (NH₃)₅]²⁺

Question 7.
Using IUPAC norms write the systematic names of the following:
(i) [CO(NH₃)₆]Cl₃
(ii) [Pt(NH₃)₂Cl(NH₂CH₃)]Cl
(iii) [Ti(H₂O)₆]³⁺
(iv) [CO(NH₃)₄Cl(NO₂)]CI
(v) [Mn(H₂O)₆]²⁺
(vi) [NiCl₄]^2-
(vii) [Ni(NH₃)₆]Cl₂
(viii) [Co(en)₃]³⁺
(ix) [Ni(CO)₄]
Answer:
(i) Hexaamminecobalt(III) chloride
(ii) Diamminechlorido(methylamine) platinum(II) chloride
(iii) Hexaquatitanium(III) ion
(iv) Tetraamminechloridonitrito-N-Cobalt(III) chloride
(v) Hexaquamanganese(II) ion
(vi) Tetrachloridonickelate(II) ion
(vii) Hexamminenickel(II) chloride
(viii) Tris(ethane-1, 2-diamine) cobalt(III) ion
(ix) Tetracarbonylnickel(O)

Question 8.
List various types of isomerism possible for coordination compounds giving an example of each.
Answer:
Two principal types of isomerism are known among coordination compounds :
(A) Sterioisomerism,
(B) Structural isomerism.
Each of which can be further sub-divided as :
(A) Stereoisomerism
(i) Geometrical isomerism : It arises in heteroleptic complexes due to different possible geometric arrangements of the ligands.
Example: Pt[(NH₃)₂Cl₂]

(ii) Optical isomerism : It is common in octahedral complexes involving didentate ligands.
Example : [Pt Cl₂(en) ₂]²⁺

Optical isomers (d and l) of cis-[PtCl₂(en)₂]²⁺

(B) Structural isomerism
(i) Linkage isomerism.
Example: [Co(NH₃)₅ (NO₂)]Cl₂
(ii) Coordination isomerism.
Example: [Co(NH₃)₆] [Cr(CN)₆]
(iii) Ionisation isomerism.
Example: [Co(NH₃)₅SO₄]Br and [CO(NH₃)₅ Br]SO₄
(iv) Solvate isomerism.
Example : [Cr(H₂O)₆] Cl₃ (violet) its solvate isomer
[Cr(H₂O)₅Cl]Cl₂. H₂O (grey-green)

Question 9.
How many geometrical isomers are possible in the following coordination entities? ’
(i) [Cr(C₂O₄)₃]^3-
(ii) [Co(NH₃)₃Cl₃]
Answer:
(i) [Cr(C₂O₄)₃]^3-,
No geometric isomer is possible as it is a bidentate ligand.
(ii) [CO(NH₃)₃Cl₃] .
Two geometrical isomers are possible.

Question 10.
Draw the structures of optical isomers of:
(i) [Cr(C₂O₄)₃]^3-
(ii) [PtCl₂(en)₂]²⁺
(iii) [Cr(NH₃)₂ Cl₂ (en)]⁺
Answer:
(i) [Cr(C₂O₄)₃]^3-

Question 11.
Draw all the isomers (geometrical and optical) of:
(i) [CoCl₂ (en)₂]⁺
(ii) [Co(NH₃)Cl(en)₂]²⁺
(iii) [Co(NH₃)₂Cl₂(en)]⁺
Answer:
(i) [CoCl₂ (en)₂]⁺

Question 12.
Write all the geometrical isomers of [Pt(NH₃)(Br)(Cl)(py)] and how many of these will exhibit optical isomers?
Answer:
Three isomers are possible as follows :

Isomers of this type do not show any optical isomerism. Optical isomerism rarely occurs in square planar or tetrahedral complexes and that too when they contain unsymmetrical chelating ligand.

Question 13.
Aqueous copper sulphate solution (blue in colour) gives :
(i) a green precipitate with aqueous potassium fluoride, and
(ii) a bright green solution with aqueous potassium chloride. Explain these experimental results.
Answer:
Aqueous copper sulphate exists as [Cu(H₂O)₄]SO₄. It is a labile complex. The blue colour of the solution is due to [Cu(H₂O)₄]²⁺ ions,

(i) When KF is added, the weak H₂O ligands are replaced by F^– ligands forming [CuF₄]^2- ions, which is a green precipitate.

(ii) When KCl is added, Cl^– ligands replace the weak H₂O ligands forming [CuCl₄]^2- ion, which has bright green colour.

Question 14.
What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H₂S (g) is passed through this solution?
Answer:
K₂[Cu(CN)₄] is formed when excess of aqueous KCN is added to an aqueous solution of CuSO₄.

As CN^–ions are strong ligands the complex is very stable. It is not replaced by S^2- ions when H₂S gas is passed through the solution and thus no precipitate of CuS is obtained.

Question 15.
Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:
(i) [Fe(CN)₆]^4-
(ii) [FeFe₆]^3-
(iii) [Co(C₂O₄)₃]^3-
(iv) [CoF₆]^3-
Answer:
(i) [Fe(CN)₆]^4-
In the above coordination complex, iron exists in the +2 oxidation state.
Fe = [Ar] 3d⁶ 4s²
Outer configuration of Fe²⁺ = 3d⁶ 4s⁰
Orbitals of Fe²⁺ ion:

As CN^– is a strong field ligand, it causes the pairing of the unpaired 3d electrons.

Since, there are six ligands around the central metal ion, the most feasible hybridisation is d²sp³. d²sp³ hybridised orbitals of Fe²⁺ are :

6 electron pairs from CN ions occupy the six hybrid d²sp³ orbitals.
Then,

Hence, the geometry of the complex is octahedral and the complex is diamagnetic (as there are no unpaired electrons).

(ii) [FeF₆]^3-
In this complex, the oxidation state of Fe is + 3.
Fe³⁺ = 3d⁵ 4s⁰
Orbitals of Fe³⁺ ion:

There are 6F^– ions. Thus, it will undergo d²sp³ or sp³d² hybridisation. As F^– is a weak field ligand, it does not cause the pairing of the electrons in the 3d orbital. Hence, the most feasible hybridisation is sp³d². sp³d² hybridised orbitals of Fe are:

Hence, the geometry of the complex is found to be octahedral.

(iii) [Co(C₂O₄)₃]^3-
Cobalt exists in the + 3 oxidation state in the given complex.
Outer configuration of Co = 3d₇ 4s₂
Co₃₊ = 3d₆4s₀
Orbitals of Co₃₊ ion:

Oxalate is a weak field ligand. Therefore, it cannot cause the pairing of the 3d electrons. As there are 6 ligands, hybridisation has to be either sp³d² or d²sp³ hybridisation. sp³d² hybridisation of Co³⁺.

The 6 electron pairs from the 3 oxalate ions (oxalate anion is a bidentate ligand) occupy these sp³d² orbitals.

Hence, the geometry of the complex is found to be octahedral.

(iv) [CoF₂]^3-
Cobalt exists in the + 3 oxidation state.
Orbitals of Co³⁺ ion:

Again, fluoride ion is a weak field ligand. It cannot cause the pairing of the 3d electrons. As a result, the Co³⁺ ion will undergo sp³d² hybridisation.
sp³d² hybridised orbitals of Co³⁺ ion are :

Hence, the geometry of complex is octahedral, 6 electron pants.

Question 16.
Draw figure to show the splitting of d-orbitals in an octahedral crystal field.
Answer:

Question 17.
What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
Answer:
The arrangement of ligands in order of their increasing field strengths, i.e., increasing crystal field splitting energy (CFSE) values is called spectrochemical series.

The ligands with a small value of CFSE (△₀) are called weak field ligands whereas those with a large value of CFSE are called strong field ligands.

Question 18.
What is crystal field splitting energy? How does the magnitude of △₀ decide the actual configuration of d-orbitals in a coordi-nation entity?
Answer:
When ligands approach a transition metal ion, the d-orbitals split into two sets, one with lower energy and the other with higher energy. The difference of energy between the two sets of orbitals is called crystal field splitting energy (△₀) in case of octahedral field.

If △₀ < P, (pairing energy), the 4th electron enters one of the e_g orbitals giving the configuration \(t_{2 g}^{3} e_{g}^{1}\), thereby forming high spin complexes.

Such ligands for which A 0 < P are called weak field ligands.
If △₀ > P, the 4th electron pairs up in one of the t2g orbitals giving the configuration \(t_{2 g}^{4} e_{g}^{0}\), thus forming low spin complexes. Such ligands for which △₀ > P are called strong field ligands.

Question 19.
[Cr(NH₃)₆]³⁺ is paramagnetic while [Ni(CN)₄]^2- is diamagnetic. Explain why?
Answer:
Cr is in the +3 oxidation state i.e., d³ configuration. Also, NH₃ is a weak field ligand that does not cause the pairing of the electrons in the orbital.

Therefore, it undergoes d²sp³ hybridisation and the electrons in the 3d orbitals remain unpaired. Hence, it is paramagnetic in nature.
In [Ni(CN)₄]^2-, Ni exists in the + 2 oxidation state i. e., d⁸ configuration.

CN^– is a strong field ligand. It causes the pairing of the 3d electrons. Then, Ni²⁺ undergoes dsp² hybridisation.

As there are no unpaired electrons, it is diamagnetic.

Question 20.
A solution of [Ni(H₂O)₆]²⁺ is green but a solution of [Ni(CN)₄]^2- is colourless. Explain.
Answer:
In [Ni(H₂O)6]²⁺, \(\mathrm{H}_{2} \ddot{\mathrm{O}}\) is a weak field ligand. Therefore, there are unpaired electrons in Ni²⁺. In this complex, the d electrons from the lower energy level can be excited to the higher energy level i. e., the possibility of d-d transition is present. Hence, [Ni(H₂O)₆]²⁺ is coloured.

In [Ni(CN)₄]²⁺, the electrons are all paired as CN^– is a strong field ligand. Therefore, d-d transition is not possible in [Ni(CN)₄]^2-. Hence, it is colourless.

Question 21.
[Fe(CN)₆]^4- and [Fe(H₂O)₆]²⁺ are of different colours in dilute solutions. Why?
Answer:
In both the complex compounds, Fe is in +2 oxidation state with configuration 3d⁶, i.e., it has four unpaired electrons. In the presence of weak H₂O ligands, the unpaired electrons do not pair up. But in the presence of strong ligand CN^– they get paired up. Then no unpaired electron is left. Due to this, difference in the number of unpaired electrons, both complex ions have different colours.

Question 22.
Discuss the nature of bonding in metal carbonyls.
Answer:
The metal carbon in metal carbonyls possesses both CT and π character. The ligand to metal is CT bond and metal to ligand is π bond. The effect of CT bond strengthens the rcbond and vice-versa. This is called synergic effect. This unique synergic provides stability to metal carbonyls.

Question 23.
Give the oxidation state, d-orbital occupation and coordination number of the central metal ion in the following complexes:
(i) K₃[CO(C₂O₄)₃]
(ii) cis-[Cr(en)₂Cl₂]Cl
(iii) (NH₄)₂[CoF₄]
(iv) [Mn(H₂O)₆]S0₄
Solution:
(i) K₃[CO(C₂O₄)₃]
The central metal ion is Co.
The oxidation state can be given as :
(+1) × 3 + × + (- 2) × 3 = 0
x – 6 = -3 ⇒ x = + 3
The d orbital occupation for Co³⁺ is \(t_{2 g}^{6} e g^{0}\).
(as \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) is strong field ligand)
Coordination number of Co = 3 × denticity of C₂O₄
= 3 × 2 (as \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) is a bidentate ligand) = 6

(ii) cis-[Cr(en)₂Cl₂]Cl
The central metal ion is Cr.
The oxidation state can be given as:
x + 2(0) + 2(-1) + (-1) = 0
x – 2 – 1 = 0
x = + 3
The d orbital occupation for Cr³⁺ is \(t_{2 g}^{3}\).
Coordination number of Cr
= 2 × denticity of en + 2
= 2 × 2 + 2 = 6

(iii) (NH₄)₂[CoF₄]
The central metal ion is Co.
The oxidation state can be given as:
(+1) × 2 + × + (-1) × 4 = 0
x – 4 = -2
x = + 2
The d orbital occupation for Co²⁺ is d⁷ or \(t_{2 g}^{5} e_{g}^{2}\). (as F^– is a weak ligand)
Coordination number of Co = 4

(iv) [Mn(H₂O)₆]S0₄
The central metal ion is Mn.
The oxidation state can be given as:
x + (0) × 6 + (- 2) = 0
x = + 2
The d orbital occupation for Mn is d⁵ or [latext_{2 g}^{3} e_{g}^{2}][/latex].
Coordination number of Mn = 6

Question 24.
Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex:
(i) K[Cr(H₂O)₂(C₂O₄)₂] 3H₂O
(ii) [Co(NH₃)₅Cl]Cl₂
(iii) CrCl₃(py)₃
(iv) Cs[FeCl₄]
(v) K₄[Mn(CN)₆]
Answer:
(i) K[Cr(H₂O)₂ (C₂O₄)2 ] ∙ 3H₂O
IUPAC name : Potassium diaquadioxalatochromate (III) hydrate.
Oxidation state of chromium
+1 + x + (0) × 2 + (- 2) × 2 + 3(0) = 0
+ 1 + x – 4 = 0
x = + 3
Electronic configuration of Cr⁺³= 3d³ = (\(t_{2 g}^{3} e_{g}^{0}\))
Coordination number = 6
Shape : Octahedral


Magnetic moment (μ) = \(\sqrt{n(n+2)}\)
= \(\sqrt{3(3+2)}\)
= \(\sqrt{15}\) = 3.87 BM

(ii) [Co(NH₃)₅Cl]Cl₂
IUPAC name : Pentaammine chlorido cobalt(III) chloride
Oxidation state of Co
x + (0)5 + (-1) + (-1) × 2 = 0
x – 3 =0
x = + 3
Coordination number = 6
Shape: Octahedral.
Electronic configuration of Co³⁺ = 3d⁶ = \(t_{2 g}^{6} e_{g}^{0}\)
The complex does not exhibit geometrical as well as optical isomerism.
Magnetic Moment (μ) = \(\sqrt{n(n+2)}\)BM = \(\sqrt{0(0+2)}\) BM = 0 BM

(iii) CrCl₃(py)₃
IUPAC name : Trichlorido tripyridine chromium (III) Oxidation state of Cr
x + (-1) × 3 + (0)3 = 0
x = + 3
Electronic configuration of Cr = 3d³ = (\(t_{2 g}^{3} e_{g}^{0}\))
Coordination number = 6
Shape : Octahedral
Stereochemistry

Both isomers are optically active. Therefore, a total of 4 isomers exist.
Magnetic moment (μ) = \(\sqrt{n(n+2)}\) = \(\sqrt{3(3+2)}\)
= \(\sqrt{15}\) = 3.87 BM

(iv) Cs[FeCl₄]
IUPAC name : Caesium tetrachlorido ferrate (III)
Oxidation state of Fe
+ 1 + x + (-1) × 4= 0
x – 3 = 0
x = + 3
Electronic configuration of Fe = 3d⁵(\(t_{2 g}^{3} e_{g}^{2}\))
Coordination number = 4
Shape : Tetrahedral
The complex does not exhibit geometrical or optical isomerism, (stereo isomerism).
Magnetic moment (μ) = \(\sqrt{n(n+2)}\)
= \(\sqrt{5(5+2)}\)
= \(\sqrt{35}\) = 5.92 BM

(v) K₄[Mn(CN)₆]
IUPAC name : Potassium hexacyanomanganate(II)
Oxidation state of Mn
(+1) × 4 + x + (-1) × 6 = 0
x – 2 = 0
x = + 2
Electronic configuration of Mn = 3d⁵ (\(t_{2 g}^{5} e_{g}^{0}\))
Coordination number = 6
Shape : Octahedral.
The complex does not exhibit stereo isomerism.
Magnetic moment (μ) = \(\sqrt{n(n+2)}\)
= \(\sqrt{1(1+2)}\)
= \(\sqrt{3}\)
= 1.732 BM

Question 25.
What is meant by stability of a coordination compound in solution? State the factors which govern the stability of complexes.
Answer:
The stability of a coordination compound in solution refers to the degree of association between the two species involved in the state of equilibrium. The stability of the coordination compound is measured in term of magnitude of stability or formation of equilibrium constant.
M + 4L → ML₄
K = \(\frac{\left[\mathrm{ML}_{4}\right]}{[\mathrm{M}][\mathrm{L}]^{4}}\)
Larger the stability constant, the higher is the proportion of ML₄ that exists in solution.

Factors on which stability of the complex depends are as follows :

1. Charge on the central metal ion : Greater the charge on the central metal ion, greater is the stability of the complex.
2. Nature of the metal ion : Groups 3 to 6 and inner transition element form stable complexes when donor atoms of the ligands are N, O and F. The element after group 6 of the transition metals which have d-orbitals (e.g., Rh, Pd, Ag, Au, Hg, etc.) form stable complexes when the donor atoms of the ligands are heavier members of N, O and F family.
3. Basic nature of the ligand : Greater the basic strength of the ligand, greater is the stability of the complex.
4. Chelate effect: Presence of chelate rings in the complex increases its stability. It is called chelate effect. It is maximum for the 5- and 6- membered rings.
5. Effect of multidentate cyclic ligands : If the ligands happen to be multidentate and cyclic without any steric effect, the stability of the complex is further increased.

Question 26.
What is meant by chelate effect? Give an example.
Answer:
When a didentate or a polydentate ligand contains donor atoms positioned in such a way that when they coordinate with the central metal ion, a five or a six-membered ring is formed, the effect is called chelate effect. Example, [PtCl₂(en)].

Question 27.
Discuss briefly giving an example in each case the role of coordination compounds in:
(i) biological systems
(ii) medicinal chemistry
(iii) analytical chemistry
(iv) extraction/metallurgy of metals
Answer:
(i) Role of coordination compounds in biological systems :

• Haemoglobin, the oxygen carrier in blood, is a complex of Fe²⁺ with porphyrin.
• The pigment chlorophyll in plants, responsible for photosynthesis, is a complex of Mg²⁺ with porphyrin.
• Vitamin B₁₂ (cyanocobalamine) the antipemicious anaemia factor, is a complex of cobalt.

(ii) Role of coordination compounds in medicinal chemistry :

• The platinum complex cis-[Pt(NH₃)₂Cl₂] (cis-platin) is used in the treatment of cancer.
• EDTA complex of calcium is used in the treatment of lead poisoning. Ca-EDTA is a weak complex; when it is administered, calcium in the complex is replaced by the lead present in the body and is eliminated in the urine.
• The excess of copper and iron present in animal system are removed by the chelating ligands D-penicillamine and desferroxime B via the formation of complexes.

(iii) Role of coordination compounds in analytical chemistry :
Complex formation is frequently encountered in qualitative and quantitative chemical analysis.
(a) Qualitative analysis
I. Detection of Cu²⁺ is based on the formation of a blue tetraammine copper (II) ion.

II. Ni²⁺ is detected by the formation of a red complex with dimethyl glyoxime (DMG).

III. The separation of Ag⁺ and Hg²⁺ in group I is based on the fact that while AgCl dissolves in NH₃, forming a soluble complex, Hg₂Cl₂ forms an insoluble black substance.

(b) Quantitative analysis : Gravimetric estimation of Ni²⁺ is carried out by precipitating Ni²⁺ as red nickel dimethyl glyoxime complex in the presence of ammonia.

EDTA is used in the complexometric determination of several metal ions such as Ca²⁺, Zn²⁺, Fe²⁺, Co²⁺, Ni²⁺ etc.

(iv) Role of coordination compounds in extraction/metallurgy of metals : Extraction of various metals from their ore involves complex formation. For example, silver and gold are extracted from their ore by forming cyanide complex.

Purification of some metals can be achieved through complex formation. For example in Mond process, impure nickel is converted into [Ni(CO)₄] which is decomposed to yield pure nickel.

Question 28.
How many ions are produced from the complex Co(NH₃)₆ Cl₂ in solution?
(i) 6
(ii) 4
(iii) 3
(iv) 2
Answer:
The correct option is (iii)
Coordination number of cobalt = 6. It ionises in the solution as

Hence, 3 ion are produced.

Question 29.
Amongst the following ions, which one has the highest magnetic moment value?
(i)[Cr(H₂O)₆]³⁺
(ii)[Fe(H₂O)₆]²⁺
(iii) [Zn(H₂O)₆]²⁺
Answer:
The oxidation state are: Cr (III), Fe (II) and Zn (II).
Electronic configuration of Cr³⁺ = 3d³, unpaired electrons = 3
Electronic configuration of Fe²⁺ = 3d⁶, unpaired electrons = 4
Electronic configuration of Z²⁺ = 3d¹⁰, unpaired electrons = 0
As μ = \(\sqrt{n(n+2)}\), therefore, (ii) has the highest magnetic moment.

Question 30.
The oxidation number of cobalt in K[Co(CO)₄] is
(i) +1
(ii) +3
(iii) -1
(iv) -3
Solution:
Oxidation number of Co : K[Co(CO)₄]
x+ (4 × 0) = -1; x = -1
Thus, correct answer is (iii).

Question 31.
Amongst the following, the most stable complex is
(i) [Fe(H₂O)₆]³⁺
(ii) [Fe(NH₃)₆]³⁺
(iii) [Fe(C₂O₄)₃]^3-
(iv) [FeCl₆]^3-
Answer:
In all these complexes, Fe is in +3 oxidation state. However, the complex (iii) is a chelate because three \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) ions acts as the chelating ligands. Thus, the most stable complex is [Fe(C₂O₄)₃]^3-. Thus, correct answer is (iii).

Question 32.
What will be the correct order for the wavelengths of absorption in the visible region of the following:
[Ni(NO₂)₆]^4-, [Ni(NH₃)₆]²⁺, [Ni(H₂O)₆]²⁺
Answer:
As metal ion is fixed, the increasing CFSE values of the ligands from the spectrochemical series are in the order :
H₂O < NH₃ < \(\mathrm{NO}_{2}^{-}\)
Hence, the energies absorbed for excitation will be in the order :
[Ni(H₂O)₆]²⁺ < [Ni(NH₃)₆]²⁺ < [Ni(NO₂)₆]^4-
As E = \(\frac{h c}{\lambda}\), therefore, the wavelengths absorbed will be in the opposite order,
[Ni(NO₂)₆]^4- < [Ni(NH₃)₆]²⁺ < [Ni(H₂O)₆]²⁺

Chemistry Guide for Class 12 PSEB Coordination Compounds Textbook Questions and Answers

Question 1.
Write the formulas for the following coordination compounds :
(i) Tetraamminediaquacobalt (III) chloride
(ii) Potassiumtetracyanidonickelate(II)
(iii) Tris(ethane-l,2-diammine)chromium(III) chloride
(iv) Amminebromidochloridonitrito-N-platinate(II)
(v) Dichloridobis(ethane-l,2-diammine) platinum (IV) nitrate
(vi) Iron(III)hexacyanidoferrate(II).
Answer:
(i) [Co(NH₃)₄(H₂O)₂]Cl₃
(ii) K₂[Ni(CN)₄
(iii) (Cr(en)₃]Cl₃
(iv) [Pt(NH₃)BrCl(NO₂)]^–
(v) [PtCl₂(en)₂] (NO₃)₂
(vi) Fe₄[Fe(CN)₆]₃

Question 2.
Write the IUPAC names of the following coordination compounds:
(i) [CO(NH₃)₆]Cl₃
(ii) [CO(NH₃)₅Cl]Cl₂
(iii) K₃[Fe(CN)₆]
(iv) K₃[Fe(C₂O₄)₃]
(v) K₂[PdCl₄]
(vi) [Pt(NH₃)₂Cl(NH₂CH₃)]Cl
Answer:
(i) Hexaamminecobalt(III)chloride
(ii) Pentaamminechloridocobalt(III)chloride
(iii) Potassiumhexacyanoferrate(III)
(iv) Potassiumtrioxalatoferrate (III)
(v) Potassiumtetrachloridopalladate (II)
(vi) Diamminechloridomethylamine platinum(II) chloride.

Question 3.
Indicate the types of isomerism exhibited by the following complexes and draw the structures of these isomers :
(i) K[Cr(H₂O)₂](C₂O₄)₂]
(ii) [Co(en)₃]Cl₃
(iii) [CO(NH₃)₅(NO₂)](NO₃)₂
(iv) [Pt(NH₃)(H₂O)Cl₂]
Answer:
(i) (a) Both geometrical isomer (cis and traits):

(b) Cis-isomer of this compound can exist as pair of optical is :

(ii) Complex will exist as optical isomers:

The compound will show ionisation as well as linkage isomerism.

(iii) Ionisation isomer :
[Co(NH₃)₅(NO₂)](NO₃)₂,
[Co(NH₃)₅ (NO)₃] (NO₂) (NO₃)
Linkage isomers :
[Co(NH₃)₅ (NO₂)](NO₃)₂;
[CO(NH₃)₅ (ONO)](NO₃)₂

(iv) Geometrical isomerism (cis and trans) :

Question 4.
Give evidence that [Co(NH₃)₅Cl]SO₄ and [Co(NH₃)₅SO₄]Cl are ionisation isomers.
Answer:
When they are dissolved in water, they give different ions in the solution which can be tested by adding AgNO₃ solution and BaCl₂ solution. If Cl^– dons are the counter ions, a white precipitate will be obtained with AgNO₃ solution. If \(\mathrm{SO}_{4}^{2-}\) ions are the counter ions, a white precipitate will be obtained with BaCl₂ solution.

Question 5.
Explain on the basis of valence bond theory that [Ni(CN)₄]^2- ion with square planar structure is diamagnetic and the [Ni(Cl)₂₄]^2- ion with tetrahedral geometry is paramagnetic.
Answer:
Nickel in [Ni(CN)₄]^2- is in the +2 oxidation state. The formation of [[Ni(CN)₄]^2- may be explained through hybridisation as follows :
Ni atom in the ground state

Since no unpaired electrons is present, the square planar complex is diamagnetic. In [Ni(CN)₄]^2-, Cl^– is a weak field ligand. It is, therefore, unable to pair up the unpaired electrons of the 3d orbital. Hence, the hybridisation involved is sp3 and the shape is tetrahedral. Since all the electrons are unpaired, it is paramagnetic

Question 6.
[Ni(CN)₄]^2- is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral. Why?
Answer:
In [Ni(CO)₄] Ni is in zero oxidation state whereas in [NiCl₄]^2-, it is in
+ 2 oxidation state. In the presence of strong ligand, CO ligand, the unpaired d electrons of Ni pair up but Cl^– being a weak ligand is unable to pair up the unpaired electrons.

Question 7.
[Fe(H₂O)₆]³⁺ is strongly paramagnetic whereas [Fe(CN)₆]^3- is weakly paramagnetic. Explain.
Answer:
In presence of CN^– (a strong ligand), the 3d⁵ electrons pair up leaving only one unpaired electron. The hybridisation is d²sp³ forming an inner orbital complex. In the presence of H₂O (a weak ligand), 3d electrons do not pair up. The hybridisation is sp³d² forming an outer orbital complex containing five unpaired electrons. Hence, it is strongly paramagnetic.

Question 8.
Explain [Co(NH₃)₆]³⁺ is an inner orbital complex whereas [Ni(NH₃)₆]²⁺ is an outer orbital complex.
Answer:
In [CO(NH₃)₆]³⁺, CO is in +3 oxidation state and has d⁶ electrons. In the presence of NH₃, the 3d electrons pair up leaving two d-orbitals empty to be involved in d²sp³ hybridisation forming inner orbital complex. In [Ni(NH₃)₆]²⁺, Ni is in +2 oxidation state and has d⁸ configuration. The hybridisation involved is sp³d², forming the outer orbital complex.

Question 9.
Predict the number of unpaired electrons in the square planar [Pt(CN)₄]^2- ion.
Answer:
₇₈Pt lies in group 10 with the configuration 5d⁹6s¹. Thus Pt²⁺ has the configuration :

For square planar shape, the hybridisation is dsp². Hence, the unpaired electrons in 5d orbital pair up to make one d orbital empty for dsp² hybridisation.
Thus there is no unpaired electron.

Question 10.
The hexaquomanganese(II) ion contains five impaired electrons, while the hexacyano ion contains only one unpaired electron. Explain using crystal field theory.
Answer:
Mn in the + 2 oxidation state has the configuration 3d⁵. In the presence of H₂O a weak ligand, the distribution of these five electrons is \(t_{2 g}^{3} e_{g}^{2}\)
i.e., all the electrons remain unpaired

However, in the presence of CN^– the distribution of these electrons is \(\), i.e., two t_2g orbitals contain paired electrons while the third t_2g orbital contains one unpaired electron.

Question 11.
Calculate the overall complex dissociation equilibrium constant for the \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) ion, given that β₄ for this complex is 2. 1 × 10¹³.
Solution:
The overall complex dissociation equilibrium constant is the reciprocal of the overall stability constant, β₄.
∴ \(\frac{1}{\beta_{4}}\) = \(\frac{1}{2.1 \times 10^{13}}\)
∴ = 4.7 × 10^-14

Punjab State Board PSEB 12th Class English Book Solutions English Grammar Translation Exercise Questions and Answers, Notes.

PSEB 12th Class English Grammar Translation

Some Common Proverbs

Exercise 1 (Solved)

Translate the following sentences into English:

Answer:
1. Every dog has his day.
2. He is completely a yes-man.
3. He has his own axe to grind.
4. No new taxes have been imposed upon us.
5. People (or subjects) have revolted against the king.

Exercise 2 (Solved)

Translate the following sentences into English:

Answer:
1. His poverty is an open secret.
2. He is at the root of this mischief.
3. My prosperity. is an eye-sore to him.
4. Ram and Sham are sworn enemies.
5. Where there is a will, there is a way.

Exercise 1 (For Practice)

Translate the following sentences into Hindi/Punjabi:
1. He works hard from dawn to dusk.
2. Most of our politicians are very corrupt.
3. Selfishness is the order of the day.
4. You must reduce your flab¹.
5. Spoken English is gaining importance these days.
6. Spoken English does not mean repeating in the parrot-fashion some crammed sentences.
7. Those who speak English fluently get easily noticed by others.
8. He is the only bread-earner² of the family.
9. Shaleenta is a very good Kathak dancer.
10. He is working as an executive³ in a bank.
11. My colleague in my college is a very good singer.
12. I feel pleased to meet you.
13. What is the nature of your job?
Hints:
1. ਸੀਟਾਧਾ, ਮੋਟਾਪਾ 2. ਹੇਜ਼ੀ ਸਾਰੇ ਭਾਗ, ਰੋਜ਼ੀ ਕਮਾਉਣ ਵਾਲਾ, 3. ਜਿਵੇਕ, ਨਿਦੇਸ਼ਕ ।

Note : Some of the sentences are very easy. It should not be difficult for many of you to translate them in simple Hindi and Punjabi. Hereafter we shall give a few more exercises with meanings in Hindi and Punjabi for your convenience.

Exercise 2 (For Practice)

Translate the following sentences into Hindi/Punjabi:
1. Gandhiji is known for his successful efforts¹ to liberate² India.
2. The lady is very mean in money matters³.
3. She did not allow her husband to spend much on himself.
4. This story tells us about the greed of well-to-do⁴ people.
5. Guru Gobind Singh was a great son of India who led India to sublime⁵ heights of glory⁶.
6. Why is this enormous⁷ increase in population taking place ?
7. Drug-addiction⁸ is a major social evil of modern times.
8. Nations become great by self-confidence⁹.
9. Mankind has undoubtedly progressed since medieval¹⁰ times.
10. Vidyasagar was a very generous¹¹ and charitable¹² man.
Hints:

Exercise 3 (For Practice)

Translate the following sentences into Hindi/Punjabi:
1. Everyone knows the events¹ of Buddha’s life.
2. Newton lived to a ripe age² and earned a great renown³.
3. Books are a wonderful blessing⁴.
4. Ishwara Chandra realised that he must go to England in oder to complete his scientific education.
5. Guru Teg Bahadur was the youngest son of Guru Hargobind Sahib.
6. Science is an angel⁵ in peace but a devil⁶ in war.
7. Wars of today are more dreadful⁷ than those of yesterday.
8. He stumbled⁸ at every step.
9. Silence prevailed⁹ all round the hospital.
10. It has been raining continuously¹⁰ for three hours.
11. It has been drizzling¹¹ for an hour.
Hints:

Exercise 4 (For Practice)

Translate the following sentences into Hindi/Punjabi:
1. This story is a typical example¹ of the storywriter’s work as a story.
2. He failed to find a man before whom he could express the feelings of his heart².
3. He wants to share his grief³ with somebody.
4. This story points out⁴ the truth that poverty dehumanises⁵ a man.
5. I am bereft⁶ of money at this time.
6. I want to know your opinion about etiquette⁷.
7. The teacher punished the students for being noisy in the class.
8. Some shopkeepers fleeces⁸ their customers.
9. A spendthrift lacks balance and perspective⁹.
10. He was reluctant¹⁰ to take the examination.
Hints:

Exercise 5 (For Practice)

Translate the following sentences into Hindi/Punjabi :
1. I have lost the novel¹ that you gave me.
2. She dressed in an elegant² manner.
3. We eat so that we may live.
4. God’s will be done³.
5. Prevention is better than cure⁴.
6. A bird in the hand is worth two in the bush⁵.
7. He is on the wrong side of sixty⁶.
8. We should not look down upon⁷ the poor.
9. I am at a loss to know⁸ what to do.
Hints:

Exercise 6 (For Practice)

Translate the following sentences into Hindi/Punjabi:
1. Trees give shade for the benefit of others.
2. Machines have become the slaves¹ of modern man.
3. A child should be trained to love the reading of books.
4. Politicians worship the rising sun².
5. Our body is like a machine.
6. Newton was a very kind-hearted³ man.
7. He never lost his temper⁴.
8. One should never read anybody’s letter without permission.
9. Children explode crackers on Diwali.
10. The life of modern cities is tiresome⁵.
Hints:

Punjab State Board PSEB 12th Class English Book Solutions English Grammar Use of Non-finites (Infinitives, Gerunds and Participles) Exercise Questions and Answers, Notes.

PSEB 12th Class English Grammar Use of Non-finites (Infinitives, Gerunds and Participles)

We divide verbs into two types. One of the types of verb is the finite verb. A verb that changes its form according to person, tense, mood is called a Finite Verb.

Read the examples given below:
Neeru runs fast.
They run fast.
She is running fast.
We have run fast.

The verbs used in the above example are finite verbs because they change according to person, number and tense.

Now have a look at the sentences given below:

1. She wants to be prosperous.
2. They wanted to be famous. .
3. You wanted to be rich and famous.
4. She made me realise my mistake.

The verbs be and realise in the above sentences do not change according to tense, number and person. Such verbs are called non-finite verbs.

Now read the following sentences:
Rashmi eats to live.
Reshma eats to live.
Rajan eats to live.
A girl eats to live.
These girls eat to live.
You eat to live.
They eat to live.

The verbs used in the above sentences/examples are finite verbs because they change according to person, number and tense.

In the above examples containing eat and to live, eat is governed by person, number and tense. So it is finite verb.

In all other sentences to live is not governed by person and tense. That is an infinite verb.

Three kinds of verbs are there in the infinite verb.

The Infinitive

1. In simple language infinitive means the use of to before the verb.
Examples : I want to go. I want to hit you.
I shall watch Rekha dance.
He made me leave the room.

There are some verbs before which we do not use to. These verbs are make, observe, bid, notice, make, feel, need, dare, see, hear.

2. I need to warn you.
Do you need to leave this place just now ?
He dares me to kill the mongoose ?
If need and dare are used as main verb, to has to be used after these verbs.

3. We do not use to before did, may, do, shall, might, can, could.
He could ride a car.
We shall laugh,
I might leave for Jalandhar.

4. There are some verbs which take an infinitive with the verb.
Promise, refuse, wish, want, dare, fail, decide, agree, hope, desire, expect, ready, glad, happy.
Sentences : I promise to help you. He refused to obey me. I wish to see you tonight.

We use infinitive after some adjectives:
Ready, glad, happy, eager, easy, anxious, hard, able etc.
I am ready to leave for Gurdaspur.
I am glad to see my friend.
I am happy to meet you.
I am eager to know the result on my examination.
It is easy to solve this problem.
I am anxious to see my father.
It is hard to solve this riddle.
He was able to get the job.

Forms of Infinitive

The Infinitive has two forms:

(i) Bare Infinitive
Or
Infinitive without to.
run, fight

(ii) to + Infinitive
to repair
to go

Use of To + Infinitive

We use to + infinitive in the following ways:
1. As a verb
(a) To walk is a good exercise
To smoke is very harmful.

(b) I want to learn French. (Object of the transitive verb want)
I know how to cook pasta. (Object)

(c) The best thing is to help yourself. (Complement to the linking verb is)
Her target was to become an actress. (Complement to the linking verb was)

It is easier to say than to do. (After the dummy subject ‘it’)

2. As an adverb to modify a verb or an adjective:

He went to see the Chairman. (modifies the verb went)
They stood up to protest. (modifies the verb stood)
She is anxious to resign. (modifies the adjective anxious)
He is hard to please. (modifies the adjective hard)

3. As an adjective to qualify a noun:
It is time to depart (qualifies the noun time)
We have nothing to offer. (qualifies nothing)

4. As an object complement or an object to a preposition.
He is about to retire.
We saw him run.
This shop is about to shut.
I felt something fall on my shoulder.

5. As an adjunct :
The captain was the last to abandon the ship.
Kindly give me some water to drink.

6. As part of too + adjective/adverb + infinitive.
He is too weak to walk.
He doesn’t have money enough to buy a flat.

7. As an absolute to modify the whole sentence.
To be candid with you, you are unfit for the job.
To be brief, you have cheated all of us.

Use the Bare Infinitive

8. We do not use to (bare infinitive) before some verbs like watch, bid, see, let, make, help, hear, feel, behold etc.
I shall watch you write an apology.
I heard her sing.
Mummy helped her daughter do her homework.
The Chairman made me wait.
I like you to be with me.
Let me know the truth.

9. With had better, had rather.
You had better stay where you are.
You had rather patch up with your friend.

10. With some prepositions like except, but, than etc.
He does nothing but play.
I would rather give up.
He does nothing except crib.
I would rather die than beg.

11. Passive form of the infinitive (to + be + past participle)
Women like to be flattered by men.
It is an insult to be maltreated like that.

Exercise 1

Fill in the blanks using the verbs within brackets (with or without to):

1. He made me ………… (to laugh)
2. He is too weak ………… (to walk)
3- Raju let me ………… his car. (to drive)
4. He must ………… them from going, (to prevent)
5. You had better ………… (to go)
6. The mango is ………… (eat)
7. You are too young ………… (to understand)
8. There is nothing ………… (to say)
9. We watched them ………… (to work)
10. Let him ………… this work (to do)
Answer:
1. laugh
2. to walk
3. drive
4. prevent
5. go
6. to eat
7. to understand
8. to be said
9. work
10. do.

Exercise 2

Fill up the blanks with infinitive:

1. There is no one ………… this claim, (challenge)
2. He did not let me ………… my work in peace, (do)
3. He was sure ………… a scholarship, (get)
4. The teacher asked the students ………… silence, (maintain)
5. The doctor advised me ………… regular exercise, (to take)
6. They did nothing but ………… (to think)
7. You had rather ………… there, (to go)
8. Everyone wants ………… peacefully, (to live)
9. My father allowed me ………… (to go)
10. She expects ………… (to succeed)
Answer:
1. to challenge
2. do
3. to get
4. to maintain
5. to take
6. think
7. go
8. to live
9. to go
10. to succeed.

The Gerund

A gerund is that form of the verb which ends in V-ing and has the force of a noun. That is why a gerund is called a verbal noun.

Use of Gerund

1. As a Subject
Running is good for health.
Swimming is useful for reducing weight.
Seeing is believing
Reading makes a man complete.
Writing makes a man perfect.

2. As an object
She likes painting.
Rekha loves dancing.

3. As a complement to a Linking Verb
My first love is singing.
My favourite sport is swimming.

4. As an object of preposition
I am tired of sitting all day long.
I am fond of fishing.
He is addicted to gambling.

5. As part of a Noun Phrase
Watching snowfall is soothing to the eyes.
It is foolish catching fire.

Note : Addicted to, look forward to are followed by gerund -ing.
He is addicted to drinking.
He is looking forward to meeting his friend.

The Participle

A participle is that form of verb which partakes of the nature both of a verb and of an adjective.

Kinds of Participle

Participle is of two kinds:

Present Participle	Past Participle
V1 + ing
go + ing	V3 gone
eat + ing	V3 eaten (eat, ate, eaten) V3
V1 is the first form of the verb.	V3 is the third form of the verb.

Use of Participle:

1. As an adjective :
I will buy a talking clock.
This house has no running water.

2. As an object complement:
I found the conditions soothing.
She found the baby weeping.

3. As an adverb :
She went away smiling.
The player left the ground abusing.
The girl stood whistling.

4. As part of an Adjective Phrase:
The lady doctor living on the ground floor is very charming.
The man managing the event is my uncle.

5. As a Participle Phrase :
Placing his hand on the Gita, he swore that he would speak the truth.
Taking her bag, the lady walked out of the Mall.

Difference between a Gerund and a Participle:
Throwing a grenade at the C.R.P.F. picket, the terrorist ran away. (Present Participle)
Throwing stones at other people’s houses is an anti-social act. (Gerund)
Swimming in the fast flowing water, Atul saved a drowning boy. (Present Participle)
Jogging is a good exercise. (Gerund)

Use of Past Participle:

1. As an Adjective:
The injured man was carried to the trauma centre.
He is a gone case.

2. As an Object Complement:
I found the door shut.
The murder of the leader left us shocked.

3. As an Adverb:
He left the gym totally exhausted.
Nervous, he kept on moving to and fro.

4. As part of a Participle Phrase:
I saw a dog crushed under a truck.
The landlady found her house burgled from the front and the back.

Exercise 1

Do as directed:

Question 1.
………… (hear) a noise, I turned round. (Fill up the blank with a participle)
Answer:
Hearing a noise, I turned round.

Question 2.
I saw a storm ………… (approach). (Fill up the blank with a participle)
Answer:
I saw a storm approaching.

Question 3.
(Hunt) deer is not allowed in this area. (Fill up the blank with a gerund)
Answer:
Hunting deer is not allowed in this area.

Question 4.
Children love ………… make mud houses. (Fill up the blank with a gerund)
Answer:
Children love making mud houses.

Question 5.
………… (toil) is the lot of mankind. (Fill up the blank with an infinitive)
Answer:
To toil or toiling is the lot of mankind.

Exercise 2

Do as directed:

Question 1.
Combine the following sentences using an infinitive:
(i) He went to Amritsar.
(ii) He wanted to visit the Golden Temple.
Answer:
He went to Amritsar to visit the Golden Temple.

Question 2.
Combine the following sentences using the participle:
(i) I speak the truth.
(ii) I am not afraid of speaking of it.
Answer:
I am not afraid of speaking the truth.

Question 3.
A ……….. candle fell off the table, (burn) (Fill up the blank with a participle)
Answer:
A burning candle fell off the table.

Question 4.
He left the tap (run). (Fill up the blank with a participle)
Answer:
He left the tap running.

Question 5.
Combine the following sentences using a participle :
He had resolved on a certain course.
He acted with vigour.
Answer:
Having resolved on a course, he acted with vigour.

Question 6.
Combine the following sentences using a participle:
They had no fodder.
They could give the cow nothing to eat.
Answer:
Having no fodder, they could give the cow nothing to eat.

Question 7.
Success is not merely ………….. (win) applause. (Fill up the blank with a gerund)
Answer:
Success is not merely winning applause.

Question 8.
………….. (amass) wealth often ruins the health. (Fill up the blank with a gerund)
Answer:
Amassing wealth often ruins the health.

Exercise 3

Do as directed:

Question 1.
The ability …………. (laugh) is peculiar to mankind. (Fill the blank with an infinitive)
Answer:
The ability to laugh is peculiar to mankind.

Question 2.
Can you hope (count) the stars ? (Fill up the blank with an infinitive)
Answer:
Can you hope to count the stars ?

Question 3.
Combine the following sentences using a participle:
(i) I call a spade a spade.
(ii) I am not afraid of it.
Answer:
I am not afraid of calling a spade a spade.

Question 4.
The man seems …………. (worry) (Fill up the blank with a participle)
Answer:
The man seems worried.

Question 5.
We had a drink of the ………….. (sparkle) water. (Fill up the blank with a participle)
Answer:
We had a drink of the sparkling water.

6. Combine the following sentences using a participle:
(i) The stable door was open.
(ii) The horse was stolen.
Answer:
The stable door having been opened, the horse was stolen.

Question 7.
Combine the following sentences using a participle:
(i) We met a man.
(ii) He was carrying a log of wood.
Answer:
We met a man carrying a log of wood.

Question 8.
We were prevented from …………. (enter) the house. (Fill up the blank by using a gerund)
Answer:
We were prevented from entering the house.

Question 9.
We heard her …………… (sing) at the function. (Fill up the blank with a gerund)
Answer:
We heard her singing at the function.

Exercise 4 (Textual)

Do as directed:

Question 1.
He is slow, ……………. (forgive) (Fill up the blank with an infinitive)
Answer:
He is slow to forgive.

Question 2.
I am sorry …………… (hear) this. (Fill up the blank with an infinitive)
Answer:
I am sorry to hear this.

Question 3.
Combine the following sentences using an infinitive:
(i) He collects old stamps even at great expense.
(ii) It is his hobby.
Answer:
It is his hobby to collect old stamps even at great expense.

Question 4.
…………… (Run) water is not always fit for drinking. (Fill up the blank with a participle)
Answer:
Running water is not always fit for drinking.

Question 5.
………… (carry) by the wind, seeds are scattered far and wide. (Fill up the blank with a participle)
Answer:
Having been carried by the wind, seeds are scattered far and wide.

Question 6.
Combine the following sentences using a participle:
(i) He was dissatisfied.
(ii) He resigned his job.
Answer:
Having been dissatisfied, he resigned his job.

Question 7.
Combine the following sentences using a participle :
(i) We met a girl.
(ii) She was carrying a basket of flowers.
Answer:
We met a girl carrying a basket of flowers.

Question 8.
He is fond of ……….. (swim) (Fill up the blank with a gerund)
Answer:
He is fond of swimming.

Question 9.
Are you afraid of his …………… (hear) you ? (Fill up the blank with a gerund)
Answer:
Are you afraid of his hearing you ?

Question 10.
I have come ……………. (see) you. (Fill up the blank with an infinitive)
Answer:
I have come to see you.

Exercise 5 (Textual)

Question 1.
Combine the following sentences using an infinitive:
(i) He has five children.
(ii) He must provide for them.
Answer:
He must provide for them.

Question 2.
He has five children to provide for.
(i) He wants to earn his livelihood.
(ii) He works hard for this reason.
Answer:
He works hard to earn his livelihood.

Question 3.
(i) I saw him (enter) the house. (Fill up the blank with a participle.)
Answer:
I saw him entering the house.

Question 4.
He played a …………. (lose) game. (Fill up the blank with a participle)
Answer:
He played a losing game.

Question 5.
Combine the following sentences using a participle:
(i) He staggered back.
(it) He sank to the ground.
Answer:
He staggered back sinking to the ground.

Question 6.
She is very keen to …………. modelling, (take up) (Fill up the blank with an infinitive)
Answer:
She is very keen to take up modelling.

Question 7.
What she hates most is ……………. (smoke) (Fill up the blank with a gerund)
Answer:
What she hates most is smoking.

Question 8.
He objected to …………. money on cosmetics, (spend) (Fill up the blank with a gerund)
Answer:
He objected to spending money on cosmetics.

Question 9.
It is a penal offence ……………. bribe a public servant. (Fill up the blank with an infinitive)
Answer:
It is a penal offence to bribe a public servant.

Question 10.
The boys are anxious ………….. (learn) (Fill up the blank with an infinitive)
Answer:
to learn.

Exercise 6 (Textual)

Question 1.
Combine the following sentences using participles:
(i) The strikers held a meeting.
(ii) They wished to discuss the terms of the employers.
Answer:
The strikers held a meeting wishing to discuss the terms of the employers.

Question 2.
(i) The robber took out a knife.
(ii) He wanted to frighten the old man.
Answer:
The robber took out a knife wanting to frighten the old man.

Question 3.
…………. (consider) the facts, he received scant justice. (Fill up the blank with a participle)
Answer:
Considering the facts, he received scant justice.

Question 4.
Combine the following sentences using a participle:
(i) He walked away.
(ii) He was whistling.
Answer:
He walked away whistling.

Question 5.
My hair needs …………….. (cut) (Fill up the blanks with a gerund)
Answer:
My hair needs cutting.

Question 6.
I saw him …………….. (cross) the road. (Fill up the blank with a gerund)
Answer:
I saw him crossing the road.

Exercise 7 (Textual)

Combine each of the following pairs of sentences into one sentence by using infinitives:
Question 1.
I am learning Sanskrit. I want to study the Gita.
Answer:
I am learning Sanskrit to study the Gita.

Question 2.
He labours hard. He wants to succeed.
Answer:
He labours hard to succeed.

Question 3.
I am learning English. I want to study the works of Shakespeare.
Answer:
I am learning English to study the works of Shakespeare.

Question 4.
Children are going to the field. They want to play.
Answer:
Children are going to the field to play.

Question 5.
I have come here. I would like to see you.
Answer:
I have come here to see you.

Exercise 8 (Textual)

Fill up the blanks with one of these: (Infinitive, Gerund, Participle)

Question 1.
He has a …………… look, (surprise)
Answer:
surprised (Participle)

Question 2.
My father is a …………… man. (retire)
Answer:
retired (Participle)

Question 3.
They promised …………… me. (help)
Answer:
ro help (Infinitive)

Question 4.
…………… is pleasant, (play)
Answer:
Playing (Gerund)

Question 5.
…………… on the footpath is safe, (walk)
Answer:
Walking on the footpath is safe. (Gerund)

Question 6.
He gave me a pen …………… with, (write)
Answer:
He gave me a pen to write with. (Infinitive)

Question 7.
He gave up …………… (drink)
Answer:
He gave up drinking. (Gerund)

Miscellaneous Exercises

Infinitive

Exercise 1

Fill in the blanks with a infinitive:

1. …………… (waste) time is a folly.
2. …………… (err) is human.
3. I find it sensible …………… (remain) silent.
4. She seems …………… (be) happy.
5. …………… (toil) is the lot of mankind.
6. You need …………… (do) a lot.
7. I gave him a chance …………… (reconsider) his stand.
8. You are advised not …………… (mislead) your friends.
9. The game was about …………… (begin).
10. Let him leave the room.
11. Make him realise his mistake.
Answer:
1. To waste
2. To err
3. to remain
4. to be
5. To toil
6. to do
7. to reconsider
8. to mislead
9. to begin
10. To is not to be used in these sentences.
11. To is not to be used in these sentences.

Exercise 2

Use the following pair of sentences by using an infinitive:
1. The poor man had live children.
He must provide for them.
2. Turn to the left.
You will find my house.
3. My teacher will learn about my success.
He will be delighted.
4. This box is very big.
He can’t lift it.
5. I had no money.
I could buy no clothes.
Answer:
1. The poor man had ro provide tor five children.
2. Turn to the left for my house.
Or
Turn to the left to find rny house.
3. My teacher will be delighted to learn about my success.
4. This box is too big to lift.
5. I had no money to buy clothes.

Gerund

Exercise 1

Complete the following sentences with correct use oi gerund form of the verb:

1. He is good ar …………… (dance).
2. He is crazy about …………… (sing).
3. He doesn’t like …………… (play; cards.
4. I am afraid of …………… (swim) in the canal.
5. I am always interested in …………… (make) ftiends.
6. He is scared of …………… (travel! by air.
7. I suggest …………… (do) sonic m ire sums.
8. They insisted on …………… (cook) the dinner at home.
9. I thanked him for …………… (fix) my door.
10. I praised him for …………… (help) me in distress.
Answer:
1. dancing
2. playing
3. playing
4. swimming
5. making
6. travelling
7. doing
8. cooking
9. fixing
10. helping.

Participle

Exercise 1

Fill in the blanks with Present Participle:

1. The man …………… (drive) the car is my uncle.
2. Lorries …………… (come) over the bridge have to be careful of the wind.
3. Who was the girl …………… (wear) the red dress ?
4. Students …………… (submit) their essays late will lose ten marks.
5. The …………… (bud) flowers looked very lovely.
6. The …………… (run) bus rammed against the wall.
7. …………… (hear) the noise, he rushed to the spot.
8. They told us an …………… (amuse) account of their journey.
9. I saw the lion …………… (approach) us.
10. He watched them …………… (fight) over trigger.
Answer:
1. driving
2. coming
3. wearing
4. submitting
5. budding
6. running
7. Hearing
8. amusing
9. approaching
10. fighting.

Punjab State Board PSEB 12th Class English Book Solutions English Letter Writing Business Letters Exercise Questions and Answers, Notes.

PSEB 12th Class English Letter Writing Business Letters

1. Applications for Different Jobs

1. Write an application for the post of a Steno-typist.

14, Civil Lines,
Amritsar.
August 6, 20 ………….

The Advertiser,
Post Box No. 313,
The Tribune,
Chandigarh.
Sir,

This is in response to your advertisement in ‘The Tribune’ dated 4th August 20 for the post of a Steno-typist in your office. I beg to offer myself as a candidate for the same. As regards my qualifications, they are detailed below:
I passed my BA. Examination in 20 ………. from D.A.V. College, Amritsar. I was placed in the second division. After my graduation, I joined a commercial college where I learnt both short-hand and type-writing. My speed in short-hand is 75 w.p.m. and my speed in type-writing is 50 w.p.m. I have been working as a steno-typist with Sehgal & Co., Amritsar for the last five years. Besides short-hand and type-writing, I am also well conversant with other official work and can speak fluently English, Hindi and Punjabi.

I am still in the service of the same firm. I am seeking better opportunities which my present employer, with his limited business, cannot supply. He has no objection to my seeking a better job somewhere else.

I am a youngman of 25 with a sound physique and an impressive personality.

I enclose a certificate from my present employer in regard to my character, experience and ability. Should you appoint me to the post, it would be my sincerest effort to give you every satisfaction.

Yours faithfully,
H.M.L. Sood

2. Reply to above.

Messrs Datta & Co.,
130, Park Lane,
New Delhi-15.
August 11, 20……

Mr. H.M.L. Sood,
14 Civil Lines,
Amritsar.

Dear Sir,

We have your application of the 6th August, for the post of a steno-typist in our office. We are glad to inform you that we have decided to appoint you for the post.

Your appointment will be on a temporary basis for three months. If your work and conduct are found satisfactory, you will be made permanent. Your salary will be Rs 12000/- per month during the temporary period and Rs 12450/- per month with a yearly increment of Rs 450/- on your being made permanent. Your promotion depends entirely on your hard work, honesty and sincerity.

If the appointment is acceptable to you, please let us know by return of post the date when you can join.

Yours faithfully,
R.M. Sinha,
Manager

3. Write an application for the job of an Asstt. Manager in a large firm – Messrs Ghooki and Sons – selling electrical appliances. Give details of your qualifications and experience and mention the minimum salary acceptable to you.

510, Church Street,
Chandigarh.
July, 28, 20……….

Messrs Ghooki and Sons,
R-120, Connaught Place,
New Delhi-1.
Dear Sirs,

In response to your advertisement in ‘The Times of India’ dated July 26, 20…. for the post of an Asstt. Manager in your firm engaged in selling electrical appliances, I beg to offer my services for the same. As desired by you, I give below a few particulars about myself.

I passed my M.A. (Economics) from St. Stephen’s College, New Delhi in 20… After my Post-Graduation, I passed M.B.A. course in 20…. from Punjab University, Chandigarh. I was placed in the first division. Since then I have been working in Mumbai Electricals in the capacity of an Asstt. Manager. I am 30 years of age. I have an experience of six years. I can plan and execute sales independently. I have an unblemished record and my present employer wants me to stay on. But I wish to come to my home town that is, Delhi.

At present I am drawing ? 25000/- p.m. and have a free furnished accommodation. I can join your concern at a minimum salary of ^ 35500/- plus a free furnished accommodation and a car at the company’s expense.

I am prepared to attend an interview only at your expense. Copies of my certificates and testimonials are enclosed herewith for your kind perusal.

Yours faithfully,
Shankar Chauhan

4. Reply to Application No. 3.

Messrs Ghooki & Sons,
R-120, Connaught Place,
New Delhi-I.
July 31, 20 ………..

Sh. Shankar Chauhan,
510, Church Street,
Chandigarh.
Dear Sir,

In reply to your application dated 28th July, for the post of an Asstt. Manager, we are glad to inform you that we are ready to offer you the job at the salary demanded by you. We shall also give you free furnished accommodation and a car at company’s expense. However, you will have to sign a contract to serve in our firm for a minimum period of 4 years.

If the above terms are acceptable to you, kindly make it convenient to call at our office on 10th August, 20 ………… at 10 A.M. to settle other details of your appointment.

Yours faithfully,
R.P. Ghooki,
Managing Director

5. Write an application for the post of a P.A. to the Technical Manager of Calico Textiles. The post requires knowledge of short-hand typing and textiles of all varieties. Quote references.

14, Patel Road,
New Delhi-33.
May 26, 20 ……….

The Technical Manager,
Calico Textiles,
Amritsar.
Dear Sir,

In response to your advertisement in ‘The Times of India dated May 22, 20…. for the post of a P.A., I beg to offer myself as a candidate for the same.

I passed my Higher Secondary Examination in the first division. I passed B. Com. Final in 20 with Hons, in Secretarial Practice. I was immediately employed by Mohindra Textiles, New Delhi and have been working there to the satisfaction of my superiors. I am seeking better opportunities which my present employers with their limited range of business cannot supply. They have no objection to my seeking a better job somewhere.

I am a young man of 26 with a pleasing personality. I am capable of doing hard work. I enclose a certificate from my present employers in regard to my character, experience and ability.

Should you appoint me to the post, it would be my sincerest effort to give you every satisfaction. Once again I assure you that I will do my duty honestly and conscientiously.

Yours faithfully,
Sunil Sharma

6. Reply to above.

The Technical Manager,
Calico Textiles,
Amritsar.
May 30, 20 …….

Mr. Sunil Sharma,
14, Patel Road,
New Delhi-33.
Dear Sir,

In reply to your application for the post of a P.A. dated May 26, 20 …………. I should be glad if you call at my office on Monday morning next so that I may discuss details concerning the post for which you have applied.

The initial salary offered is Rs 12400/- per month, the engagement being terminable at one month’s notice on either side.

Yours faithfully,
S.M. Kakkar

2. Placing Orders

7. Placing Order for Steel Furniture.

Messrs Raja Sodhi &C Co.,
(Steel Furniture Dealers)
Prasad Road,
Ludhiana.
October 3, 20 ……….

Ref. No. 602/…. RS
The Manager,
Messrs Mahesh & Co.,
Rani Jhansi Road,
New Delhi-11.
Dear Sir,

We are in receipt of your quotation October, 1, 20… and thank you for the same. Please arrange to supply us the following goods as early as possible:
12 Nos…. Steel Almirah, size 84⊄⊄ at Rs 700 Per Piece.
12 Nos…. Manager’s Table, size 60⊄⊄ at Rs 400 Per Piece.
Sales and any other tax, if required, will be paid extra.

As desired by you, we are enclosing herewith our cheque No. STB/610073 dated October 3, 20….. for Rs 10,000/- only as advance and balance will be paid as soon as the goods are delivered.

In this connection, please note that if you make your terms liberal, we confirm, we shall be able to give you a good turnover per year. Please see what you can do.
Thanking you,

Yours faithfully,
Ram Chand
Manager

Enel. : One cheque.
No. : STB/610073.
Date : October 3, 20 …..
Amount : Rs 10,000/- only.
Back : State Bank of India, New Delhi-1.

8. Reply to above.

Messrs Mahesh & Co.,
Rani Jhansi Road,
New Delhi-11.
October 10, 20 …….

The Manager,
Messrs Raja Sodhi & Co.
(Steel Furniture Dealers)
Prasad Road,
Ludhiana.
Dear Sir,

We thank you for your order of the 3rd October, 20… and are pleased to state that the goods required have been despatched by passenger train this morning.

We trust that the goods will arrive in excellent condition and their quality will induce your good self to send further orders.

Yours faithfully,
K. Mahesh,
Manager

Enel. : Invoice
R/R No. : Nd/16003
Dated : October 10, 20….

9. Placing Order for Books.

R. Chand & Co.,
(Booksellers)
College Road,
Patiala.
July 18, 20 ……..

Messrs Malhotra Publishers,
Railway Road,
Jalandhar City.
Dear Sirs,

We are in receipt of your letter No. B-605/78 K dated 8th July, 20 alongwith a list of your publications. We thank you for the same.
Please supply us immediately:
40 Copies AS YOU LIKE IT by Prof. Ashok @ Rs 19/- each.
40 Copies MBD English Guide, TDC, III @ Rs 45/- each.
30 Copies MBD Economics Refresher B.Com. II, @ Rs 26/- each.
40 Copies MBD History Refresher TDC 11 @ Rs 12/- each 25 Copies
MBD Refresher English Guide, B.Com. 1, @ Rs 16/- each.

As desired by you, we are enclosing herewith our cheque No. SKB/K 10032 dated July 18, 20…… for Rs 1500/- only as advance and the balance will be paid against delivery.

Our next order will be forwarded as and when necessary.

Yours faithfully,
Ravinder Ghai,
Proprietor

Enel. : One cheque.
No. : SKB/K10032.
Date : July 18, 20 …….
Amount : Rs 1500/-
Bank : State Bank of India,
Patiala.

10. Acknowledging a Big Order.

Messrs Malhotra Publishers,
Railway Road,
Jalandhar City.
July 12, 20……

Ref. : MBD/MA-16 TR.
Messrs Bose & Co.,
University Road,
New Delhi-16.
Dear Sirs,

We thank you for your order No. 818. dated July 6, 20 ….. for 300 copies of MBD ENGLISH LITERATURE SERIES, “AS YOU LIKE IT” by Prof. Ashok @ Rs 90/- each copy. The order has been booked for execution and the books will be despatched by goods train on 14th July.

This is the first occasion we have the pleasure to execute your order. We heartily welcome you to our list of will-wishers and patrons. We assure you that you will always find our books satisfactory and hope this order will lead to an enduring connection with you.

Yours faithfully,
Balwant Sharma
Manager

11. Placing Order for Sports Goods.

“SPORTS CORNER”
14, Lodhi Market,
New Delhi-21.
July 6, 20 ……

Ref. No. Sc/506.
Messrs Kohli & Co.,
Sports Goods Manufacturers,
Industrial Estate,
Jalandhar City.
Dear Sirs,

I am in receipt of your quotation dated July 2, 20 …… and thank you for the same. Please supply us immediately the following items:
6 dozen Cricket balls @ Rs 120/- per dozen.
20 Cricket Bats (Durrani) @ Rs 75/- per bat.
50 pairs of Batting Gloves @ Rs 10/- per pair,
8 dozen Hockey balls (Ashok) @ Rs 100/- per dozen.
5 dozen Hockey Sticks (Champion) @ Rs 400/- per dozen.
Sales and any other tax, if leviable, will be paid extra.

As desired by you. I am enclosing herewith my cheque No. PB/c 70078 dated July 6, 20… for Rs 2,500/- only as advance and balance will be paid against delivery.
Thanking you,

Yours faithfully,
J.L. Mehta
Manager

Enel. : One cheque.
No. : PB/c 70078
Date : July 6, 20…
Amount : Rs 2,500/- only.
Bank : Punjab National Bank, New Delhi-1.

12. Reply to Letter No. 11.

Messrs Kohli & Co.,
Sports Goods Manufacturers,
Industrial Estate,
Jalandhar City.
July 9, 20 …….

The Manager,
“Sports Corner”,
14, Lodhi Market,
New Delhi-21.
Dear Sirs,

We are indebted to you for your order of the 6th July, 20 ……. and pleased to state that the goods required have been despatched by passenger train this morning.

We trust that the goods will arrive in excellent condition and their quality will induce you to send further orders.

Yours faithfully,
Satish Kohli
Sales Manager

Enel. : Invoice
R/R No. : JK/622532
Date : July 2, 20 ……..

13. Placing order for Servicing of Typewriting Machines.

Michael Limited,
(Commission Agents & Gen. Order Suppliers)
150, Park Road,
Kolkata-500025.
August 12, 20 ………

Ref. No. 1832/ML-630
Messrs Baliram & Co.
18, Church Lane,
New Delhi-32.
Dear Sirs,

Thank you for your quotation dated August 8, 20… for servicing of our different Typewriting Machines.

We do hereby accept your quotation under reference and request you to start work from the next week.

Regarding payment, though you did not mention anything in your quotation, we propose to make payments bi-monthly (i.e. 6 times in a year). Hence you are to submit your bill after completion of servicing of every two months.

Thanking you,

Yours faithfully,
R.S. Singh Purchase
Manager

14. Asking for details of Order.

Renu Hosiery Goods,
Industrial Estate,
Jalandhar City.
October 6, 20 …….

Messrs Dimple Readymade Garments,
106, Main Market,
New Delhi-13.
Dear Sirs,

We thank you for your Order No. 512 October 4, 20… for 5 dozen All-wool Pullovers, Catalogue No. 12 A, @ Rs 80/- each. We should like to inform you that you did not write the specifications in regard to size and colour. Surely, you would like to make your own choice and do not want us to send you size and colour of our selection.

We shall be glad if you kindly let us know your choice.

Yours faithfully,
V.K. Nanda
Sales Manager

3. Complaints And Adjustments

15. Complaint for Non-execution of an Order.

Upkar Traders,
6-Gurudwara Road,
Amritsar.
December 10, 20 …….

Ref. No. UT/103/………
Messrs Aggarwal & Sons,
15/162-E, Rampur Road,
Agra.
Dear Sirs,

Please refer to our order No. UT/503/ ………. dated November 2, 20 ……… for 200 Tins of Ghee (Lotus Brand).

We note with regret that the goods ordered were not delivered till this day, in spite of the fact that a prompt delivery was guaranteed and the order obtained on the strength of guarantee. The reasons are, however, beyond our knowledge.

This is not the first time a delay in delivery has occurred. Any way, we shall like to point out that business on these conditions can’t be continued for long.

We hope that this letter will cause you to settle the problem finally. Your immediate reply is expected.

Thanking you,

Yours faithfully,
Rajinder K. Singh
Purchase Officer

16. A Negative Reply to Letter No. 15.

Messrs Aggarwal & Sons,
15/162-E, Rampura Road,
Agra.
December 13, 20 ……..

Ref. No. A/s-603213-G,
The Purchase Officer,
Upkar Traders,
6-Gurudwara Road,
Amritsar.
Dear Sir,

We are in receipt of your letter No. UT/1003/ ……….. dated December 10, 20 ……….. and thank you for the same.

Please note that since there are some technical difficulties in our factory, we are not in a position to execute the order this time. We request you to procure the same from some other source.

We hope that you will realise that it is due to circumstances beyond our control. We do sincerely hope that you will not mind anything. We shall inform you as soon as the position becomes normal.

We are extremely sorry for this but are helpless. Thanking you in anticipation and expecting your best cooperation always.

Yours faithfully,
R.N. Aggarwal
Proprietor

17. An Affirmative Reply.

Messrs Aggarwal & Sons,
15/162-E, Rampura Road,
Agra.
December 13, 20 ………

Ref. No. A/s-603213-G,
The Purchase Officer,
Upkar Traders,
6-Gurudwara Road,
Amritsar.
Dear Sir,

We are in receipt of your letter No. UT/1003/ ……….. dated December 10,20 …………. and thank you for the same.

We are really sorry for the delay in executing your order due to a strike by the workers in our factory. Now the things are settled and we shall send the goods within two’ or three days.

We hope you will realise that the delay is due to circumstances beyond our control. We shall try our best to execute all your future orders in time.

Yours faithfully,
R.H. Aggarwal
Proprietor

18. Complaint for Express Bill.

District Sports Officer,
Govinda Stadium,
Amritsar.
July 10, 20……….

Ref. No. D50/6-CR.
M/s R.P. Kalra & Co.,
Sports Goods Manufacturers,
16, Industrial Estate,
Jalandhar City.
Dear Sir,

We are in receipt of the hockey sticks sent under cover of your invoice No. RPC/340 ……… dated June, 8, 20 ………. and thank you for the same.

In this connection, kindly note that in your invoice, you have charged Rs 26/- per stick while you quoted Rs 24/- per stick.

Please rectify the mistake and either send us the excess amount charged by M.O. or issue us a C.N. for the same. Prompt action is expected.

Yours sincerely,
Swami Talwar
D.S.O.

19. Reply to Above.

M/S R.P. Kalra & Co.,
Sports Goods Manufacturers,
16, Industrial Estate,
Jalandhar city.
July 13, 20 ……..

Ref. No. RPK/32-
The District Sports Officer,
Govind Stadium,
Amritsar.
Dear Sir,

We are in receipt of your letter No. DSO/6-CR dated July 10, 20…. and note the contents with thanks. The mistake took place due to the inadvertence of our billing clerk for which we are extremely sorry.

Enclosed please find our Credit Note No. RPK/6-88 dated July 13, 20…….. for the excess amount which may be adjusted with your next order.
Thanking you in anticipation and assuring you of our best co-operation and service. We remain,

Yours faithfully,
K.L. Sharma
Sales Manager.

20. Complaint for sending wrong books.

“Book-Shop”
University Campus,
Patiala.
July 26, 20…….

Ref. No. BS/102 ……….
Messrs Manjit Bros.,
Educational Publishers,
Jalandhar City.
Dear Sirs,

I am in receipt of the parcel sent under cover of your invoice No. MB 4078 dated July 24, 20…….. (for Rs 5200/-) and thank you for the same.

But on opening the parcel, I find that you have sent us the Punjabi Edition intead of Hindi Edition.

I presume due to the inadvertence of your packer, this mistake has taken place. I am sending the parcel back. I request you to replace the same immediately. I do hope that you shall do the needful as soon as my letter reaches you.

Yours faithfully,
K.C. Sachdeva
Manager

21. Reply to Above Letter of Complaint.

Messrs Manjit Bros.,
Educational Publishers,
Jalandhar City.
July 28, 20…….

Ref. No. MB/607……
The Manager,
“Book-Shop”
University Campus,
Patiala.
Dear Sir,
We are in receipt of your letter No. BS/102 ……….. dated July 26,20 …… and noted the contents with thanks.

We are sending you the Hindi Edition by the next available Passenger Train. On receipt of the parcel, please do drop a line to us.

We are extremely sorry for the inconvenience caused to you due to this. We assure you that necessary care will be taken at the time of execution of all your future orders. Please never mind for this.

Thanking you in anticipation and assuring you of our best services always.

Yours faithfully,
D.K. Joshi
(Partner)
For Manjit Bros.

22. Complaint for Defective Goods.

Rehman & Sons,
Shawl Dealers,
Manik Road,
Amritsar.
November 10, 20 ……..

The Manager,
Messrs Bal Chand Lai Chand,
The Mall,
Jalandhar.
Dear Sir,

We thank you for the prompt execution of our Order No. RS/265-D, dated October 21, 20 ……

Unfortunately, on opening the case, we find twenty shawls torn which are unsaleable. There are six shawls whose finish is decidedly bad. In no way they appear like high-priced goods, a fact which is very much against our business ethics.

We are unable to offer such shawls to our customers and therefore returning them in the hope that you will let us have goods of a better quality.

Yours faithfully,
Kadir Rahman
(Partner)

23. Reply to Above.

Messrs Bal Chand Lai Chand,
The Mall,
Jalandhar.
November 18, 20 ……..

Ref. No. BCLC/……….
Messrs Rehman & Sons,
Shawl Dealers,
Manik Road,
Amritsar.
Dear Sirs,

We are in receipt of your letter dated November 10, 20……. complaining about the defective goods supplied by us. We are extremely sorry that we have not been able to supply goods to your satisfaction. Generally, our men thoroughly check the goods before they are despatched to our clients, but as some of our shawls have been found defective and inferior, we are ready to replace them at our cost.

We shall forward you a fresh supply of shawls within a few days and we trust that these will be to your full satisfaction.

Yours faithfully,
H.D. Sharma
Manager

24. You recently travelled on an old public transport bus which broke down a couple of times on the way. Write a letter to the manager of the company giving details of the poor condition of the bus and the harassment that it caused to the passengers. Also make a few suggestions to improve the service

To
The Manager,
Malhi Bus Service Private Limited,
Jalandhar Bus Stop.
Jalandhar.
Sir,

I wish to bring to your kind notice that I had a very bitter experience while travelling by one of the buses of your company BTK 2534 plying between Batala and Jalandhar on 12th January, 2012, I was in a hurry to reach Jalandhar. After buying a ticket for my journey, I boarded the bus. The bus left Batala at ten o’clock. As I took my seat, I found the seat cover torn. Some of the windows had no panes. The floor or deck of the bus was in a wretched shape. The iron plates were broken and twisted. Technically it could be called more of scrap than a roadworthy bus. It appeared to be a very aged bus.

After leaving Batala, the bus began to slow down after covering about two kilometres. Then it broke down and stopped near Achal Sahib. The driver and the conductor tried to repair it but it showed no sign of getting re-started. I got the impression that the bus was not at all roadworthy even when it was going to start from the Batala bus stop. Probably the conductor and the driver were in league with each other to dupe the passengers. After struggling to start, the bus showed signs of moving. But it began to run at a slower speed. All the passengers felt uncertain about the bus reaching the destination. We had already wasted 45 minutes at Achal Sahib.

Thereafter the bus came to a grinding halt at Baba Bakala. All the passengers got down. They asked the conductor to return their fare as they wanted to take other buses to reach Jalandhar. The conductor refused to refund the fare. He wanted the passengers to wait for the bus to re-start. Another forty-five minutes were wasted. Someone brought a mechanic.

He tried his level best to start the bus. The bus moved at a very slow speed. It broke down again at Beas. We had a similar experience at Subhanpur. It was 4.30 p.m.. by the time we reached Jalandhar. A journey of one and a half hours was completed within six hours.

It is a reflection on your competence as the manager of the company. You are requested to remove such a useless bus from plying. It is nothing short of deliberate cheating. You should ask the driver and the conductor why they sent such a bus to transport passengers. The public expects a proper value of their money.

Yours truly,
Gurtej Singh Dhillon
and ten others.

4. Inquiries, Quotations & Replies

25. Letter of Inquiry concerning the Status of the Firm.

Raman Manufacturing Co. Ltd.
Industrial Estate,
Jalandhar City.
December 4, 20……

Mr. D.R Azad,
5/31, Nehru Road,
Delhi.
Dear Sir,

We have been requested by M/s Singh Bros., Civil Lines, Delhi, to forward twenty pieces of Steel Almirahs for which they propose to pay by means of a three months’ bill. As the amount involved is more than Rs 10,000/-, we are naturally a bit reluctant to allow credit without some assurance as to their financial standing.

We shall feel obliged if you give us the detailed information regarding their financial status and reputation. Please do inform us whether you consider it proper to grant the said credit to them.

We thank you in advance for any other information which you can give us.

Yours faithfully,
K.R. Mani
Sales Manager

26. A favourable Reply to the above letter.

5/31, Nehru Road,
Delhi.
December 10, 20 …….

The Sales Manager,
Messrs Raman Manufacturing Co. Ltd.
Industrial Estate,
Jalandhar City.
Dear Sir,

I am pleased to be of service to you in ascertaining the financial status and reputation of M/s Singh Bros, Civil Lines, Delhi, about whom you enquired on December 4, 20……

I am glad to inform you that the said firm enjoys the fullest respect and reputation in local commercial circles. The firm is known to be possessed of a considerable amount of capital.

There seems to be no reason for withholding credit to the extent you mention, and you should have no hesitation in doing business with the firm on the terms suggested.

Yours faithfully,
D.P. Azad

27. An Unfavourable Reply.

5/31-Nehru Road,
Delhi.
December, 10, 20……….

The Sales Manager,
Messrs Raman Manufacturing Co. Ltd.
Jalandhar City.
Dear Sir,

In reply to your inquiry of the 4th December concerning M/s Singh Bros., Civil Lines, Delhi, I could recommend a policy of caution. Though the said firm is an excellent business organisation with a wide circle of customers, yet their operation hardly warrants an allowance of credit to the extent of Rs 10,000/- You should hesitate to accept the conditions they suggest. As far as I can see, payment of half this sum i.e. Rs 5000/- in cash would be advisable.

I hope that you will recognise the importance of keeping the information strictly private.

Yours faithfully,
D.R Azad

28. Asking for Quotation of Pencils.

Pandit Brothers,
Gandhi Road,
Ludhiana.
April 10, 20 ……..

Messrs Jain Bros.,
Pencil Manufacturers,
10, Industrial Area,
New Delhi-28.
Dear Sirs,

We are very much interested to purchase various brands of pencils manufactured by you, in large quantities regularly every month.

We shall be glad if you send us the details of the pencils manufactured by you quoting prices and other terms.

We assure you that we shall be able to give you good sales provided your terms and conditions suit us.

We hope you will facilitate business by quoting us the lowest possible rates. Your immediate action will be highly appreciated.

Yours faithfully,
S.M. Pandit

29. Reply to Above.

Messrs Jain Bros.,
Pencil Manufacturers,
10, Industrial Area,
New Delhi-28.
April 14, 20 ………

Ref. No. JB/16032-D.
Mr. S.M. Pandit,
Pandit Brothers,
Gandhi Road,
Ludhiana.
Dear Sir,

We are in receipt of your letter dated April 10, 20 ………. and thank you for the same. We are sending herewith our catalogue containing quotations for quantities from our existing stocks. You will also find the terms of payment indicated therein. But packing charges will be extra.

We like to add here that we shall be very glad to offer you a special discount of 2½% over and above our usual trade terms if your monthly requirement is 200 grosses (assorted).

Please let us have your standing order as early as possible which will always receive our most careful and prompt attention.

Yours faithfully,
P.K. Jain
Managing Director