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NCERT Solutions For Class 12 Chemistry Chapter 14 Biomolecules

Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 14 Biomolecules:

Section Name	Topic Name
14	Biomolecules
14.1	Carbohydrates
14.2	Proteins
14.3	Enzymes
14.4	Vitamins
14.5	Nucleic Acids
14.6	Hormones

NCERT INTEXT QUESTIONS

14.1. Glucose or sucrose are soluble in water but cyclohexane and benzene (simple six membred ring compounds) are insoluble in water Explain.
Ans: The .solubility of a solute in a given solvent follows the rule ‘ Like dissolves like’.Glucose contains five and sucrose contains eight -OH groups. These -OH groups form H-bonds with water. As a result of this extensive intermoleeular H-bonding, glucose and sucrose are soluble in water.On the other hand, benzene and cyclohexane do not contain -OH bonds and hence do not form H-bonds with water. Moreover, they are non-polar molecules and hence do not dissolve in polar water molecules.

14.2. What are the expected products of hydrolysis of lactose?
Ans: Lactose being a disaccharide gives two molecules of monosaccharides Le. one molecule each of D-(+) – glucose and D-(+)-galactbse.

14.3. How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?
Ans: The cyclic hemiacetal form of glucose contains an -OH group at C-l which gets hydrolysed in aqueous solution to produce open chain aldehydic form which then reacts with NH₂OH -to form corresponding oxime. Thus, glucose contains an aldehydic group. However, when glucose is reacted with acetic anhydride, the -OH group at C-l along with the other -OH groups at C-2, C-3, C-4 and C-6 form a pentaacetate.
Since the penta acetate of1 glucose does not contain a free -OH group at C-l, it cannot get hydrolysed in aqueous solution to produce open chain aldehydic form and hence glucose pentaacetate does not react with NH₂OH to form glucose oxime. The reactions are shown as:

14.4. The melting points and solubility in water of a-amino acids are generally higher than those of corresponding haloacids. Explain.
Ans: a-amino acids as we all know, are dipolar in nature (\(\overset { + }{ N }\)H₃-CHR-COO^– ) and have strong dipolar interactions. As a result, these are high melting solids. These are also involved in intermolecular hydrogen bonding with the  molecules of water and are therefore, water soluble. On the contrary, the haloacids RCH(X)COOH are not dipolar like a-amino acids. Moreover, only the carboxyl group of haloacids are involved in hydrogen bonding with the molecules of water and not the halogen atoms. These have therefore, comparatively less melting points and are also soluble in water to smaller extent.

14.5. Where does the water present in the egg go after boiling the egg?
Ans: When egg is boiled, proteins first undergo denaturation and then coagulation and the water present in the egg gets absorbed in coagulated protein, probably through H- bonding

14.6. Why cannot Vitamin C be stored in our body?
Ans: Vitamin C cannot be stored in the body because it is water soluble and is, therefore, easily excreted in urine.

14.7. Which products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
Ans: Upon hydrolysis, nucleotide from DNA would form 2-deoxyribose and phosphoric acid along-with thymine.

14.8. When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?
Ans: A DNA molecule has two strands in which the four complementary bases pair each other, i.e., cytosine (C) always pair with guanine (G) while thymine (T) always pairs with adenine (A). Thus, when a DNA molecule is hydrolysed, the molar amounts of cytosine is always equal to that of guanine and that of adenine is always equal to thymine.In RNA, there is no relationship between the quantities of four bases (C, G, A and U) obtained, therefore, the base pairing principle, i.e. A pairs with U and C pairs with G is not followed. Therefore, unlike DNA, RNA has a single strand.

NCERT EXERCISES

14.1. What are monosaccharides ?
Ans: Monosaccharides are carbohydrates Which cannot be hydrolysed to smaller molecules.Their general formula is (CH₂O)n Where n=3-7 These are of two types: Those which contain an aldehyde group (-CHO) are called aldoses and those which contain a keto (C=O) group are called ketoses.
They are further classified as trioses , tetroses ,pentoses , hexoses and heptoses according as they contain 3,4,5,6, and 7 carbon atoms respectively.For example.

14.2. What are reducing sugars?
Ans: Reducing sugars are those which can act as reducing agents. They contain in them a reducing group which may be aldehydic (-CHO) or ketonic (>C=0) group. The characteristic reactions of reducing sugars are with Tollen’s reagent and Fehling solution. Non-reducing sugars donot give these reactions. For example, glucose, fructose, lactose etc. are reducing sugars. Sucrose is regarded as a non-reducing sugar because both glucose and fructose are linked through their aldehydic and ketonic groups by glycosidic linkage. Since these groups are not free, sucrose is a non-reducing sugar.

14.3. Write two main functions of carbohydrates in plants.
Ans: Two major functions of carbohydrates in plants are following
(a)Structural material for plant cell walls: The polysaccharide cellulose acts as the chief structural material of the plant cell walls.
(b)Reserve food material: The polysaccharide starch is the major reserve food material in the plants. It is stored in seeds and act as the reserve food material for the tiny plant till it is capable of making its own food by photosynthesis.

14.4. Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.
Ans: Monosaccharides: Ribose, 2-deoxyribose, galactose and fructose. Disaccharides: Maltose and lactose.

14.5. What do you understand by the term glycosidic linkage?
Ans: The ethereal or oxide linkage through which two monosaccharide units are joined together by the loss of a water molecule to form a molecule of disaccharide is called the glycosidic linkage. The glycosidic linkage in maltpse molecule is shown below:

14.6. What is glycogen? How is it different from starch?
Ans: The carbohydrates are stored in animal body as glycogen. It is also called animal starch and its structure is similar toamylopectin which means that it is a branched chain polymer of α-D-glucose units in which the chain is formed by C₁ – C₄ glycosidic linkage whereas branching occurs by the formation of C₁– C₆ glycosidic linkage. One main difference between glycogen and amylopectin is the length of the chain. In amylopectin, the chain consists of 20 – 25 α – D – glucose molecules whereas in glycogen, there are 10 -14 molecules of α – D – glucose present. Glycogen is more branched than amylopectin. It is present mainly in liver, muscles and also in brain. Glycogen gets converted into glucose when the body needs it with the help of certain enzymes present in the body. Glycogen has also been found to be present in yeast and fungi.

Starch is a major source of carbohydrates which are very much essential to the human body since they supply energy to the body. It occurs as granules mainly in seeds, fruits, tubers and also in the roots of the plants. The chief commercial sources of starch are wheat, maize, rice, potatoes etc.

14.7. What are the hydrolysis products of (i) sucrose, and (ii) lactose?
Ans: Both sucrose and lactose are disaccharides. Sucrose on hydrolysis gives one molecule each of glucose and fructose but lactose on hydrolysis gives one molecule each of glucose and galactose.

14.8. What is the basic structural difference between starch and cellulose?
Ans: Starch consists of amylose and amylopectin. Amylose is a linear polymer of α-D-glucose while cellulose is a linear polymer of β -D- glucose. In amylose, C -1 of one glucose unit is connected to C – 4 of the other through α-glycosidic linkage. However in cellulose, C – 1 of one glucose unit is connected to C-4 of the other through β – glycosidic linkage. Amylopectin on the other hand has highly branched structure.

14.9. What happens when D-glucose is treated with . the following reagents.
(i) HI
(ii) Bromine water
(iii) HNO₃
Ans:

14.10. Enumerate the reactions of D-glucose which cannot be explained with open chain structure. (C.B.S.E. Delhi 2008, C.B.S.E. Sample Paper 2011)
Ans:
Open chain structure of D-glucose contains a free aldehydic group (- CHO). However, it does not give the following reactions:

• D(+) glucose does not react with 2, 4 D.N.P.
• D(+) glucose does not react with NaHSO₃.
• D(+) glucose does not restore the pink colour to Schiff’s reagent.
• Penia acetyl glucose formed by carrying acetylation with acetic anhydride does not react with hydroxyl amine
  (NH₂OH) which is the characteristic reaction of all aldehydes.
• D( +) glucose is found to exist in two different crystalline forms which are named as α and β. Both these forms have actually been isolated. For example, α form with m.p. 419 K is obtained by the crystallisation of the saturated solution of glucose prepared at 303 K. Similarly, β-form with m.p. 423 K is isolated by carrying out the crystallisation of the saturated solution of glucose prepared at 371 K. Apart from that the a-form has a specific rotation (α) equal to + 112° while the β- form has specific rotation (α) equal to + 19°.

In the light of the limitations stated above, Tollen stated that an open chain structure for D(+) glucose is probably not practicable. He proposed a cyclic structure which is a hemiacetal structure. In this structure, the aldehydic (CHO) group
is involved in the form of a ring with the -OH group attached to C₅ carbon. It is a six membered ring, often called ô-
oxide ring. The ring structure accounts for the two isomeric forms a and shown below.

14.11. What are essential and non-essential amino acids? Give two examples of each type.
Ans: α-Amino acids which are needed for good health and proper growth of human beings but are not synthesized by the human body are called- essential amino acids. For example, valine, leucine, phenylalanine, etc. On the other hand, α-amino acids which are needed for health and growth of human beings and are synthesized by the human body are called non-essential amino acids. For example, glycine, alanine, aspartic acid etc.

14.12. Define the following as related to proteins:
(i) Peptide linkage
(ii) Primary structure
(iii) Denaturation
Ans: (i) Peptide bond: Proteins are condensation polymers of α-amino acids in which the same or different α-amino acids are joined by peptide bonds. Chemically, a peptide bond is an amide linkage formed between – COOH group of one α-amino acid and -NH-, group of the other α-amino acid by lo;ss of a molecule of water. For example,

(ii) Primary structure: Proteins may contain one or more polypeptide chains. Each . polypeptide chain has a large number of α-amino acids which are linked to one another in a specific manner. The specific sequence in which the various amino acids present in a protein linked to one another is called its primary structure. Any change in the sequence of α-amino acids creates a different protein.

(iii) Denaturation: Each protein in the biological system has a unique three-dimensional structure and has specific biologicalactivity. This is called native form of a protein. When a protein in its native form is subjected to a physical change such as change in temperature or a chemical change like change in pH, etc., hydrogen bonds gets broken. As a result, soluble forms of proteins such as globular proteins undergo coagulation or precipitation to give fibrous proteins which are insoluble in water. This coagulation also results in loss of biological activity of the proteins and this loss in biological activity, is called denaturation. During denaturation, 2° and 3° structures of proteins are destroyed but 1° structure remains intact.
The most common example of denaturation of proteins is the coagulation of albumin present in the white of an egg. When the egg is boiled hard, the soluble globular protein present in it is denatured and is converted into insoluble fibrous protein.

14.13. What are the common types of secondary structure of proteins?
Ans:
Secondary structure of protein refers to the shape in which a long polypeptide chain can exist. These are found to exist in two types :

• α-helix structure
• β-pleated sheet structure.

Secondary Structure of Proteins:
The long, flexible peptide chains of proteins are folded into the relatively rigid regular conformations called the
secondary structure. It refers to the conformation which the polvpeptide chains assume as a result of hydrogen bonding
between the > C= O and > N-H groups of different peptide bonds.
The type of secondary structure a protein will acquire, in general depends upon the size of the R-group. If the size of the
R-groups are quite large, the protein will acquire ct-helix structure. If on the other hand, the size of the R-groups are relatively
smaller, the protein will acquire a β – flat sheet structure.

(a) α-Helix structure: If the size of the R-groups are quite large, the hydrogen bonding occurs between > C = O group
of one amino acid unit and the > N-H group of the fourth amino acid unit within the same chain. As such the polypeptide
chain coils up into a spiral structure called right handed ct- helix structure. This type of structure is adopted by most of the
fibrous structural proteins such as those present in wool, hair and muscles. These proteins are elastic i.e., they can be
stretched. During this process, the weak hydrogen bonds causing the a – helix are broken. This tends to increase the length of
the helix like a spring. On releasing the tension, the hydrogen bonds are reformed, giving back the original helical shape.

(b) β—Flat sheet or β—Pleated sheet structure: If R-groups are relatively small, the peptide chains lie side by side in a zig
zag manner with alternate R-groups on the same side situated at fixed distances apart. The two such neighbouring chains are held together by intermolecular hydrogen bonds. A number of such chains can be inter-bonded and this results in the formation of a flat sheet structure These chains may contract or bend a little in order to accommodate moderate sized R-groups. As a result, the sheet bends into parallel folds to form pleated sheet structure known as β – pleated sheet structure. These sheets are then stacked one above the other like the pages of a book to form a three dimensional structure. The protein fibrion in silk fibre has a β – pleated sheet structure. The characteristic mechanical properties of silk can easily be explained on the basis of its β – sheet structure. For example, silk is non-elastic since stretching leads to pulling the peptide covalent bonds. On the other hand, it can be bent easily like a stack of pages because during this process, the sheets slide over each other.

14.14. What types of bonding helps in stabilising the α-helix structure of proteins?
Ans: α-helix structure of proteins is stabilised through hydrogen bonding. (a) α -Helix structure. If the size of the R-groups are quite large, the hydrogen bonding occurs between > C = O group of one amino acid unit and the > N- H group of the fourth amino acid unit within the same chain. As such the polypeptide chain coils up into a spiral structure called right handed a—helix structure. This type of structure is adopted by most of the fibrous structural proteins such as those present in wool, hair and muscles. These proteins are elastic i.e., they can be stretched. During this process, the weak hydrogen bonds causing the α-helix are broken. This tends to increase the length of the helix like a spring. On releasing the tension, the hydrogen bonds are reformed, giving back the original helical shape.

14.15: Differentiate between globular and fibrous proteins.
Ans. (i) Fibrous proteins: These proteins consist of linear thread like molecules which tend to lie side by side (parallel) to form fibres. The polypeptide chains in them are held together usually at many points by hydrogen bonds and some disulphide bonds. As a result,intermolecular forces of attraction are very’ strong and hence fibrous proteins are insoluble in water. Further, these proteins are stable to moderate changes in temperature and pH. Fibrous proteins serve as the chief structural material of animal tissues.For example, keratin in skin, hair, nails and wool, collagen in tendons, fibrosis in silk and myosin in muscles.
(ii) Globular proteins: The polypeptide chain in these proteins is folded around itself in such a way so as to give the entire protein molecule an almost spheroidal shape. The folding takes place in such a manner that hydrophobic (non-polar) parts are pushed inwards and hydrophilic (polar) parts are pushed outwards. As a result, water molecules interact strongly with the polar groups and hence globular protein are water soluble. As compared to fibrous proteins, these are very sensitive to small changes of temperature and pH. This class of proteins include all enzymes, many hormones such as insulin from pancreas, thyroglobulin from thyroid gland, etc.

14.16. How do you explain the amphoteric behaviour of amino acids?
Ans: Amino acids contain an acidic (carboxyl group) and basic (amino group) group in the same molecule. In aqueous solution, they neutralize each other. The carboxyl group loses a proton while the amino group accepts it. As a result, a dipolar or zwitter ion is formed.

In zwitter ionjc form, a-amino acid show amphoteric behaviour as they react with both acids and bases.

14.17. What are enzymes?
Ans: Enzymes are biological catalyst. Each biological reaction requires a different enzyme. Thus, as compared to conventional catalyst enzymes are very specific and efficient in their action. Each type of enzyme has its own specific optimum conditions of concentration, pH and temperature at which it works best.

14.18. What is the effect of denaturation on the structure of proteins?
Ans: Denaturation of proteins is done either by change in temperature (upon heating) or by bringing a change in the pH of the medium. As a result, the hydrogen bonding is disturbed and the proteins lose their biological activity i.e., their nature changes. During the denaturation, both the tertiary and secondary structures of proteins are destroyed while the primary structures remain intact.

14.19. How are vitamins classified? Name the vitamin responsible for the coagulation of blood.
Ans: Vitamins are classified into two groups depending upon their solubility in water or fat: (i) Water soluble vitamins: These include vitamin B-complex (B₁, B₂, B₅, i.e., nicotinic acid,B₆, B₁₂, pantothenic acid, biotin, i.e., vitamin H and folic acid) and vitamin C.
(ii) Fat soluble vitamins: These include vitamins A, D, E and K. They are stored in liver and adipose (fat storing) tissues. Vitamin K is responsible for coagulation of blood.

14.20. Why are vitamin A and vitamin C essential to us? Give their important sources.
Ans: Vitamin A is essential for us because its deficiency causes xerophthalmia (hardening of cornea of eye) and night blindness.
Sources: Fish liver oil, carrots, butter, milk, etc. Vitamin C is essential for us because its deficiency causes scurvy (bleeding of gums) and pyorrhea (loosening and bleeding of teeth). Sources: Citrous fruits, amla, green leafy vegetables etc.

14.21. What are nucleic acids ? Mention their two important functions.
Ans: Nucleic acids are biomolecules which are found in the nuclei of all living cell in form of nucleoproteins or chromosomes (proteins contains nucleic acids as the prosthetic group).

Nucleic acids are of two types: deoxyribonucleic acid (DNA) and ribonucleic acid.(RNA).
The two main functions of nucleic acids are:
(a) DNA is responsible for transmission of hereditary effects from one generation to another. This is due to its unique property of replication, during cell division and two identical DNA strands are transferred to the daughter cells.
(b) DNA and RNA are responsible for synthesis of all proteins needed for the growth and maintenance of our body. Actually the proteins are synthesized by various RNA molecules (r-RNA, m-RNA) and t-RNA) in the cell but the message for the synthesis of a particular protein is coded in DNA.

14.22. What is the difference between a nucleoside and a nucleotide?
Ans: A nucleoside contains only two basic components of nucleic acids i.e., a pentose sugar and a nitrogenous base. It is formed when 1- position of pyrimidine (cytosine, thiamine or uracil) or 9-position of purine (guanine or adenine) base is attached to C -1 of sugar (ribose or deoxyribose) by a β-linkage. Nucleic acids are also called polynucleotides since the repeating structural unit of nucleic acids is a nucleotide.
A nucleotide contains all the three basic . components of nucleic acids, i.e., a phosphoric acid group, a pentose sugar and a nitrogenous base. These are obtained by esterification of C₅, – OH group of the pentose sugar by phosphoric acid.

14.23. The two strands in DNA are not identical but are complementary. Explain.
Ans: The two strands in DNA molecule are held together by hydrogen bonds between purine base of one strand and pyrimidine base of the other and vice versa. Because of different sizes and geometries of the bases, the only possible pairing in DNA are G (guanine) and C (cytosine) through three H-bonds, (i.e.,C = G) and between A (adenine) and T (thiamine) through two H-bonds (i.e., A = T). Due to this base -pairing principle, the sequence of bases in one strand automatically fixes the sequence of bases in the other strand. Thus, the two strands are complimentary and not identical.

14.24. Write the important structural and functional differences between DNA and RNA.
Ans:

14.25. What are the different types of RNA found in the cell?
Ans: There are three types of RNA:
(a) Ribosomal RNA (r RNA)
(b) Messenger RNA (m RNA)
(c) Transfer RNA (t RNA)

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NCERT Solutions For Class 12 Chemistry Chapter 16 Chemistry in Everyday Life

Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 16 Chemistry in Everyday Life:

Section Name	Topic Name
16	Chemistry in Everyday Life
16.1	Drugs and their Classification
16.2	Drug-Target Interaction
16.3	Therapeutic Action of Different Classes of Drugs
16.4	Chemicals in Food
16.5	Cleansing Agents

NCERT INTEXT QUESTIONS

16.1. Sleeping pills are recommended by doctors to the patients suffering from sleeplessness but it is not advisable to take its doses without consultation with the doctor. Why?
Ans: Most of drugs taken in doses higher than recommended may produce harmful effects and act as poison and cause even death. Therefore, a doctor must always be consulted before taking the drug.

16.2. “Ranitidine is an antacid” With reference to which classification, has this statement been given?
Ans: Ranitidine is labelled as antacid since it is quite effective in neutralising the excess of acidity in the stomach. It is sold in the market under the trade name Zintac.

16.3. Why do we require artificial sweetening agents?
Ans: To reduce calorie intake and to protect teeth from decaying, we need artificial sweeteners.

16.4. Write the chemical equation for preparing sodium soap from glyceryl oleate and glyceryl palmitate. Structures of these compounds are given below:
(i) (C₁₅H₃₁COO)₃C₃H₅-Glyceryl palmitate
(ii) (C₁₇H₃₂COO)₃C₃H₅-Glyceryl oleate
Ans:

16.5. Following type of non-ionic detergents are present in liquid detergents, emulsifying agents and wetting agents. Label the hydrophilic and hydrophobic part in the molecule. Identify the functional group (s) present in the molecule.

Ans:

Functional groups present in the detergent molecule are:
(i)ether
(ii)1°alcoholic group

NCERT EXERCISES

16.1. Why do we need to classify drugs in different ways?
Ans: Drugs are classified in following different ways:
(a) Based on pharmacological effect.
(b) Based on action on a particular biochemical process.
(c) Based on chemical structure.
(d) Based on molecular targets.
Each classification has its own usefulness.
(а) Classification based on pharmacological effect is useful for doctors because it provides them the whole range of drugs available for the treatment of a particular disease.
(b) Classification based on action on a particular biochemical proc*ess is useful for choosing the correct compound for designing the synthesis of a desired drug.
(c) Classification based on chemical structure helps us to design the synthesis of a number of structurally similar compounds having different substituents and then choosing the drug having least toxicity.
(d) Classification on the basis of molecular targets is useful for medical chemists so that they can design a drug which is most effective for a particular receptor site.

16.2. Explain the following as used in medicinal chemistry
(a) Lead compounds
(b) Target molecules or drug targets.
Ans:
(a) Lead compounds are the compounds which are effective in different drugs. They have specific chemical formulas and may be extracted either from natural sources (plants and animals) or may be synthesised in the laboratory.

(b) Target molecules or drug targets. An enzyme (E) functions by combining with the reactant (called substrate) denoted as ‘S’ to form an activated complex known as enzyme-substrate complex (E-S). The complex dissociates to form product and releases the enzyme for carrying out further activity.

16.3. Name the macro molecules that are chosen as drug targets.
Ans: Proteins, carbohydrates, lipids and nucleic acids are chosen as drug targets.

16.4. Why the medicines should not be taken without consulting doctors?
Ans: No doubt medicines are panacea for most of the body ailments. But their wrong choice and overdose can cause havoc and may even prove to be fatal. Therefore, it is of utmost importance that the medicines should not be given without consulting doctors.

16.5. Define the term chemotherapy.
Ans: It is the branch of chemistry that deals with the treatment of diseases by using chemicals as medicines.

16.6. Which forces are involved in holding the drugs to the active site of enzymes?
Ans: The following forces are involved in holding the drugs to the active site of enzymes:
(a) Hydrogen bonding
(b) Ionic bonding
(c) Dipole-dipole interactions
(d) van der Waals interactions

16.7. Antacids and antiallergic drugs interfere with the function of histamines but do not interfere with the function of each other. Explain.
Ans: They donot interfere with the functioning of each other because they work on different receptors in the body.Histamine stimulates the secretion of pepsin and hydrochloric acid in the stomach. The drug cimetidine (antacid) was designed to prevent the interaction of histamine with the receptors present in the stomach wall. This resulted in release of lesser amount of acid. Antacid and antiallergic drugs work on different receptors.

16.8. Low level of noradrenaline is the cause of depression. What type of drugs are needed to cure this problem? Name two drugs.
Ans: In case of low level of neurotransmitter, . noradrenaline, tranquilizer (antidepressant) drugs are required because low levels of noradrenaline leads to depression. These drugs inhibit the enzymes which catalyse the degradation of noradrenaline. If the enzyme is inhibited, noradrenaline is slowly metabolized and can activate its receptor for longer periods of time thereby reducing depression. Two important drugs are iproniazid and phenylzine.

16.9. What is meant by the term broad spectrum antibiotics? Explain.
Ans: Broad spectrum antibiotics are effective against several different types or wide range of harmful bacteria. For example, tetracycline, chloramphenicol and of loxacin. Chloramphenicol can be used in case of typhoid, dysentry, acute fever, urinary infections, meningitis and pneumonia.

16.10. How do antiseptics differ from disinfectants ? Give one example of each.
Ans: Many times, the same substance can act as an antiseptic as well as disinfectant by changing the concentration of the solution used. For example, a 0.2 per cent solution of phenol acts as an antiseptic and its 1 percent solution is a disinfectant. Chlorine is used in India for making water fit for drinking at a concentration of 0.2 to 0.4 ppm (parts per million). Low concentration of sulphur dioxide is used for sterilizing squashes for preservation. A few points of distinction between antiseptics and disinfectants are listed.

Antiseptics	Disinfectants
1. Can kill or prevent the growth of micro-organisms.	1.Can kill micro-organisms.
2. Do not harm the living tissues. Therefore, these can be applied to the skin.	2. Toxic to the living tissues. Therefore, these cannot be applied to the skin.
3. These are used for the dressing of wounds, ulcers and in the treatment of diseased skin.	3. These are used for disinfecting floors, toilets, drains, instruments etc.

16.11. Why are cimetidine and ranitidine better antacids than sodium hydrogencarbonate or magnesium or aluminium hydroxide?
Ans: If excess of NaHCO₃ or Mg(OH)₂ or Al(OH)₃ is used, it makes the stomach alkaline and thus triggers the release of even more HCl which may cause ulcer in the stomach. In contrast, cimetidine and ranitidine prevent the interaction of histamine with the receptor cells in the stomach wall and thus release of HCl will be less as histamine stimulates the secretion of acid.

16.12. Name a substance which can be used as an antiseptic as well as disinfectant.
Ans: 0.2% solution of phenol acts as antiseptic while 1% solution acts as a disinfectant.

16.13. What are the main constituents of dettol?
Ans: Chloroxylenol .and α-terpineol in a suitable solvent.

16.14. What is tincture of iodine? What is its use?
Ans: 2-3% solution of iodine in alcohol and water is called tincture of iodine. It is a powerful antiseptic. It is applied on wounds.

16.15. What are food preservatives?
Ans: Preservation has a major role in the food products. Chemically preserved squashes and crushes can be kept for a fairly long time even after opening the seal of bottle.
A preservative may be defined as the substance which is capable of inhibiting or arresting the process of fermentation, acidification or any other decomposition of food. Salting i.e. addition of table salt is a well known method for food preservation and was applied in ancient times for preserving raw mangoes, tamarind, meat, fish etc. Sugar syrup can also act as a preservative. Vinegar is a useful preservative for pickles. Apart from these, sulphur dioxide and benzoic acid can be employed for the preservation of food. The major source of sulphur dioxide is potassium metabisulphite (K₂S₂O₅). It is fairly stable in neutral and alkaline medium but gets decomposed by weak acids such as carbonic, citric, tartane and malic acids. Benzoic acid is used either as such or in the form of sodium benzoate. However, sulphur dioxide has a better preservative action than sodium benzoate against bacteria and moulds. It also retards the development of yeast in juice but fails to arrest their multiplication once the number has reached a high value. Sorne salts of sorbic acid and propionic acid are also being used these days for the preservation of the food.
The use of preservatives must be properly controlled as their indiscriminate use is likely to be harmful. The preservative should not be injurious to health and should be also non-irritant.

16.16. Why is the use of aspartame limited to cold foods and drinks?
Ans: This is because it decomposes at baking or cooking temperatures and hence can be used only in cold foods and drinks as an artificial sweetener

16.17. What are artificial sweetening agents? Give two examples.
Ans: Artificial sweeteners are chemical substances which are sweet in taste but do not add any calories to our body. They are excreted as such through urine. For example, saccharin, aspartame, alitame etc.

16.18. Name the sweetening agent used in the preparation of sweets for a diabetic patient.
Ans: Saccharine, aspartame or alitame may be used in the preparation of sweets for a diabetic patient.

16.19. What problem arises in using alitame as artificial sweetener?
Ans: Alitame is a high potency artificial sweetener.Therefore, it is difficult to control the sweetness of the food to which it is added.

16.20. How are synthetic detergents better than soaps?
Ans: They can be used in hard water as well as in acidic solution. The reason being that sulphonic acids and their calcium and magnesium salts are soluble in water thus they do not form curdy white precipitate with hard water but the fatty acids and their calcium and magnesium salts of soaps are insoluble. Detergents also works in slightly acidic solution due to formation of soluble alkyl hydrogen sulphates. Soaps react with acidic solution to form insoluble fatty acids.

16.21. Explain the following terms with suitable examples:
(i) cationic detergents (ii) anionic detergents and (iii) non-ionic detergents
Ans: (i) Cationic detergents: These are quaternary ammonium salts, chlorides, acetates, bromides etc containing one or more long chain alkyl groups. For example, cetyltrimethyl ammonium chloride.
(ii) Anionic detergents are called so because a large part of their molecules are anions. ‘These are of two types:
(a) Sodium alkyl sulphates: For example, sodium lauryl sulphate, C₁₁H₂₃CH₂OSO₃ Na⁺.
(b) Sodium alkylbenzenesulphonates.Vor example, sodium 4-(l-dodecyl) benzenesu Iphphonate (SDS).

(iii) Neutral or non-ionic detergents: These are esters of high molecular mass alcohols with fatty acids. These can also be obtained by treatment of long chain alcohols by with excess of ethylene oxide in presence of a base. For example, polyethylene glycol stearate,CH₃(CH₂)₁₆COO (CH₂CH₂O)₁₁ CH₂CH₂OH Polyethylene glycol stearate.

16.22. What are biodegradable and non-biodegradable detergents? Give one example of each.
Ans: Detergents having straight chain hydrocarbons are easily degraded (or decomposed) by microorganisms and hence are called biodegradable detergents while detergents containing branched hydrocarbon chains are not easily degraded by the microorganisms find hence are called non-biodegradable detergents. Consequently, non-biodegradable detergents accumulate in rivers and water ways thereby causing severe water pollution. Examples of biodegradable detergents are sodium lauryl sulphate, sodium 4-(-l-dodecyl) benzenesulphonate and sodium 4-(2-dodecyl) benzenesulphonate.
Examples of non-biodegradable detergents is sodium 4-(1, 3,5,7 – tetramethyloctyl) benzene sulphonate.

16.23. Why do soaps not work in hard water? (C.B.S.E. Outside Delhi 2009, 2011)
Ans: Soaps are water soluble sodium or potassium salts of higher fatty acids like palmitic acid (C₁₅H₃₁COOH), oleic acid (C₁₇H₃₃COOH) and stearic acid (C₁₇H₃₅COOH). Hard water contains certain calcium and magnesium salts which combine with soaps to form corresponding magnesium compounds. These being insoluble, get separated as curdy white precipitates resulting in wastage of soap.

16.24. Can you use soaps and synthetic detergents to check the hardness of water?
Ans: Soaps get precipitated as insoluble calcium and magnesium soaps in hard water but detergents do not. Therefore, soaps but not synthetic detergents can be used to check the hardness of water.

16.25. Explain the cleansing action of soaps.
Ans: Cleansing action of soaps : Soaps contain two parts, a large hydrocarbon which is a hydrophobic (water repelling) and a negative charged head, which is hydrophillic (water attracting). In solution water molecules being polar in nature, surround the ions & not the organic part of the molecule. When a soap is dissolved in water the molecules gather together as clusters, called micelles. The tails stick inwards & the head outwards. The hydrocarbon tail attaches itself to oily dirt. When water is agitated, the oily dirt tends to lift off from the dirty surface & dissociates into fragments. The solution now contains small globules of oil surrounded detergent molecules. The negatively charged heads present in water prevent the small globules from coming together and form aggregates. Thus the oily dirt is removed from the object.

16.26. If water contains dissolved calcium hydrogencarbonate, out of soaps and synthetic detergents, which one will you use for cleaning clothes?
Ans: Calcium hydrogencarbonate makes water hard. Therefore, soap cannot be used because it gets precipitated in hard water. On the other hand, a synthetic detergent does not precipitate in hard water because its calcium salt is also soluble in water. Therefore, synthetic detergents can be used for cleaning clothes in hard water.

16.27. Label the hydrophilic and hydrophobic parts in the following compounds.
(i)cCH₃(CH₂)₁₀CH₂OSO₃ ^–Na⁺
(ii) CH₃(CH₂)₁₅ -N+(CH₃)₃Br^–
(iii) CH₃(CH₂)₁₆C00(CH₂CH₂O)₁₁CH₂CH₂OH
Ans:

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NCERT Solutions For Class 12 Chemistry Chapter 8 The d and f Block Elements

Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 8 The d and f Block Elements:

Section Name	Topic Name
8	The d – and f – Block Elements
8.1	Position in the Periodic Table
8.2	Electronic Configurations of the d-Block Elements
8.3	General Properties of the Transition Elements (d-Block)
8.4	Some Important Compounds of Transition Elements
8.5	The Lanthanoids
8.6	The Actinoids
8.7	Some Applications of d – and f -Block Elements

NCERT Solutions CBSE Sample Papers ChemistryClass 12 Chemistry

NCERT IN TEXT QUESTIONS 

8.1. Silver atom has completely filled d orbitals (4d¹⁰) in its ground state. How can you say that it is a transition element?
Ans: The outer electronic configuration of Ag (Z=47) is 4d¹⁰5s¹. It shows+1 and + 2 O.S. (in AgO and AgF₂). And in + 2 O.S., the electronic configuration is d⁹ i.e, d-subshell is incompletely filled. Hence, it is a transition element.

8.2. In the series Sc(Z = 21) to (Z = 30), the enthalpy of atomisation of zinc is the lowest i.e., 126 kJ mol^-1. Why?
Ans: The enthalpy of atomisation is directly linked with the stability of the crystal lattice and also the strength of the metallic bond. In case of zinc (3d¹⁰4s² configuration), no electrons from the 3d-orbitals are involved in the formation of metallic bonds since all the orbitals are filled. However, in all other elements belonging to 3d series one or more d-electrons are involved in the metallic bonds. This means that the metallic bonds are quite weak in zinc and it has therefore, lowest enthalpy of atomisation in the 3d series.

8.3. Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?
Ans: Manganese (Z = 25) shows maximum number of O.S. This is because its outer EC is 3d⁵4s². As 3d and 4s are close in energy, it has maximum number of e^-1 s to loose or share. Hence, it shows O.S. from +2 to +7 which is the maximum number.

8.4.

Ans.

8.5. How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements?
Ans:  There is a irregularity in the IE’s of 3d-series due to alternation of energies of 4s and 3d orbitals when an e^-1 is removed. Thus, there is a reorganisation energy accompanying ionization. This results into release of exchange energy which increases as the number of e^-1 s increases in the dn configuration. Cr has low 1st IE because loss of 1 e- gives stable EC (3d⁶). Zn has very high IE because e~ has to be removed from 4s orbital of the stable configuration (3d¹⁰ 4s²) After the loss of one e^–, removal of 2nd e^–, becomes difficult. Hence, 2nd IE’s are higher and in general, increase from left to right. However, Cr and Cu show much higher values because 2nd e^– has to be removed from stable configuration of Cr⁺ (3d⁵) and Cu⁺ (3d¹⁰)

8.6. Why is the highest oxidation state of a metal exhibited by its fluoride and oxide only? (C.B.S.E. Delhi 2010)
Ans: Both fluorine and oxygen have very high electronegativity values. They can oxidise the metals to the highest oxidation state. As a result, the highest oxidation states are shown by the fluorides and oxides of the metals; transition metals in particular.

8.7.Which is a stronger reducing agent Cr²⁺ or Fe²⁺ and why?
Ans: Cr²⁺ is a stronger reducing agent than Fe²⁺. This is because E°(Cr³⁺/Cr²⁺) is negative (- 0.41V) whereas E°(Fe³⁺/Fe²⁺) is positive (+ 0.77 V). Thus, Cr²⁺ is easily oxidised to Fe³⁺ but Fe²⁺ cannot be easily oxidised to Fe³⁺.

8.8.Calculate the ‘spin only’ magnetic moment of M²⁺(aq) ion (Z = 27).
Ans:

8.9.Explain why Cu⁺ ion is not stable in aqueous solutions?
Ans: Cu⁺ (aq) is not stable, while Cu²⁺ (aq) is stable. This is becuase Δ_hydH of Cu²⁺(aq) is much higher than that of Cu+(aq) and hence it compensates for the 2nd IE of Cu. Thus, many Cu(I) compounds are unstable in aqueous solution and undergo disproportionation as follows :
2 Cu⁺ —–> Cu²⁺ + Cu

8.10. Actinoid contraction is greater from element to element than lanthanoid contraction. Why? (C.B.S.E. Sample Paper 2011, Jharkhand Board 2010)
Ans: The decrease or contraction in atomic radii, as well as ionic radii in actinoid elements (actinoid contraction), is more as compared to lanthanoid contraction because 5/ electrons have more poor shielding effect as compared to 4f electrons. Therefore, the effect of increased nuclear charge leading to contraction in size is more in case of actinoid elements.

NCERT EXERCISES

8.1. Write down the electronic configuration of (i) Cr³⁺ (ii) Pm³⁺ (iii) Cu⁺ (iv) Ce⁴⁺(v) Co²⁺ (vi) Lu²⁺(vii) Mn²⁺ (viii) Th4+.
Sol: (i) Cr³⁺ = [Ar]¹⁸3d³
(ii)Pm³⁺ = [Xe]⁵⁴ 4f⁴
(iii)Cu⁺ = [Ar]¹⁸ 3d¹⁰
(iv)Ce⁴⁺ = [Xe]⁵⁴
(v)Co²⁺ = [Ar]¹⁸ 3d⁷
(vi)Lu²⁺ = [Xe]⁵⁴ 4f¹⁴ 5d¹
(vii) Mn²⁺ = [Ar]¹⁸ 3d⁵ (viii)Th⁴⁺= [Rn]⁸⁶

8.2. Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation to their+3 state?
Sol: Electronic configuration of Mn²⁺ is 3d⁵. This is a half-filled configuration and hence stable. Therefore, third ionization enthalpy is’very high, i. e., third electron cannot be lost easily. Electronic configuration of Fe²⁺ is 3d⁶. It can lose one electron easily to achieve a stable configuration 3d⁵.

8.3. Explain briefly how+2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?
Sol: Here after losing 2 electrons from j-orbitals, the 3d-orbital gets gradually occupied with increase in atomic number. Since the number of unpaired electrons in 3d orbital increases, the stability of the cations (M²⁺) increases from Sc²⁺ to Mn²⁺.

8.4. To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.
Sol: In the first series of transition elements, the oxidation states which lead to exactly half-filled or completely filled d-orbitals are more stable. For example, Mn (Z = 25) has electronic configuration [Ar] 3d⁵ 4 s². It shows oxidation states + 2 to + 7 but Mn (II) is most stable because of half-filled configuration [Ar] 3d⁵. Similarly Sc³⁺ is more stable then Sc+ and Fe³⁺ is more stable than Fe²⁺ due to half filled it f-orbitals.

8.5. What may be the stable oxidation state of the transition element with the following delectron configurations in the ground state of their atoms: 3d³,3d⁵, 3d⁸ and 3d⁴?
Sol: (a) 3d³ 4s¹ = + 5.
(b) 3d⁵ 4s² = + 2, + 7,3d⁵ 4s¹ =+6.
(c)3d⁸4s² = + 2.
(d)3d⁴4s² = 3d⁵ 4s¹ = + 6(and + 3).

8.6. Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.
Sol: Cr₂0₇^2- and Cr0₄^2- (Group number = Oxidation state of Cr = 6).
Mn0₄^–  (Group number = Oxidation state of Mn = 7).

8.7. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?
Sol: Lanthanoid Contraction : In the lanthanoids , the electrons are getting filled in the 4f-subshell. On moving from left to right, the nuclear charge increases and this increase is expected to be compensated by the increase in the magnitude of shielding effect by the 4 f- electrons However,
the f-electrons have very poor shielding effect. Consequently, the atomic and ionic radii decrease from left to right and this is knwon as lanthanoid contraction.
Consequences of lanthanoid Contraction
(a)Separation Lanthanoids: All the lanthanoids have quite similar properties and due to this reason they are difficult to separate.
(b)Variation in basic strength of hydroxides: Due to lanthanoid contraction, size of M³⁺ ions decreases and thus there is a corresponding increase in the covalent character in M—OH bond. Thus basic character of oxides and hydroxides decreases from La(OH)₃ to Lu(OH)₃.
(c)Similarity in the atomic sizes of the elements of second and third transition series present in the same group. The difference in the value of atomic radii of Y and La is quite, large as compared to the difference in the value of Zr and Hf. This is because of the lanthanoid contraction.
(d)Variation in standard reduciton potential: Due to lanthanoid contraction there is a small but steady increase in the standard reduction potential (E°) for the reduction process.
M³⁺ (aq) + 3e^– —–> 4 M(aq)
(e)Variation in physical properties like melting point, boiling point, hardness etc.

8.8. What are the characteristics of the transition . elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?
Sol: General characteristics of transition elements.
(i)Electronic configuration – (n -1) d^1-10 ns^1-2
(ii)Metallic character – With the exceptions of Zn, Cd and Hg, they have typical metallic structures.
(iii)Atomic and ionic size-ions of same charge in a given series show progressive decrease in radius with increasing atomic number.
(iv)Oxidation state-Variable; ranging from+2 to +7.
(v)Paramagnetism – The ions with unpaired electrons are paramagnetic.
(vi)Ionisation enthalpy – Increases with increase in charge.
Formation of coloured ions – Due to presence of unpaired electrons.
(viii) Formation of complex compounds – Due to small size and high charge density of metal ions.
(ix)They possess catalj^c properties – Due to
their ability to adopt multiple oxidation states. .
(x)Formation of interstitial compounds.
(xi)Alloy formation.
They are called transition elements due to their incompletely filled d-orbitals in ground state or in any stable oxidation state and they are placed between s and p- block elements. Zn, Cd and Hg have fully filled d- orbitals in their ground state hence may not be regarded as the transition elements.

8.9. In what way are the electronic configuration of the transition elements different from non-transition elements?
Sol: Electronic configuration of transition elements : (n – 1)d^1-10 ns^1-2. Electronic configuration of non-transition elements : ns^1-2 or ns²np^1-6. From comparison, it is quite evident that the transition elements have incomplete d-orbitals (s- orbitals in some cases) while the non-transition elements have no d-orbitals present in the valence shells of their atoms. This is responsible for the difference in the characteristics of the elements belonging to these classess of elements.

8.10. What are the different oxidation states exhibited by the lanthanoids?
Sol: Lanthanides exhibits + 2, + 3 and + 4 oxidation states. The most common oxidation state of lanthanoids is +3.

8.11. Explain giving reasons:
(i)Transition metals and many of their compounds show paramagnetic behaviour.
(ii)The enthalpies of atomisation of the transition metals are high.
(iii)The transition metals generally form coloured compounds.
(iv)Transition metals and their many compounds act as good catalyst
Sol: (i) Magnetic properties: Transition elements and many of their compounds are paramagnetic, i.e., they are weakly attracted by a magnetic field. This is due to the presence of unpaired electrons in atoms, ions or molecules. The paramagnetic character increases as the number of . unpaired electrons increases. The paramagnetic character is measured in terms of magnetic moment and is given by
\(\mu =\sqrt { n(n+2) }\) where n – number of unpaired electrons.
(ii) Because of large number of unpaired electrons in d-orbitals of their atoms they have stronger interatomic intefactions and hence stronger metallic bonding between atoms resulting in higher enthalpies of atomisation.
(iii) Formation of coloured compounds (both in solid state as well as in aqueous solution) is another very common characteristics of transition metals. This is due to absorption of some radiation from visible light to cause d-d transition of electrons in transition metal atom. The d-orbitals do not have same energy and under the influence of ligands, the d-orbitals split into two sets of orbitals having different energies; transition of electrons can take place from one set of d-orbitals to another set within the same sub-shell. Such transitions are called d-d transitions. The energy difference for these d-d transitions fall in the visible region. When white light is incident on compounds of transition metals, they absorb a particular frequency and remaining colours are emitted imparting a characteristic colour to the complex. Zn²⁺ and Ti⁴⁺ salts are white because they do not absorb any radiation in visible region.
(iv)Catalytic properties: Many of transition metals and their compounds act as catalyst in variety of reactions, e.g., finely divided iron in manufacture of NH₃ by Haber’s process, V₂O₅ or Pt in manufacture of H₂S0₄ by Contact process, etc.). The catalytic activity is due to following two reasons.
(a)The ability of transition metal ion to pass ” easily from one oxidation state to another
and thus providing a new path to reaction with lower activation energy.
(b)The surface of transition metal acts as very good adsorbent and thus provides increased concentration of reactants on their surface causing the reaction to occur.

8.12. What are interstitial compounds? Why are such compounds well known for transition metals?
Sol: Transition metals form large number of interstitial compounds. They are able to entrap small atoms of elements like H, G, N, B, etc., in their crystal lattice and even can make weak bonds with them.
Due to formation of interstitial compounds, their malleability and ductility decreases and tensile . strength increases. Steel and cast iron are hard in comparison to wrought iron due to the presence of trapped carbon atoms in interstitial spaces.

8.13. How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.
Sol: The transition metals show a number of variable oxidation states due to the participation of (n – 1) d electrons in addition to ns electrons in the bond formation. They therefore, exhibit a large number of variable oxidation states. On the other hand, the non-transition metals generally belonging to s-block do not show variable oxidation states because by the loss of valence s-electrons, they acquire the configuration of the nearest noble gas elements.

In the p-block the lower oxidation states are favoured by the heavier members (due to inert pair effect), the opposite is true in the groups of d-block. For example, in group 6, Mo(VI) and W(VI) are found to be more stable than Cr(VI). Thus Cr(VI) in the form of dichromate in acidic medium is a strong oxidising agent, whereas MoO₃ and WO₃ are not.

8.14. Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?
Sol: Potassium dichromate is prepared from chromate, which in turn is obtained by the fusion of chromite ore (FeCr₂O₃) with sodium or potassium carbonate in free excess of air. The reaction with sodium carbonate occurs as follows:

The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate, Na₂Cr,0₇.2H₂0 can be crystallised.

Sodium dichromate is more soluble than potassium dichromate. The latter is therefore, prepared by treating the solution of sodium dichromate with potassium chloride.

Orange crystals of potassium dichromate crystallise out. The chromates and dichromates depending upon p^H of the solution. If p^H of potassium dichromate is increased it is converted to yellow potassium chromate.

8.15. Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:
(i)iodide
(ii)iron (II) solution and
(iii)H₂S
Sol: K₂Gr₂0₇is a powerful oxidising agent. In dilute sulphuric acid medium the oxidation state of Cr changes from+6 to + 3. The oxidising action can be represented as follows:

8.16. Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron (II) ions (ii) S0₂ and (iii) oxalic acid? Write the ionic, equations for the reactions.
Sol: Potassium permanganate (KMn04) is prepared by the fusion of a mixture of pyrolusite (Mn0₂),potassiufn hydroxide and oxygen, first green coloured potassium manganate is formed. 2MnO₂ + 4KOH + 0₂ —> 2K₂Mn0₄+2H₂0 The potassium manganate is extracted by water, which then undergoes disproportionation in neutral or acidic solution to give potassium permanganate.

8.17. For M²⁺/M and M³⁺/M²⁺ systems the E° values
for some metals are as follows:
Cr²⁺/Cr   –> -0.9 V
Mn²⁺/Mn  –> -1.2V
Fe²⁺/Fe     –> -0.4 V
Cr³⁺/Cr²⁺  –> -0.4 V
Mn³⁺/Mn²⁺   –>+ 1.5V
Fe³⁺/Fe²⁺   –>+ 0.8V
(ii)the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.
Sol: (i) Cr³⁺/Cr²⁺ has negative reduction potential. Hence, Cr³⁺ cannot be reduced to Cr²⁺. Mn³⁺/Mn²⁺ has a large positive reduction potential. Hence, Mn³⁺ can be easily reduced to Mn²⁺. Fe³⁺/Fe²⁺ has small positive reduction potential. Hence, Fe³⁺ is more stable than Mn³⁺ but less stable than Cr³⁺.
(ii)From the E° values, the order of oxidation of the metal to the divalent cation is : Mn > Cr > Fe.

8.18. Predict which of the following will be coloured in aqueous solution?
Ti³⁺, V³⁺, Cu⁺, Sc³⁺, Mn²⁺, Fe³⁺, Co²⁺.
Sol: Only those ions will be coloured which have incomplete d-orbitals. The ions with either empty or filled d-orbitals are colourless. Keeping this in view, the coloured ions among the given list are :
Ti³⁺(3d¹), V³⁺(3d²), Mn²⁺(3d⁵), Fe³⁺(3d⁵), Co²⁺ (3d⁷)
Sc³⁺ (3d°) and Cu⁺ (3d¹⁰) ions are colourless.

8.19. Compare the stability of +2 oxidation state for the elements of the first transition series.
Sol: In general, the stability of +2 oxidation state in first transition series decreases from left to right due to increase in the sum of first and second ionisation energies. However Mn²⁺ is more stable due to half filled d-orbitals (3d⁵) and Zn²⁺ is more stable due to completely filled d-orbitals (3d¹⁰).

8.20. Compare the chemistry of actinoids with that of the lanthanoids with special reference to
(i)electronic configuration,
(ii)atomic and ionic sizes and
(iii)oxidation state
(iv)chemical reactivity.
Sol: (i) Electronic configuration: The general electronic configuration of lanthanoids is [Xe]⁵⁴ 4f^1-14 5d^0-1  6s² and that of actinoids is [Rn]⁸⁶ 5f^0-14 6d^0-1  7s², lanthanoids . belong to 4 f series whereas actinoids belong to 5f-series.
(ii) Atomic and ionic sizes: Both lanthanoids and actinoids show decrease in size of their atoms or ions in + 3 oxidation state as we go from left to right. In lanthanoids, the decrease is called lanthanoid contraction whereas in actinoids, it is called actinoid contraction. The contratibn is greater from element to element in actinodes due to poorer shielding by 5f electrons.
(iii)Oxidation state: Lanthanoids show limited oxidation states (+ 2, + 3, + 4) out of which + 3 is most common whereas actinoids show +3, +4, +5, +6, +7 oxidation states.This is because of large energy gap between 4f 5d and 6s orbitals. However, actinoids show a large number of oxidation states because of small energy ap- between 5f 6d and Is orbitals.
(iv) Chemical reactivity: The earlier members
of the lanthanoids series are quite reactive similar to calcium but, with increase in atomic number, they behave more like aluminium. The metals combine with hydrogen when . gently heated in the gas. Carbides, Ln₃C, Ln₂C₃ and LnC₂ are formed when the metals are heated with carbon. They liberate hydrogen from dilute acid and burn in halogens to form halides. They form oxides M₂0₃ and hydroxides M(OH)₃.
Actinoids are highly reactive metals, especially when finely divided. The action of boiling water on them gives a mixture of oxide and hydride and combination with most non-metals take place at moderate temperatures. HCl attacks all metals but most are slightly affected by nitric acid owing to the formation of protective oxide layers, alkalis have no action. Actinoids are more reactive than lanthanoids due to bigger atomic size and lower ionisation energy.

8.21. How would you account for the following:
(i) Of the d⁴ species, Cr²⁺ is strongly reducing while manganese (III) is strongly oxidizing.
(ii) Cobalt (II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.
(iii) The d¹ configuration is very unstable in ions.
Sol: (i) E° value for Cr³⁺/Cr²⁺ is negative (-0-41 V) whereas E° values for Mn³⁺/Mn²⁺is positive (+1.57 V). Hence, Cr²⁺ ion can easily undergo oxidation to give Cr³⁺ ion and, therefore, act as strong reducing agent whereas Mn³⁺ can easily undergo’ reduction to give Mn²⁺ and hence act as an oxidizing agent.
(ii) Co (III) has .greater tendency to form coordination complexes than Co (II). Hence, in the presence of ligands, Co (II) changes to Co (III), i.e., is easily oxidized.
(iii) The ions with dx configuration have the tendency to lose the only electron present in d-subshell to acquire stable d° configuration. Hence, they are unstable and undergo oxidation or disproportionation.

8.22. What is meant by disproportionation? Give two examples of disproportionation reaction in aqueous solution
Sol: Disproportionation reactions are those in which the same substance undergoes oxidation as well as reduction, i.e., oxidation number of an element increases as well as decreases to form two different products.

8.23. Which metal in the first transition metal series exhibits +1 oxidation state most frequently and why?
Sol: Cu with configuration [Ar] 4s¹3d¹⁰ exhibits +1 oxidation state and forms Cu⁺ ion because by losing one electron, the cation or positive ion acquires a stable configuration of d-orbitals (3d¹⁰).

8.24. Calculate the number of unpaired electrons in the following gaseous ions : Mn³⁺, Cr³⁺, V³⁺ and Ti³⁺. Which one of these is the most stable in aqueous solution.
Sol: Mn³⁺ = 3d¹ = 4 unpaired electrons, Cr³⁺ = 3d³ = 3 electrons,V³⁺ = 3d² = 2 electrons, Ti³⁺=3d¹ = l electron.Out of these, Cr³⁺ is most stable in aqueous solution because of half-filled t_2g level.

8.25. Give examples and suggest reasons for the following features of the transition metal chemistry:
(i) The lowest oxide of transition metal is basic the highest is amphoteric/ acidic.
(ii) A transition metal exhibits highest oxidation state ih oxides and fluorides.
(iii) The highest oxidation state is exhibited in oxoanions of a metal.
Sol: (i) The lower oxide of transition metal is basic because the metal atom has low oxidation state whereas higher once are acidic due to high oxidation state. For example, MnO is basic whereas Mn₂O₇is acidic. Oxides in lower oxidation state are ionic hence basic. Oxides in higher oxidation state are covalent hence acidic
(ii) A transition metal exhibits higher oxidation states in oxides and fluorides because oxygen and fluorine are highly electronegative elements, small in size and strongest oxidising agents. For example, osmium shows an oxidation states of + 6 in O₅F₆and vanadium shows an oxidation states of + 5 in V₂O₅.
(iii) Oxo metal anions have highest oxidation state, e.g., Cr in Cr₂0₇^2- has an. oxidation state of + 6 whereas Mn in Mn0₄^– has an oxidation state of + 7. This is again due to the combination of the metal with oxygen, which is highly electronegative and oxidizing agent.

8.26. Indicate the steps in the preparation of:
(i)K₂Cr₂0₇from chromite ore
(ii)KMn0₄ from pyrolusite ore.
Sol:

8.27. What are alloys? Name an alloy which contains some lanthanoid metals. Mention its uses.
Sol: An alloy is a homogeneous mixture of different metals or metals and non-metals.
Misch metal is an alloy of cerium (Ce). lanthanum (La), neodymium (Nd), iron (Fe) and traces of carbon, sulphur, aluminium etc. It is used in making parts of jet engines.

8.28. What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements: 29,59,74,95,102,104.
Sol: The f-block elements in which the. last electron enters into f-sub shell-are called inner-transition elements. These include lanthanoids (Z=58 to 71) and actinoids (Z=90 to 103). Thus, the elements with atomic numbers 59,95 and 102 are the? inner transition elements.

8.29. The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements.
Sol: Lanthanoids show limited number of oxidation state, viz, + 2, + 3 and + 4 (out of which + 3 is most common). This is because of large energy gap between 4f 5d and 6s subshells. The dominant oxidation state of actinoids is also + 3 but they show a number of other oxidation states also. For example, uranium (Z=92) and plutonium (Z – 94), show + 3, + 4, + 5 and + 6, neptunium (Z = 94) shows + 3, +4, + 5 and + 7, etc. This is because of the small energy difference between. 5f, 6d and 7s orbitals of the actinoids.

8.30. Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element
Sol: Last actinoid=Lawrencium (Z = 103)
Electronic configuration = [Rn]⁸⁶ 5f¹⁴ 6d¹ 7s² Possible oxidation state = + 3.

8.31 Use Hund’s rule to derive the electronic configuration of Ce³⁺ ion, and calculate its magnetic moment on the basis of ‘spin-only’ formula.
Sol.

8.32. Name the members of the lanthanoid series which exhibit +4 oxidation state and those which exhibit +2 oxidation state. Try to co-relate this type of behaviour with the electronic configuration of these elements.
Sol: +4 oxidation state : ₅₈Ce, ₅₉Pr, ₆₅Tb
+ 2 oxidation state : ₆₀Nd, ₆₂Sm, ₆₃Eu, ₆₉Tm, ₇₀Yb.
In general +2 oxidation state is exhibited by the elements with configuration 5d⁰6s² so that two electrons may be easily lost. Similarly +4 oxidation state is shown by the elements which after losing four electrons acquire configuration either close to 4f⁰ or 4f⁷.

8.33. Compare the chemistry of actinoids with that of lanthanoids with reference to:
(i)Electronic configuration
(ii)Oxidation states
(iii)Chemical reactivity
Sol: (i)Electronic configuration : In lanthanoids 4f- orbitals are progressively filled whereas in actinoids 5f-orbitals are progressively filled.
(ii)Oxidation states : Lanthanoids shows +3 oxidation state. Some elements shows +2 and +4 oxidation state also. Actinoids shows +3, +4, +5 +6, +7 oxidation states. Although +3 and +4 are most common.
(iii)Chemical reactivity : Actinoids are more reactive than lanthanoids due to bigger atomic size and lower ionisation energy.

8.34. Write the electronic configurations of the elements with the atomic numbers 61,91,101 and 109.
Sol: Z=61 (Promethium, Pm) [Xe]⁵⁴4f⁵5d⁰ 6s²
Z = 91 (Protactinium, Pa) => [Rn]⁸⁶ 5f² 6d¹ 7s²
Z = 101 (Mendelevium, Md)=> [Rn]⁸⁶ 5f¹³ 6d⁰ 7s²
Z = 109 (Meitnerium, Mt) [Rn]⁸⁶ 5f¹⁴ 6d⁷ 7s²

8.35. Com pare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:
(i)electronic configurations
(ii)oxidation states
(iii)ionisation enthalpies and
(iv)atomic sizes
Sol: (i) Electronic configuration: The elements in the same vertical column generally have similar electronic configuration. First transition series shows only two exceptions, i.e., Cr = 3d⁵  4s¹  and Cu = 3d¹⁰ 4s¹. But second transition series shows more exceptions, i.e., Y = 4d¹ 5s², Nb = 4d¹ , 5s¹ , Mo=4d⁵  5s¹ , Ru=4d¹  5s¹ , Rh=4d⁸  5s¹ , Pd , =4d¹⁰ 5s°, Ag=4d¹⁰ 5s¹ . In third transition, there are two exceptions, i.e„ Pt = 5d⁹  6s¹  and Au = 5d¹⁰ 6s¹ .
Thus in the same vertical column, in a number of cases, the electronic configuration of the elements of three series are not similar.
(ii) Oxidation states: The elements in the same vertical column generally show similar oxidation states. The number of oxidation states shown by the elements in the middle of each series is maximum and minimum at the extreme ends.
(iii)Ionization enthalpies: The first ionization enthalpies in each series generally increases gradually as we more from left to right though some exceptions are observed in each series. The first ionization enthalpies of some elements in the second (4d) series are higher while some of them have lower value than the elements of 3d series in the same vertical column. However, the first ionization enthalpies of third (5d) series are higher than those of 3d and Ad series. This is because of weak shielding of nucleus by 4f-electrons in the 5d series.
(iv)Atomic sizes: In general, ions of the same charge or atoms in a given series show progressively decrease in radius with increasing atomic number though the decrease is quite small. But the size of the atoms of the Ad series is larger then the corresponding elements of the 3d series whereas size of elements of the 5d-series nearly the same as those of Ad series because of lanthanoid contraction.

8.36. Write down the number of 3d electrons in each of the following ions:Ti²⁺, V²⁺, Cr³⁺, Mn²⁺, Fe²⁺, Fe²⁺, Co²⁺, Ni²⁺ and Cu²⁺. Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).
Sol:

8.37. Comment on the statement that elements of first transition series possess many properties different from those of the heavier transition elements.
Sol: The heavier transition elements belong to fourth (Ad) and fifth (5d) and sixth (6d) transition series. Their properties are expected to be different from the elements belonging to the first (3d) series due to the following reasons :
(i) The atomic radii of the elements belonging to Ad and 5d series are more due to greater number of electron shells. However, the difference in Ad and 5d transition elements are comparatively less because of lanthanoid contraction.
(ii) Because of stronger inter atomic bonding, the m.p. and b.p. of the elements of Ad and 5d series are higher.
(iii) Ionisation enthalpies are expected to decrease as we move from one series to the other. However, the values for the elements of 5d series are higher as compared to the elements belonging to the other two series due to lanthanoid contraction.
Actually atomic size decreases on account of it and effective nuclear charge increases. As a result, there is an increase in ionisation energy in case of 3d elements.

8.38. What can be inferred from the magnetic moment values of the following complex species?

Sol:

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