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Explore the highly sought-after NCERT Solutions for Class 12 Science Chemistry Chapter 5: Biomolecules. These step-by-step solutions offer clear explanations, making them a favorite among Chemistry students. Whether you’re completing homework assignments or gearing up for exams, the Biomolecules Solutions for Class 12 Science provide a valuable resource. All questions and answers from Chapter 5 of the NCERT Book for Class 12 Science Chemistry are readily available here for free, ensuring a convenient and effective study experience. Page No 412: Question 14.1: Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six membered ring compounds) are insoluble in water. Explain. ANSWER: A glucose molecule contains five −OH groups while a sucrose molecule contains eight −OH groups. Thus, glucose and sucrose undergo extensive H-bonding with water. Hence, these are soluble in water. But cyclohexane and benzene do not contain −OH groups. Hence, they cannot undergo H-bonding with water and as a result, are insoluble in water. Page No 412: Question 14.2: What are the expected products of hydrolysis of lactose? ANSWER: Lactose is composed of β-D galactose and β-D glucose. Thus, on hydrolysis, it gives β-D galactose and β-D glucose. Page No 412: Question 14.3: How do you explain the absence of aldehyde group in the pentaacetate of D-glucose? ANSWER: D-glucose reacts with hydroxylamine (NH2OH) to form an oxime because of the presence of aldehydic (−CHO) group or carbonyl carbon. This happens as the cyclic structure of glucose forms an open chain structure in an aqueous medium, which then reacts with NH2OH to give an oxime. But pentaacetate of D-glucose does not react with NH2OH. This is because pentaacetate does not form an open chain structure. Page No 417: Question 14.4: The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain. ANSWER: Both acidic (carboxyl) as well as basic (amino) groups are present in the same molecule of amino acids. In aqueous solutions, the carboxyl group can lose a proton and the amino group can accept a proton, thus giving rise to a dipolar ion known as a zwitter ion. Due to this dipolar behaviour, they have strong electrostatic interactions within them and with water. But halo-acids do not exhibit such dipolar behaviour. For this reason, the melting points and the solubility of amino acids in water is higher than those of the corresponding halo-acids. Page No 417: Question 14.5: Where does the water present in the egg go after boiling the egg? ANSWER: When an egg is boiled, the proteins present inside the egg get denatured and coagulate. After boiling the egg, the water present in it is absorbed by the coagulated protein through H-bonding. Page No 422: Question 14.6: Why cannot vitamin C be stored in our body? ANSWER: Vitamin C cannot be stored in our body because it is water soluble. As a result, it is readily excreted in the urine. Page No 422: Question 14.7: What products would be formed when a nucleotide from DNA containing thymine is hydrolysed? ANSWER: When a nucleotide from the DNA containing thymine is hydrolyzed, thymine β-D-2-deoxyribose and phosphoric acid are obtained as products. Page No 422: Question 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? ANSWER: A DNA molecule is double-stranded in which the pairing of bases occurs. Adenine always pairs with thymine, while cytosine always pairs with guanine. Therefore, on hydrolysis of DNA, the quantity of adenine produced is equal to that of thymine and similarly, the quantity of cytosine is equal to that of guanine. But when RNA is hydrolyzed, there is no relationship among the quantities of the different bases obtained. Hence, RNA is single-stranded. Page No 423: Question 14.1: What are monosaccharides? ANSWER: Monosaccharides are carbohydrates that cannot be hydrolysed further to give simpler units of polyhydroxy aldehyde or ketone. Monosaccharides are classified on the bases of number of carbon atoms and the functional group present in them. Monosaccharides containing an aldehyde group are known as aldoses and those containing a keto group are known as ketoses. Monosaccharides are further classified as trioses, tetroses, pentoses, hexoses, and heptoses according to the number of carbon atoms they contain. For example, a ketose containing 3 carbon atoms is called ketotriose and an aldose containing 3 carbon atoms is called aldotriose. Page No 423: Question 14.2: What are reducing sugars? ANSWER: Reducing sugars are carbohydrates that reduce Fehling’s solution and Tollen’s reagent. All monosaccharides and disaccharides, excluding sucrose, are reducing sugars. Page No 423: Question 14.3: Write two main functions of carbohydrates in plants. ANSWER: Two main functions of carbohydrates in plants are: (i) Polysaccharides such as starch serve as storage molecules. (ii) Cellulose, a polysaccharide, is used to build the cell wall. Page No 423: Question 14.4: Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose ANSWER: Monosaccharides: Ribose, 2-deoxyribose, galactose, fructose Disaccharides: Maltose, lactose Page No 423: Question 14.5: What do you understand by the term glycosidic linkage? ANSWER: Glycosidic linkage refers to the linkage formed between two monosaccharide units through an oxygen atom by the loss of a water molecule. For example, in a sucrose molecule, two monosaccharide units, ∝-glucose and β-fructose, are joined together by a glycosidic linkage. Page No 423: Question 14.6: What is glycogen? How is it different from starch? ANSWER: Glycogen is a carbohydrate (polysaccharide). In animals, carbohydrates are stored as glycogen. Starch is a carbohydrate consisting of two components − amylose (15 − 20%) and amylopectin (80 − 85%). However, glycogen consists of only one component whose structure is similar to amylopectin. Also, glycogen is more branched than amylopectin. Page No 423: Question 14.7: What are the hydrolysis products of (i) sucrose and (ii) lactose? ANSWER: (i) On hydrolysis, sucrose gives one molecule of ∝-D glucose and one molecule of β- D-fructose. (ii) The hydrolysis of lactose gives β-D-galactose and β-D-glucose. Page No 423: Question 14.8: What is the basic structural difference between starch and cellulose? ANSWER: Starch consists of two components …

NCERT Solutions for Class 12 Science Chemistry Chapter 5 – Biomolecules Read More »

Explore the comprehensive NCERT solutions for Class 12 Science Chemistry Chapter 4 on Amines, complete with easy-to-follow step-by-step explanations. Widely favored among Class 12 Science students, these Chemistry Amines Solutions are invaluable for efficiently tackling homework assignments and gearing up for exams. Free of charge, you can access all questions and answers from Chapter 4 of the NCERT Book for Class 12 Science Chemistry, ensuring a convenient resource for your academic needs. Page No 384: Question 13.1: Classify the following amines as primary, secondary or tertiary: (i) (ii) (iii) (C2H5)2CHNH2 (iv) (C2H5)2NH ANSWER: Primary: (i) and (iii) Secondary: (iv) Tertiary: (ii) Page No 384: Question 13.2: (i) Write structures of different isomeric amines corresponding to the molecular formula, C4H11N (ii) Write IUPAC names of all the isomers. (iii) What type of isomerism is exhibited by different pairs of amines? ANSWER: (i), (ii) The structures and their IUPAC names of different isomeric amines corresponding to the molecular formula, C4H11N are given below: (a) CH3-CH2-CH2-CH2-NH2 Butanamine (10) (b)  Butan-2-amine (10) (c)  2-Methylpropanamine (10) (d) 2-Methylpropan-2-amine (10) (e) CH3-CH2-CH2-NH-CH3 N-Methylpropanamine (20) (f) CH3-CH2-NH-CH2-CH3 N-Ethylethanamine (20) (g) N-Methylpropan-2-amine (20) (h) N,N-Dimethylethanamine (3°) (iii) The pairs (a) and (b) and (e) and (g) exhibit position isomerism. The pairs (a) and (c); (a) and (d); (b) and (c); (b) and (d) exhibit chain isomerism. The pairs (e) and (f) and (f) and (g) exhibit metamerism. All primary amines exhibit functional isomerism with secondary and tertiary amines and vice-versa. Page No 387: Question 13.3: How will you convert? (i) Benzene into aniline (ii) Benzene into N, N-dimethylaniline (iii) Cl−(CH2)4−Cl into hexan-1, 6-diamine? ANSWER: (i) (ii) (iii) Page No 396: Question 13.4: Arrange the following in increasing order of their basic strength: (i) C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2 and (C2H5)2NH (ii) C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2 (iii) CH3NH2, (CH3)2NH, (CH3)3N, C6H5NH2, C6H5CH2NH2. ANSWER: (i) Considering the inductive effect of alkyl groups, NH3, C2H5NH2, and (C2H5)2NH can be arranged in the increasing order of their basic strengths as: Again, C6H5NH2 has proton acceptability less than NH3. Thus, we have: Due to the −I effect of C6H5 group, the electron density on the N-atom in C6H5CH2NH2 is lower than that on the N-atom in C2H5NH2, but more than that in NH3. Therefore, the given compounds can be arranged in the order of their basic strengths as: (ii) Considering the inductive effect and the steric hindrance of the alkyl groups, C2H5NH2, (C2 H5)2NH2, and their basic strengths as follows: Again, due to the −R effect of C6H5 group, the electron density on the N atom in C6H5 NH2 is lower than that on the N atom in C2H5NH2. Therefore, the basicity of C6H5NH2 is lower than that of C2H5NH2. Hence, the given compounds can be arranged in the increasing order of their basic strengths as follows: (iii) Considering the inductive effect and the steric hindrance of alkyl groups, CH3NH2, (CH3)2NH, and (CH3)3N can be arranged in the increasing order of their basic strengths as: In C6H5NH2, N is directly attached to the benzene ring. Thus, the lone pair of electrons on the N−atom is delocalized over the benzene ring. In C6H5CH2NH2, N is not directly attached to the benzene ring. Thus, its lone pair is not delocalized over the benzene ring. Therefore, the electrons on the N atom are more easily available for protonation in C6H5CH2NH2 than in C6H5NH2 i.e., C6H5CH2­NH2 is more basic than C6H5NH2. Again, due to the −I effect of C6H5 group, the electron density on the N−atom in C6H5CH2NH2 is lower than that on the N−atom in (CH3)3N. Therefore, (CH3)3N is more basic than C6H5CH2NH2. Thus, the given compounds can be arranged in the increasing order of their basic strengths as follows. Page No 396: Question 13.5: Complete the following acid-base reactions and name the products: (i) CH3CH2CH2NH2 + HCl  (ii) (C2H5)3N + HCl  ANSWER: (i)  (ii) (C2H5)3N Triethylamine+ HCl → (C2H5)3N+HCl−TriethylammoniumchlorideC2H53N Triethylamine+ HCl → C2H53N+HCl-Triethylammoniumchloride Page No 396: Question 13.6: Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution. ANSWER: Aniline reacts with methyl iodide to produce N, N-dimethylaniline. With excess methyl iodide, in the presence of Na2CO3solution, N, N-dimethylaniline produces N, N, N−trimethylanilinium carbonate. Page No 396: Question 13.7: Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained. ANSWER: Page No 396: Question 13.8: Write structures of different isomers corresponding to the molecular formula, C3H9N. Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid. ANSWER: The structures of different isomers corresponding to the molecular formula, C3H9N are given below: (a)  Propan-1-amine (10) (b) Propan-2-amine (10) (c) (d) N,N-Dimethylmethanamine (30) 10amines, (a) propan-1-amine, and (b) Propan-2-amine will liberate nitrogen gas on treatment with nitrous acid. Page No 399: Question 13.9: Convert (i) 3-Methylaniline into 3-nitrotoluene. (ii) Aniline into 1,3,5-tribromobenzene. ANSWER: (i) (ii) Page No 400: Question 13.1: Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines. (i) (CH3)2 CHNH2 (ii) CH3(CH2)2NH2 (iii) CH3NHCH(CH3)2 (iv) (CH3)3CNH2 (v) C6H5NHCH3(vi) (CH3CH2)2NCH3 (vii) m−BrC6H4NH2 ANSWER: (i) 1-Methylethanamine (10 amine) (ii) Propan-1-amine (10 amine) (iii) N−Methyl-2-methylethanamine (20 amine) (iv) 2-Methylpropan-2-amine (10 amine) (v) N−Methylbenzamine or N-methylaniline (20 amine) (vi) N-Ethyl-N-methylethanamine (30 amine) (vii) 3-Bromobenzenamine or 3-bromoaniline (10 amine) Page No 400: Question 13.2: Give one chemical test to distinguish between the following pairs of compounds. (i) Methylamine and dimethylamine (ii) Secondary and tertiary amines (iii) Ethylamine and aniline (iv) Aniline and benzylamine (v) Aniline and N-methylaniline. ANSWER: (i) Methylamine and dimethylamine can be distinguished by the carbylamine test. Carbylamine test: Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form foul-smelling isocyanides or carbylamines. Methylamine (being an aliphatic primary amine) gives a positive carbylamine test, but dimethylamine does not. (ii) Secondary and tertiary amines can be distinguished by allowing them to react with Hinsberg’s reagent (benzenesulphonyl chloride, C6H5SO2Cl). Secondary amines react with Hinsberg’s reagent to form a product that is insoluble in an alkali. For example, N, N−diethylamine reacts with Hinsberg’s reagent to form N, N−diethylbenzenesulphonamide, which is insoluble in an alkali. Tertiary amines, however, do not react with Hinsberg’s reagent. (iii) Ethylamine and aniline can be distinguished using the azo-dye test. A dye is obtained when aromatic amines react with HNO2 (NaNO2 + dil.HCl) at 0-5°C, followed by a reaction with the alkaline solution of 2-naphthol. The dye is usually yellow, red, or orange in colour. Aliphatic amines give a brisk effervescence due (to the evolution of N2 gas) under similar conditions. ​ (iv) Aniline …

NCERT Solutions for Class 12 Science Chemistry Chapter 4 – Amines Read More »

Find comprehensive NCERT Solutions for Class 12 Science Chemistry Chapter 3 on Aldehydes, Ketones, and Carboxylic Acids. These step-by-step explanations are highly favored by Chemistry students for quick completion of homework and effective exam preparation. All answers to the questions in NCERT Book of Class 12 Science Chemistry Chapter 3 are available here at no cost, making them a valuable resource for students. Page No 353: Question 12.1: Write the structures of the following compounds. (i) α-Methoxypropionaldehyde (ii) 3-Hydroxybutanal (iii) 2-Hydroxycyclopentane carbaldehyde (iv) 4-Oxopentanal (v) Di-sec-butyl ketone (vi) 4-Fluoroacetophenone ANSWER: (i) (ii) (iii) (iv) (v) (vi) Page No 356: Question 12.2: Write the structures of products of the following reactions; (i) (ii) (iii) (iv) ANSWER: (iv) Page No 358: Question 12.3: Arrange the following compounds in increasing order of their boiling points. CH3CHO, CH3CH2OH, CH3OCH3, CH3CH2CH3 ANSWER: The molecular masses of the given compounds are in the range 44 to 46. CH3CH2OH undergoes extensive intermolecular H-bonding, resulting in the association of molecules. Therefore, it has the highest boiling point. CH3CHO is more polar than CH3OCH3 and so CH3CHO has stronger intermolecular dipole − dipole attraction than CH3OCH3⋅ CH3CH2CH3 has only weak van der Waals force. Thus, the arrangement of the given compounds in the increasing order of their boiling points is given by: CH3CH2CH3 < CH3OCH3 < CH3CHO < CH3CH2OH Page No 365: Question 12.4: Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions. (i)Ethanal, Propanal, Propanone, Butanone. (ii)Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone. Hint:Consider steric effect and electronic effect. ANSWER: (i) The +I effect of the alkyl group increases in the order: Ethanal < Propanal < Propanone < Butanone The electron density at the carbonyl carbon increases with the increase in the +I effect. As a result, the chances of attack by a nucleophile decrease. Hence, the increasing order of the reactivities of the given carbonyl compounds in nucleophilic addition reactions is: Butanone < Propanone < Propanal < Ethanal (ii) The +I effect is more in ketone than in aldehyde. Hence, acetophenone is the least reactive in nucleophilic addition reactions. Among aldehydes, the +I effect is the highest in p-tolualdehyde because of the presence of the electron-donating −CH3 group and the lowest in p-nitrobezaldehyde because of the presence of the electron-withdrawing −NO2 group. Hence, the increasing order of the reactivities of the given compounds is: Acetophenone < p-tolualdehyde < Benzaldehyde< p-Nitrobenzaldehyde Page No 365: Question 12.5: Predict the products of the following reactions: (i) (ii) (iii) (iv) ANSWER: (i) (ii) (iii) (iv) Page No 367: Question 12.6: Give the IUPAC names of the following compounds: (i) PhCH2CH2COOH (ii) (CH3)2C=CHCOOH (iii)  (iv)  ANSWER: (i) 3-Phenylpropanoic acid (ii) 3-Methylbut-2-enoic acid (iii) 2-Methylcyclopentanecarboxylic acid (iv)2,4,6-Trinitrobenzoic acid Page No 370: Question 12.7: Show how each of the following compounds can be converted to benzoic acid. (i) Ethylbenzene (ii) Acetophenone (iii) Bromobenzene (iv) Phenylethene (Styrene) ANSWER: (i) (ii) (iii) (iv) Page No 376: Question 12.8: Which acid of each pair shown here would you expect to be stronger? (i) CH3CO2H or CH2FCO2H (ii)CH2FCO2H or CH2ClCO2H (iii) CH2FCH2CH2CO2H or CH3CHFCH2CO2H (iv) ANSWER: (i) The +I effect of −CH3 group increases the electron density on the O−H bond. Therefore, release of proton becomes difficult. On the other hand, the −I effect of F decreases the electron density on the O−H bond. Therefore, proton can be released easily. Hence, CH2FCO2H is a stronger acid than CH3CO2H. (ii) F has stronger −I effect than Cl. Therefore, CH2FCO2H can release proton more easily than CH2ClCO2H. Hence, CH2FCO2H is stronger acid than CH2ClCO2H. (iii) Inductive effect decreases with increase in distance. Hence, the +I effect of F in CH3CHFCH2CO2H is more than it is in CH2FCH2CH2CO2H. Hence, CH3CHFCH2CO2H is stronger acid than CH2FCH2CH2CO2H. (iv) Due to the −I effect of F, it is easier to release proton in the case of compound (A). However, in the case of compound (B), release of proton is difficult due to the +I effect of −CH3 group. Hence, (A) is a stronger acid than (B). Page No 377: Question 12.1: What is meant by the following terms? Give an example of the reaction in each case. (i) Cyanohydrin (ii) Acetal (iii) Semicarbazone (iv) Aldol (v) Hemiacetal (vi) Oxime (vii) Ketal (vii) Imine (ix) 2,4-DNP-derivative (x) Schiff’s base ANSWER: (i) Cyanohydrin: Cyanohydrins are organic compounds having the formula RR′C(OH)CN, where R and R′ can be alkyl or aryl groups. Aldehydes and ketones react with hydrogen cyanide (HCN) in the presence of excess sodium cyanide (NaCN) as a catalyst to field cyanohydrin. These reactions are known as cyanohydrin reactions. Cyanohydrins are useful synthetic intermediates. (ii) Acetal: Acetals are gem−dialkoxy alkanes in which two alkoxy groups are present on the terminal carbon atom. One bond is connected to an alkyl group while the other is connected to a hydrogen atom. When aldehydes are treated with two equivalents of a monohydric alcohol in the presence of dry HCl gas, hemiacetals are produced that further react with one more molecule of alcohol to yield acetal. (iii) Semicarbarbazone: Semicarbazones are derivatives of aldehydes and ketones produced by the condensation reaction between a ketone or aldehyde and semicarbazide. Semicarbazones are useful for identification and characterization of aldehydes and ketones. (iv) Aldol: A β-hydroxy aldehyde or ketone is known as an aldol. It is produced by the condensation reaction of two molecules of the same or one molecule each of two different aldehydes or ketones in the presence of a base. (v) Hemiacetal: Hemiacetals are α−alkoxyalcohols General structure of a hemiacetal Aldehyde reacts with one molecule of a monohydric alcohol in the presence of dry HCl gas. (vi) Oxime: Oximes are a class of organic compounds having the general formula RR′CNOH, where R is an organic side chain and R′ is either hydrogen or an organic side chain. If R′ is H, then it is known as aldoxime and if R′ is an organic side chain, it is known as ketoxime. On treatment with hydroxylamine in a weakly acidic medium, aldehydes or ketones form oximes. (vii) Ketal: Ketals are gem−dialkoxyalkanes in which two alkoxy groups are present on the same carbon atom within the chain. The other two bonds of the carbon atom are connected to two alkyl groups. Ketones react with ethylene glycol in the presence of dry HCl gas to give a cyclic product known as ethylene glycol ketals. (viii) Imine: Imines are chemical compounds containing a carbon …

NCERT Solutions for Class 12 Science Chemistry Chapter 3 – Aldehydes, Ketones And Carboxylic Acids Read More »

Explore the comprehensive NCERT solutions for Class 12 Science Chemistry Chapter 2 – “Alcohols, Phenols, and Ethers” with straightforward, step-by-step explanations. These solutions have gained immense popularity among Chemistry students in class 12 Science, serving as valuable resources for completing homework efficiently and preparing for exams. All the questions and answers from the NCERT Book of Class 12 Science Chemistry Chapter 2 are available here at no cost, providing a convenient and accessible aid for students. Page No 317: Question 11.1: Classify the following as primary, secondary and tertiary alcohols: (i) (ii)  (iii)  (iv) (v) (vi) ANSWER: Primary alcohol → (i), (ii), (iii) Secondary alcohol → (iv), (v) Tertiary alcohol → (vi) Page No 317: Question 11.2: Identify allylic alcohols in the above examples. ANSWER: The alcohols given in (ii) and (vi) are allylic alcohols. Page No 320: Question 11.3: Name the following compounds according to IUPAC system. (i) (ii) (iii) (iv) (v) ANSWER: (i) 3-Chloromethyl-2-isopropylpentan-1-ol (ii) 2, 5-Dimethylhexane-1, 3-diol (iii) 3-Bromocyclohexanol (iv) Hex-1-en-3-ol (v) 2-Bromo-3-methylbut-2-en-1-ol Page No 325: Question 11.4: Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal? (i) (ii) ANSWER: (i) (ii) Page No 325: Question 11.5: Write structures of the products of the following reactions: (i) (ii) (iii) ANSWER: (i)   (ii) (iii) Page No 335: Question 11.6: Give structures of the products you would expect when each of the following alcohol reacts with (a) HCl-ZnCl2 (b) HBr and (c) SOCl2. (i) Butan-1-ol (ii) 2-Methylbutan-2-ol ANSWER: (a) (i) Primary alcohols do not react appreciably with Lucas’ reagent (HCl-ZnCl2) at room temperature. (ii) Tertiary alcohols react immediately with Lucas’ reagent. (b) (i) (ii) (c) (i) (ii) Page No 335: Question 11.7: Predict the major product of acid catalysed dehydration of (i) 1-Methylcyclohexanol and (ii) Butan-1-ol ANSWER: (ii) Page No 335: Question 11.8: Ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions. ANSWER: Resonance structure of the phenoxide ion Resonance structures of p-nitrophenoxide ion Resonance structures of o-nitrophenoxide ion It can be observed that the presence of nitro groups increases the stability of phenoxide ion. Page No 335: Question 11.9: Write the equations involved in the following reactions: (i) Reimer-Tiemann reaction (ii) Kolbe’s reaction ANSWER: Page No 342: Question 11.10: Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol. ANSWER: In Williamson synthesis, an alkyl halide reacts with an alkoxide ion. Also, it is an SN2 reaction. In the reaction, alkyl halides should be primary having the least steric hindrance. Hence, an alkyl halide is obtained from ethanol and alkoxide ion from 3-methylpentan-2-ol. Page No 342: Question 11.11: Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene and why? (i) (ii) ANSWER: Set (ii) is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene. In set (i), sodium methoxide (CH3ONa) is a strong nucleophile as well as a strong base. Hence, an elimination reaction predominates over a substitution reaction. Page No 343: Question 11.12: Predict the products of the following reactions: (i)  (ii) (iii) (iv)  ANSWER: (i) (ii) (iii) (iv) Page No 344: Question 11.1: Write IUPAC names of the following compounds: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x)  (xi)  (xii) ANSWER: (i) 2, 2, 4-Trimethylpentan-3-ol (ii) 5-Ethylheptane-2, 4-diol (iii) Butane-2, 3-diol (iv) Propane-1, 2, 3-triol (v) 2-Methylphenol (vi) 4-Methylphenol (vii) 2, 5-Dimethylphenol (viii) 2, 6-Dimethylphenol (ix) 1-Methoxy-2-methylpropane (x) Ethoxybenzene (xi) 1-Phenoxyheptane (xii) 2-Ethoxybutane Page No 344: Question 11.2: Write structures of the compounds whose IUPAC names are as follows: (i) 2-Methylbutan-2-ol (ii) 1-Phenylpropan-2-ol (iii) 3,5-Dimethylhexane −1, 3, 5-triol (iv) 2,3 − Diethylphenol (v) 1 − Ethoxypropane (vi) 2-Ethoxy-3-methylpentane (vii) Cyclohexylmethanol (viii) 3-Cyclohexylpentan-3-ol (ix) Cyclopent-3-en-1-ol (x) 3-Chloromethylpentan-1-ol. ANSWER: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) Page No 344: Question 11.3: (i) Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names. (ii) Classify the isomers of alcohols in question 11.3 (i) as primary, secondary and tertiary alcohols. ANSWER: (i) The structures of all isomeric alcohols of molecular formula, C5H12O are shown below: (a)  Pentan-1-ol (1°) (b) 2-Methylbutan-1-ol (1°) (c) 3-Methylbutan-1-ol (1°) (d) 2, 2-Dimethylpropan-1-ol (1°) (e) Pentan-2-ol (2°) (f) 3-Methylbutan-2-ol (2°) (g) Pentan-3-ol (2°) (h) 2-Methylbutan-2-ol (3°) (ii) Primary alcohol: Pentan-1-ol; 2-Methylbutan-1-ol; 3-Methylbutan-1-ol; 2, 2−Dimethylpropan-1-ol Secondary alcohol: Pentan-2-ol; 3-Methylbutan-2-ol; Pentan-3-ol Tertiary alcohol: 2-methylbutan-2-ol Page No 344: Question 11.4: Explain why propanol has higher boiling point than that of the hydrocarbon, butane? ANSWER: Propanol undergoes intermolecular H-bonding because of the presence of −OH group. On the other hand, butane does not Therefore, extra energy is required to break hydrogen bonds. For this reason, propanol has a higher boiling point than hydrocarbon butane. Page No 344: Question 11.5: Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact. ANSWER: Alcohols form H-bonds with water due to the presence of −OH group. However, hydrocarbons cannot form H-bonds with water. As a result, alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Page No 344: Question 11.6: What is meant by hydroboration-oxidation reaction? Illustrate it with an example. ANSWER: The addition of borane followed by oxidation is known as the hydroboration-oxidation reaction. For example, propan-1-ol is produced by the hydroboration-oxidation reaction of propene. In this reaction, propene reacts with diborane (BH3)2 to form trialkyl borane as an addition product. This addition product is oxidized to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide. Page No 344: Question 11.7: Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O. ANSWER: Page No 344: Question 11.8: While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason. ANSWER: Intramolecular H-bonding is present in o-nitrophenol. In p-nitrophenol, the molecules are strongly associated due to the presence of intermolecular bonding. Hence, o-nitrophenol is steam volatile. Page No 344: Question 11.9: Give the equations of reactions for the preparation of phenol from cumene. ANSWER: To prepare phenol, cumene is first oxidized in the presence of air of cumene hydro-peroxide. Then, cumene hydroxide is treated with dilute acid to prepare phenol and acetone as by-products. Page No 344: Question 11.10: Write chemical reaction for the preparation of phenol from chlorobenzene. ANSWER: …

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Here, you can find NCERT solutions for Class 12 Science Chemistry Chapter 1 – Haloalkanes and Haloarenes. These solutions offer clear and straightforward explanations, making them highly favored among Chemistry students. Whether you need assistance with homework or preparing for exams, these solutions for Haloalkanes and Haloarenes are a valuable resource. All questions and answers from the NCERT Book of Class 12 Science Chemistry Chapter 1 are available here for free, providing a convenient and effective way to complete assignments and study for examinations. Page No 285: Question 10.1: Write structures of the following compounds: (i) 2-Chloro-3-methylpentane (ii) 1-Chloro-4-ethylcyclohexane (iii) 4-tert. Butyl-3-iodoheptane (iv) 1,4-Dibromobut-2-ene (v) 1-Bromo-4-sec. butyl-2-methylbenzene ANSWER: (i) 2-Chloro-3-methyl pentane (ii) 1-Chloro-4-ethylcyclohexane (iii) 4- tert-Butyl-3-iodoheptane (iv) 1, 4-Dibromobut-2-ene (v) 1-Bromo-4-sec-butyl-2-methylbenzene Page No 289: Question 10.2: Why is sulphuric acid not used during the reaction of alcohols with KI? ANSWER: In the presence of sulphuric acid (H2SO4), KI produces HI Since  is an oxidizing agent, it oxidizes HI (produced in the reaction to I2). As a result, the reaction between alcohol and HI to produce alkyl iodide cannot occur. Therefore, sulphuric acid is not used during the reaction of alcohols with KI. Instead, a non-oxidizing acid such as H3PO4 is used. Page No 289: Question 10.3: Write structures of different dihalogen derivatives of propane. ANSWER: There are four different dihalogen derivatives of propane. The structures of these derivatives are shown below. (i) 1, 1-Dibromopropane (ii) 2, 2-Dibromopropane (iii) 1, 2-Dibromopropane (iv) 1, 3-Dibromopropane Page No 289: Question 10.4: Among the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical chlorination yields (i) A single monochloride. (ii) Three isomeric monochlorides. (iii) Four isomeric monochlorides. ANSWER: (i) To have a single monochloride, there should be only one type of H-atom in the isomer of the alkane of the molecular formula C5H12. This is because, replacement of any H-atom leads to the formation of the same product. The isomer is neopentane. Neopentane (ii) To have three isomeric monochlorides, the isomer of the alkane of the molecular formula C5H12 should contain three different types of H-atoms. Therefore, the isomer is n-pentane. It can be observed that there are three types of H atoms labelled as a, b and c in n-pentane. (iii) To have four isomeric monochlorides, the isomer of the alkane of the molecular formula C5H12 should contain four different types of H-atoms. Therefore, the isomer is 2-methylbutane. It can be observed that there are four types of H-atoms labelled as a, b, c, and d in 2-methylbutane. Page No 289: Question 10.5: Draw the structures of major monohalo products in each of the following reactions: (i) (ii) (iii) (iv) (v) (vi) ANSWER: (i) (ii) (iii) (iv) (v) (vi) Page No 291: Question 10.6: Arrange each set of compounds in order of increasing boiling points. (i) Bromomethane, Bromoform, Chloromethane, Dibromomethane. (ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane. ANSWER: (i) For alkyl halides containing the same alkyl group, the boiling point increases with an increase in the atomic mass of the halogen atom. Since the atomic mass of Br is greater than that of Cl, the boiling point of bromomethane is higher than that of chloromethane. Further, for alkyl halides containing the same alkyl group, the boiling point increases with an increase in the number of halides. Therefore, the boiling point of Dibromomethane is higher than that of chloromethane and bromomethane, but lower than that of bromoform. Hence, the given set of compounds can be arranged in the order of their increasing boiling points as: Chloromethane < Bromomethane < Dibromomethane < Bromoform. (ii) For alkyl halides containing the same halide, the boiling point increases with an increase in the size of the alkyl group. Thus, the boiling point of 1-chlorobutane is higher than that of isopropyl chloride and 1-chloropropane. Further, the boiling point decreases with an increase in branching in the chain. Thus, the boiling point of isopropyl alcohol is lower than that of 1-chloropropane. Hence, the given set of compounds can be arranged in the increasing order of their boiling points as: Isopropyl chloride < 1-Chloropropane < 1-Chlorobutane Page No 307: Question 10.7: Which alkyl halide from the following pairs would you expect to react more rapidly by an SN2 mechanism? Explain your answer. (i) (ii) (iii) ANSWER: (i) 2-bromobutane is a 2° alkylhalide whereas 1-bromobutane is a 1° alkyl halide. The approaching of nucleophile is more hindered in 2-bromobutane than in 1-bromobutane. Therefore, 1-bromobutane reacts more rapidly than 2-bromobutane by an SN2 mechanism. (ii) 2-Bromobutane is 2° alkylhalide whereas 2-bromo-2-methylpropane is 3° alkyl halide. Therefore, greater numbers of substituents are present in 3° alkyl halide than in 2° alkyl halide to hinder the approaching nucleophile. Hence, 2-bromobutane reacts more rapidly than 2-bromo-2-methylpropane by an SN2 mechanism. (iii) Both the alkyl halides are primary. However, the substituent −CH3 is at a greater distance to the carbon atom linked to Br in 1-bromo-3-methylbutane than in 1-bromo-2-methylbutane. Therefore, the approaching nucleophile is less hindered in case of the former than in case of the latter. Hence, the former reacts faster than the latter by SN2 mechanism. Page No 307: Question 10.8: In the following pairs of halogen compounds, which compound undergoes faster SN1 reaction? (i) (ii) ANSWER: (i) SN1 reaction proceeds via the formation of carbocation. The alkyl halide (I) is 3° while (II) is 2°. Therefore, (I) forms 3° carbocation while (II) forms 2° carbocation. Greater the stability of the carbocation, faster is the rate of SN1 reaction. Since 3° carbocation is more stable than 2° carbocation. (I), i.e. 2−chloro-2-methylpropane, undergoes faster SN1 reaction than (II) i.e., 3-chloropentane. (ii) The alkyl halide (I) is 2° while (II) is 1°. 2° carbocation is more stable than 1° carbocation. Therefore, (I), 2−chloroheptane, undergoes faster SN1 reaction than (II), 1-chlorohexane. Page No 307: Question 10.9: Identify A, B, C, D, E, R and R1 in the following: ANSWER: Since D of D2O gets attached to the carbon atom to which MgBr is attached, C is Therefore, the compound R − Br is When an alkyl halide is treated with Na in the presence of ether, a hydrocarbon containing double the number of carbon atoms as present in the original halide is obtained as product. This is known as Wurtz …

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Here, you can find comprehensive NCERT solutions for Class 12 Science Chemistry Chapter 9 on Coordination Compounds. These solutions include easy-to-follow, step-by-step explanations. Widely appreciated by class 12 Science students, these Chemistry Coordination Compounds Solutions are invaluable for efficiently completing homework assignments and preparing for exams. All the questions and answers from the NCERT Book of Class 12 Science Chemistry Chapter 9 are available here at no cost. Page No 244: Question 9.1: Write the formulas for the following coordination compounds: (i) Tetraamminediaquacobalt(III) chloride (ii) Potassium tetracyanonickelate(II) (iii) Tris(ethane−1,2−diamine) chromium(III) chloride (iv) Amminebromidochloridonitrito-N-platinate(II) (v) Dichloridobis(ethane−1,2−diamine)platinum(IV) nitrate (vi) Iron(III) hexacyanoferrate(II) ANSWER: (i)  (ii)  (iii)  (vi)  (v)  (vi)  Page No 244: Question 9.2: 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) Potassium hexacyanoferrate(III) (iv) Potassium trioxalatoferrate(III) (v) Potassium tetrachloridopalladate(II) (vi) Diamminechlorido(methylamine)platinum(II) chloride Page No 247: Question 9.3: Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers: ANSWER: Trans-isomer is optically inactive. On the other hand, cis-isomer is optically active. (ii) Two optical isomers for  exist. Two optical isomers are possible for this structure. (iii)  A pair of optical isomers: It can also show linkage isomerism. and It can also show ionization isomerism. (iv) Geometrical (cis-, trans-) isomers of can exist. Page No 247: Question 9.4: Give evidence that [Co(NH3)5Cl]SO4 and [Co(NH3)5SO4]Cl are ionization isomers. ANSWER: When ionization isomers are dissolved in water, they ionize to give different ions. These ions then react differently with different reagents to give different products. Page No 254: Question 9.5: Explain on the basis of valence bond theory that [Ni(CN)4]2− ion with square planar structure is diamagnetic and the [NiCl4]2− ion with tetrahedral geometry is paramagnetic. ANSWER: Ni is in the +2 oxidation state i.e., in d8 configuration. There are 4 CN− ions. Thus, it can either have a tetrahedral geometry or square planar geometry. Since CN− ion is a strong field ligand, it causes the pairing of unpaired 3d electrons. It now undergoes dsp2 hybridization. Since all electrons are paired, it is diamagnetic. In case of [NiCl4]2−, Cl− ion is a weak field ligand. Therefore, it does not lead to the pairing of unpaired 3d electrons. Therefore, it undergoes sp3 hybridization. Since there are 2 unpaired electrons in this case, it is paramagnetic in nature. Page No 254: Question 9.6: [NiCl4]2− is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why? ANSWER: Though both [NiCl4]2− and [Ni(CO)4] are tetrahedral, their magnetic characters are different. This is due to a difference in the nature of ligands. Cl− is a weak field ligand and it does not cause the pairing of unpaired 3d electrons. Hence, [NiCl4]2− is paramagnetic. In Ni(CO)4, Ni is in the zero oxidation state i.e., it has a configuration of 3d8 4s2. But CO is a strong field ligand. Therefore, it causes the pairing of unpaired 3d electrons. Also, it causes the 4s electrons to shift to the 3d orbital, thereby giving rise to sp3 hybridization. Since no unpaired electrons are present in this case, [Ni(CO)4] is diamagnetic. Page No 254: Question 9.7: [Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3− is weakly paramagnetic. Explain. ANSWER: In both and , Fe exists in the +3 oxidation state i.e., in d5 configuration. Since CN− is a strong field ligand, it causes the pairing of unpaired electrons. Therefore, there is only one unpaired electron left in the d-orbital. Therefore, On the other hand, H2O is a weak field ligand. Therefore, it cannot cause the pairing of electrons. This means that the number of unpaired electrons is 5. Therefore, Thus, it is evident that is strongly paramagnetic, while is weakly paramagnetic. Page No 254: Question 9.8: Explain [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital complex. ANSWER: Oxidation state of cobalt = +3 Oxidation state of Ni = +2 Electronic configuration of cobalt = d6 Electronic configuration of nickel = d8 NH3 being a strong field ligand causes the pairing. Therefore, Cobalt can undergo d2sp3 hybridization.Hence, it is an inner orbital complex. If NH3 causes the pairing, then only one 3d orbital is empty. Thus, it cannot undergo d2sp3hybridization. Therefore, it undergoes sp3d2 hybridization.Hence, it forms an outer orbital complex. Page No 254: Question 9.9: Predict the number of unpaired electrons in the square planar [Pt(CN)4]2− ion. ANSWER: In this complex, Pt is in the +2 state. It forms a square planar structure. This means that it undergoes dsp2 hybridization. Now, the electronic configuration of Pd(+2) is 5d8. CN− being a strong field ligand causes the pairing of unpaired electrons. Hence, there are no unpaired electrons in Page No 254: Question 9.10: The hexaquo manganese(II) ion contains five unpaired electrons, while the hexacyanoion contains only one unpaired electron. Explain using Crystal Field Theory. ANSWER: Mn is in the +2 oxidation state. Mn is in the +2 oxidation state. The electronic configuration is d5. The electronic configuration is d5. The crystal field is octahedral. Water is a weak field ligand. Therefore, the arrangement of the electrons in is t2g3eg2. The crystal field is octahedral. Cyanide is a strong field ligand. Therefore, the arrangement of the electrons in  isT2g5eg0.     Hence, hexaaquo manganese (II) ion has five unpaired electrons, while hexacyano ion has only one unpaired electron. Page No 256: Question 9.11: Calculate the overall complex dissociation equilibrium constant for the Cu(NH3)42+ ion, given that β4 for this complex is 2.1 × 1013. ANSWER: β4 = 2.1 × 1013 The overall complex dissociation equilibrium constant is the reciprocal of the overall stability constant, β4. Page No 258: Question 9.1: Explain the bonding in coordination compounds in terms of Werner’s postulates. ANSWER: Werner’s postulates explain the bonding in coordination compounds as follows: (i) A metal exhibits two types of valencies namely, primary and secondary valencies. Primary valencies are satisfied by negative ions while secondary valencies are satisfied by both negative and neutral ions. (In modern terminology, the primary valency corresponds to the oxidation number of the metal ion, whereas the secondary valency refers to the coordination number of the metal ion. (ii) A metal ion has a definite number of secondary valencies around the central atom. Also, these valencies project in a specific direction in the space assigned to the definite geometry of the coordination compound. (iii) Primary valencies are usually ionizable, while secondary valencies are non-ionizable. Page No 258: Question 9.2: FeSO4 solution mixed with …

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Explore the NCERT Solutions for Class 12 Science Chemistry Chapter 8: “The D And F Block Elements,” featuring clear and concise step-by-step explanations. Widely embraced by class 12 Science students, these solutions prove invaluable for completing homework efficiently and preparing for exams. Free access to all questions and answers from the NCERT Book of class 12 Science Chemistry Chapter 8 is available here, making it a convenient resource for students. Page No 212: Question 8.1: Silver atom has completely filled d orbitals (4d10) in its ground state. How can you say that it is a transition element? ANSWER: Ag has a completely filled 4d orbital (4d10 5s1) in its ground state. Now, silver displays two oxidation states (+1 and +2). In the +1 oxidation state, an electron is removed from the s-orboital. However, in the +2 oxidation state, an electron is removed from the d-orbital. Thus, the d-orbital now becomes incomplete (4d9). Hence, it is a transition element. Page No 215: Question 8.2: In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomization of zinc is the lowest, i.e., 126 kJ mol−1. Why? ANSWER: The extent of metallic bonding an element undergoes decides the enthalpy of atomization. The more extensive the metallic bonding of an element, the more will be its enthalpy of atomization. In all transition metals (except Zn, electronic configuration: 3d10 4s2), there are some unpaired electrons that account for their stronger metallic bonding. Due to the absence of these unpaired electrons, the inter-atomic electronic bonding is the weakest in Zn and as a result, it has the least enthalpy of atomization. Page No 217: Question 8.3: Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why? ANSWER: Mn (Z = 25) = 3d5 4s2 Mn has the maximum number of unpaired electrons present in the d-subshell (5 electrons). Hence, Mn exhibits the largest number of oxidation states, ranging from +2 to +7. Page No 217: Question 8.4: The Eθ(M2+/M) value for copper is positive (+0.34V). What is possibly the reason for this? (Hint: consider its high ΔaHθ and low ΔhydHθ) ANSWER: The Eθ(M2+/M) value of a metal depends on the energy changes involved in the following: 1. Sublimation: The energy required for converting one mole of an atom from the solid state to the gaseous state. 2. Ionization: The energy required to take out electrons from one mole of atoms in the gaseous state. 3. Hydration: The energy released when one mole of ions are hydrated. Now, copper has a high energy of atomization and low hydration energy. Hence, the Eθ(M2+/M) value for copper is positive. Page No 219: Question 8.5: How would you account for the irregular variation of ionization enthalpies (first and second) in the first series of the transition elements? ANSWER: Ionization enthalpies are found to increase in the given series due to a continuous filling of the inner d-orbitals. The irregular variations of ionization enthalpies can be attributed to the extra stability of configurations such as d0, d5, d10. Since these states are exceptionally stable, their ionization enthalpies are very high. In case of first ionization energy, Cr has low ionization energy. This is because after losing one electron, it attains the stable configuration (3d5). On the other hand, Zn has exceptionally high first ionization energy as an electron has to be removed from stable and fully-filled orbitals (3d10 4s2). Second ionization energies are higher than the first since it becomes difficult to remove an electron when an electron has already been taken out. Also, elements like Cr and Cu have exceptionally high second ionization energies as after losing the first electron, they have attained the stable configuration (Cr+: 3d5 and Cu+: 3d10). Hence, taking out one electron more from this stable configuration will require a lot of energy. Page No 220: Question 8.6: Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only? ANSWER: Both oxide and fluoride ions are highly electronegative and have a very small size. Due to these properties, they are able to oxidize the metal to its highest oxidation state. Page No 220: Question 8.7: Which is a stronger reducing agent Cr2+ or Fe2+ and why? ANSWER: The following reactions are involved when Cr2+ and Fe2+ act as reducing agents. Cr2+ Cr3+ Fe2+ Fe3+ The value is −0.41 V and  is +0.77 V. This means that Cr2+ can be easily oxidized to Cr3+, but Fe2+ does not get oxidized to Fe3+ easily. Therefore, Cr2+ is a better reducing agent that Fe3+. Page No 222: Question 8.8: Calculate the ‘spin only’ magnetic moment of M2+(aq) ion (Z = 27). ANSWER: Z = 27  [Ar] 3d7 4s2  M2+ = [Ar] 3d7 3d7 =  i.e., 3 unpaired electrons n = 3 μ ≈ 4 BM Page No 224: Question 8.9: Explain why Cu+ ion is not stable in aqueous solutions? ANSWER: In an aqueous medium, Cu2+ is more stable than Cu+. This is because although energy is required to remove one electron from Cu+ to Cu2+, high hydration energy of Cu2+ compensates for it. Therefore, Cu+ ion in an aqueous solution is unstable. It disproportionates to give Cu2+ and Cu. Page No 232: Question 8.10: Actinoid contraction is greater from element to element than lanthanoid contraction. Why? ANSWER: In actinoids, 5f orbitals are filled. These 5f orbitals have a poorer shielding effect than 4f orbitals (in lanthanoids). Thus, the effective nuclear charge experienced by electrons in valence shells in case of actinoids is much more that that experienced by lanthanoids. Hence, the size contraction in actinoids is greater as compared to that in lanthanoids. Page No 234: Question 8.1: Write down the electronic configuration of: (i) Cr3++ (iii) Cu+(v) Co2+ (vii) Mn2+ (ii) Pm3+(iv) Ce4+ (vi) Lu2+(viii) Th4+ ANSWER: (i) Cr3+: 1s2 2s2 2p6 3s2 3p6 3d3 Or, [Ar]18 3d3 (ii) Pm3+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f4 Or, [Xe]54 3d3 (iii) Cu+: 1s2 2s2 2p6 3s2 3p6 3d10 Or, [Ar]18 3d10 (iv) Ce4+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 Or, [Xe]54 (v) Co2+: 1s2 2s2 2p6 3s2 3p6 3d7 Or, [Ar]18 3d7 (vi) Lu2+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f14 5d1 Or, [Xe]54 2f14 3d3 (vii) Mn2+: 1s2 2s2 2p6 3s2 3p6 3d5 Or, [Ar]18 3d5 (viii) Th4+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f14 5s2 5p6 5d10 6s2 6s6 Or, [Rn]86 Page No 234: Question 8.2: Why are Mn2+compounds more stable than Fe2+towards oxidation to their +3 state? ANSWER: Electronic configuration of Mn2+ is [Ar]18 3d5. Electronic configuration of Fe2+ is [Ar]18 3d6. It is known that half-filled and fully-filled orbitals are more stable. Therefore, Mn in (+2) state has a stable d5 configuration. This is the reason Mn2+ shows resistance to oxidation to Mn3+. Also, Fe2+ has 3d6 configuration and by losing one …

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Find comprehensive solutions for Class 12 Science Chemistry Chapter 7 – “The P Block Elements” in the NCERT book. These step-by-step explanations are highly sought after by Chemistry students, aiding in homework completion and exam preparation. The popularity of these solutions among class 12 Science students is attributed to their effectiveness. Access free answers to all questions from the NCERT Book of Class 12 Science Chemistry Chapter 7, ensuring a valuable resource for your studies. Page No 169: Question 7.1: Why are pentahalides more covalent than trihalides? ANSWER: In pentahalides, the oxidation state is +5 and in trihalides, the oxidation state is +3. Since the metal ion with a high charge has more polarizing power, pentahalides are more covalent than trihalides. Page No 169: Question 7.2: Why is BiH3 the strongest reducing agent amongst all the hydrides of Group 15 elements? ANSWER: As we move down a group, the atomic size increases and the stability of the hydrides of group 15 elements decreases. Since the stability of hydrides decreases on moving from NH3 to BiH3, the reducing character of the hydrides increases on moving from NH3 to BiH3. Page No 170: Question 7.3: Why is N2 less reactive at room temperature? ANSWER: The two N atoms in N2 are bonded to each other by very strong triple covalent bonds. The bond dissociation energy of this bond is very high. As a result, N2 is less reactive at room temperature. Page No 172: Question 7.4: Mention the conditions required to maximise the yield of ammonia. ANSWER: Ammonia is prepared using the Haber’s process. The yield of ammonia can be maximized under the following conditions: (i) High pressure (∼ 200 atm) (ii) A temperature of ∼700 K (iii) Use of a catalyst such as iron oxide mixed with small amounts of K2O and Al2O3 Page No 172: Question 7.5: How does ammonia react with a solution of Cu2+? ANSWER: NH3 acts as a Lewis base. It donates its electron pair and forms a linkage with metal ion. Page No 173: Question 7.6: What is the covalence of nitrogen in N2O5? ANSWER: From the structure of N2O5, it is evident that the covalence of nitrogen is 4. Page No 177: Question 7.7: Bond angle in is higher than that in PH3. Why? ANSWER: In PH3, P is sp3 hybridized. Three orbitals are involved in bonding with three hydrogen atoms and the fourth one contains a lone pair. As lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion, the tetrahedral shape associated with sp3 bonding is changed to pyramidal. PH3 combines with a proton to form  in which the lone pair is absent. Due to the absence of lone pair in, there is no lone pair-bond pair repulsion. Hence, the bond angle in is higher than the bond angle in PH3. Page No 177: Question 7.8: What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO2? ANSWER: White phosphorous dissolves in boiling NaOH solution (in a CO2 atmosphere) to give phosphine, PH3. Page No 178: Question 7.9: What happens when PCl5 is heated? ANSWER: All the bonds that are present in PCl5 are not similar. It has three equatorial and two axial bonds. The equatorial bonds are stronger than the axial ones. Therefore, when PCl5 is heated strongly, it decomposes to form PCl3. Page No 178: Question 7.10: Write a balanced equation for the hydrolytic reaction of PCl5 in heavy water. ANSWER: Page No 180: Question 7.11: What is the basicity of H3PO4? ANSWER: H3PO4 Since there are three OH groups present in H3PO4, its basicity is three i.e., it is a tribasic acid. Page No 180: Question 7.12: What happens when H3PO3 is heated? ANSWER: H3PO3,on heating, undergoes disproportionation reaction to form PH3 and H3PO4. The oxidation numbers of P in H3PO3,PH3, and H3PO4 are +3, −3, and +5 respectively. As the oxidation number of the same element is decreasing and increasing during a particular reaction, the reaction is a disproportionation reaction. Page No 183: Question 7.13: List the important sources of sulphur. ANSWER: Sulphur mainly exists in combined form in the earth’s crust primarily as sulphates [gypsum (CaSO4.2H2O), Epsom salt (MgSO4.7H2O), baryte (BaSO4)] and sulphides [(galena (PbS), zinc blends (ZnS), copper pyrites (CuFeS2)]. Page No 183: Question 7.14: Write the order of thermal stability of the hydrides of Group 16 elements. ANSWER: The thermal stability of hydrides decreases on moving down the group. This is due to a decrease in the bond dissociation enthalpy (H−E) of hydrides on moving down the group. Therefore, Page No 183: Question 7.15: Why is H2O a liquid and H2S a gas? ANSWER: H2O has oxygen as the central atom. Oxygen has smaller size and higher electronegativity as compared to sulphur. Therefore, there is extensive hydrogen bonding in H2O, which is absent in H2S. Molecules of H2S are held together only by weak van der Waal’s forces of attraction. Hence, H2O exists as a liquid while H2S as a gas. Page No 185: Question 7.16: Which of the following does not react with oxygen directly? Zn, Ti, Pt, Fe ANSWER: Pt is a noble metal and does not react very easily. All other elements, Zn, Ti, Fe, are quite reactive. Hence, oxygen does not react with platinum (Pt) directly. Page No 185: Question 7.17: Complete the following reactions: (i) C2H4 + O2→ (ii) 4Al + 3O2→ ANSWER: (i)  (ii)  Page No 187: Question 7.18: Why does O3 act as a powerful oxidising agent? ANSWER: Ozone is not a very stable compound under normal conditions and decomposes readily on heating to give a molecule of oxygen and nascent oxygen. Nascent oxygen, being a free radical, is very reactive. Therefore, ozone acts as a powerful oxidising agent. Page No 187: Question 7.19: How is O3 estimated quantitatively? ANSWER: Quantitatively, ozone can be estimated with the help of potassium iodide. When ozone is made to react with potassium iodide solution buffered with a borate buffer (pH 9.2), iodine is liberated. This liberated iodine can be titrated against a standard solution of sodium thiosulphate using starch as an indicator. The reactions involved in the process are given below. Page No 189: Question 7.20: What happens when sulphur dioxide is …

NCERT Solutions for Class 12 Science Chemistry Chapter 7 – The P Block Elements Read More »

Here, you will find NCERT solutions for Class 12 Science Chemistry Chapter 6 – “General Principles And Processes Of Isolation Of Elements” with clear and straightforward explanations. These solutions are widely favored by Class 12 Science students studying Chemistry. They serve as a valuable resource for completing homework efficiently and preparing for exams. All the questions and answers from Chapter 6 of the NCERT Book for Class 12 Science Chemistry are available here at no cost. Page No 150: Question 6.1: Which of the ores mentioned in Table 6.1 can be concentrated by magnetic separation method? ANSWER: If the ore or the gangue can be attracted by the magnetic field, then the ore can be concentrated by the process of magnetic separation. Among the ores mentioned in table 6.1, the ores of iron such as haematite (Fe2O3), magnetite (Fe3O4), siderite (FeCO3), and iron pyrites (FeS2) can be separated by the process of magnetic separation. Page No 150: Question 6.2: What is the significance of leaching in the extraction of aluminium? ANSWER: In the extraction of aluminium, the significance of leaching is to concentrate pure alumina (Al2O3) from bauxite ore. Bauxite usually contains silica, iron oxide, and titanium oxide as impurities. In the process of leaching, alumina is concentrated by digesting the powdered ore with a concentrated solution of NaOH at 473-523 K and 35-36 bar. Under these conditions, alumina (Al2O3) dissolves as sodium meta-aluminate and silica (SiO2) dissolves as sodium silicate leaving the impurities behind. The impurities are then filtered and the solution is neutralized by passing CO2 gas. In this process, hydrated Al2O3 gets precipitated and sodium silicate remains in the solution. Precipitation is induced by seeding the solution with freshly prepared samples of hydrated Al2O3. Hydrated alumina thus obtained is filtered, dried, and heated to give back pure alumina (Al2O3). Page No 157: Question 6.3: The reaction, is thermodynamically feasible as is apparent from the Gibbs energy value. Why does it not take place at room temperature? ANSWER: The change in Gibbs energy is related to the equilibrium constant, K as . At room temperature, all reactants and products of the given reaction are in the solid state. As a result, equilibrium does not exist between the reactants and the products. Hence, the reaction does not take place at room temperature. However, at a higher temperature, chromium melts and the reaction takes place. We also know that according to the equation Increasing the temperature increases the value of making the value of  more and more negative. Therefore, the reaction becomes more and more feasible as the temperature is increased. Page No 157: Question 6.4: Is it true that under certain conditions, Mg can reduce SiO2 and Si can reduce MgO? What are those conditions? ANSWER: The temperature range in which  is lesser than, Mg can reduce SiO2 to Si. On the other hand, the temperatures range in which  is less than, Si can reduce MgO to Mg. The temperature at which ΔfG curves of these two substances intersect is 1966 K. Thus, at temperatures less than 1966 K, Mg can reduce SiO2 and above 1966 K, Si can reduce MgO. Page No 163: Question 6.1: Copper can be extracted by hydrometallurgy but not zinc. Explain. ANSWER: The reduction potentials of zinc and iron are lower than that of copper. In hydrometallurgy, zinc and iron can be used to displace copper from their solution. But to displace zinc, more reactive metals i.e., metals having lower reduction potentials than zinc such as Mg, Ca, K, etc. are required. But all these metals react with water with the evolution of H2 gas. As a result, these metals cannot be used in hydrometallurgy to extract zinc. Hence, copper can be extracted by hydrometallurgy but not zinc. Page No 163: Question 6.2: What is the role of depressant in froth floatation process? ANSWER: In the froth floatation process, the role of the depressants is to separate two sulphide ores by selectively preventing one ore from forming froth. For example, to separate two sulphide ores (ZnS and Pbs), NaCN is used as a depressant which selectively allows PbS to come with froth, but prevents ZnS from coming to froth. This happens because NaCN reacts with ZnS to form Na2[Zn(CN)4]. Page No 163: Question 6.3: Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction? ANSWER: The Gibbs free energy of formation (ΔfG) of Cu2S is less than that of and. Therefore, H2 and C cannot reduce Cu2S to Cu. On the other hand, the Gibbs free energy of formation of  is greater than that of. Hence, C can reduce Cu2O to Cu. Hence, the extraction of copper from its pyrite ore is difficult than from its oxide ore through reduction. Page No 163: Question 6.4: Explain: (i) Zone refining (ii) Column chromatography. ANSWER: (i) Zone refining: This method is based on the principle that impurities are more soluble in the molten state of metal (the melt) than in the solid state. In the process of zone refining, a circular mobile heater is fixed at one end of a rod of impure metal. As the heater moves, the molten zone of the rod also moves with it. As a result, pure metal crystallizes out of the melt and the impurities pass onto the adjacent molten zone. This process is repeated several times, which leads to the segregation of impurities at one end of the rod. Then, the end with the impurities is cut off. Silicon, boron, gallium, indium etc. can be purified by this process. (ii) Column chromatography: Column chromatography is a technique used to separate different components of a mixture. It is a very useful technique used for the purification of elements available in minute quantities. It is also used to remove the impurities that are not very different in chemical properties from the element to be purified. Chromatography is based on the principle that different components of a mixture are differently adsorbed on an adsorbent. In chromatography, there are two phases: mobile phase and stationary phase. The stationary phase is immobile and immiscible. Al2O3 column …

NCERT Solutions for Class 12 Science Chemistry Chapter 6 – General Principles And Processes Of Isolation Of Elements Read More »

Find comprehensive NCERT Solutions for Class 12 Science Chemistry Chapter 5 on Surface Chemistry, featuring clear, step-by-step explanations. These solutions are widely favored by Class 12 Science students as they offer a convenient way to complete homework assignments and prepare for exams. All questions and answers from Chapter 5 of the NCERT Book for Class 12 Science Chemistry are available here at no cost, providing valuable assistance to students in their studies. Page No 127: Question 5.1: Write any two characteristics of Chemisorption. ANSWER: 1. Chemisorption is highly specific in nature. It occurs only if there is a possibility of chemical bonding between the adsorbent and the adsorbate. 2. Like physisorption, chemisorption also increases with an increase in the surface area of the adsorbent. Page No 127: Question 5.2: Why does physisorption decrease with the increase of temperature? ANSWER: Physisorption is exothermic in nature. Therefore, in accordance with Le-Chateliere’s principle, it decreases with an increase in temperature. This means that physisorption occurs more readily at a lower temperature. Page No 127: Question 5.3: Why are powdered substances more effective adsorbents than their crystalline forms? ANSWER: Powdered substances are more effective adsorbents than their crystalline forms because when a substance is powdered, its surface area increases and physisorption is directly proportional to the surface area of the adsorbent. Page No 133: Question 5.4: Why is it necessary to remove CO when ammonia is obtained by Haber’s process? ANSWER: It is important to remove CO in the synthesis of ammonia as CO adversely affects the activity of the iron catalyst, used in Haber’s process. Page No 133: Question 5.5: Why is the ester hydrolysis slow in the beginning and becomes faster after sometime? ANSWER: Ester hydrolysis can be represented as: The acid produced in the reaction acts as a catalyst and makes the reaction faster. Substances that act as catalysts in the same reaction in which they are obtained as products are known as autocatalysts. Page No 133: Question 5.6: What is the role of desorption in the process of catalysis? ANSWER: The role of desorption in the process of catalysis is to make the surface of the solid catalyst free for the fresh adsorption of the reactants on the surface. Page No 145: Question 5.7: What modification can you suggest in the Hardy-Schulze law? ANSWER: Hardy-Schulze law states that ‘the greater the valence of the flocculating ion added, the greater is its power to cause precipitation.’ This law takes into consideration only the charge carried by an ion, not its size. The smaller the size of an ion, the more will be its polarising power. Thus, Hardy-Schulze law can be modified in terms of the polarising power of the flocculating ion. Thus, the modified Hardy-Schulze law can be stated as ‘the greater the polarising power of the flocculating ion added, the greater is its power to cause precipitation.’ Page No 145: Question 5.8: Why is it essential to wash the precipitate with water before estimating it quantitatively? ANSWER: When a substance gets precipitated, some ions that combine to form the precipitate get adsorbed on the surface of the precipitate. Therefore, it becomes important to wash the precipitate before estimating it quantitatively in order to remove these adsorbed ions or other such impurities. Page No 146: Question 5.1: Distinguish between the meaning of the terms adsorption and absorption. Give one example of each. ANSWER: Adsorption is a surface phenomenon of accumulation of molecules of a substance at the surface rather than in the bulk of a solid or liquid. The substance that gets adsorbed is called the ‘adsorbate’ and the substance on whose surface the adsorption takes place is called the ‘adsorbent’. Here, the concentration of the adsorbate on the surface of the adsorbent increases. In adsorption, the substance gets concentrated at the surface only. It does not penetrate through the surface to the bulk of the solid or liquid. For example, when we dip a chalk stick into an ink solution, only its surface becomes coloured. If we break the chalk stick, it will be found to be white from inside. On the other hand, the process of absorption is a bulk phenomenon. In absorption, the substance gets uniformly distributed throughout the bulk of the solid or liquid. Page No 146: Question 5.2: What is the difference between physisorption and chemisorption? ANSWER: Physisorption Chemisorption 1. In this type of adsorption, the adsorbate is attached to the surface of the adsorbent with weak van der Waal’s forces of attraction. In this type of adsorption, strong chemical bonds are formed between the adsorbate and the surface of the adsorbent. 2. No new compound is formed in the process. New compounds are formed at the surface of the adsorbent. 3. It is generally found to be reversible in nature. It is usually irreversible in nature. 4. Enthalpy of adsorption is low as weak van der Waal’s forces of attraction are involved. The values lie in the range of 20-40 kJ mol−1. Enthalpy of adsorption is high as chemical bonds are formed. The values lie in the range of 40-400 kJ mol−1. 5. It is favoured by low temperature conditions. It is favoured by high temperature conditions. 6. It is an example of multi-layer adsorption It is an example of mono-layer adsorption. Page No 146: Question 5.3: Give reason why a finely divided substance is more effective as an adsorbent. ANSWER: Adsorption is a surface phenomenon. Therefore, adsorption is directly proportional to the surface area. A finely divided substance has a large surface area. Both physisorption and chemisorption increase with an increase in the surface area. Hence, a finely divided substance behaves as a good adsorbent. Page No 146: Question 5.4: What are the factors which influence the adsorption of a gas on a solid? ANSWER: There are various factors that affect the rate of adsorption of a gas on a solid surface. (1) Nature of the gas: Easily liquefiable gases such as NH3, HCl etc. are adsorbed to a great extent in comparison to gases such as H2, …

NCERT Solutions for Class 12 Science Chemistry Chapter 5 – Surface Chemistry Read More »