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In This Post we are  providing Chapter-13 AMINES NCERT MOST IMPORTANT QUESTIONS for Class 12 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON AMINES

Question 1.
Write the main products of the following reactions:

Answer:

Answer:

Answer:

Question 2.
Write the main products of the following reactions:

Answer:

Answer:

Answer:

Question 3.
Account for the following:
(i) Primary amines (R-NH₂) have a higher boiling point than tertiary amines (R₃N).
(ii) Aniline does not undergo Friedel — Crafts reaction.
(iii) (CH₃)₂NH is more basic than (CH₃)₃N in an aqueous solution.
OR
Give the structures of A, B, and C in the following reactions:

Answer:
(i) Primary amines (RNH₂) have two hydrogen atoms on the N atom and therefore, form intermolecular hydrogen bonding.

Tertiary amines (R₃N) do not have hydrogen atoms on the N atom and therefore, these do not form hydrogen bonds. As a result of hydrogen bonding in primary amines, they have higher boiling points than tertiary amines of comparable molecular mass. For example, b.p. of n-butylamine is 351 K while that of tert-butylamine is 319 K.

(ii) Aniline being a Lewis base reacts with Lewis acid such as AlCl₃ to form a salt.

As a result, N of aniline acquires +ve charge and hence it acts as a strong deactivating group for electrophilic substitution reactions. Hence aniline does not undergo Friedel Crafts reaction.

(iii) Due to the presence of lone pair of electrons on the N atom, amines are basic in nature. The methyl group is the electron releasing group (+I inductive effect) and therefore, it increases the electron density on the N atom, and therefore, basic character increases, so that (CH₃)₃N should be more basic than (CH₃)₂NH. But tertiary ammonium ion formed from tertiary amines is less hydrated than secondary ammonium ion formed from secondary amine. Therefore, (CH₃)₃N has less tendency to form ammonium ion, and consequently, it is less basic than (CH₃)₂NH. Thus, (CH₃)₂NH is more basic than (CH₃)₃N due to the combined effect of inductive effect and hydration effect.
OR

Question 4.
Give the structures of A, B, and C in the following reactions:

OR
How will you convert the following:
(i) Nitrobenzene into aniline
(ii) Ethanoic acid into methanamine
(iii) Aniline into N-phenylethylamine
(Write the chemical equations involved.) (CBSE Delhi 2014)
Answer:

OR

Question 5.
Write chemical equations for the following conversions:
(i) Nitrobenzene to benzoic acid.
Answer:

(ii) Benzyl chloride to 2-phenytethanamine.
Answer:

(iii) Aniline to benzyl alcohol. (CBSE Delhi 2012)
Answer:

Question 12.
(a) Identify ‘A’ and ‘B’ in the following reaction:

Answer:

(b) Why is ethylamine soluble in water whereas aniline is not?
Answer:
Ethylamine dissolves in water due to intermolecular hydrogen bonding as shown below:

However, because of the large hydrophobic part (i.e. hydrocarbon part) of aniline, the extent of hydrogen bonding is less and therefore, aniline is insoluble in water.

Question 6.
(a) Identify X and Y in the following:

Answer:
X =
benzene diazonium ch(orlde,
Y =
 Cyanobenzene

(b) Amino group is o, p-directing for aromatic electrophilic substitution reactions. Why does aniline on nitration give m-nitroaniline? (CBSE 2019C)
Answer:
Under strongly acidic conditions of nitration, most of the aniline is converted into anilinium ion having an NH₃⁺ group. This group is an m-directing group, therefore, m-nitro aniline is also obtained along with o- and p-products.

Question 14.
Give the structures of A, B, and C in the following reactions: (CBSE Delhi 2013)

Answer:

Answer:

Question 7.
Do as directed:
(i) Arrange the following compounds in the increasing order of their basic strength in an aqueous solution:
CH₃NH₃, (CH₃)₃N, (CH₃)₂NH.
Answer:
(CH₃)₃N < CH₃NH₂ < (CH₃)₂ NH

(ii) Identify ‘A’ and ‘B’:

Answer:
A: C6H5N2+ Cl- B: C6H5OH

(iii) Write the equation of carbylamine reaction. (CBSE 2018C)
Answer:

Question 8.
(i) Illustrate the following reactions giving a suitable example in each case:
(a) Hoffmann bromamide degradation reaction
(b) Diazotisation
(c) Gabriel phthalimide synthesis
(ii) Distinguish between the following pairs of compounds:
(a) Aniline and N-methylaniline
(b) (CH₃)₂NH and (CH₃)₃N
OR
(i) Write the structures of main products when benzene diazonium chloride (C₆H₅N₂⁺Cl^–) reacts with the following reagents:
(a) CuCN/KCN
(b) H₂0
(c) CH₃CH₂OH
(ii) Arrange the following:
(a) C₂H₅NH₂, C₂H₅OH, (CH₃)₃N – in the increasing order of their boiling point.
(b) Aniline, p-nitroaniline, p-methyl aniline – in the increasing order of their basic strength. (CBSE Delhi 2015)
Answer:
(i) (a) Hoffmann bromamide degradation reaction: Primary amines can be prepared from amides by treatment with Br₂ and KOH solution. The amine formed contains one carbon atom less than the parent amide.

(b) Diazotisation: The reaction of aniline or other aromatic amines, with nitrous acid at 0-5 °C to form diazonium salts is called diazotization. Nitrous acid needed for this reaction is prepared in situ by the action of dil. HCl on NaNO₂.

(c) Gabriel’s phthalimide synthesis. This method is used for preparing only primary amines. In this method, phthalimide is treated with alcoholic KOH to give potassium phthalimide, which is treated with an alkyl halide or benzyl halide to form N-alkyl or aryl phthalimide. The hydrolysis of N-alkyl phthalimide with 20% HCl under pressure or refluxing with NaOH gives primary amine.

Phthalic acid can again be converted into phthalimide and is used again and again. This method is very useful because it gives pure amines. Aryl halides cannot be converted to arylamines by Gabriel synthesis because they do not undergo nucleophilic substitution with potassium phthalimide.

(ii) (a) Add an alcoholic solution of KOH and CHCl3 to the compounds. Aniline gives the foul smell of isocyanide whereas N-methyl aniline does not give a foul smell.

(b) When treated with Hinsberg’s reagent (benzene sulphonyl chloride, C₆H₅SO₂CI), dimethylamine, (CH₃)₂NH gives precipitate which is insoluble in aqueous KOH.

(CH₃)₃N does not react with Hinsberg’s reagent.
Or

(ii) (a) (CH₃)₂ N < C₂H₅NH₂ < C₂H₅OH
(b) p-nitroaniline < aniline < p-methylaniline

Question 9.
Give the IUPAC names of the following compounds:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

(CBSE Delhi 2017)
Answer:

(CBSE Delhi 2017)
Answer:

Answer:

Question 10.
Complete the following reactions:

Answer:

Answer:

Answer:

Answer:

Answer:

In This Post we are  providing Chapter-14 BIOMOLECULES NCERT MOST IMPORTANT QUESTIONS for Class 12 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON BIOMOLECULES

Question 1.
Define the following terms:
(i) Glycosidic linkage
Answer:
Glycosidic linkage: The condensation of hydroxyl groups of two monosaccharides to form a link between them is called glycosidic linkage.

(ii) Invert sugar
Answer:
Invert sugar: The sugar which on hydrolysis with dilute adds or enzymes gives mixture having specific rotation opposite to the original is called invert sugar. For example, sucrose is inverted sugar.

(iii) Oligosaccharides
Answer:
Oligosaccharides: These are the carbohydrates that give two to ten monosaccharide molecules on hydrolysis. These are further classified as disaccharides, trisaccharides, tetrasaccharides, etc. depending upon the number of monosaccharide units present in their molecules. For example, Disaccharides: Sucrose, lactose, maltose. All these have the molecular

Formula C₁₂H₂₂O₁₁.
Trisaccharide: Raffinose (C₁₈H₃₂O₁₆).
Tetrasaccharides: Stachyose (C₂₄H₄₂O₂₁ ).

Question 2.
Define the following terms:
(i) Nucleotide
Answer:
Nucleotide: A unit formed by the combination of a nitrogen-containing heterocyclic base, a pentose sugar and a phosphoric acid group.

(ii) Anomers
Answer:
Anomers: The anomers are the isomers formed due to the change in the configuration of the -OH group at C-1 of glucose. For example, α-and β-forms of glucose are anomers.

(iii) Essential amino acids.
Answer:
Essential amino acids: The amino acids which cannot be made by our bodies and must be supplied in our diet for the growth of the body are called essential amino acids.

Question 3.
Differentiate between the following:
(a) Fibrous protein and Globular protein
Answer:
Difference between fibrous protein and globútar protein:

Fibrous	Globular
(i) The polypeptide chains run parallel and are held together by hydrogen and disulphide bonds.	1. The polypeptide chains colt around to give a spherical shape.
(ii) These are insoluble in water.	2. These are soluble in water.
(iii) For example, keratin in hair	3. For example, albumin in egg.

(b) Essential amino acids and Non-essential amino acids
Answer:
Difference between essential amino acids and non-essential amino acids:

Essential amino acids	Non-essential amino acids
These are not synthesìsed in our body and must be supplied in the diet.	These are synthesised in our body and not required in our diet.
For example, valine.	For example, alanine

(C) Amylose and Amylopectin (CBSE AI 2019)
Answer:
Difference between amylose and amylopectin:

Amylose	Amylopectin
It consists of branched polymeric chains of α – D – glucose.	It consists of a long straight chain of α – D – glucose.
It is water-insoluble.	It is water-soluble.

Question 4.
What is essentially the difference between the α-form of D-glucose and β-form of D-glucose? Explain.
Answer:
α-form and β-form of glucose differ in the orientation of —H and —OH groups around the C₁ atom. The isomer having the -OH group on the right is called α-D-glucose while the one having -OH group on the left is called β-D-glucose. Such pairs of optical isomers which differ in the configuration only around C₁ are called anomers. The structures of these two may be shown below:

These two forms are crystalline and have different melting points and optical rotations. For example, α-form of glucose has m.p. 419 K and |α|_D = +111° and the β-form of glucose has m.p. 423 K and |α|_D = +19.2°.

Question 5.
Explain the following:
(i) Amino acids behave like salts rather than simple amines or carboxylic acids. (CBSE 2018C)
Answer:
Due to the formation of zwitterion.

(ii) The two strands of DNA are complementary to each other.
Answer:
The two strands are complementary to each other because the hydrogen bonds are formed between specific pairs of bases.

(iii) Reaction of glucose indicates that the carbonyl group is present as an aldehydic group in the open structure of glucose.
Answer:

Glucose gets oxidised to gluconic acid on reaction with a mild oxidising agent like Bromine water.

Question 6.
What is essentially the difference between α-glucose and β-glucose? What is meant by the pyranose structure of glucose?
Answer:
α-form and β-form of glucose differ in the orientation of -H and -OH groups around C, atom. The isomer having the -OH group on the right is called α-D-glucose while the one having -OH group on the left is called β-D-glucose. Such pairs of optical isomers which differ in the configuration only around C₁ are called anomers. The structures of these two may be shown below:

These two forms are crystalline and have different melting points and optical rotations. For example, α-form of glucose has m.p. 419 K and |α|_D = + 111° and the β-form of glucose has m.p. 423 K and |α|_D = + 19.2°.

The α-D-glucose and β-D-glucose can be drawn in a simple six-membered ring form called pyranose structures. These resemble pyran which is a six-membered heterocyclic ring containing five carbon atoms and one oxygen atom.

These are known as pyranose structures and are shown below:

Question 7.
Which sugar is called invert sugar? Why is it called so?
Answer:
Sucrose is called invert sugar. The sugar obtained from sugar beet is a colourless, crystalline and sweet substance. It is very soluble in water and its aqueous solution is dextrorotatory having [α]_D = + 66.5°.

On hydrolysis with dilute acids or enzyme invertase, cane sugar gives an equimolar mixture of D-(+)-glucose and D-(-)-fructose.

So, sucrose is dextrorotatory but after hydrolysis, gives dextrorotatory glucose and laevorotatory fructose. D-(-)-fructose has a greater specific rotation than D-(+)- glucose. Therefore, the resultant solution upon hydrolysis is laevorotatory in nature with a specific rotation of (-39.9°). Since there is a change in the sign of rotation from Dextro before hydrolysis to Laevo after hydrolysis, the reaction is called Inversion reaction and the mixture (glucose and fructose) is called invert sugar.

Question 8.
How do you explain the presence of an aldehydic group in a glucose molecule?
Answer:
Glucose reacts with hydroxylamine to form a monoxime and adds one molecule of hydrogen cyanide to give cyanohydrin.

Therefore, it contains a carbonyl group which can be an aldehyde or a ketone. On mild oxidation with bromine water, glucose gives gluconic acid which is a carboxylic acid-containing six carbon atoms.

This indicates that the carbonyl group present in glucose is an aldehydic group.

Question 9.
Define the following with an example of each: (CBSE 2018)
(i) Polysaccharides
(ii) Denatured protein
(iii) Essential amino acids
OR
(i) Write the product when D-glucose reacts with the cone. HNO₃.
(ii) Amino acids show amphoteric behaviour. Why?
(iii) Write one difference between a-helix and p-pleated structures of proteins.
Answer:
(i) Carbohydrates that give a large number of monosaccharide units on hydrolysis or a large number of monosaccharide units joined together by glycosidic linkage, e.g. starch, glycogen, cellulose.
(ii) Proteins that lose their biological activity or proteins in which secondary and tertiary structures are destroyed, e.g. curdling of milk.
(iii) Amino acids cannot be synthesised in the body. e.g. Valine / Leucine

OR
(i) Saccharic acid / COOH-(CHOH)₄-COOH
(ii) Due to the presence of carboxyl and amino group in the same molecule or due to formation of zwitterion or dipolar ion.
(iii) a-helix has intramolecular hydrogen bonding while p-pleated has intermolecular hydrogen bonding / a-helix results due to regular coiling of polypeptide chains while in p-pleated all polypeptide chains are stretched and arranged side by side.

Question 10.
Describe the term D- and L-configuration used for sugars with examples.
Answer:
The sugars are divided into two families: the D-family and L-family which have definite configurations. These configurations are represented with respect to glyceraldehyde as the standard. The glyceraldehyde may be presented in two forms:

The D-configuration has —OH attached to the carbon adjacent to —CH₂OH on the right while L-configuration has —OH attached to the carbon adjacent to —CH₂OH on left.

The sugars are calLed D- or L- depending upon whether the configuration of the molecule is related to D-glyceraldehyde or L-glyceraldehyde. It has been found that all naturally occurring sugars beLong to D-series, e.g. D-glucose, D-ribose and D-fructose.

In This Post we are  providing Chapter-12 ALDEHYDES, KETONES AND CARBOXYLIC ACIDS NCERT MOST IMPORTANT QUESTIONS for Class 12 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON ALDEHYDES, KETONES AND CARBOXYLIC ACIDS

Question 1.
(a) Write the structures of A and B in the following reactions:

(b) Distinguish between:
(i) C₆H₅— COCH₃ and C₆H₅— CHO
(ii) CH₃COOH and HCOOH
(c) Arrange the following in the increasing order of their boiling points:
CH₃CHO, CH₃COOH, CH₃CH₂OH
Answer:

(b) (i) Benzaldehyde and acetophenone :
By Iodoform test: Acetophenone being a methyl ketone on treatment with I₂ and NaOH (NaOI) undergoes iodoform test to give yellow ppt. of iodoform on heating whereas benzaldehyde does not.

(ii) CH₃COOH (Acetic acid) and HCOOH (Formic acid). Formic acid is the only acid which contains aldehydic group and thus shows reactions with Tollen’s reagent (silver nitrate) and Fehling’s solution which Acetic acid does not show.
Tollen’s Test:
Add ammonical solution of silver nitrate to both the compounds, HCOOH gives silver mirror but CH₃COOH does not

(c) CH₃CHO < CH₃CH₂OH < CH₃COOH

Question 2.
(a) Write the chemical reaction involved in Wolff-Kishner reduction.
(b) Arrange the following in the increasing order of their reactivity towards nucleophilic addition reaction:
C₆H₅COCH₃, CH₃— CHO, CH₃COCH₃
(c) Why carboxylic acid does not give reactions of carbonyl group?
(d) Write the product in the following reaction

(e) A and B are two functional isomers of compound C₃H₆O. On heating with NaOH and I₂, isomer B forms yellow precipitate of iodoform whereas isomer A does not form any precipitate. Write the formulae of A and B.
Answer:
(a) Wolff-Kishner reduction reaction : The reduction of aldehydes and ketones to the corresponding hydrocarbons by heating them with hydrazine and KOH or potassium tert-butoxide in a high boiling solvent like ethylene glycol is called Wolff-Kishner reduction.

(b) C₆H₅COCH₃ < CH₃COCH₃ < CH₂CHO
(c) The carboxylic carbon is less electrophilic than carbonyl carbon because of the possible resonance structure.

(e) The given compound has molecular formula C3H60. One of its functional isomer i.e., B shows iodoform test which can be only shown by compounds having methyl ketone so the compound B will be Acetone or 2-propanone. Its functional isomer A will be propanal.

Question 3.
(a) Write the product(s) in the following :

(b) Give simple tests to distinguish the following pairs of compounds :
(i) Ethanal and Propanal
(ii) Benzaldehyde and Acetophenone
(iii) Benzoic acid and Ethyl benzoate
Answer:

(b) (i) On heating with NaOH and I₂, ethanal forms yellow ppt of CHI₃ whereas propanal can not.
CH₃CHO + 3I₂ → NaOH -4 CHI₃ + 3NaI + HCOONa + 3H₂O
(ii) On heating with NaOH and I2, aceptophenone forms yellow ppt of CHI3 whereas benzaldehyde does not.
C₆H₅COCH₃ + 3NaOI → C₆H₅COONa + CHI₃4 + 2NaOH
(iii) On adding NaHCO,, benzoic acid produces brisk effervescence of C02 gas whereas ethylbenzoate does not.

Question 4.
(a) Give reasons :
(i) CH₃—CHO is more reactive than CH₃COCH₃ towards HCN.
(ii) 4-nitrobenzoic acid is more acidic than benzoic acid.
(b) Describe the following :
(i) Acetylation (ii) Cannizzaro reaction (iii) Cross aldol condensation
Answer:
(a) (i) Because carbonyl carbon of CH₃—CHO is more electrophilic than CH₃COCH₃ due to only one electron donating CH₃– group.
(ii) Because of electron withdrawing nature of -NO₂ group.
(b) (i) Acetylation : Introduction of an acetyl group/CH3CO- by heating an organic compound with acetyl chloride/acetic anhydride.

(ii) Cannizzaro reaction : Aldehydes having no a-hydrogen atom when treated with cone. NaOH, undergoes self-oxidation and self-reduction simultaneously

(iii) Cross Aldol Condensation : When aldol condensation is carried out between two different aldehydes or ketones, it is called cross aldol condenstation.

Question 5.
(a) Complete the following equations :

Answer:

(b) (i) On adding NaHCO₃, CH₃COOH produces brisk effervescence of CO₂ gas whereas phenol does not.
(ii) On heating with Tollen’s reagent, CH₃CHO forms silver mirror whereas CH₃COCH₃ does not.

Question 6.
(a) What is meant by the following terms? Give an example of the reaction in each case.
(i) Aldol (ii) Semicarbazone
(b) Complete the following :

Answer:
(a) (i) Two molecules of aldehyde and ketones containing a-hydrogen atom react in the presence of
aqueous alkali giving product known as Aldol.
Example :

Question 7.
Write the product(s) in the following reactions.

Answer:

Question 8.
(a) Write the product(s) in the following reactions:

(b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Butanal and Butan-2-one (ii) Benzoic acid and Phenol
Answer:

(b) (i) Pollen’s reagent test. Add ammoniacal solution of sliver nitrate (Tollen’s Reagent) in both the solutions. Butanal gives silver mirror whereas Butan-2-one does not. Therefore Butanal gives Tollen’s test.
(ii) Ferric chloride test. Add neutral FeCl₃ in both the solutions, phenol reacts with neutral FeCl₃ to form an iron-phenol complex giving violet colour but benzoic acid does not.

Question 9.
(a) Write the reactions involved in the following:
(i) Etard reaction (ii) Stephen reduction
(b) How will you convert the following in not more than two steps:
(i) Benzoic acid to Benzaldehyde (ii) Acetophenone to Benzoic acid
(iii) Ethanoic acid to 2-Hydroxyethanoic acid (All India 2017)
Answer:
(a) (i) Etard reaction

(ii) Stephen reduction:

(b) (i) Benzoic acid to Benzaldehyde

(ii) Acetophenone to Benzoic acid

Question 10.
(a) How will you convert:
(i) Benzene to acetophenone (ii) Propanone to 2-Methylpropan-2-ol
(b) Give reasons :
(i) Electrophilic substitution in benzoic acid takes place at meta position.
(ii) Carboxylic acids are higher boiling liquids than aldehydes, ketones and alcohols of comparable molecular masses.
(iii) Propanal is more reactive than propanone in nucleophilic addition reactions. (Comptt. Delhi 2017)
Answer:
(i) Benzene to acetophenone

(b) (i) Because -COOH group is electron withdrawing group and deactivates the benzene ring. As a result of this ortho and para position acquires positive charge but only meta does not, so electrophile can attack on rneta position.
(ii) Because -COOH group of carboxylic acids is capable to do intermolecular hydrogen bonding forming a dimer while alcohols, aldehydes and ketones can not.

(iii) Because of smaller +1 effect of one alkyl group in propanal as compared to larger + I effect ol 2 alkyl groups of propanone, the magnitude of positive charge on the carbonyl carbon is more in propanal than propanone.

In This Post we are  providing Chapter-10 HALOALKANES AND HALOARENES NCERT MOST IMPORTANT QUESTIONS for Class 12 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON HALOALKANES AND HALOARENES

Question 1- Among the following pairs which one undergoes S_N2 substitution reaction faster. State reasons.  

(i) 

(ii) 

Answer: (i) It undergoes S_N2 reaction faster as it is a primary halide. 

(ii) Due to its large size, Iodine is a better leaving group, therefore, in the presence of an incoming nucleophile, it will be released at a faster rate. 

Question 2- Among the following pairs which one undergoes S_N1 substitution reaction faster. State reasons.  

(i) 

(ii) 

Answer: i) Tertiary halide reacts faster than secondary halide because of the greater stability of tert-carbocation.          

(ii) Because of the greater stability of secondary carbocation than primary.

Question 3- How can the following conversions be carried out? 

(i) Methyl magnesium bromide to 2-methyl- propan-2-ol

(ii) Benzyl chloride to benzyl alcohol

(iii) 2-Bromopropane to 1-bromopropane

(iv) Benzene to p-chloronitrobenzene

Answer: (i) 

(ii)

(iii)

(iv)

Question 4- Although chlorine is an electron-withdrawing group, yet it is ortho-, para- directing in electrophilic aromatic substitution reactions. Why?

Answer: Chlorine withdraws electrons through inductive effect and releases electrons through resonance. Through inductive effect, chlorine destabilised the intermediate carbocation formed during the electrophilic substitution. Through resonance, halogen tends to stabilise the carbocation and the effect is more pronounced at the ortho- and para- positions. The inductive effect is stronger than resonance and causes net electron withdrawal and thus causes net deactivation. The resonance effect tends to oppose the inductive effect for the attack at the ortho- and para positions and hence makes the deactivation less for ortho- and para attack. Reactivity is thus controlled by the stronger inductive effect and orientation is controlled by resonance effect.

Question 5- State reasons for: 

(i) p-dichlorobenzene has a higher melting point than its o- and m-isomers. 

(ii) (±)-Butan-2-ol is optically inactive. 

(iii) The C–Cl bond length in chlorobenzene is shorter than that in CH₃–Cl.

(iv) Chloroform is stored in closed dark brown bottles.

(v) Alkyl halides, though polar, are immiscible with water.

Answer: (i) p-isomers are comparatively more symmetrical, fits closely in the crystal lattice. Therefore, they require more heat to break the strong forces of attraction. As a result, p-isomers have a higher melting point than o- and m-isomers.

(ii)  In a racemic mix, one type of rotation is cancelled by another. Therefore, (±)-Butan-2-ol is optically inactive. 

(iii)  The C–Cl bond length in chlorobenzene is shorter than that in CH₃–Cl as in haloalkanes, the halogen atom is attached to sp³ hybridized carbon whereas in haloarenes it is attached to sp² hybridized carbon whose size is smaller than sp³ orbital carbon.

(iv) Chloroform is stored in closed dark brown bottles because chloroform gets slowly oxidised by air in the presence of light and forms an extremely poisonous gas (carbonyl chloride), popularly known as phosgene. 

(v) Alkyl halides are polar molecules that are held together by dipole-dipole interaction. The molecules of H₂O are held together by H- bonds. The forces of attraction between water and alkyl halide molecules are weaker than the existing forces of attraction between alkyl halide-alkyl halide molecules and water-water molecules.

Question 6-  Identify chiral and achiral molecules in each of the following pair of compounds:

(i) 

(ii)

(iii)

Answer: (i) 

(ii)

(iii)

Question 7- Define: 

(i) ambident nucleophiles

(ii) racemic mixture

(iii) Sandmeyer’s reaction

Answer: (i) Groups like cyanides and nitrites possess two nucleophilic centres and are called ambident nucleophiles. Cyanide group is a hybrid of two contributing structures and can act as a nucleophile in two different ways [VC≡N ↔ :C=NV], i.e., linking through carbon atoms resulting in alkyl cyanides and through nitrogen atoms leading to isocyanides. 

(ii) A mixture containing two enantiomers in equal proportions will have zero optical rotation, as the rotation due to one isomer will be cancelled by the rotation due to the other isomer. Such a mixture is known as a racemic mixture or racemic modification. A racemic mixture is represented by prefixing dl or (±) before the name, for example (±) butan-2-ol. The process of conversion of enantiomer into a racemic mixture is known as racemisation.

(iii) When a primary aromatic amine, dissolved or suspended in cold aqueous mineral acid, is treated with sodium nitrite, a diazonium salt is formed. Mixing the solution of the freshly prepared diazonium salt with cuprous chloride or cuprous bromide results in the replacement of the diazonium group by –Cl or –Br. Replacement of the diazonium group by iodine does not require the presence of cuprous halide and is done simply by shaking the diazonium salt with potassium iodide.

Question 8- Write the structures for the following: 

(i)  4-tert-Butyl-3-iodoheptane

(ii) 4-Bromo-3-methylpent-2-ene

(iii) p-Bromochlorobenzene

(iv) 1-Chloro-4-ethylcyclohexane

Answer: (i) 

(ii) 

(iii)

(iv)

Question 9- Give reasons: 

(i) Grignard’s reagents should be prepared under anhydrous conditions.

(ii) n-Butyl bromide has a higher boiling point than f-butyl bromide.

Answer: (i) Grignard’s reagents react with alcohol, water, amines etc. to form the corresponding hydrocarbon and thus, must be prepared under anhydrous conditions. 

R-MgX + HOH → RH + Mg(OH)X

(ii) n-Butyl bromide has a higher boiling point than f-butyl bromide as it has a larger surface area. Therefore,  it has more Van der Waals’ forces.

Question 10- What happens when: 

(i) Chlorobenzene is treated with Cl₂/FeCl₃

(ii) Ethyl chloride is treated with AgNO₂

(iii) 2-bromopentane is treated with alcoholic KOH

Answer: (i)

(ii) 

(iii)

In This Post we are  providing Chapter-11 ALCOHOLS, PHENOLS AND ETHERS NCERT MOST IMPORTANT QUESTIONS for Class 12 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON ALCOHOLS, PHENOLS AND ETHERS

Question 1.
What happens when
(a) Sodium phenoxide is treated with CH₃Cl?
Answer:

(b) CH₂ = CH – CH₂ – OH is oxidised by PCC?
Answer:

(c) Phenol is treated with CH₃COCI/anhydrous AlCl₃?
Write chemical equations in support of your answer.
Answer:

Question 2.
(a) How will you convert the following:
(i) Phenol to benzoquinone
Answer:

(ii) Propanone to 2-methyl propane-2-ol
Answer:

(b) Why does propanol have a higher boiling point than that butane? (CBSE 2019C)
Answer:
The molecules of propanol are held together by intermolecular hydrogen bonding while butane molecules have only weak van der Waals forces of attraction. Since hydrogen bonds are stronger than van der Waals forces, therefore, propanol has a higher boiling point than butane.

Question 3.
Identify the product formed when propan-1 -ol is treated with a cone. H₂S0₄ at 413 K. Write the mechanism involved for the above reaction.
Answer:
(a) 1-Propoxypropane is formed.
Mechanism involved:
Step 1: Formation of protonated alcohol. Propanol gets protonated in the presence of H⁺.

Step 2: Nucleophilic attack. Due to the presence of a +ve charge on the oxygen atom, the carbon of the CH₂ part becomes electron deficient. As a result, a nucleophilic attack by another alcohol molecule (unprotonated) occurs on the protonated alcohol with the elimination of a molecule of water.

Step 3: Deprotonation. Oxonium ion loses a proton to form an ether.

Question 4.
Write the equations involved in the following reactions:
(i) Reimer-Tiemann reaction
Answer:
Reimer-Tiemann reaction: When phenol is refluxed with chloroform in the presence of aqueous caustic alkali at 340 K, an aldehydic group (CHO) gets introduced in the ring at a position ortho to the phenolic group. Ortho hydroxy benzaldehyde or salicylaldehyde is formed as the product of the reaction.

(ii) Williamson’s ether synthesis
Answer:
Williamson’s ether synthesis. This is used to prepare symmetrical and unsymmetrical ethers by treating alkyl halide with either sodium alkoxide or sodium phenoxide.

Aryl halides cannot be used for the preparation of alkyl-aryl ethers because of their low reactivity.

Question 5.
Draw the structure and name the product formed if the following alcohols are oxidised. Assume that an excess of the oxidising agent is used.
(i) CH3CH2CH2CH2OH
Answer:

(ii) 2-butenol
Answer:

(iii) 2-methyl-1-propanol
Answer:

Question 6.
Write the main product(s) in each of the following reactions:

Answer:

Answer:

Answer:

Question 7.
Give reasons for the following:
(i) Protonation of phenols Is difficult whereas ethanol easily undergoes protonation.
Answer:
In phenols, the oxygen atom acquires a partial positive charge due to resonance and therefore, it is not easily protonated. On the other hand, in ethanol, the alkyl group is an electron-releasing group and increases the electron density on 0 atoms. Therefore, ethanol is easily protonated.

(ii) Bolting point of ethanol is higher than that of dimethyl ether.
Answer:
The molecules of ethanol are held together by intermolecular hydrogen bonding while dimethyl molecules have only weak van der Waals forces of attractions. Since hydrogen bonds are stronger than van der Waals forces, ethanol has a higher boiling point than dimethyl ether.

(iii) Anisole on reaction with Hl gives phenol and CH₃-I as main products and not iodobenzene and CH₃OH.
Answer:
This is because, during the reaction, the attack of halide ion occurs to the protonated anisole, i.e. methyl phenyl oxonium ion, which is formed during the protonation of anisole. Due to steric hindrance of the bulky phenyl group, the attack preferably occurs to the alkyl group forming methyl iodide and phenol.

Question 8.
How do you convert the following: (CBSE Delhi 2015, Delhi)
(i) Phenol to anisole
(ii) Propan-2-ol to 2-methylpropan-2-ol
(iii) Aniline to phenol?
OR
(i) Write the mechanism of the following reaction:
2CH₃CH₂OH CH₃CH₂—O—CH₂CH₃
(ii) Write the equation involved In the acetylation of salicylic acid.
Answer:
(i) Phenol to anisole

(iii) Aniline to phenol

Or
(i) The mechanism for the formation of ether from ethanol at 413 K is a nucleophilic bimolecular reaction as given below:
(a) Ethyl alcohol gets protonated in the presence of H⁺

(b) Due to the presence of a +ve charge on the oxygen atom, the carbon of CH₂ part of CH₃CH₂ becomes electron deficient. As a result, nucleophilic attack by another alcohol molecule (unprotonated) occurs on the protonated alcohol with the elimination of a

(c) Oxonium ion loses a proton to form an ether.

Question 9.
Give reasons for the following:
(i) o-nitrophenol is more acidic than o-methoxyphenyl.
Answer:
This is because —NO₂ (nitro group) is an electron-withdrawing group and will increase the +ve charge on oxygen to make it more acidic. On the other hand, the -OCH₃ group is an electron¬releasing group and will decrease +ve charge on oxygen making it less acidic as O-H bond will not break easily.

(ii) Butan-1-oi has a higher boiling point than diethyl ether.
Answer:
Butan-1 -ol has intermolecular hydrogen bonding between their molecules. Therefore, it exists as associated molecules and large amount of energy is required to break these bonds and hence, its boiling point is high. But diethyl ether does not show any association by intermolecular hydrogen bonding. Hence, its boiling point is low.

(iii) (CH₃)₃C – O – CH₃ on reaction with HI gives (CH₃)₃C – I and CH₃ – OH as the main products and not (CH₃)₃C – OH and CH₃ – I.
Answer:
The reaction:

gives (CH₃)₃C – I and CH₃OH as the main products and not (CH₃)₃COH and CH₃I. This is because the reaction occurs by the S_N1 mechanism and the formation of products is governed by the stability of carbocation formed from the cleavage of the C-0 bond in the protonated ether. Since tert. butyl carbocation, (CH₃)₃C⁺ is more stable than methyl carbocation, CH₃, the cleavage of C-0 gives a more stable carbocation, [(CH₃)₃C]⁺ and methanol. Then, iodide ion, I^– attacks this tert. butyl carbocation to form tert. butyl iodide.

Question 10.
(i) Write the mechanism of the following reaction:

Answer:
HBr → H⁺ + Br^–
(a) H+ attacks oxygen of O-H to form protonated alcohol

(b) Protonated alcohol Loses a molecule of water to form a carbocation.

(C) Br attacks the carbocation to form bromoatkane
CH₃CH₂+ + Br^– → CH₃CH₂Br (Bromoethane)

In This Post we are  providing Chapter- 7 THE p-BLOCK ELEMENTS NCERT MOST IMPORTANT QUESTIONS for Class 12 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON THE p-BLOCK ELEMENTS

Question 1.
(a) Draw the molecular structures of following compounds:
(i) XeF₆
(ii) H₂S₂O₈
Answer:
(i)

(ii)

(b) Explain the following observations:
(i) The molecules NH₃ and NF₃ have dipole moments which are of opposite directions.
(ii) All the bonds in PCl₅ molecule are not equivalent.
(iii) Sulphur in vapour state exhibits paramagnetism.
Answer:
(i) Both NH₃ and NF₃ have pyramidal shape with one lone pair on N atom.

The lone pair on N is in opposite direction to the N-F bond moments and therefore, it has very low dipole moment (about 0.234 D). But ammonia has high dipole moment because its lone pair is in the same direction as the N-H bond moments.

(ii) PCl₅ has trigonal bipyramidal structure in which there are three P-Cl equatorial bonds and two P-Cl axial bonds. The two axial bonds are being repelled by three bond pairs at 90° while the three equatorial bonds are being repelled by two bond pairs at 90°. Therefore, axial bonds are repelled more by bond pairs than equatorial bonds and hence are larger (219 pm).

(iii) In vapour state, sulphur partly exists as S₂ molecule and S₂ molecule like O₂ has two unpaired electrons in anti-bonding π* molecular orbitals. Therefore, it is paramagnetic.

OR

(a) Complete the following chemical equations:
(i) XeF₄ + SbF₅ →
(ii) Cl₂ + F₂ (excess ) →
Answer:
(i) XeF₄ + SbF₅ → XeF₄.SbF₅ → [XeF₃]⁺ [SbF₆]^–

(ii)

(b) Explain each of the following:
(i) Nitrogen is much less reactive than phosphorus.
(ii) The stability of +5 oxidation state decreases down group 15.
(iii) The bond angles (O – N – O) are not of the same value in NO₂^– and NO₂⁺. (CBSE Delhi 2012)
Answer:
(i) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N = N) and it is non-polar in character.
Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol^-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

(ii) The stablity of +5 oxidation state decreases down the group because of inert pair effect. Therefore, the +5 oxidation state of Bi is less stable than that of Sb.

(iii) In NO₂, there is one electron on N while in NO₂⁺ there is no electron and in NO₂⁺, there is a lone pair of electrons on N as shown below:

NO₂⁺ molecule is linear and has bond angle of 180°. The repulsion by single electron is less as compared to repulsion by a lone pair of electrons. Therefore, bond pairs in NO₂^– are forces more closer than in NO₂.

Question 2.
(a) Draw the structure of the following molecule: H₃PO₂
Answer:
(a)

(b) Explain the following observations:
(i) Nitrogen is much less reactive than phosphorus.
(ii) Despite having greater polarity, hydrogen fluoride boils at a lower temperature than water.
(iii) Sulphur has greater tendency for catenation than oxygen in the same group.
Answer:
(i) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N = N) and it is nonpolar in character.
Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol^-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

(ii) Though electronegativity of F is more than O, yet water forms more extensive hydrogen bonding than HF. In H₂O, each oxygen is tetrahedrally surrounded by two covalent bonds and two hydrogen bonds, (-O….H). In HF there is only one hydrogen bond (-F….H).

(iii) Sulphur has strong tendency for catenation, i.e. forming bonds with itself in comparison to oxygen. This is because of stronger S-S single bond than O-O single bond.

OR

(a) Draw the structures of the following molecules:
(i) N₂O₅
(ii) HClO₄
Answer:

(a)
(i)

(ii)

(b) Explain the following observations:
(i) H₂S is more acidic than H₂O.
(ii) Fluorine does not exhibit any positive oxidation state.
(iii) Helium forms no real chemical compound. (CBSE 2012)
Answer:
(i) The size of S is more than that of O. Therefore, the distance between S and H, i.e. S-H bond length, is more than O-H bond length. As a result, the bond dissociation enthalpy of S-H will be less and it will be easier to break the bond in H₂S than O-H bond in water. Therefore, H₂S will be more acidic than H₂O.

(ii) Fluorine is the most electronegative element and therefore, it shows oxidation state of – 1 only. It does not show any positive oxidation state.

(iii) Helium does not form compounds because it has very high ionisation enthalpy and smallest size. Its electron gain enthalpy is also almost zero.

Question 3.
Complete the following reactions
(i) NaOH + Cl₂ →
hot and conc.
(ii) NH₃ + Cl₂ (excess) →
(iii) NaNO₂ + HCl →
(iv) F₂(g) + H₂O(l) →
(v) K₂CO₃ + HCl →
(vi) Cl₂ + H₂O →
(vii) F₂ + 2Cl^– →
Answer:
(i) 6NaOH + 3Cl₂ → 5NaCl + NaClO₃ + 3H₂O
hot and conc.
(ii) NH₃ + 3Cl₂ (excess) → NCl₃ + 3HCl
(iii) 2NaNO₂ + 2HCl → 2NaCl + H₂O + NO₂ + NO
(iv) 2F₂(g) + 2H₂O(l) → 4H⁺(aq) + 4F^– (aq) + O₂(g)
(v) K₂CO₃ + 2HCl → 2KCl + CO₂ + H₂O
(vi) 2Cl₂ + 2H₂O → 4HCl + O₂
(vii) F₂ + 2Cl^– → 2F^– + Cl₂

Question 4.
Complete the following reactions:
(i) XeF₄ + SbF₅ →
(ii)

(iii) XeF₂ + H₂O →
(iv) XeF₄ + H₂O →
(v) XeF₆ + H₂O →
(vi) XeF₆ + 2H₂O →
(vii) XeF₆ + 3H₂O →
Answer:
(i) XeF₄ + SbF₅ → [XeF₃]⁺ [SbF₆]^–
(ii)

(iii) XeF₂ + H₂O → 2Xe + 4HF + O₂
(iv) XeF₄ + H₂O → 4Xe + 2XeO₃ + 24HF + 3O₂
(v) XeF₆ + H₂O → XeOF₄ + 2HF
(vi) XeF₆ + 2H₂O → XeO₂F₂ + 4HF
(vii) XeF₆ + 3H₂O → XeO₃ + 6HF

Question 6.
(a) Account for the following:
(i) Ozone is thermodynamically unstable.
(ii) Solid PCl₅ is ionic in nature.
(iii) Fluorine forms only one oxoacid HOF.
Answer:
(a) (i) Ozone is thermodynamically unstable with respect to oxygen because it results in liberation of heat (ΔH is -ve) and increase in entropy ΔS is +ve). These two factors reinforce each other resulting in negative ΔG (ΔG = ΔH – TΔS) for its conversion to oxygen.

(ii) PCl₅ has trigonal bipyramidal structure and is not very stable. It splits up into more stable tetrahedral and octahedral structures which are stable as
PCl₅ ⇌ [PCl₄]⁺ [PCl₆]^–
Therefore, it exists as ionic.

(iii) Due to small size and high electronegativity, fluorine cannot act as central atom in higher oxoacids.

(b) Draw the structure of
(i) BrF₅
(ii) XeF₄
Answer:
(i)

Square Pyramidal

(ii)

OR

(i) Compare the oxidising action of F₂ and Cl₂ by considering parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
Answer:
F₂ is stronger oxidising agent than Cl₂. This can be explained on the basis of bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
The process of oxidising behaviour may be expressed as:

The overall tendency for the change (i.e. oxidising behaviour) depends upon the net effect of three steps. As energy is required to dissociate or convert molecular halogen into atomic halogen, the enthalpy change for this step is positive. On the other hand, energy is released in steps (II) and (III), therefore, enthalpy change for these steps is negative.

Now although fluorine has less negative electron gain enthalpy, yet it is stronger oxidising agent because of low enthalpy of dissociation and very high enthalpy of hydration. In other words, large amount of energy released in step (III) and lesser amount of energy required in step (I) overweigh the smaller energy released in step (II) for fluorine. As a result, the AH overall is more negative for F₂ than for Cl₂. Hence, F₂ is stronger oxidising agent than Cl₂.

(ii) Write the conditions to maximise the yield of H₂SO₄ by contact process.
Answer:
Conditions for maximum yield of H₂SO₄ by Contact Process:
(a) Low temperature (optimum temperature 720 K)
(b) High pressure (optimum pressure 2 bar)
(c) Presence of catalyst (V₂O₅ catalyst).

(iii) Arrange the following in the increasing order of property mentioned:
(a) H₃PO₃, H₃PO₄, H₃PO₂ (Reducing character)
(b) NH₃, PH₃, ASH₃, SbH₃, BiH₃ (Base strength) (CBSE 2016)
Answer:
(a) H₃PO₂ > H₃PO₃ > H₃PO₄
(b) NH₃ > PH₃ > ASH₃ > SbH₃ > BiH₃

Question 5.
(a) Account for the following:
(i) Acidic character increases from HF to HI.
(ii) There is large difference between the melting and boiling points of oxygen and sulphur.
(iii) Nitrogen does not form pentahalide.
Answer:
(a) (i) In gaseous state, hydrogen halides are covalent. But in aqueous solution, they ionise and behave as acids. The acidic strength of these acids decreases in the order:
HI > HBr > HCl > HF
Thus, HF is the weakest acid and HI is the strongest acid among these hydrogen halides.

The above order of acidic strength is reverse of that expected on the basis of electronegativity. Fluorine is the most electronegative halogen, therefore, the electronegativity difference will be maximum in HF and should decrease gradually as we move towards iodine through chlorine and bromine.

Thus, HF should be most ionic in nature and consequently it should be strongest acid. Although many factors contribute towards the relative acidic strengths, the major factor is the bond dissociation energy. The bond dissociation energy decreases from HF to HI so that HF has maximum bond dissociation energy and HI has the lowest value.

Since H-I bond is weakest, it can be dissociated into H⁺ and I^– ions readily while HF can be dissociated with maximum difficulty. Thus, HI is the strongest acid while HF is the weakest acid among the hydrogen halides.

(ii) Oxygen molecule is held by weak van der Waals forces because of the small size and high electronegativity of oxygen. On the other hand, sulphur molecules do not exist as S₂ but form polyatomic molecules having eight atoms per molecule (S₈) linked by single bonds. Therefore, S atoms are strongly held together by intermolecular forces and its melting point is higher than that of oxygen. Hence, there is large difference in melting and boiling points of oxygen and sulphur.

(iii) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N ≡ N) and it is non-polar in character. Due to the presence of a triple bond, it has very high bond dissociation enthalpy (941.4 kJ mol^-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

(b) Draw the structures of the following:
(i) CIF₃
(ii) XeF₄
Answer:
(i)

T – Shaped molecule

(ii)

Square planar molecule

OR

(i) Which allotrope of phosphorus is more reactive and why?
Answer:
White phosphorus is most reactive of all the allotropes because it is unstable due to angular strain on P₄ molecule with bond angle of 60°.

(ii) How are the supersonic jet aeroplanes responsible for the depletion of ozone layers?
Answer:
Nitrogen oxide emitted from exhausts of supersonic jet aeroplanes readily combines with ozone to form nitrogen dioxide and diatomic oxygen. Since supersonic jets fly in the stratosphere near ozone layer, they are responsible for the depletion of ozone layer.

(iii) F₂ has lower bond dissociation enthalpy than Cl₂. Why?
Answer:
Because of small size of fluorineatoms, there are strong electron-electron repulsions between the lone pairs of electrons on F atoms. Hence, bond dissociation enthalpy of F₂ is lower than that of Cl₂.

(iv) Which noble gas is used in filling balloons for meteorological observations?
Answer:
Helium is used for filling balloons for meteorological observations because it is non-inflammable.

(v) Complete the equation: (CBSE 2015)
XeF₂ + PF₅ →
Answer:
XeF₂ + PF₅ → [XeF]⁺ [PF₆]^–

Question 6.
(a) Account for the following:
(i) Interhalogens are more reactive than pure halogens.
(ii) N₂ is less reactive at room temperature.
(iii) Reducing character increases from NH₃ to BiH₃.
Answer:
(i) Interhalogen compounds are more reactive than component halogens.
This is because covalent bond between dissimilar atoms in interhalogen compounds is polar and weaker than between similar atoms in halogens (except F-F). This is due to the fact that the overlapping of orbitals of dissimilar atoms is less effective than the overlapping of orbitals of similar atoms.

(ii) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N ≡ N) and it is non-polar in character. Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol^-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

(iii) The reducing character of hydrides of group 15 depends upon the stability of the hydride. On going down the group, the size of the central atom increases and therefore, its tendency to form stable covalent bond with small hydrogen atom decreases.

The greater unstability of a hydride, the greater is its reducing character. Since the stability of the group 15 hydrides decreases from NH₃ to BiH₃, hence reducing character increases.

(b) Draw the structures of the following:
(i) H₄P₂O₇ (Pyrophosphoric acid)
(ii) XeF₄
Answer:
(i)

(ii)

OR

(a) Which poisonous gas is evolved when white phosphorus is heated with conc. NaOH solution? Write the chemical equation involved.
Answer:
Phosphine, PH₃
P₄ + 3NaOH + 3H₂O → 3NaH₂PO₃ + PH₃

(b) Which noble gas has the lowest boiling point?
Answer:
Helium

(c) Fluorine is a stronger oxidising agent than chlorine. Why?
Answer:
Fluorine has lower bond dissociation enthalpy of F-F bond than Cl-Cl bond of Cl₂ and high enthalpy of hydration because of smaller size of F ion. As a result it has greater tendency to accept electron in solution and is stronger oxidising agent than chlorine.

(d) What happens when H₃PO₃ is heated?
Answer:

(e) Complete the equation:
PbS + O₃ → (CBSE 2015)
Answer:
PbS + 4O₃ → PbSO₄ + 4O₂

Question 7.
(a) Account for the following observations:
(i) SF₄ is easily hydrolysed whereas SF₆ is not easily hydrolysed.
(ii) Chlorine water is a powerful bleaching agent.
(iii) Bi(V) is a stronger oxidising agent than Sb(V).
Answer:
(a) (f) S atom in SF₄ is not sterically protected as it is surrounded by only four F atoms, so attack of water molecules can take place easily. In contrast, S atom in SF₆ is protected by six F atoms. Thus attack by water molecules cannot take place easily.

(ii) Chlorine water produces nascent oxygen (causes oxidation) which is responsible for bleaching action.
Cl₂ + H₂O → 2HCl + O

(iii) Due to inert pair effect Bi(V) can accept a pair of electrons to form more stable Bi (III) (+3 oxidation state of Bi is more stable than its +5 oxidation state).

(b) What happens when
(i) White phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO₂.
(ii) XeF₆ undergoes partial hydrolysis.
(Give the chemical equations involved).
Answer:
(i) Phosphorus undergoes disproportionation reaction to form phosphine gas.
P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂

(ii) On partial hydrolysis, XeF6 gives oxyfluoride XeOF4 and HF.
XeF₆ + H₂O → XeOF₄ + 2HF

OR

(a) What inspired N.Bartlett for carrying out reaction between Xe and PtF₆?
Answer:
N. Bartlett first prepared a red compound O₂⁺PtF₆^–. He then realised that the first ionisation enthalpy of molecular oxygen was almost identical with Xenon. So he carried out reaction between Xe and PtF₆.

(b) Arrange the following in the order of property indicated against each set:
(i) F₂, I₂, Br₂, Cl₂ (increasing bond dissociation enthalpy)
(ii) NH₃, ASH₃, SbH₃, BiH₃, PH₃ (decreasing base strength)
Answer:
(i) I₂ < F₂ < Br₂ < Cl₂
(ii) NH₃ > PH₃ > ASH₃ > SbH₃ > BiH₃

(c) Complete the following equations:
(i) Cl₂ + NaOH (cold and dilute) →
(ii) Fe³⁺ + SO₂ + H₂O →
(CBSE Sample Paper 2018)
Answer:
(i) 2NaOH + Cl₂ → NaCl + NaOCl + H₂O
(ii) 2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺ + SO₄^2- + 4H⁺

Question 8.
(a) Give reasons:
(i) H₃PO₃ undergoes disproportionation reaction but H₃PO₄ does not.
(ii) When Cl₂ reacts with excess of F₂, ClF₃ is formed and not FCl₃.
(iii) Dioxygen is a gas while Sulphur is a solid at room temperature.
Answer:
(i) In +3 oxidation state phosphorus tends to disproportionate to higher and lower oxidation states / Oxidation state of P in H₃PO₃ is +3 so it undergoes disproportionation but in H₃PO₄ it is +5 which is the highest oxidation state.

(ii) F cannot show positive oxidation state as it has highest electronegativity/ Because Fluorine cannot expand its covalency / As Fluorine is a small sized atom, it cannot pack three large sized Cl atoms around it.

(iii) Oxygen has multiple bonding due to pπ-pπ bonding whereas sulphur shows catenation. Oxygen is diatomic therefore held by weak intermolecular force while sulphur is polyatomic and held by strong intermolecular forces.

(b) Draw the structures of the following:
(i) XeF₄
(ii) HClO₃
Answer:
(i)

(ii)

OR

(a) When concentrated sulphuric acid was added to an unknown salt present in a test tube a brown gas (A) was evolved. This gas Intensified when copper turnings were added to this test tube. On cooling, the gas (A) changed into a colourless solid (B).
(i) Identify (A) and (B).
(ii) Write the structures of (A) and (B).
(iii) Why does gas (A) change to solid on cooling?
Answer:
(i) A = NO₂, B = N₂O₄
(ii)

(iii) NO₂ contains an odd electron. So it dimerises to give N₂O₄.

(b) Arrange the following In the decreasing order of their reducing character:
HF, HCl, HBr, HI
Answer:
HI > HBr > HCl > HF

Complete the following reaction:
XeF₄ + SbF₅ →, (CBSE 2018)
Answer:
(C) XeF₄ + SbF₅ → [XeF₃] [SbF₆]

Question 9.
(i) What happens when
(a) chlorine gas reacts with cold and dilute solution of NaOH?
(b) XeF₂ undergoes hydrolysis?
Answer:
(a) 2NaOH + Cl₂ → NaCl + NaOCl + H₂O
(cold and dilute)
(b) 2XeF₂(s) + 2H₂O (l) → 2Xe (g) + 4HF(aq) + O₂(S)

(ii) Assign suitable reasons for the following:
(a) SF₆ Is inert towards hydrolysis.
(b) H₃PO₃ is diprotic.
(C) Out of noble gases only Xenon is known to form established chemical compounds.
Answer:
(a) Sulphur is sterically protected by six F atoms, hence does not allow the water molecules to attack.

(b) It contains only two ionisable H-atoms which are present as -OH groups, thus behaves as dibasic acid.

(c) Xe has least ionisation energy among the noble gases and hence it forms chemical compounds particularly with O₂ and F₂.

OR

(i) Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F₂ and Cl₂.
Answer:
F₂ is stronger oxidising agent than Cl₂. This can be explained on the basis of bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
The process of oxidising behaviour may be expressed as:

The overall tendency for the change (i.e. oxidising behaviour) depends upon the net effect of three steps. As energy is required to dissociate or convert molecular halogen into atomic halogen, the enthalpy change for this step is positive. On the other hand, energy is released in steps (II) and (III), therefore, enthalpy change for these steps is negative.

Now although fluorine has less negative electron gain enthalpy, yet it is stronger oxidising agent because of low enthalpy of dissociation and very high enthalpy of hydration. In other words, large amount of energy released in step (III) and lesser amount of energy required in step (I) overweigh the smaller energy released in step (II) for fluorine. As a result, the AH overall is more negative for F₂ than for Cl₂.
Hence, F₂ is stronger oxidising agent than Cl₂.

(ii) Complete the following reactions :
(a) Cu + HNO₃(dilute) →
(b) Fe³⁺ + SO₂ + H₂O →
(c) XeF₄ + O₂F₂ → (CBSE 2018)
Answer:
(a) 3Cu + 8 HNO₃ (dilute) → 3Cu(NO₃)₂ + 2NO + 4H₃O
(b) 2Fe³⁺ + SO₃ + 2H₃O → 2 Fe²⁺ + SO₄^2- + 4H⁺
(c) XeF₄ + O₂F₂ → XeF₆ + O₂

Question 10.
A crystalline solid ‘A’ burns in air to form a gas ‘B’ which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified KMnO₄ (aq.) solution and reduces Fe³⁺ to Fe²⁺. Identify ‘A’ and ‘B’ and write the reactions involved.
Answer:
A = S₈ / Sulphur
S₈ + 8 O₂ → 8SO₂ / S + O₂ → SO₂
B = SO₂

Decolourises KMnO₄
2KMnO₄ + 5 SO₂ + 2H₄O → 2H₂SO₄ + 2MnSO₄ + K₂SO₄ / 2MnO₄^– + 5SO₂ + 2H₂O → 4H⁺ + 2Mn²⁺ + 5SO₄^2-
Reduces Fe³⁺ to Fe²⁺
2Fe³⁺ + SO₂ + 2 H₂O → 2 Fe²⁺ + SO₄^2- + 4H⁺

OR

Answer the following:
(a) Arrange the following hydrides of Group 16 elements in the decreasing order of their acidic strength:
H₂O, H₂S, H₂Se, H₂Te
Answer:
H₂Te > H₂Se > H₂S > H₂O

(b) Which one of PCl₄⁺ and PCl₄^– is not likely to exist and why?
Answer:
PCl^4- is not likely to exist because lone pair on P in PCl₃ can be donated to Cl⁺ and not to Cl^–. Phosphorus has 10e^– which cannot be accommodated in sp³ orbitals.

(c) Which aliotrope of sulphur is thermally stable at room temperature?
Answer:
Rhombic sulphur is thermally stable at room temperature. Its melting point is 385.8 K. All other varieties of sulphur change into this form on standing. It has low thermal and electrical conductivity.

(d) Write the formula of a compound of phosphorus which is obtained when cone. HNO₃ oxidises P₄.
Answer:
HNO₃ oxidises phosphorus to phosphoric acid. H₃PO₄ is formed
P₄ + 2OHNO₃ → 4H₃PO₄ + 2ONO₂ + 4H₂O

(e) Why does PCl₃ fume in moisture?
Answer:
PCl₃ fumes in moist air and reacts with water violently to form phosphorus acid. It hydrolyses in the presence of moisture to give fumes of HCl.
PCl₃ + 3H₂O → H₃PO₃ + 3HCl

In This Post we are  providing Chapter- 6 THE d AND f-BLOCK ELEMENTS NCERT MOST IMPORTANT QUESTIONS for Class 12 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON THE d AND f-BLOCK ELEMENTS

Question 1.
Explain the following:
(i) Out of Sc³⁺, Co²⁺, and Cr³⁺ ions, only Sc³⁺ is colorless in aqueous solutions.
(Atomic no.: Co = 27; Sc = 21 and Cr = 24)
(iI) The E°Cu²⁺/Cu for copper metal is positive (+0.34), unlike the remaining members of the first transition series
(iiI) La(OH)₃ is more basic than Lu(OH)₃. (CBSE Sample paper 2018)
Answer:
(i) Co²⁺: [Ar]3d⁷, Sc³⁺: [Ar]3d°
Cr³⁺: [Ar]3d³
Co²⁺ and Cr³⁺ have unpaired electrons. Thus, they are colored in an aqueous solution. Sc³⁺ has no unpaired electron. Thus it is colorless.
(ii) Metal copper has a high enthalpy of atomization and enthalpy of ionization. Therefore the high energy required to convert Cu(s) to Cu²⁺(aq) is not balanced by its hydration enthalpy.
(iii) Due to lanthanoid contraction the size of lanthanoid ion decreases regularly with the increase in atomic size. Thus covalent character between lanthanoid ion and OH’ increases from La³⁺ to Lu³⁺. Thus the basic character of hydroxides decreases from La(OH)₃ to Lu(OH)₃

Question 2.
(a) Complete the following chemical reactions:
(i) 2MnO₄^–+ 5 NO₂^–+ 6 H⁺ →
(ii) 3Mn0₄^2- + 4 H⁺ →
Answer:
(i) 5NO₂– + 2MnO4^– + 6H+ → 2Mn²⁺ + 5NO₃^– + 3H₂O
(ii) 3MnO₄^2-+ 4H+ → 2MnO₄^– + MnO₂ + 2H₂O

(b) Name a member of the lanthanoid series which shows a +4 oxidation state. (CBSE 2019C)
Answer:
Cerium / Ce

Question 3.
Give reasons:
(i) E° value for Mn³⁺/Mn²⁺ couple is much more positive than that for Fe³⁺/Fe²⁺.
Answer:
The comparatively high value for Mn shows that Mn²⁺(d⁵) is particularly stable/Much larger third ionization energy of Mn (where the required change is from d⁵ to d⁴).

(ii) Iron has a higher enthalpy of atomization than copper.
Answer:
Due to the higher number of unpaired electrons.

(iii) Sc³⁺ is colorless in an aqueous solution whereas Ti³⁺ is colored. (CBSE 2018)
Answer:
The absence of unpaired d-electron in Sc³⁺ whereas in Ti³⁺ there is one unpaired electron or Ti³⁺ shows the d-d transition.

Question 4.
(i) Complete the following chemical equations for reactions:
(a) MnO₄^– (aq) + S₂O₃2- (aq) + H₂O (I) →
(b) Cr₂O₇^– (aq) + H₂S(g) + H⁺(aq) →
Answer:
(a) 8MnO₄^– (aq) + 3S₂O₃2-  (aq) + H₂O (l) → 8MnO₂ (s) + 6SO₄2- (aq) + 2OH- (aq)
(b) Cr₂O₇^2- (aq) + 3H₂S(g) + 8H⁺ (aq) → 2Cr³⁺(aq) + 3S(s) + 7H₂O

(ii) Give an explanation for each of the following observations:
(a) The gradual decrease in size (actinoid contraction) from element to element is greater among the actinoids than that among the lanthanoids (lanthanoid contraction).
Answer:
(a) This is because of relatively poor shielding by 5f electrons in actinoids in comparison with shielding of 4f electrons in lanthanoids.

(b) The greatest number of oxidation states are exhibited by the members in the middle of a transition series.
Answer:
This is because, in the middle of the transition series, the maximum number of electrons are available for sharing with others. The small number of oxidation states at the extreme left side is due to a lesser number of electrons to lose or share. On the other hand, at the extreme right-hand side, due to a large number of electrons, only a few orbitals are available in which the electrons can share with others for higher valence.

(c) With the same d-orbital configuration (d⁴) Cr²⁺ ion is a reducing agent but the Mn³⁺ ion is an oxidizing agent. (CBSE Delhi 2009)
Answer:
Cr²⁺ is reducing as its configuration changes from d⁴ to d³, the latter having a half-filled t_2g level. On the other hand, the change from Mn³⁺ to Mn²⁺ results in the half-filled (d⁵) configuration which has extra stability.

Question 5.
(i) Complete the following chemical reaction equations:
(a) Fe²⁺ (aq) + MnO^–4 (aq) + H⁺ (aq) →
(b) Cr₂O₇^2- (aq) + I^– (aq) + H⁺ (aq) →
Answer:
(a) 5Fe2+ (aq) + MnO4- (aq) + 8H+ (aq) → Mn²⁺ (aq) + 5Fe³⁺ (aq) + 4H₂O(l)
(b) Cr₂O₇^2- (aq) + 6I^– (aq) + 14H+ (aq) → 2Cr³⁺(aq) + 3I₂ (s) + 7H₂O(l)

(ii) Explain the following observations:
(a) Transition elements are known to form many interstitial compounds.
Answer:
Transition metals form interstitial compounds. Transition metals have a unique character to form interstitial compounds with small non-metallic elements such as hydrogen, boron, carbon, and nitrogen. The small atoms of these non-metallic elements (H, B, C, N, etc.) fit into the vacant spaces of the lattices of the transition metal atoms. As a result of the filling up of the interstitial spaces, the transition metals become rigid and hard.

These interstitial compounds have similar chemical properties as the parent metals but have different physical properties, particularly, density, hardness, and conductivity. For example, steel and cast iron are hard because of the formation of interstitial compounds with carbon. These interstitial compounds have a variable composition and cannot be expressed by a simple formula. Therefore, these are called nonstoichiometric compounds.

(b) With the same d⁴ d-orbital configuration Cr²⁺ ion is reducing while the Mn³⁺ ion is oxidizing.
Answer:
Cr²⁺ is reducing as its configuration changes from d⁴ to d³, the latter having a half-filled t_2g level. On the other hand, the change from Mn³⁺ to Mn²⁺ results in the half-filled (d⁵) configuration which has extra stability.

(c) The enthalpies of atomization of the transition elements are quite high. (CBSE Delhi 2009)
Answer:
The transition metals have strong interactions of electrons in the outermost orbitals and therefore, have high melting and boiling points. These suggest that the atoms of these elements are held together by strong forces and have high enthalpy of atomization.

Question 6.
(i) Complete the following chemical equations:
(a) MnO₄^– (aq) + S₂O₃^2- (aq) + H₂O(l) →
(b) Cr₂O₇^2- (aq) + Fe²⁺ (aq) + H⁺ (aq) →
Answer:
(a) 8MnO₄^– + 3S₂O₃^2- + H₂O → 8MnO₂ + 6SO₄^2-+ 2OH-
(b) Cr₂O₇^2- + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ +6Fe³⁺ + 7H₂O

(ii) Explain the following observations:
(a) La3+ (Z = 57) and Lu3+ (Z = 71) do not show any colour in solutions.
Answer:
In La3+ there are no f-electrons and in Lu³⁺ the 4f sub-shell is complete (4/14). Therefore, there are no unpaired electrons and consequently, d-d transitions are not possible. Hence, La³⁺ and LU³⁺ do not show any color in solutions.

(b) Among the divalent cations in the first series of transition elements, manganese exhibits the maximum paramagnetism.
Answer:
Among the divalent cations in the first
transition series, Mn (Z = 25) exhibits the maximum paramagnetism because it has the maximum number of unpaired electrons (3d⁵): five unpaired electrons.

(c) Cu⁺ ion is not known in aqueous solutions. (CBSE 2010)
Answer:
In an aqueous solution, Cu⁺ undergoes disproportionation changing to Cu²⁺ ion.
2Cu⁺ → Cu²⁺ + Cu

Question 7.
(i) Give reasons for the following:
(a) Mn³⁺ is a good oxidizing agent.
(b) E°M²⁺/M values are not regular for first-row transition metals (3d series).
(c) Although ‘F’ is more electronegative than ‘O’, the highest Mn fluoride is MnF4, whereas the highest oxide is Mn207.
(ii) Complete the following equations:
(a) 2CrO₄^2- + 2H⁺ →
(b) KMnO₄ 
OR
(i) (a) Why do transition elements show variable oxidation states?
(b) Name the element showing a maximum number of oxidation states among the first series of transition metals from Sc(Z = 21) to Zn (Z = 30).
(c) Name the element which shows only the +3 oxidation state.
(ii) What is lanthanoid contraction? Name an important alloy that contains some of the lanthanoid metals. (CBSE2013)
Answer:
(i) (a) Mn³⁺ (3d⁴) on changing to Mn²⁺ (3d⁵) becomes stable, half-filled configuration has extra stability. Therefore, Mn³⁺ can be easily reduced and acts as a good oxidizing agent.

(b) E°(M²⁺/M) values are not regular in the first transition series metals because of irregular variation of ionization enthalpies (IE₁ + IE₂) and the sublimation energies.

(c) Among transition elements, the bonds formed in +2 and +3 oxidation states are mostly ionic. The compounds formed in higher oxidation states are generally formed by sharing of d-electrons. Therefore, Mn can form MnO4- which has multiple bonds also, while fluorine cannot form multiple bonds.

(ii) (a) 2CrO₄^2- + 2H⁺ → Cr₂O₇^2- + H₂O
(b) 2KMnO₄  K₂MnO₄ + MnO₂ + O₂
OR
(i) (a) The transition elements show variable oxidation states because their atoms can lose a different number of electrons. This is due to the participation of inner (n -1). d-electrons in addition to outer ns electrons because the energies of the ns and (n – 1) d-subshells are almost equal.

(b) Manganese

(c) Scandium

(ii) The steady decrease in atomic and ionic sizes of lanthanide elements with increasing atomic numbers is called lanthanide contraction. In the lanthanoids, there is a regular decrease in the size of atoms and ions with an increase in atomic number. For example, the ionic radii decrease from Ce³⁺ (111 pm) to Lu³⁺ (93 pm).

Cause of lanthanoid contraction. As we move through the lanthanoid series, 4f-electrons are being added, one at each step. The mutual shielding effect of electrons is very little, even smaller than that of d-electrons. This is due to the shape of f-orbitals. The nuclear charge, however, increases by one at each step. Hence, the inward pull experienced by the 4f-electrons increases. This causes a reduction in the size of the entire 4f shell. The sum of the successive reductions gives the total lanthanoid contraction. The important alloy is mischmetal.

Question 8.
(i) How do you prepare:
(a) K₂MnO₄ from MnO₂?
(b) Na₂Cr₂O₇ from Na₂CrO₄?
(ii) Account for the following:
(a) Mn²⁺ is more stable than Fe²⁺ towards oxidation to +3 state.
(b) The enthalpy of atomization is lowest for Zn in the 3d series of the transition elements.
(c) Actinoid elements show a wide range of oxidation states.
OR
(a) Name the element of 3d transition series which shows a maximum number of oxidation states. Why does it show so?
(b) Which transition metal of 3d series has a positive E°(M²⁺/M) value and why?
(c) Out of Cr³⁺ and Mn³⁺, which is a stronger oxidizing agent and why?
(d) Name a member of the lanthanoid series which is well known to exhibit a +2 oxidation state.
(e) Complete the following equation:
Mn0₄^– + 8H⁺ +5e^– → (CBSE Delhi 2014)
Answer:
(i) (a) When Mn0₂ is fused with caustic potash (KOH) or potassium carbonate in the presence of air, a green mass of potassium manganate is formed.
2MnO₂ + 4KOH + O₂  4 2K₂MnO₄ + 2H₂O
Or
2MnO₂ + 2K₂CO₂ + O₂ → 2K₂MnO₄ + 2CO₂.

(b) Na2Cr04 is acidified with dilute sulphuric acid to get Na₂Cr₂O₇.
2Na₂CrO₄ + H₂SO₄ → Na₂Cr₂O₇ + Na₂SO₄ + H₂O

(ii) (a) Electronic configuration of Mn²⁺ is [Ar]Bd⁵ which is half-filled and hence is stable. Therefore, the third ionization enthalpy is very high, i.e. the third electron cannot be easily removed. In the case of Fe²⁺, the electronic configuration is 3d⁶. Therefore, Fe²⁺ can easily lose one electron to acquire a 3d⁵ stable configuration. Thus, Mn²⁺ is more stable than Fe²⁺ towards oxidation to + 3 states.

(b) The high enthalpies of atomization of transition elements are due to the participation of electrons (n – 1) d-orbitals in addition to ns electrons in the interatomic metallic bonding. In the case of zinc, no electrons from 3d-orbitals are involved in the formation of metallic bonds. On the other hand, in all other metals of 3d series electrons from d-orbitals are always involved in the formation of metallic bonds.

(c) Lanthanoids show a limited number of oxidation states, such as +2, +3, and +4 (+3 is the principal oxidation state). This is because of the large energy gap between 5d and 4/ subshells. On the other hand, actinoids also show a principal oxidation state of +3 but show a number of other oxidation states also. For example, uranium (Z = 92) exhibits oxidation states of +3, +4, +5 and +6 and neptunium (Z = 94) shows oxidation states of +3, +4, +5, +6 and +7. This is because of the small energy difference between 5f and 6d orbitals.
OR
(a) Manganese (Z = 25) shows the largest number of oxidation states because it has the maximum number of unpaired electrons. It shows oxidation states from +2 to +7, i.e. +2, +3, +4, +5, +6 and +7.

(b) The E°(M²⁺|M) value for copper is positive and this shows that it is the least reactive metal among the elements of the first transition series. This is because copper has a high enthalpy of atomization and enthalpy of ionization. Therefore, the high energy required to convert Cu(s) to Cu²⁺(aq) is not balanced by its hydration enthalpy.

(c) Mn³⁺ is a stronger oxidizing agent than Cr³⁺ because Mn³⁺(3d⁴) changes to stable half-filled (3d⁵) configuration while in the case of Cr³⁺ (3d⁵) is stable (half-filled) and it cannot be readily changed to Cr²⁺ (3d⁴).

(d) Europium.

(e) MnO₄^– + 8H⁺ + 5e^– → Mn²⁺ + 4H₂O.

Question 9.
(a) Account for the following:
(i) Manganese shows the maximum number of oxidation states in 3d series.
(ii) E° value for Mn³⁺/Mn²⁺ couple is much more positive than that for Cr³⁺/Cr²⁺.
(iii) Ti⁴⁺ is colorless whereas V⁴⁺ is colored in an aqueous solution.
(b) Write the chemical equations for the preparation of KMn04 from Mn02. Why does the purple color of acidified permanganate solution decolorize when it oxidizes Fe²⁺ to Fe³⁺?
OR
(a) Write one difference between transition elements and p-block elements with reference to variability of oxidation states.
(b) Why do transition metals exhibit higher enthalpies of atomization?
(c) Name an element of the lanthanoid series which is well known to show a +4 oxidation state. Is it a strong oxidizing agent or a reducing agent?
(d) What is lanthanoid contraction? Write its one consequence.
(e) Write the ionic equation showing the oxidation of Fe(ll) salt by acidified dichromate solution. (CBSE Al 2019)
Answer:
(a) (i) Manganese has the electronic configuration: 3d⁵ 4s². It has a maximum number of unpaired electrons and hence shows maximum oxidation states.
(ii) Mn²⁺ has 3d⁵ electronic configuration. It is stable because of the half-filled configuration. Therefore, Mn3 easily gets reduced to Mn²⁺. Thus, E°(M³⁺/Mn²⁺) is positive. On the other hand, Cr³⁺ is more stable in the +3 oxidation state due to stable t2g3 configuration.
(iii) Ti⁴⁺ (3d°) does not have any d-electrons. Therefore, there are no d-d transitions whereas V⁴⁺, (3d¹) has one electron in d-subshell and d-d transitions are possible and hence it is colored.

(b) 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
3K₂MnO₄ + 4HCL → 2KMnO + MnO₂ + 2H₂O

When acidified KMnO₄ oxidizes Fe2+ to Fe3+ its purple color gets decolorized due to the formation of Mn2+ from
MnO₄– ion.
MnO₄– + 5Fe²⁺ + 8W → Mn²⁺ + 5Fe³⁺ + 4H₂O
Or
(a) Transition elements, in general, show variable oxidation states which differ by 1 unit, whereas p-block elements show variable oxidation states which differ by 2 units.

(b) Transition eLements have unpaired d-electron and therefore have strong metallic bonding between atoms. Hence, they have high enthalpies of atomization.

(c) Cerium. It is a strong oxidizing agent.

(d) The regular decrease In atomic and ionic radii of the Lanthanides with increasing atomic number is catted Lanthanoid contraction.

Consequence: 5d series elements have almost the same size as the 4d series.

(e) 6Fe²⁺ + Cr₂O₇^2- + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O

Question 10.
(i) Account for the following:
(a) Mn shows the highest oxidation state of +7 with oxygen but with fluorine, it shows the highest oxidation state of +4.
(b) Zirconium and Hafnium exhibit similar properties.
(c) Transition metals act as catalysts.
(ii) Complete the following equations:
(a) 2MnO₂ + 4KOH + O₂ 
(b) Cr₂O₇^2- + 14H⁺ + 6I^–
Or
The elements of 3d transition series are given as:
Sc Ti V Cr Mn Fe Co Ni Cu Zn
Answer the following:
(a) Write the element which is not regarded as a transition element. Give reason.
(b) Which element has the highest m.p.?
(c) Write the element which can show an oxidation state of +1.
(d) Which element is a strong oxidizing agent in the +3 oxidation state and why? (CBSE 2016)
Answer:
(i) (a) Manganese shows the highest oxidation state of +7 with oxygen but +4 with fluorine. This is because oxygen has a tendency to form multiple bonds and hence stabilize the high oxidation state.
(b) Due to lanthanoid contraction, Zr and Hf show similar properties.
(c) The transition metals act as catalysts. This is due to their ability to show multiple oxidation states.

(ii) (a) 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
(b) Cr₂O₇^2- + 14H+ + 6I^– → 2Cr³⁺ + 7H₂O + 3I₂
Or
(a) Zn because of not having partially filled d-orbitals in its ground state or ionic state.
(b) Chromium, Cr
(c) Copper, Cu
(d) Mn, because Mn²⁺ has extra stability due to half-filled d-subshell.
Mn: [Ar] 3d⁵ 4s²; Mn²⁺: [Ar]3d⁵
Mn³⁺ has four electrons (3d⁴) in 3d subshell and requires only one electron to get a half-filled 3d subshell and therefore, acts as a strong oxidizing agent.

In This Post we are  providing Chapter-9 COORDINATION COMPOUNDS NCERT MOST IMPORTANT QUESTIONS for Class 12 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON COORDINATION COMPOUNDS

Question 1.
When a coordination compound NiCl₂.6H₂0 is mixed with AgNO₃, 2 moles of AgCI are precipitated per mole of the compound. Write
(i) Structural formula of the complex
(ii) IUPAC name of the complex
Answer:
For one mole of the compound, 2 moles of AgCI are precipitated with AgNO₃, which indicates 2 ionisable Cl” ions in the complex.

1. Structural formula: [Ni (H₂0)₆]Cl₂
2. IUPAC name: Hexaaquanickel (II) chloride

Question 2.
Write IUPAC name of the complex [Cr(NH₃)₄Cl₂]⁺. Draw structures of geometrical isomers for this complex.
OR
Using IUPAC norms write the formulae for the following:
(i) Pentaamminenitrito-O-cobalt(III) chloride
(ii) Potassium tetracyanidonickelate(II) (CBSE Delhi 2019)
Answer:
IUPAC name: Tetramminedichloridochromi um(III) ion.
Geometrical isomers:

OR
(i) [CO(NH₃)₅(ONO)]Cl₂
(ii) K₂[Ni(CN)₄]

Question 3.
Write the hybridisation and magnetic character of the following complexes:
(i) [Fe(H₂O)₆]²⁺
(ii) [Fe(CO)₅] (Atomic no. of Fe = 26)
Answer:
[Fe(H₂0)6]²⁺
Hybridisation: sp³d²
Magnetic character: Paramagnetic due to 4 unpaired electrons.
Fe(C0)₅
Hybridisation: dsp³

Question 4.
Draw one of the geometrical isomers of the complex [Pt(en)₂Cl₂]²⁺ which is optically inactive. Also, write the name of this entity according to the IUPAC nomenclature.
Or
Discuss the bonding in the coordination entity [Co(NH₃)₆ ]³⁺ on the basis of valence bond theory. Also, comment on the geometry and spin of the given entity. (Atomic no. of Co= 27) (CBSE Sample Paper 2019)
Answer:

IUPAC Name of the entity:
Dichloridobis(ethane – 1,2-diamine) platinum(IV) ion
Or

Question 5.
(a) Draw the structures of geometrical isomers of [Fe(NH₃)₂ (CN)₄]^–
(b) [NiCl₄]^2- is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral. Why? [Atomic number of Ni = 28]
OR
Define the following:
(a) Ambidentate ligands
(b) Spectra chemical series
(C) Heteroleptic complexes
Answer:

(b) In the presence of strong field ligand CO, the unpaired d-electrons of Ni pair up so [Ni(CO)₄] is diamagnetic but Cl^– being a weak ligand is unable to pair up the unpaired electrons, so [Ni(Cl₄)]^2- is paramagnetic.

and

OR
Answer:
(a) Ambidentate ligands: The monodentate ligands which can coordinate with the central atom through more than one site are called ambidentate ligands.
These ligands contain more than one coordinating atoms in their molecules.
For example, NO₂ can coordinate to the metal atom through N or 0 as

Similarly, CN^– can coordinate through C or N as

Thiocyanato (SCN) can coordinate through S or N

(b) The arrangement of ligands in the increasing order of crystal field splitting is called spectrochemical series. This is shown below:
I^– < Br^– < SCN^– < Cl^– < F^– < OH^– < Ox^2- < O^2- < H₂0 < NCS^– < py = NH₃ <en <NO₂^– < CN^– < CO.

Weak field ligands are those which cause less crystal field splitting. These form high spin complexes. For example, Cl^–, F^–, etc.

Strong field ligands are those which cause greater crystal field splitting. These form low spin complexes. For example, CN^–, NO₂^–, CO.

(c) The complexes in which the metal is bound to more than one kind of donor groups (ligands) are called heteroleptic complexes. Some common examples of heteroleptic complexes are [NiCl₂(H₂O)₄], [CoCl₂(NH₃)₄]⁺, etc.

Question 6.
Write the structures and names of all the stereoisomers of the following compounds:
(i) [Co(en)₃]Cl₃
Answer:

(ii) [Pt(NH₃)₂Cl₂]
Answer:

(iii) [Fe(NH₃)4Cl₂]Cl (CBSE 2011)
Answer:

Question 7.
Write the name, stereochemistry and magnetic behaviour of the following:
(i) K₄[Mn(CN)₆]
(ii) [Co(NH₃)₂ C1]Cl₂
(iii) K₂[Ni(CN)₄] (CBSE Delhi 2011)
Answer:

Complex	Name	Stereochemistry	Magnetic behaviour
(i) K4[Mn(CN)6]	Potassium hexacyanomanganate (I)	Octahedral	Paramagnetic
(ii) [Co(NH3)5Cl]Cl2	Pentaamminechtorido cobalt (III) chloride	Octahedral	Diamagnetic
(iii) K2[Ni(CN)4)	Potassium tetracyanonicketate (II)	Square planar	Diamagnetic

Question 8.
For the complex [Fe(en)₂Cl₂]Cl identify the following:
(i) Oxidation number of iron
Answer:
III.

(ii) Hybrid orbitals and shape of the complex
Answer:
d²sp³ hybridisation, octahedral

(iii) Magnetic behaviour of the complex
Answer:
Paramagnetic due to one unpaired electron

(iv) Number of its geometric isomers
Answer:
Two

(v) Whether there may be optical isomer also
Answer:
One optical, an isomer of cls-geometrical isomer.

(vi) Name the complex. (CBSE 2011)
Answer:
Dichloridobis(ethylenediamine) iron (III) chloride.

Question 9.
Give the formula of each of the following coordination entities:
(i) CO³⁺ ion is bound to one Cl^–, one NH₃ molecule and two bidentate ethylene diamine (en) molecules.
(ii) Ni²⁺ ¡on is bound to two water molecules and two oxalate ions. Write the name and magnetic behaviour of each of the above coordination entitles. (At. nos. CO = 27, Ni = 28) (CBSE 2012)
Answer:
(i) [Co(NH₃) (Cl) (en)₂]²⁺ Amminechtoriðo bls-(ethane -1, 2-diamine) cobalt (III) ion Co(27):

Since there are no unpaired electrons,
complex is diamagnetic.

(ii) [Ni(H₂O)₂ (C₂O₄)₂]^2- Diaquadioxatatonickelate (II) ion Ni(28):

The complex has two unpaired electrons, therefore, it will be paramagnetic.

Question 10.
State a reason for each of the following situations:
(i) CO²⁺ is easily oxidised to CO³⁺ in presence of a strong ligand.
Answer:
The configuration of CO²⁺ is t_2g⁶ e_g¹ and for CO³⁺, it is t_2g⁶. The crystal field stabilisation energy is more than compensated for the third ionisation enthalpy. Therefore, CO²⁺ is easily oxidised to CO³⁺ in the presence of a strong ligand.

(ii) CO is a stronger complexing reagent than NH₃.
Answer:
CO is a stronger complexing ligand than NH₃ because it contains both σ and π character and can form a back bond (M → CO) also. Therefore, CO forms a stronger bond with the metal. It is also called a strong field ligand.

(iii) The molecular shape of Ni(CO)₄ is not the same as that of [Ni(CN)₄]^2-. (CBSE Delhi 2012)
Answer:
The molecular shape of [Ni(C0)₄] is tetrahedral because this complex nickel involves sp³ hybridisation. In [Ni(CN)₄]^2-, nickel involves dsp² and its shape is square planer.

In This Post we are  providing Chapter- 4 CHEMICAL KINETICS NCERT MOST IMPORTANT QUESTIONS for Class 12 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON CHEMICAL KINETICS

Question 1.
What is meant by rate of a reaction? Differentiate between average rate and instantaneous rate of a reaction.
Answer:
Rate of reaction: It is the change in concentration of the reactants or products in a unit time. Average rate : Average rate depends upon the change in concentration of reactants or products and the time taken for the change to occur. R → P

Instantaneous rate: It is defined as the rate of change in concentration of any one of the reactant or product at a particular moment of time.

Question 2.
(a) What is the physical significance of energy of activation ? Explain with diagram.
(b) In general, it is observed that the rate of a chemical reaction doubles with every 10 degree rise in temperature. If the generalization holds good for the reaction in the temperature range of 295 K to 305 K, what would be the value of activation energy for this reaction ?
[R = 8.314 J mol^-1 K^-1]
Answer:
(a) The minimum extra amount of energy absorbed by the reactant molecules so that their energy becomes equal to threshold value is called activation energy. Less is the activation energy, faster is the reaction or greater is the activation energy, slower is the reaction

Question 3.
(a) A reaction is second order in A and first order in B.
(i) Write the differential rate equation,
(ii) How is the rate affected on increasing the concentration of A three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled?
(b) A first order reaction takes 40 minutes for 30% decomposition. Calculate t_1/2 for this reaction. (Given log 1.428 = 0.1548)
Answer:
(a) (i) Differential rate equation :
dxdt = K [A]²[B]
(ii) When concentration of A is increased to three times, the rate of reaction becomes 9 times
r = K[3A]²B ∴ r = 9KA²B i.e. = 9 times
(iii) r = K[2A]²[2B] ∴ r = 8KA²B i.e. = 8 times

(b) Given : Time, t = 40 minutes, t =?
Let a = 100, ∴ x = 30% of 100 = 30
Using the formula :

Question 4.
(a) For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction. .
(b) Rate constant ‘k’ of a reaction varies with temperature ‘T’ according to the equation:
log k = log A – Ea2.303R(1T)
where E_a is the activation energy. When a graph is plotted for log k vs. 1T¯¯¯¯,a straight line with a slope of – 4250 K is obtained. Calculate ‘E_a‘ for the reaction. (R = 8.314 JK^-1 mol^-1)
Answer:

∴ E_a = 2.303 × 8.314 (JK^-1 mol^-1) × 4250 K
= 81.375 J mol^-1 or 81.375 kJ mol^-1

Question 5.
(a) The decomposition of A into products has a value of K as 4.5 × 10³ s^-1 at 10°C and energy of activation 60 kj mol-1. At what temperature would K be 1.5 × 10⁴ s^-1?
(b) (i) If half life period of a first order reaction is x and 3/4,^th life period of the same reaction is y, how are x and y related to each other?
(ii) In some cases it is found that a large number of colliding molecules have energy more than threshold energy, yet the reaction is slow. Why?
Answer:
(a) Given : K₁ = 4.5 × 10³ s^-1,
T₁ = 10K + 273K = 283K
K₂ = 1.5 × 10⁴ s^-1, T₂ = ?
E_a = 60 KJ mol^-1
Using formula :

∴ Temperature, T₂ will be = 297° – 273° = 24° C

(b) (i) t_1/2 = 0.693K (For first order reaction)
t_3/4 = K ⇒ t_3/4 = 1.3864K
According to condition
(The value 1.3864 is double of 0.693)
From the above equation it is clear that
t_3/4 = 2t_1/2 ∴ y = 2X
(ii) It is due to improper orientation of the colliding molecules at the time of collision.

Question 6.
(a) A first order reaction takes 100 minutes for completion of 60% of the reaction. Find the time when 90% of the reaction will be completed.
(b) With the help of diagram explain the role of activated complex in a reaction. (Comptt. Delhi 2013)
Answer:
(a) For the first order reaction,

(b) In order that the reactants may change into products, they have to cross an energy barrier as shown in the diagram

This diagram is obained by plotting potential energy vs. reaction coordinate. It is believed that when the reactant molecules absorb energy, their bonds are loosened and new bonds are formed between them. The intermediate complex thus formed is called activated complex. It is unstable and immediately dissociates to form the stable products.

Question 7.
For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained :

t/s	0	30	60
[CH3COOCH3]/mol L-1	0.60	0.30	0.15

(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
(ii) Calculate the average rate of reaction between the time interval 30 to 60 seconds. (Given log 2 = 0.3010, log 4 = 0.6021)
Answer:

As k is constant in both the readings, hence it is a pseudo first order reaction.
(ii) Rate = – Δ[R] /Δt, Average rate between 30 to 60 seconds
= −(0.15−0.30)60−30=0.1530
= 0.5 × 10^-2 mol L^-1 sec^-1

Question 8.
(a) For a reaction A + B → P, the rate is given by Rate = k[A] [B]²
(i) How is the rate of reaction affected if the concentration of B is doubled?
(ii) What is the overall order of reaction if A is present in large excess?
(b) A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction.
(log 2 = 0.3010)
Answer:
(a) For the reaction A + B → P rate is given
by Rate = k[A]¹[B]²
(i) r₁ = k[A]1 [B]2
r₂ = k[ A]¹[2B]² =
r₂ = k[A]¹ [2B]²=4k[A]¹ [B]²
r₁ = 4r₂, rate will increase four times of actual rate.

(ii) When A is present in large amount, order w.r.t. A is zero.
Hence overall order = 0 + 2 = 2, second order reaction.

Question 9.
For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained:

t/s	0	10	20
[CH3COOCH3]/mol L-1	0.10	0.05	0.025

(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
(ii) Calculate the average rate of reaction between the time interval 10 to 20 seconds. (Given : log 2 = 0.3010, log 4 = 0.6021)
Answer:

As k₁ and k₂ are equal, hence pseudo rate constant is same.
It follows the pseudo first order reaction.
(ii) Average rate of reaction between 10 to 20 seconds
= −Δ[R]Δt=−(0.025−0.05)(20−10)=0.02510
= 0.0025 mol lit^-1 sec^-1

Question 10.
(a) For a reaction A + B → P, the rate is given by Rate = k[A] [B]²
(i) How is the rate of reaction affected if the concentration of B is doubled?
(ii) What is the overall order of reaction if A is present in large excess?
(b) A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction.
Answer:
(a) For the reaction A + B → P
rate is given by Rate = k[A]¹[B]²
(i) r₁ = k[A]¹ [B]²
r₂= k[A]¹ [2B]² = 4k[A]¹ [B]²
r₁ = 4r₂ (rate of reaction becomes 4 times)

(ii) When A is present in large amounts, order w.r.t. A is zero.
Hence overall order = 0 + 2 = 2

In This Post we are  providing Chapter- 6 GENERAL PRINCIPLES AND PROCESSESS OF ISOLATION OF ELEMENTS NCERT MOST IMPORTANT QUESTIONS for Class 12 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON GENERAL PRINCIPLES AND PROCESSESS OF ISOLATION OF ELEMENTS

1. How are gold and silver extracted?
Ans. Gold and silver are extracted by leaching the metal with CN^–. The metal is later recovered by displacement method in which zinc acts as reducing agent.


2. Give two examples of metal refined by
 a) Distillation
 b) Liquation
 c) Electrolytic refining
Ans. a) Distillation – Zinc and Mercury
b) Liquation – Tin and Antimony
c) Electrolytic refining – Copper and Zinc
3. Explain electrolytic refining of copper.   
Ans. Electrolytic refining of copper-
In this method impure copper acts as anode and a strip of the same metal in the pure form is used as cathode. The electrolyte is acidified solution of copper sulphate. The net result is the transfer of copper in pure form from the anode to cathode.


Impurities from the blister copper like antimony, selenium, tellurium, silver, gold and platinum deposit as anode mud.
4. Write a short note on Mond’s process.
Ans. Mond’s Process- In this process, nickel is heated in a stream of carbon monoxide to give a volatile complex, nickel tetra carbonyl.

The carbonyl is heated to higher temperature

5. Which method is used for refining of zirconium? Explain.
Ans. Zirconium and Titanium are refined by van Ankle process, Here the crude metal is heated in an evacuated vessel with iodine.

The metal iodide is decomposed on a tungsten filament, electrically heated to about 1800K and pure metal is deposited on the filament.

6. What is the principle behind chromatography? Name some types of chromatographic techniques.
Ans. The principle behind chromatography is that different components of a mixture are differently adsorbed on an adsorbent. Some of the chromatographic techniques are paper chromatography, column chromatography, gas chromatography etc.
7. Explain the extraction of copper?   
Ans. The sulphide ores of copper are roasted to give oxides:

The oxide can then be easily reduced to metallic copper using coke.

The impurities like iron oxide are removed as slag by reacting with, added as flux.

8. What is the role of depressant in froth floatation process?
Ans. 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 .

9.  Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction?
Ans. The Gibbs free energy of formation of is less than that of and. Therefore, and C cannot reduce to Cu.
On the other hand, the Gibbs free energy of formation of is greater than that of CO. Hence, C can reduce  to Cu.

Hence, the extraction of copper from its pyrite ore is difficult than from its oxide ore through reduction.
10. State the role of silica in the metallurgy of copper.
Ans. During the roasting of pyrite ore, a mixture of FeO and is obtained.



The role of silica in the metallurgy of copper is to remove the iron oxide obtained during the process of roasting as ‘slag’. If the sulphide ore of copper contains iron, then silica is added as flux before roasting. Then, FeO combines with silica to form iron silicate, (slag).