Previous Year Question Paper of Chemistry Class 12 CBSE with Solution PDF (PYQ)

The Chemistry subject is fascinating and that contains many equations and chemical reactions. With so many chemical formulas and concepts present in the class 12 NCERT books, students get confused among them. Due to this, there is always a chance that students may skip some important concepts and study the not-so-important concepts. That is why, it is important to know the types of questions and concepts that are asked frequently in the board exam.

Chemistry PYQ Class 12

To help students in their preparation for the Chemistry board exam, below we are providing students with the Class 12 previous year question papers. Along with the question papers, our expert faculty at Adda247 have also solved the entire question papers of the previous year’s chemistry PYQ Class 12 as per the latest marking scheme.

Chemistry Class 12 Previous Year Question Paper with Solutions

Class 12 Chemistry Previous Year Question Paper plays important role in preparation. With the help of the Class 12 Chemistry Previous Year Question Paper, students can understand the types of questions asked in class 12 chemistry and they can give shape to their preparation by analyzing the Class 12 Chemistry Previous Year Question Paper. On this page, we have given Class 12 Chemistry Previous Year Question Paper with solutions. Students must solve the Class 12 Chemistry Previous Year Question Paper given on this page to pass their CBSE Class 12 Exam 2024 with flying colours.

Class 12 Biology Important Questions With Solutions

Previous Year Question Paper of Chemistry Class PYQ of Chemistry Class 12 with Solutions – 2021

Section – A

Question 1.
Write balanced chemical equations for the following processes:
(a) Cl₂ is passed through slaked lime.
(b) SO₂ gas is passed through an aqueous solution of Fe (III) salt. [2]
OR
(a) Write two poisonous gases prepared from chlorine gas.
(b) Why does the Cu²⁺ solution give a blue colour on reaction with ammonia?
Answer:
(a) Cl₂ is passed through slaked lime to give bleaching powder [Ca(OCl)₂]
2Ca(OH)₂ + 2Cl₂ → Ca(OCl)₂ + CaCl₂ + 2H₂O
(b) When SO2 gas is passed through a Fe(III) aqueous solution, Fe(III) is reduced to Fe(II) ion:
2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺ + SO₄^2- + 4H⁺
OR
(a) Two poisonous gases prepared from chlorine – Phosgene (COCl₂) and tear gas (CCl₃NO₂).
(b) Nitrogen in ammonia has a lone pair of electrons, which makes it a Lewis base. It donates the electron pair and forms linkage with metal ions-

Question 2.
Give reasons:
(a) Cooking is faster in a pressure cooker than in cooking the pan.
(b) Red Blood Cells (RBC) shrink when placed in saline water but swell in distilled water. [2]
Answer:
(a) Boiling points increase in increasing the pressure in case of liquids. Water used for cooking attains a higher temperature than the usual boiling temperature inside the pressure cooker due to the existing high pressure inside the pressure cooker vessel. This leads to a faster flow of water inside the vegetables or grains etc. resulting in faster cooking of food in a pressure cooker than in the cooking pan.

(b) Red blood cells shrink when placed in saline water because of exosmosis, i.e., water comes out from the cell to surrounding (more concentrated) to equate the concentration. Whereas, when placed in distilled water concentration within the cell becomes more than the surrounding, hence water comes inside and endosmosis takes place to equate the concentrations.

Question 3.
Define the order of the reaction. Predict the order of reaction in the given graphs:

where [R]₀ is the initial concentration of reactant and t_1/2 is a half-life. [2]
Answer:
It is defined as the sum of powers to which the concentration terms are raised in the rate law equation.
(a) In this graph, as t_1/2 is independent of initial reactant concentration, it is a first-order reaction.
(b) In this graph, as tin is directly proportional to the initial concentration of reactant hence, it is a zero-order reaction.

Question 4.
When FeCr₂O₄ is fused with Na₂CO₃ in the presence of air it gives a yellow solution of compound (A). Compound (A) on acidification gives compound (B). Compound (B) on reaction with KCl forms an orange coloured compound (C). An acidified solution of compound (C) oxidises Na₂O₃ to (D). Identify (A), (B), (C) and (D). [2]
Answer:

Question 5.
Write IUPAC name of the complex [Co(en)₂(NO₂)Cl]⁺. What type of structural isomerism is shown by this complex? [2]
OR
Using IUPAC norms, write the formulae for the following complexes:
(a) Hexaaquachromium (III) chloride
(b) Sodium trioxalatoferrate (III)
Answer:
IUPAC name of [Co(en)₂(NO₂)Cl]⁺ is Chlorobis(ethane-1, 2-diamine)nitrocobalt(III).
This compound shows geometrical isomerism.
OR
(a) Hexaaquachromium(III) chloride- [Cr(H₂O)₆]Cl₃
(b) Sodium trioxalatoferrate(III)- Na₃[Fe(C₂O₄)₃]

Question 6.
(a) Although both [NiCl₄]^2- and [Ni(CO)₄] have sp³ hybridisation yet [NiCl₄]^2- is paramagnetic and [Ni(CO)₄] is diamagnetic. Give reason. (Atomic no. of Ni = 28).
(b) Write the electronic configuration of d⁵ on the basis of crystal field theory when
(i) ∆₀ < P and (ii) ∆₀ > P [2]
Answer:
(a) [NiCl₄]^2- is a high spin complex and there are two impaired electrons with 3d8 electronic configuration of a central metal atom, hence it is paramagnetic. Whereas in [Ni(CO)₄] Ni is in zero oxidation state and contains no unpaired electrons, hence it is diamagnetic in nature.
(b) (i) Electronic configuration of d⁵ when ∆o < P is given as t_2g³ e_g²
(ii) Electronic configuration of d⁵ when ∆o > P is given as t_2g⁵ e_g⁰

Question 7.
Write structures of main compounds A and B in each of the following reactions: [2]

Question 8.
Arrange the following in decreasing order of basic character: [1]
C₆H₅NH₂, (CH₃)₃N, C₂H₅NH₂
Answer:
Decreasing order of basic character:
CH₃CH₂NH₂ > (CH₃)₃N > C₂H₅NH₂

Question 9.
What type of colloid is formed when a solid is dispersed in a liquid ? Give an example. [1]
Answer:
Sols are formed when a solid is dispersed in a liquid.
Example – Paints.

Question 10.
Out of Chlorobenzene and Cyclohexyl chloride, which one is more reactive towards nucleophilic substitution reaction and why?
Answer:
Cyclohexyl chloride is more reactive towards nucleophilic substitution reaction because the carbon bearing the chlorine atom is deficient in electron and seeks a nucleophile. In Chlorobenzene the carbon bearing the halogen is a part of an aromatic ring and is electron-rich due to the electron density in the ring.

Question 11.
What is the basic structural difference between starch and cellulose? [1]
OR
Write the products obtained after hydrolysis of DNA.
Answer:
Starch consists of two components- amylose and amylopectin. Amylose is a long linear chain of α-D -(+)-glucose units joined by C₁-C₄ glycosidic linkage (α-link). Amylopectin is a branched-chain polymer of α-D-glucose units, in which the chain is formed by C₁-C₄ glycosidic linkage and the branching occurs by C₁-C₆ glycosidic linkage. On the other hand, cellulose is a straight-chain polysaccharide of β-D-glucose units joined by C₁-C₄ glycosidic linkage (β-link).
OR
Hydrolysis of DNA yields a pentose sugar (β-D-2deoxyribose), phosphoric acid and nitrogen-containing heterocyclic compounds called bases (Adenine, Guanine, Cytosine and Thymine).

Question 12.
(a) The conductivity of 0.001 mol L^-1 acetic acid is 4.95 × 10^-5 S cm^-1. Calculate the dissociation constant if ∧⁰_m for acetic acid is 390.5 S cm² mol^-1.
(b) Write Nest equation for the reaction at 25°C:
2Al(s) + 3Cu²⁺ (aq) → 2 Al³⁺ (aq) + 3Cu (s) (c)
What are secondary batteries? Give an example. [5]
OR
(a) Represent the cell in which the following reaction takes place:
2Al (s) + 3 Ni²⁺ (0.1M) → 2Al³⁺ (0.01M) + 3 Ni (s)
Calculate its emf if E⁰_cell = 1.41 V.
(b) How does molar conductivity vary with increase in concentration for strong electrolyte and weak electrolyte? How can you obtain limiting molar conductivity (∧m0) for weak electrolyte?
Answer:
(a) Conductivity ∧_m of a solution is given by the following equation:
∧_m = kc,
where k is dissociation constant and c is the concentration of the solution.
Here, given.
Conductivity, k = 4.95 × 10^-5 S cm^-1
Limiting molar conductivity,
∧⁰_m = 390.5 S cm² mol^-1
Concentration,
c = 0.001 mol L^-1 = 1 × 10^-3 mol L^-1
Substituting the given values in above equation
Molar conductivity,

(b) Nemst equation for the given reaction can be written as

(c) A secondary battery can be recharged after use, bypassing current through it in the opposite direction so that it can be used again.
Example: The most important secondary cell is lead storage cell. It consists of a lead anode and a grid of lead packed with lead dioxide as a cathode. A 38% solution of sulphuric acid is used as an electrolyte.
OR
(a) The cell can be represented as

(b) For strong electrolytes, the molar conductivity is increased only slightly on dilution. A strong electrolyte is completely dissociated in solution and thus, furnishes all ions for conductance. However, at higher concentrations, the dissociated ions are close to each other and thus, the interionic attractions are greater. These forces retard the motion of the ions and thus, conductivity is low. With a decrease in concentration (dilution), the ions move away from each other thereby feeling less attractive forces from the counterions. This results in an increase in molar conductivity with dilution. The molar conductivity approaches a maximum limiting value at infinite dilution designated as ∧m0.

In the case of weak electrolytes as the solution of a weak electrolyte is diluted, its ionization is increased. This results in more number of ions in solution and thus, there is an increase in molar conductivity, also there is a large increase in the value of molar conductivity with dilution, especially near-infinite dilution. However, the conductance of a weak electrolyte never approaches a limiting value. Or in other words, it is not possible to find conductance at infinite dilution (zero concentration).
So, limiting molar conductivity for weak electrolytes are obtained by using Kohlrausch law, from the limiting molar conductivities of individual ions (λ0).

Kohlrausch law of independent migration of ions states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and the cation of the electrolyte.
∧_m⁰ = λ⁰⁺ + λ⁰

Question 13.
(a) Give the equation of the following reactions:
(i) Phenol is treated with cone. HNO₃.
(ii) Propene is treated with B₂H₆ followed by H₂O₂/OH-.
(iii) Sodium t-butoxide is treated with CH₃Cl.
(b) How will you distinguish between butane-l-ol and butane-2-ol?
(c) Arrange the following in increasing order of acidity: Phenol, ethanol, water [5]
OR
(a) How can you obtain Phenol from (i) Cumene, (ii) Benzene sulphonic acid, (iii) Benzene diazonium chloride?
(b) Write the structure of the major product obtained from the denitration of 3-methyl phenol.
(c) Write the reaction involved in Kolbe’s reaction.
Answer:
(a) (i) Phenol is treated with conc. HNO3 to obtain 2,4,6-trinitrophenol picric acid.

(ii) Propene undergoes hydroboration-oxidation when treated with B₂H₆ followed by hydrogen peroxide in basic medium to give propan-1-ol.

(iii) Methyl tert-butyl ether is produced when sodium tert-butoxide is treated with methyl chloride.

(b) Butan-l-ol and Butan-2-ol can be distinguished using Lucas reagent (ZnCl²⁺HCl), where butan-2-ol would react with Lucas reagent in around 5 minutes to give a white precipitate of 2-chlorobutane, whereas butan-l-ol won’t give any reaction at room temperature.
(c) Increasing order of acidity can be given as Ethanol < water < phenol
OR
(a) (i) Phenol from cumene

(ii) Phenol from benzene sulphonic acid

(iii) Phenol from benzene diazonium chloride

(b) The combined influence of -OH and -CH₃ groups determine the position of the entering groups, also the sterically hindered positions are not substituted.

(c) In Kolbe’s reaction phenol is reacted with CO₂ in the presence of sodium hydroxide, followed by acidification, to give a carboxylic acid group on 2-position of phenol-

Question 10.
(a) Account for the following:
(i) The tendency to show – 3 oxidation state decreases from N to Bi in group 15.
(ii) Acidic character increases from H₂O to H₂Te.
(iii) F₂ is more reactive than CIF₃, whereas ClF₃ is more reactive than Cl₂.
(b) Draw the structure of (i) XeF₂ (ii) H₄P₂O₇. [5]
OR
(a) Give one example to show the anomalous reaction of fluorine.
(b) What is the structural difference between white phosphorous and red phosphorous?
(c) What happens when XeF₆ reacts with NaF?
(d) Why is H₂S a better reducing agent than H₂O?
(e) Arrange the following acids in the increasing order of their acidic character: HF, HCl, HBr and HI
Answer:
(a) (ii) Acidic character increases from H₂O to H₂Te due to decrease in E—H bond dissociation enthalpy down the group. Thus it becomes easy to lose proton going down the group.
(iii) F₂ is more reactive than ClF₃ because of the small size of fluorine atom F—F bond, bond dissociation enthalpy is low (thus is reactive).
Whereas ClF₃ is more reactive than Cl₂ because ClF₃ is an interhalogen compound with weak Cl—F bond (compared to Cl—Cl bond) due to the difference in atomic sizes (hence ineffective overlap of orbitals).
(b) (i) Structure of XeF₂ is linear.

(ii) Structure of H₄P₂O₇:

OR
(a) Fluorine reacts with cold sodium hydroxide solution to give OF₂.
2F₂ (g) + 2NaOH (aq) → 2NaF (aq) + OF₂ (g) + H₂O(l)
(c) XeFg reacts with NaF as follows:
XeF₆ + NaF → Na⁺[XeF₇]^–
(d) Ability to reduce is judged by ease with which an atom can donate its electrons to the species which is getting reduced. Now, the size of oxygen atom in H₂O is smaller than that of Sulphur atom in H₂S, due to which the lone pair of electrons on oxygen are more attracted by the oxygen nucleus, making it difficult to donate the electrons (by oxygen compared to Sulphur, while in H₂S the influence of the nucleus is less on lone pair of electrons of sulphur and hence, it can give away its electrons, easily compared to oxygen, and thus acts as a better reducing agent.
(e) The increasing order of acidic character can be written as
HF < HCl < HBr < HI

Question 11.
The following data were obtained for the reaction:
A + 2B → C

(a) Find the order of reaction with respect to A and B.
(b) Write the rate law and overall order of the reaction.
(c) Calculate the rate constant (k). [3]
Answer:
The reaction is A + 2B → C
(a) It can be seen that when the concentration of A is doubled keeping B constant, then the rate increases by a factor of 4 (from 4.2 × 10^-2 to 1.68 × 10^-1). This indicates that the rate depends on the square of the concentration of the reactant A. Also when the concentration of reactant B is made four times, keeping the concentration of reactant A constant, the reaction rate also becomes 4 times (2.4 × 10^-2 to 6.0 × 10^-3). This indicates that the rate depends on the concentration of reactant B to the first power.
(b) So, the rate equation will be:
Rate = k[A]²[B]
Overall order of reaction will be 2 + 1 = 3.
(c) Rate constant can be. calculated by putting the values given.
4.2 × 10^-2 M min^-1= k (0.2)²(0.3) M
k = 0.0420.012 = 3.5 min^-1

Question 12.
(a) Write the dispersed phase and dispersion medium of dust.
(b) Why is physisorption reversible whereas chemisorption is irreversible?
(c) A colloidal sol is prepared by the method given in the figure.
What is the charge of Agl colloidal particles formed in the test tube?
How is this sol represented? [3]

Answer:
(a) In dust, the dispersed phase is solid particles and the dispersion medium is air (gas).
(b) Physisorption occurs only because of physical attractive forces, like van der Waals forces between molecules of adsorbate and adsorbent, hence that can be reversed on the application of bigger forces but chemisorption occurs due to the chemical reaction between molecules of adsorbate and adsorbent, and hence can’t be reversed.
(c) When KI solution is added to AgNO₃ a positively charged sol results due to absorption of Ag⁺ ions from dispersion medium-AgI/ Ag⁺(p0sitively charged)

Question 13.
A solution containing 19 g per 1oo mL of K (M = 74.5 g mol^-1) is isotonic with a solution containing 3 g per 100 mL of urea (M = 60 g mol^-1). Calculate the degree of dissociation of KCl solution. Assume that both the solutions have the same temperature. [3]
Answer:
Two solutions having the same osmotic pressure at a given temperature are called isotonic solution. Now in the given problem, the KCl and urea solutions are given to be isotonic.
Osmotic pressure π is given by the equation
π = (n₂/V)RT,
where n₂ = moles of solute,
V = volume of a solution in litre.
Also, n₂ = w₂/M₂,
where W₂ = grams of solute
and M₂ = molar mass of solute.
The other given information is
The molar mass of KCl = 74.5 g mol^-1
Weight of KCl, W₂= 1.9 g, V = 100 mL
So, for KCl
n = (w₂/M₂ × V)RT
nRTKCl = 1.9/(74.5 × 100) = 2.55 × 10^-4
Now as the solutions are isotonic at the same temperature:
πRTKCl = πRTurea
Hence, substituting the values for urea:
2.55 × 10^-4 = 3/M₂ × 100
M₂ = 117.6
So, the experimentally determined molecular weight of urea is found to be as 117.6, so the degree of dissociation can be given as:
Osmotic pressure (TT) = Experimentally determined

So, Urea dimerized in the given experimental solution.

Question 14.
Write the name and principle of the method used for refining of (a) Zinc, (b) Germanium, (c) Titanium. [3]
Answer:
(a) Distillation is used for refining zinc. As zinc is a low boiling metal, the impure metal is evaporated and the pure metal is obtained as a distillate.
(b) Zone refining is used for refining Germanium. This method is based on the principle that the impurities are more soluble in the melt than in the solid-state of the metal.
(c) Titanium is refined by van Arkel method. This method is used for the removal of oxygen and nitrogen present as an impurity. The crude metal is heated in an evacuated vessel with iodine to obtain metal iodide, which volatilizes being covalent. Later this metal iodide is decomposed through electrical heating to obtain the pure metal.

Question 16.
Give reasons for the following:
(a) Transition metals form complex compounds.
(b) E⁰ values of (Zn²⁺/Zn) and (Mn²⁺/Mn) are more negative than expected.
(c) Actinoids show a wide range of oxidation states. [3]
Answer:
(a) Transition elements have partly filled d-orbitals due to which they have variable oxidation states which enables them to bind with a variety of ligands and hence form complex compounds.
(b) Oxidation of Zn to Zn²⁺ leads to a completely filled d10 configuration in Zn²⁺, making it more stable. Also, Mn/Mn²⁺ conversion leads to a half-filled stable d⁵ configuration of Mn²⁺ ion. Hence, E⁰ value for Zn/Zn²⁺ and Mn/ Mn²⁺conversion has negative values.
(c) Actinoids show a wide range of oxidation states due to their partially filled f-orbitals and they have comparable energies as well.

Question 17.
Write the structures of monomers used for getting the following polymers:
(a) Nylon-6
(b) Terylene
(c) Buna-N [3]
OR
(a) Is [CH₂-CH(C₆H₅)]_n homopolymer or copolymer? Give reason.
(b) Write the monomers of the following polymer:

(c) Write the role of benzoyl peroxide in the polymerisation of ethene.
Answer:
Structures of monomers
(a) Caprolactam is monomer of Nylon-6

Question 18.
(a) Pick out the odd one from the following on the basis of their medicinal properties:
Equanil, Seconal, Bithional, Luminal
(b) What types of detergents are used in dishwashing liquids?
(c) Why is the use of aspartame limited to cold foods? [3]
OR
Define the following terms with a suitable example of each:
(a) Antibiotics
(b) Antiseptics
(c) Anionic detergents
Answer:
(a) ‘Bithionol’ is the odd one here, as it is an antiseptic whereas others are tranquillizers.
(b) Liquid dishwashing detergents are non-ionic type.
(c) Aspartame is an artificial sweetener which is unstable at cooking temperature hence its use is limited to cold foods.
OR
(a) Antibiotics: These are the compounds (produced by microorganisms or synthetically) which either inhibit the growth of bacteria or kill bacteria. Example: Penicillin.
(b) Antiseptics: These are the chemicals used to kill or prevent the growth of microorganisms when applied to the living tissues.
Example: Soframicin.
(c) Anionic detergents: These are sodium salts of sulfonated long-chain alcohols or hydrocarbons. In these, the anionic part of the molecule is involved in the cleansing action.
Example: Sodium lauryl sulphate.

Question 19.
Among all the isomers of molecular formula C₄H₈Br, identify:
(a) the one isomer which is optically active.
(b) the one isomer which is highly reactive towards SN₂.
(c) the two isomers which give the same product on dehydrohalogenation with alcoholic KOH. [3]
Answer:
(a) 2- Bromobutane is optically active as C-2 is a chiral carbon here having all the four different groups attached to it.
(b) 1-Bromobutane being primary alkyl halide is highly reactive towards SN₂ reaction.
(c) 2-Bromo-2-methylpropane and 1-Bromo-2-methylpropane would give the same product after dehydrohalogenation.

Question 20.
Complete the following reactions: [3]
OR
How do you convert the following:
(a) N-phenylethylamine to p-bromaniline
(b) Benzene diazonium chloride to nitro-benzene
(c) Benzoic acid to aniline

Question 21.
(a) Give reasons:
(i) Benzoic acid is a stronger acid than acetic acid.
(ii) Methanal is more reactive towards nucleophilic addition reaction than ethanal.
(b) Give a simple chemical test to distinguish between propanal and propanone. [3]
Answer:
(a) (i) Benzoic acid is a stronger acid than acetic acid because the benzoate anion (the conjugate base of benzoic acid) formed after loss of H⁺ is stabilized by resonance, whereas acetate ion (CH₃COO^–) has no such extra stability. Hence, Benzoic acid has more tendency of losing proton compared to acetic acid hence more acidic.

(ii) Methanal is more reactive towards nucleophilic addition reaction than ethanal because in ethanal there is a methyl group attached to the carbonyl carbon (centre for nucleophile attack) and +1 effect of the methyl group decreases the nucleophilicity of carbonyl carbon by increasing the electron density at carbonyl carbon.

(b) Propanal and propanone can be distinguished using Tollen’s reagent by silver mirror test. Propanal being an aldehyde reacts with Tollen’s reagent to give silver deposition whereas propanone being a ketone does not give the reaction.

Question 22.
(a) What is the product of hydrolysis of maltose?
(b) What type of bonding provides stability to the α-helix structure of a protein?
(c) Name the vitamin whose deficiency causes pernicious anaemia. [3]
OR
Define the following terms:
(a) Invert sugar
(b) Native protein
(c) Nucleotide
Answer:
(a) On hydrolysis maltose gives two molecules of α-D-glucose.

(b) α-Helix structure of proteins is stabilized by hydrogen bonds between -NH group of each amino acid and -COOH group of amino acid at adjacent turn.

(c) Deficiency of Vitamin B₁₂ causes pernicious anaemia.
OR
(a) Invert sugar: It is a mixture of glucose and fructose obtained after hydrolysis of sucrose. Sucrose is dextrorotatory, but after hydrolysis gives a mixture of dextrorotatory glucose and levorotatory fructose which outweighs in magnitude and hence the whole mixture becomes levorotatory hence the mixture obtained is called invert sugar.
(b) Native protein: Protein found in a bio¬logical system with a unique three-dimensional structure and biological activity is called a native protein.
(c) Nucleotide: They are building blocks of DNA/RNA. These consist of a pentose sugar moiety attached to a nitrogenous base at V position and a phosphoric acid molecule at 5′ position.

PYQ Chemistry Class 12 with Solution – 2020

Question 1.
What happens when [3]
(a) a freshly prepared precipitate of Fe(OH)₃ is shaken with a small amount of FeCl₃ solution?
(b) persistent dialysis of a colloidal solution is carried out?
(c) an emulsion is centrifuged?
Answer:
(a) When FeCl₃ is added to a freshly prepared precipitate of Fe(OH)₃, a positively charged sol of hydrated ferric oxide is formed due to adsorption of Fe³⁺ ions.

(b) When persistent dialysis of the colloidal solution is carried out, traces of electrolytes present in the sol are removed almost completely leaving the colloids unstable and finally, coagulation takes place.

(c) Emulsions are centrifuged to separate them into constituent liquids.

Question 2.
Write the chemical reactions involved in the process of extraction of Gold. Explain the role of dilute NaCN and Zn in this process. [3]
Answer:
Extraction of gold involves leaching the metal with a dilute solution of NaCN or KCN in the presence of air (for O₂) from which the metal is obtained later by replacement method (using Zinc).
The reactions involved are:
4Au(s) + 8CN^–(aq) + 2H₂O(aq) + O₂(g) → 4[AU(CN)₂]^– (aq) + 4OH^– (aq)
2 [AU(CN)₂]^– (aq) + Zn(s) → 2Au (s) + [Zn(CN)₄]^2- (aq)

Question 3.
Give reasons:
(a) E⁰ value for Mn³⁺/Mn²⁺ couple is much more positive than that for Fe³⁺/Fe²⁺.
(b) Iron has a higher enthalpy of atomization than that of copper.
(c) Sc³⁺ is colourless in aqueous solution whereas Ti³⁺ is coloured. [3]
Answer:
(a) Mn²⁺ has a d⁵ configuration, and the extra stability of half-filled d-orbitals is compromised when another electron is taken out to give Mn³⁺, On the contrary, Fe³⁺ attains a half-filled orbital configuration when Fe²⁺ gets oxidized to Fe³⁺. Hence, the E⁰ value for Mn³⁺/ Mn²⁺ couple has more positive E⁰ value.

(b) Fe has a 3d⁶4s² outer electronic configuration whereas Cu has 3d¹⁰4s¹ configuration. Now, more the number of impaired electrons in d-orbital, more favourable are interatomic attractions and thus higher atomization enthalpies. Hence, Fe having 4 unpaired d-electrons has more enthalpy of atomization than copper having no unpaired d-electron.

(c) Sc³⁺ has a 3d⁰ configuration whereas Ti³⁺ has a 3d¹ configuration. As there are no electrons in d orbital for Sc³⁺ ion, there is no transition of electrons by absorption of energy and hence no emission in visible range imparting colour to the Sc³⁺ ion.

Question 4.
(a) Identify the chiral molecule in the following pair: [3]
(b) Write the structure of the product when chlorobenzene is treated with methyl chloride in the presence of sodium metal and dry ether.
(c) Write the structure of the alkene formed by dehydrohalogenation of 1-Bromo-1 methylcyclohexane with alcoholic KOH.
Answer:
(a) The molecule (i) is a chiral molecule.

(b) Chlorobenzene reacts with methyl chloride in the presence of sodium metal and dry ether to give toluene. This reaction is known as Wurtz-Fitting reaction.

(c) In the 1-Bromo-1-methylcyclohexane, all β-hydrogen atoms are equivalent. Thus dehydrohalogenation takes place, in the reaction of this compound with KOH.

Question 5.
(A), (B) and (C) are three non-cyclic functional isomers of a carbonyl compound with molecular formula C₄H₈O. Isomers (A) and (C) give positive Tollen’s test whereas isomer (B) does not give Tollen’s test but gives positive Iodoform test. Isomers (A) and (B) on reduction with Zn (Hg)/conc. HCl, give the same product (D).
(a) Write the structures of (A), (B), (C) and (D).
(b) Out of (A), (B) and (C) isomers, which one is least reactive towards the addition of HCN? [3]
Answer:
(a) Compound A and C give positive Tollen’s test which indicates that they are aldehydes. Compound C gives Iodoform test which means it contains a carbonyl group with a methyl group attached to the carbonyl carbon so, with formula C₄H₈O the structure of compound would be CH₃COCH₂CH₃ (Butanone).

Now upon reduction with Zn(Hg)/conc. HCl, the corresponding alkanes are obtained, so a reduction of B gives Butane (D), so the isomer A has to be a linear chain aldehyde (Butanal), giving Butane (compound D) on reduction. So, the last isomer possible is compound C, 2-Methyl propionaldehyde. The reactions involved are shown below with the structures of compounds:

(b) Out of the three isomers A, B and C, compound B’ (Butanone) would be least reactive towards the addition of HCl as the carbonyl carbon is sterically hindered and most reactive would be compound A (Butanal) towards the addition of HCN.

Question 6.
Write the structures of the main products in the following reactions: [3]
Answer:
(i) Sodium borohydride doesn’t reduce esters, so the product would be,

Question 7.

1. Why is bithional added to soap? [3]
2. What is the tincture of iodine? Write its one use.
3. Among the following, which one acts as a food preservative?
   Aspartame, Aspirin, Sodium Benzoate, Paracetamol

Answer:

1. Bithional is added to soaps to impart antiseptic properties to soap.
2. Tincture of iodine is 2-3 per cent mixture of iodine in the alcohol-water mixture. It is used as an antiseptic.
3. Sodium benzoate is used as a food preservative.

Question 8.
Define the following with an example of each: [3]
(a) Polysaccharides
(b) Denatured protein
(c) Essential amino acids
OR
(a) Write the product when D-glucose reacts with conc. HNO₃.
(b) Amino acids show amphoteric behaviour. Why?
(c) Write one difference between α-helix and β-pleated structures of proteins.
Answer:
(a) Polysaccharides: Polysaccharides are food storage materials and most commonly found carbohydrates in nature. These are the compound which is formed of a large number of monosaccharide units joined together by glycosidic linkages. Example. Starch, main storage polysaccharide of plants.

(b) Denatured protein: Proteins have a unique three-dimensional structure in their native form. If the native form of protein is subjected to any physical change (such as temperature change) or any chemical change (such as a change in pH), the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled due to which protein loses its biological activity. This is called denaturation of the protein. During denaturation 2° and 3° structures of proteins are destroyed but 1° structure remains intact. Coagulation of egg white is an example of denaturation of the protein.

(c) Essential amino acids: The amino acids which are not synthesized in our body and have to be obtained through diet are known as essential amino acids. Example: Tryptophan
OR
(a) D-Glucose gets oxidized to give saccharic acid, a dicarboxylic acid on reacting with nitric acid.

(b) Amino acids show amphoteric behaviour due to the presence of both acidic (carboxylic group) and basic (amino group) in the same molecule. So, in the basic medium, the carboxyl group can lose a proton and in acidic medium, the amino group can accept a proton.

(c) In α-helix structure the polypeptide chain forms all possible hydrogen bonds by twisting into a right-handed screw (helix) with the -NH group of each amino acid residue gets hydrogen-bonded to the -C = O of an adjacent turn of the helix (Intra. molecular bonding), whereas in β-structure all peptide chains are stretched out to nearly maximum extension and then laid side by side which are held together by intermolecular hydrogen bonds (intermolecular bonding).

Question 9.
(a) Write the formula of the following coordination compound: Iron (III) hexacyanoferrate (II)
(b) What type of isomerism is exhibited by the complex [Co(NH₃)₅ Cl]SO₄?
(c) Write the hybridisation and number of unpaired electrons in the complex [CoF₆]^3-.
(Atomic number of Co = 27) [3]
Answer:
(a) The molecular formula of Iron(III) α-cyanoferrate(II) is Fe₄[Fe(CN)₆]₃
(b) [CO(NH₃)₅Cl]SO₄ will show Ionisation isomerism and the possible isomers are [CO(NH₃)₅Cl]SO₄ and [Co (NH₃)₅SO₄]Cl
(c) Electronic configuration of Co³⁺ ion is,

Electronic configuration of sp³d² hybridized (as F is a weak field ligand) orbitals of Co³⁺, with six pairs of electrons from six F ions.

There are 4 impaired electrons in [CoF₆]₃.

Question 10.
Shyam went to a grocery shop to purchase some food items. The shopkeeper packed all the items in polythene bags and gave them to Shyam. But Shyam refused to accept the polythene bags and asked the shopkeeper to pack the items in paper bags. He informed the shopkeeper about the heavy penalty imposed by the government for using polythene bags. The shopkeeper promised that he would use paper bags in future in place of polythene bags. [4]
Answer the following:
(a) Write the values (at least two) shown by Shyam.
(b) Write one structural difference between low-density polythene and high-density polythene.
(c) Why did Shyam refuse to accept the items in polythene bags?
(d) What is a biodegradable polymer? Give an example.
Answer:
(b) Low-density polythene has a branched-chain structure, whereas the high-density polythene has a linear chain structure.
(c) Shyam refused to take the items in polythene bags as polythene is non-biodegradable neither recyclable,
(d) Biodegradable polymers contain functional groups similar to functional groups present in biopolymers, so they get degraded in the environment by certain microorganisms and thus are environment-friendly.
For example Poly β -hydroxybutyrate-co-β-hydroxy valerate (PHBV).

Question 11.
(a) Give reasons: [5]
(i) H₃PO₃ undergoes disproportionation reaction but H₃PO₄ does not.
(ii) When Cl₂ reacts with an excess of F₂, ClF₃ is formed and not FCl₃.
(iii) Dioxygen is a gas while Sulphur is a solid at room temperature.
(b) Draw the structures of the following:
(i) XeF₄
(ii) HClO₃
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?
(b) Arrange the following in the decreasing order of their reducing character: HF, HCl, HBr, HI
(c) Complete the following reaction:
XeF₄ + SbF₅ →
Answer:
(a) (i) In H₃PO₃ (orthophosphoric acid) oxidation state of phosphorus is +3 and it contains one P-H bond in addition to P = O and P-OH bonds. These type of oxoacids tend to undergo disproportionation to give orthophosphoric acid (P has +5 state) and phosphine (P has +3 state). Whereas in H₃PO₄ (orthophosphoric acid), Phosphorus is in +5 state hence no disproportionation takes place in H₃PO₄.

(ii) When Cl₂ reacts with an excess of F₂, ClF₃ is formed-and not FCl₃ because Fluorine can’t expand its valency and can show only -1 oxidation state, whereas Cl can expand its valency due to the availability of d-orbitals.

(iii) Dioxygen is a gas while sulphur is a solid at room temperature this is because sulphur have S₈ molecules and these are packed to give different crystal structure, whereas dioxygen is a diatomic molecule (O₂) and it does not have enough intermolecular attraction and thus exists in gaseous form.

(a) (i) The brown gas A is NO₂ or nitrogen dioxide. On cooling, it dimerises to N₂O₄ and solidifies as a colourless solid.

(iii) Compound A, that is, NO₂ contains the odd number of valence electrons. It behaves as a typical odd molecule. On dimerization, it is converted to stable N₂O₄ molecule with even number of electrons (thus colourless) and have better intermolecular forces to get solidified. Thus, it changes to solid on cooling.
(b) Decreasing order of reducing character: HI > HBr > HCl > HF
(c) XeF₄ + SbF₅ → [XeF₃]⁺ + [SbF₆]^–

Question 12.
(a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K: [5]
Sn(s) | Sn²⁺ (0.004 M) || H⁺ (0.020 M) | H₂(g) (1 bar) | Pt(s)
(Given: E° Sn²⁺/Sn = – 0.14V)
(b) Give reasons:
(i) On the basis of E° values, O₂ gas should be liberated at anode but it is Cl₂ gas which is liberated in the electrolysis of aqueous NaCl.
(ii) The conductivity of CH₃COOH decreases on dilution.
OR
(a) For the reaction
2AgCl(s) + H₂(g) (1 atm) → 2Ag(s) + 2H⁺ (0.1M) + 2Cl^– (0.1M), ΔG° = -43600 J at 25°C.
Calculate the e.m.f. of the cell. [log 10^-n = -n]
(b) Define fuel cell and write its two advantages.
Answer:
(a) The half cell reactions can be written as;

The reaction at the anode with a lower value of Ecell is preferred and therefore, water should get oxidized to give O₂ but on account of overpotential of oxygen, Cl^– gets oxidized preferably, liberating Cl₂ gas.
(ii) The conductivity of CH₃COOH decreases on dilution because the number of ions per unit volume that carry the current in a solution decreases on dilution.
OR
(b) Galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol etc. directly into electrical energy are called fuel cells.
Advantages of fuel cells are:

• Fuel cells produce electricity with an efficiency of about 70% compared to thermal plants whose efficiency is about 40%.
• Fuel cells are pollution-free.

Question 13.
(a) Write the reactions involved in the following: [5]
(i) Hofmann bromamide degradation reaction
(ii) Diazotisation
(iii) Gabriel phthalimide synthesis
(b) Give reasons:
(i) (CH₃)₂NH is more basic than (CH₃)₃N in an aqueous solution.
(ii) Aromatic diazonium salts are more stable than aliphatic diazonium salts.
OR
(a) Write the structures of the main products of the following reactions:
(b) Give a simple chemical test to distinguish between Aniline and N, N-dimethylaniline.
(c) Arrange the following in the increasing order of their pK_b values:
C₆H₅NH₂, C₂H₅NH₂, C₆H₅NHCH₃
Answer:
(a) (i) Hofmann bromamide degradation reaction: Acetamide can be considered for example. In this reaction, Acetamide (CH₃CONH₂) undergoes Hofmann degradation in the presence of Bromine and NaOH to give Methanamine.
CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂ + Na₂CO₃ + 2NaBr + 2H₂O

(ii) Diazotisation: The conversion of primary aromatic amines into diazonium salts is known as diazotisation.

(iii) Gabriel phthalimide synthesis: This reaction is used for the preparation of primary amines. Phthalimide on treatment with ethanolic potassium hydroxide forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis produces the corresponding primary amine.
(b) (i) (CH₃)₂ NH is more basic than (CH₃)₃N in aqueous solutions because in (CH₃)₃N the lone pair of electrons on the nitrogen atom is responsible for its basicity are quite hindered by the three methyl groups, hence are less available. Due to which it is less basic as compared to (CH₃)₂NH.

(ii) Aromatic diazonium salts are more stable than aliphatic diazonium salts because the positive charge on the nitrogen atom is stabilized by the resonance with an attached phenyl group.

(b) Aniline can be distinguished from N, N-dimethyl aniline by diazo coupling reaction. Aniline would react with benzene diazonium chloride to give a yellow dye, whereas N, N-dimethyl aniline won’t undergo this reaction.

Question 14.
CO(g) and H₂(g) react to give different products in the presence of different catalysts. Which ability of the catalyst is shown by these reactions? [1]
Answer:
CO(g) and H₂(g) react in the presence of different catalysts to give different products, this shows that the action of a catalyst is highly selective in nature.

Question 15.
Write the coordination number and oxidation state of Platinum in the complex [Pt(en)₂Cl₂]. [1]
Answer:
Coordination number: 6;
Oxidation state: +2

Question 16.
Out of chlorobenzene and benzyl chloride, which one gets easily hydrolysed by aqueous NaOH and why? [1]
Answer:
Benzyl chloride would be easily hydrolysed compared to chlorobenzene. In the given reaction condition, hydrolysis proceeds by nucleophilic substitution mechanism and the benzylic carbonium ion formed after losing the leaving group (-Cl) is better stabilized (through resonating structures) hence reacts easily.

Question 17.
Write the IUPAC name of the following: [1]
Answer:
The IUPAC name would be 3, 3- Dimethyl-pentane-2-ol.

Question 18.
Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g mol^-1) in 250 g of water.
(K_f of water = 1.86 K kg mol^-1) [2]
Answer:
Molality (m) of a given solution of Glucose:
m = [(60/180) g mol^-1/250 g] × 1000 = 1.33 mol kg^-1
Now, depression in freezing point is given by, ΔT_f = K_fm
Putting the given values,
ΔT_f = K_fm = 1.86 × 1.33 = 2.5
So, the freezing point of the solution would be = 273.15 K – 2.5 K = 270.65 K.

Question 19.
For the reaction [2]
2N₂O₅(g) → 4NO₂(g) + O₂(g)
the rate of formation of NO₂(g) is 2.8 × 10^-3 Ms^-1. Calculate the rate of disappearance of N₂O₅(g).
Answer:
Rate of reaction for the given reaction can be given as,
Rate = 1/2 {-Δ[N₂O₅]/Δt} or {-Δ[N₂O₅]/Δt} = 1/2 {[NO₂/ Δt]}
So, the rate of disappearance of N₂O₅ would be half of the rate of production of NO₂ (given 2.8 × 10^-3 Ms^-1).
So, the rate of disappearance of N₂O₅ is 1.4 × 10^-3 Ms^-1.

Question 20.
How do you convert the following? [2]
(a) Ethanal to Propanone
(b) Toluene to Benzoic acid
OR
Account for the following:
(a) Aromatic carboxylic acids do not undergo Friedel- Crafts reaction.
(b) pK_a value of 4-nitrobenzoic acid is lower than that of benzoic acid.
Answer:
(a) Conversion of ethanol to Propanone:

OR
(a) Aromatic carboxylic acids do not undergo Friedel-Crafts reaction because the carboxyl group is deactivating for electrophilic substitution reaction, secondarily, the catalyst aluminium chloride gets bonded to the carboxyl group.

(b) pK_a value of 4-Nitrobenzoic acid is lower than benzoic acid, which means 4-Nitrobenzoic acid is more acidic than the benzoic acid. Being an electron-withdrawing group, the -NO₂ group withdraws electrons towards itself resulting in ease of carboxylic proton release, hence increasing the acidity.

Question 20.
Complete and balance the following chemical equations: [2]
(a) Fe²⁺ + MnO^–₄ + H⁺ →
(b) MnO^–₄ + H₂O + I^– →
Answer:
(a) 5Fe²⁺ + MnO^–₄ + 8H⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺
(b) 2MnO^–₄ + H₂O + I^– → 2MnO₂ + 2OH^– + IO^–₃

Question 21.
Give reasons for the following: [3]
(a) Measurement of osmotic pressure method is preferred for the determination of molar masses of macromolecules such as proteins and polymers.
(b) Aquatic animals are more comfortable in cold water than in warm water.

(c) Elevation of the boiling point of 1M KCl solution is nearly double than that of 1M sugar solution.
Answer:
(a) Molar masses of macromolecules like polymers and proteins are measured through osmotic pressure method. The osmotic pressure method uses ‘molarity’ of solution (instead of molality) which has a large magnitude even for dilute solutions, given that polymers have poor solubility, osmotic pressure measurement is used for determination of their molar masses. Macromolecules such as proteins are not stable at high temperatures and because measurement of osmotic pressure is done at around room temperature, it is useful for determination of molar masses of proteins.

(b) The solubility of gases in liquids decreases on increasing the temperature. Hence, the availability of dissolved oxygen in water is more at lower temperatures hence, the aquatic animals feel more comfortable at lower temperatures than at the higher temperatures.

(c) Elevation of boiling point is a colligative property and hence depends on the number of solute particles in the solution. Now, 1 M KCl would have twice the number of solute particles, as KCl dissociates into K⁺ and Cl^–, compared to sugar solution (as sugar does not undergo any dissociation). So, the elevation of boiling point is nearly double for 1M KCl solution compared to 1M sugar solution.

Question 22.
A first-order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction. (Given: log 2 = 0.3010, log 4 = 0.6021, R = 8.314 JK^-1 mol^-1) [3]
Answer:
Rate constant for a first-order reaction is given by,\

Chemistry Class 12 PYQ with Solution – 2019

Question 1.
A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when 75% of the reaction will be completed.
Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021) [3]
Answer:
For first order reaction,

Question 2.
The following compounds are given to you: 2-Bromopentane, 2-Bromo-2-methyl butane, 1-Bromopentane
(a) Write the compound which is most reactive towards SN₂ reaction.
(b) Write the compound which is optically active.
(c) Write the compound which is most reactive towards (3-elimination reaction. [3]
Answer:
(a) 1-Bromo pentane > 2-Bromo pentane > 2-Bromo-2-methyl pentane (Reactivity towards, SN₂ reaction)
(b) 2-Bromo pentane
CH₃CH₂CH₂CHBrCH₃
This compound is most reactive towards β-elimination.

Question 3.
Write the principle of the following:

1. Zone refining
2. Froth floatation process
3. Chromatography [3]

Answer:

1. Zone refining
   • This process is used for the metals which are required in very high purity like silicon, germanium, boron, gallium, etc.
   • This method is based on the principle that the impurities are more soluble in the melt than in the solid-state of the metal.
   • In this method, impure metal is cast into a thin bar.
2. Froth floatation process
   • This method is based on the principle that difference in the wetting properties of the ore and gangue particles with water and oil.
   • This method is used for the extraction of those metals in which the ore particles are preferentially wetted by oil and gangue by water.
   • This method has been used for the concentration of sulphide ores like PbS, ZnS, CuFeS_2, etc.
3. Chromatography
   • This is a modem method of purification based on the difference in the adsorbing capacities of the metal and its impurities on a suitable adsorbent.
   • This technique is based on the principle that different components of a mixture are differently adsorbed on an adsorbent.

Question 4.
Write the structures of compounds A, B and C in the following reactions:
Question 5.
Write the structures of the monomers used for getting the following polymers:
(a) Nylon-6,6
(b) Melamine-formaldehyde polymer
(c) Buna-S [3]
Answer:
(a) Monomers of Nylon-6, 6

(b) Monomers of Melamine-formaldehyde polymer

Question 6.
Define the following: [3]
(a) Anionic detergents
(b) Limited spectrum antibiotics
(c) Antiseptics
Answer:
(a) Anionic detergents: These detergents contain an anionic hydrophilic group. These are manufactured from the long chain of alcohols. These long chain alcohols are treated with cone. H₂SO₄ to form alkyl hydrogen sulphates of high molecular mass and then are neutralized with alkali to form salts.

(b) Limited Spectrum Antibiotics: The antibiotics which are effective against single organism or disease are called limited spectrum antibiotics, example-streptomycin.

(c) Antiseptics: The chemical substances that are used to either kill or prevent the growth of micro-organisms are called antiseptics. These are not harmful to living tissues and can be safely applied on wounds, cuts, ulcers, etc., for example, Soframycin.

Question 7.
Give reasons for the following:
(a) Red phosphorus is less reactive than white phosphorus.
(b) Electron gain enthalpies of halogens are largely negative.
(c) N₂O₅ is more acidic than N₂O₃. [3]
Answer:
(b) Electron gain enthalpies of halogens are largely negative in their respective periods.
This is due to the fact that the atoms of these elements have only one electron less than the stable noble gas (ns²np⁶) configuration. Therefore, they have maximum tendency to accept an additional electron.

Question 8.
Give reasons for the following:
(a) Acetylation of aniline reduces its activation effect.
(b) CH₃NH₂ is more basic than C₆H₅NH₂.
(c) Although-NH₂ is olp directing group, yet aniline on nitration gives a significant amount of m-nitroaniline. [3]
Answer:
(a) In acetanilide, the oxygen atom of the group withdraws electrons from the NH₂ group as shown below:

As a result, the electron pair on nitrogen gets displaced to the carboxyl group. Therefore, the unshared pair of electron on nitrogen is less available for a donation of the electron to the aromatic ring.

(b) In aniline, lone pair of e~ present on ‘N’ is in conjugation with the benzene ring and become less available for protonation because of resonance.

This conjugation of lone pair of e- is not present in case of methylamine and lone pair of e^– of ‘N’ are fully available for protonation. That’s why the basicity order of aniline and methylamine is:

The reason for the formation of a large amount of m-nitroaniline is that under strongly acidic conditions, aniline gets protonated to anilinium ion (-NH₃ group). This is a deactivating group and is meta-directing in nature.

Question 9.
After watching a program on TV about the presence of carcinogens (cancer causing agents) Potassium bromate and potassium iodate in bread and other bakery products, Rupali a Class XII student decided to make others aware about the adverse effects of these carcinogens in foods. She consulted the school principal and requested him to instruct the canteen contractor to stop selling sandwiches, pizzas, burgers and other bakery products to the students. The principal took immediate action and instructed the canteen contractor to replace the bakery products with some protein and vitamin rich food like fruits, salads, sprouts, etc. The decision was welcomed by the parents and the students.

After reading the above passage, answer the following questions:
(a) What are the values (at least two) displayed by Rupali?
(b) Which polysaccharide component of carbohydrates is commonly present in bread?
(c) Write the two types of secondary structures of proteins.
(d) Give two examples of water soluble vitamins. [4]
Answer:
(b) Starch
(c) 1. α-helix structure.
2. β-pleated sheet structure.
(d) Vitamin B and Vitamin C

Question 10.
(a) Account for the following:
(i) Transition metals show variable oxidation states.
(ii) Zn, Cd, and Hg are soft metals.
(iii) E⁰ value for the Mn³⁺/Mn²⁺ couple is highly positive (+1.57 V) as compared to Cr³⁺/Cr²⁺.
(b) Write one similarity and one difference between the chemistry of lanthanoid and actinoid elements. [5]
OR
(a) Following are the transition metal ions of 3d series: Ti⁴⁺, V²⁺, Mn³⁺, Cr³⁺
(Atomic numbers: Ti = 22, V = 23, Mn = 25, Cr = 24)
Answer the following:
(i) Which ion is most stable in an aqueous solution and why?
(ii) Which ion is a strong oxidising agent and why?
(iii) Which ion is colourless and why?
(b) Complete the following equation:
(i) 2MnO₄ + 16H⁺ + 5S^2- →
(ii) KMnO₄ →
Answer
(a) (i) Transition metal ions show variable oxidation states due to the participation of (n-1) d electrons in addition to outer ns-electrons because the energies of ns and (n-1)d subshells are almost equal. As a result of which the electrons of (n-1)d and ns subshell both part in bond formation.

(ii) Zn, Cd, and Hg are soft metals because of their completely filled 3d, 4d, and 5d orbitals respectively Due to completely filled d-orbitals these metals are reluctant to form Zn-Zn, Cd-Cd, and Hg-Hg bonds.

(iii) A highly positive value of E⁰ for Mn³⁺/Mn²⁺ shows that Mn²⁺ (d⁵) is particularly stable. While low value of E⁰ for Cr³⁺/Cr²⁺ shows that Cr²⁺ (d⁴) is less stable than Cr³⁺ (d³)

(b) Similarity: In lanthanoids and actinoids both the added electron enters the antepenultimate shell 4f and 5f respectively.
Difference: Lanthanoids show a common oxidation state of +3 while actinoids show different oxidation states other than +3.
OR
(a) Ti⁴⁺ = 1s²2s²2p⁶3s²3p⁶
V²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³
Mn³⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴
Cr³⁺ = 1s² 2s² 2p⁶ 3s² 3d³
(i) Ti⁴⁺ is most stable in an aqueous solution because of full filled valence shell (3s² 3p⁶) onfiguration (noble gas configuration).
(ii) Mn³⁺ is the strong agent as it oxidises other species it will reduce itself by taking an electron and will stabilise its configuration to 3d⁵.
(iii) Ti⁴⁺ is colourless due to absence of impaired electrons (3s² 3p⁶)

Question 11.
(a) A 10% solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water if the freezing point of pure water is 273.15 K.
Given:
The molar mass of sucrose = 342g mol^-1
The molar mass of glucose = 180 g mol^-1
(b) Define the following terms:
(i) Molality (m)
(ii) Abnormal molar mass [5]
OR
(a) 30 g of urea (M = 60g mol-1) is dissolved in 846 g of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23.8 mm Hg.
(b) Write two differences between ideal solutions and non-ideal solutions.
Answer:
(a) T0 (freezing point of water) = 273.15K
Ts (freezing point of sucrose solution) = 269.15K
Weight of sucrose in solution = 10 g
Weight of glucose in solution = 10 g
The molar mass of sucrose = 342 g mol^-1
The molar mass of glucose = 180 g mol^-1
The freezing point of glucose = x
Depression in freezing point
So, the freezing point of glucose solution = 265.55 K.

(b) (i) Molality: It is the number of moles of the solute dissolved per 1000 g of the solvent. It is denoted by m.
(ii) Abnormal molar mass: Those solute that dissociates or associate in solution, show an abnormal molar mass in solution, for example, Molar mass of ethanoic acid is greater than normal molar mass.
The molar mass of KCl in solution is reduced than normal molar mass.
KCl → K⁺ + Cl^–
OR
(a) W_B = 30 g
M_B = 60 g mol^-1
W_A = 846 g
M_A = 18 g mol^-1
P⁰ = 23.8mm Hg
P_s = x
Relative lowering of vapour pressure
So, the vapour pressure of water for this solution = 23.597 mm Hg
(b)

S.No.	Ideal Solutions	Non-ideal Solutions
1.	The interactions between the components are similar to those in the pure components.	The interactions between the components are different from those of the pure components.
2.	There is no enthalpy change on mixing, ΔHmix = 0	There is enthalpy change on mixing, ΔHmix ≠ 0

Question 12.
(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
OR
(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
Answer:
(b) (i) Butanal and Butan-2-one

Thus Butanal gives silver mirror test with Tollen’s reagent whereas Butan-2-one does not.
(ii) Benzoic acid and phenol

Thus, Benzoic acid gives sodium benzoate on reaction with sodium bicarbonate whereas phenol gives no reaction with sodium bicarbonate.
OR
(a) (i) Etard reaction: The oxidation of toluene to benzaldehyde with chromyl chloride (CrO₂Cl₂) dissolved in CCl₄ or CS₂.

(ii) Stephen reaction: The partial reduction of alkyl or aryl cyanides to the corresponding aldehydes with a suspension of anhydrous SnCl₂ in ether saturated with HCl at room temperature followed by hydrolysis.

Question 13.
Write the IUPAC name of the following compound: [1]

Question 14.
What is the effect of adding a catalyst on
(a) The activation energy (E_a), and
(b) Gibbs energy (ΔG) of a reaction?
Answer:
On adding a catalyst
(a) Activation energy of the reaction decreases.
(b) Gibbs energy doesn’t change.

Question 15.
What type of colloid is formed when a liquid is dispersed in a solid? Give an example. [1]
Answer:
When a liquid is dispersed in solid, ‘gel’ colloid is formed. Examples Jelly, butter, cheese, curd, etc.

Question 16.
(a) Arrange the following compounds in the increasing order of their acid strength: [2]
p-cresol, p-nitrophenol, phenol
(b) Write the mechanism (using curved arrow notation) of the following reaction:
OR
Write the structures of the products when Butan-2-ol reacts with the following:
(a) CrO₃
(b) SOCl₂

Question 17.
Draw the structures of the following:
(a) H₂SO₃
(b) HClO₃

Question 18.
Write the name of the cell which is generally used in hearing aids. Write the reactions taking place at the anode and the cathode of this cell. [2]
Answer:
Electrolytic cells are generally used is hearing aids. At the cathode, reduction of metal takes place and at the anode, oxidation of metal takes place.
Cathode: M + e^– → M⁺
Anode: M⁺ → M + e^–

Question 19.
Using IUPAC norms write the formulae for the following:
(a) Sodium dicyanidoaurate (I)
(b) Tetraamminechloridonitrito-N-platinum (IV) sulfate [2]
Answer:
(a) Sodium dicyanoaurate (I)
Na [Au (CN)₂]
(b) Tetraammine chloridonitrito-N-platinum (IV)
Sulphate [Pt(NH₃)₄(Cl) (NO₂)]SO₄

Question 20.
(a) The cell in which the following reaction occurs:
2Fe³⁺ (aq) + 2I^– (aq) → 2Fe²⁺ (aq) + I₂ (s)
has E_cell = 0.236 V at 298 K. Calculate the standard Gibbs energy of the cell reaction.
(Given: 1F = 96,500 C mol^-1)
(b) How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours?
(Given: 1 F = 96,500 C mol^-1) [3]
Answer:
(a) ΔG = -nFE_cell = -2 × 96500 × 0.236 = -45.548 kJ/mol
(b) According to Faraday’s first law the amount of metal deposited (W).
W = i × t = 0.5 × 7200 = 3600C
1F = 96500 C mol^-1
That is e^– flows from 96500 C = 1 mol
e^– flows from 3600 C = 360096500 = 0.037 mol.
No. of electrons = 0.037 × 6.023 × 10²³
= 0.2246 × 10²³
= 22.46 × 10²¹ electrons

Question 21.
(a) What type of isomerism is shown by the complex [CO(NH₃)₅ (SCN)]²⁺?
(b) Why is [NiCl₄]^2- paramagnetic while [Ni(CN)₄]^2- is diamagnetic?
(Atomic number of Ni = 28)
(c) Why are low spin tetrahedral complexes rarely observed? [3]
Answer:
(a) Linkage isomerism
(b) [NiCl₄]^2-, Ni²⁺ = 1s²2s²2p⁶3s²3p⁶3d⁸
Cl^– is a weak field ligand.

2 electrons are impaired in [NiCl₄]^2- which provides paramagnetism to the complex.
[Ni(CN)₄]^2-
Ni²⁺ = 1s²2s²2p⁶3s²3p⁶3d⁸
CN^– is a strong, field ligand

no electron is unpaired in [Ni (CN)₄]^2- That’s why the complex is diamagnetic.
(c) In the tetrahedral complex, CFSE is very low and it is difficult for the tetrahedral complexes to exceed the pairing energy. Usually, electrons prefer to move to higher energy orbitals for pairing. Thus they usually form high spin complexes.
(CFSE) tetrahedral = 49 (CFSE)octahedral

Question 22.
Write one difference in each of the following:
(a) Multimolecular colloid and Associated colloid
(b) Coagulation and Peptization
(c) Homogeneous catalysis and Heterogeneous catalysis [3]
OR
(a) Write the dispersed phase and dispersion medium of milk.
(b) Write one similarity between physisorption and chemisorption.
(c) Write the chemical method by which Fe(OH)₃ sol is prepared from FeCl₃.
Answer:
(a) Multimolecular colloids are the colloids in which the dispersed phase consists of aggregates of atoms or molecules with molecular size less than 1 nm whereas associated colloids are the substances that are dissolved in a medium, behave as normal electrolytes at low concentration but as colloids at higher concentration.

(b) Coagulation is the process of precipitation of a colloidal solution by the addition of an excess of an electrolyte whereas peptization is the process responsible for the formation of stable dispersion of colloidal particles in the dispersion medium.

(c) Homogeneous catalysis is the one in which the phases of the reactants and the catalysts are the same whereas in heterogeneous catalysis the phases of the reactants and the catalysts are not the same.
OR
(a) Milk
Dispersed phase – Liquid
Dispersion medium – Liquid
(b) Both physisorption and chemisorption depends on the surface area. Both increases with an increase in the surface area.
(c) Fe(OH)3 sol is prepared from FeCl3 by hydrolysis method.

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