CBSE Class 11 Chemistry Question Paper Set 12 with Solutions

Answer: (d) 1 m
Moles of HCl = 18.5 ÷ 36.5 ≈ 0.507 mol; mass of solvent = 0.5 kg; molality = 0.507 ÷ 0.5 ≈ 1.01 ≈ 1 m.

Teacher's Note:
1. Molality is moles of solute per kilogram of solvent, not per kilogram of solution.
2. Convert grams of water to kilograms before dividing.

Answer: (a) 0.2
Moles of water = 36 ÷ 18 = 2; moles of glycerine = 46 ÷ 92 = 0.5; mole fraction of glycerine = 0.5 ÷ (0.5 + 2) = 0.2.

Teacher's Note:
1. Mole fraction is moles of one component divided by the total moles of all components.
2. The molar mass of glycerine is 92 g/mol, so do not use 46 as its molar mass.

Answer: (b) Fe³⁺, Mn²⁺
Fe³⁺ (26 − 3 = 23 electrons) and Mn²⁺ (25 − 2 = 23 electrons) both have the configuration [Ar] 3d⁵. The ions in the other pairs have different numbers of electrons.

Teacher's Note:
1. Count the electrons of each ion by subtracting the charge from the atomic number.
2. Remember that 4s electrons are removed before 3d electrons when a transition metal forms a cation.

Answer: (c) 14
l = 3 is the f subshell (4f). It has 2l + 1 = 7 orbitals, each holding 2 electrons, so 14 electrons.

Teacher's Note:
1. The capacity of a subshell is 2(2l + 1) electrons.
2. The value of n does not change the capacity of a subshell.

Answer: (d) 5, 0, 0, +1/2
Rb is [Kr] 5s¹, so the valence electron is in the 5s orbital: n = 5, l = 0, m = 0, s = +1/2.

Teacher's Note:
1. Write the electronic configuration first and then read off n and l of the last electron.
2. An s orbital always has l = 0 and m = 0.

Answer: (a) s
Z = 55 is caesium, [Xe] 6s¹. Its last electron enters the s subshell, so it is an s-block element (group 1).

Teacher's Note:
1. The block is decided by the subshell that receives the last electron.
2. Use the nearest noble gas (Xe, Z = 54) to find the configuration quickly.

Answer: (b) Neon
Neon has a stable, completely filled 2p subshell, so an added electron must go into the next shell and energy has to be supplied. Its electron gain enthalpy is therefore positive.

Teacher's Note:
1. Noble gases have positive electron gain enthalpies because of their stable filled configurations.
2. Oxygen and fluorine have negative values, though oxygen's second gain is positive.

Answer: (d) Ca²⁺
All four are isoelectronic with 18 electrons. In an isoelectronic series the ion with the largest nuclear charge (Ca²⁺, Z = 20) pulls the electrons in most strongly and is the smallest.

Teacher's Note:
1. Among isoelectronic species, radius decreases as atomic number increases.
2. Do not compare cations and anions of different electron counts using charge alone.

Answer: (a) 0
He₂ has configuration σ1s² σ*1s². Bond order = ½(2 − 2) = 0, so the molecule does not exist.

Teacher's Note:
1. Bond order is half the difference between bonding and antibonding electrons.
2. A bond order of zero means the molecule is not stable.

Answer: (b) sp³d
Phosphorus has 5 bond pairs and no lone pair, so it uses one s, three p and one d orbital. This gives sp³d hybridisation and a trigonal bipyramidal shape.

Teacher's Note:
1. Count the number of electron pairs around the central atom to find the hybridisation.
2. Five electron pairs always correspond to sp³d.

Answer: (c) q = 0
In an adiabatic process no heat is exchanged between the system and the surroundings, so q = 0.

Teacher's Note:
1. Adiabatic means no heat transfer, while isothermal means constant temperature.
2. Temperature can change in an adiabatic process even though q = 0.

Answer: (b) 3° < 2° < 1°
Alkyl groups push electrons (+I effect) and intensify the negative charge on the carbanion carbon, which destabilises it. So the primary carbanion is the most stable and the tertiary the least.

Teacher's Note:
1. Carbanions are the opposite of carbocations: alkyl groups destabilise them.
2. Link stability with whether the groups help to spread or concentrate charge.

Answer: (c) 2,3-dimethylbutane
2 (CH₃)₂CHBr + 2 Na (dry ether) → (CH₃)₂CH–CH(CH₃)₂ + 2 NaBr. Two isopropyl groups join to give a six-carbon chain with methyl groups at C-2 and C-3.

Teacher's Note:
1. In Wurtz reaction two alkyl groups join, so the product has twice the carbon atoms of the alkyl halide.
2. Keep the branching of the alkyl group in the product.

Answer: (b) Benzene
Three molecules of ethyne undergo cyclic polymerisation: 3 C₂H₂ → C₆H₆ (red hot iron tube, 873 K or 773 K).

Teacher's Note:
1. Cyclic polymerisation of three ethyne molecules gives a six-carbon ring.
2. Check the carbon count: 3 × 2 = 6 carbons, matching benzene.

Answer: (c) A is true but R is false.
Paraffin means little affinity; alkanes are called paraffins because they are chemically inert under ordinary conditions. So R (highly reactive) is false.

Teacher's Note:
1. The word paraffin comes from Latin parum affinis, meaning little affinity.
2. Saturated hydrocarbons have only strong, nonpolar C–C and C–H single bonds, so they resist attack.

Answer: (d) A is false but R is true.
Alkanes are nonpolar and cannot form hydrogen bonds with water, so they are insoluble in water but soluble in nonpolar solvents.

Teacher's Note:
1. Use the rule like dissolves like.
2. Water is polar and hydrogen-bonded, so it does not mix with nonpolar hydrocarbons.

Answer: (a) Both A and R are true and R is the correct explanation of A.
AlCl₃ is a Lewis acid catalyst. It reacts with R–Cl or RCOCl to form the carbocation or acylium ion (the electrophile), which attacks the benzene ring.

Teacher's Note:
1. Friedel–Crafts reactions are electrophilic substitution reactions on the benzene ring.
2. The catalyst is used precisely because it produces the attacking electrophile.

Answer: 1. (a) The longest chain with the –O– link: the smaller group CH₃O is the alkoxy group and the larger is ethane, so the name is methoxyethane.
2. (b) Longest chain has 4 carbons with a double bond between C-2 and C-3 and Cl on C-2: 2-chlorobut-2-ene.

Teacher's Note:
1. For ethers, name the smaller alkyl part as the alkoxy prefix and the larger part as the parent alkane.
2. In (b) number the chain so that the double bond gets the lowest locant, and check that the double bond is on C-2 either way before placing the substituent.

Answer: (a) Alkanes are obtained by catalytic hydrogenation (Sabatier–Senderens reaction) of alkenes or alkynes:
CH₂=CH₂ + H₂ → CH₃–CH₃ (Ni or Pt or Pd, heat)
CH≡CH + 2 H₂ → CH₃–CH₃ (Ni, 573 K)
OR
(b) Phenol is reduced by distilling it over zinc dust:
C₆H₅OH + Zn → C₆H₆ + ZnO (heat)

Teacher's Note:
1. Name the catalyst and conditions, as they carry marks along with the equation.
2. Alkynes need two moles of hydrogen to become fully saturated.

Answer: 1. At NTP, 22.4 L of a gas = 1 mole, so 11.2 L = 0.5 mol of O₂.
2. Number of molecules = 0.5 × 6.022 × 10²³ = 3.011 × 10²³ molecules.
3. Each O₂ molecule has 2 atoms, so number of atoms = 2 × 3.011 × 10²³ = 6.022 × 10²³ atoms.

Teacher's Note:
1. Use the molar volume to get moles and then multiply by Avogadro's number.
2. Do not forget that oxygen is diatomic when counting atoms.

Answer: By de Broglie's relation, λ = h / (m v).
λ = 6.626 × 10⁻³⁴ ÷ (9.11 × 10⁻³¹ × 2.05 × 10⁷)
= 6.626 × 10⁻³⁴ ÷ (1.868 × 10⁻²³)
= 3.55 × 10⁻¹¹ m (about 35.5 pm).

Teacher's Note:
1. Always write the formula, then substitute values in SI units.
2. Check the power of ten carefully when dividing numbers with exponents.

Answer: (a) Hund's rule of maximum multiplicity: pairing of electrons in the orbitals of a subshell (degenerate orbitals) does not start until each orbital has received one electron, and these single electrons have parallel spins.
(b) Cu (Z = 29): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ or [Ar] 3d¹⁰ 4s¹. This is an exception because a completely filled d subshell is extra stable.

Teacher's Note:
1. Give an example such as three electrons in a p subshell filling three orbitals singly to support Hund's rule.
2. Copper and chromium are the classic exceptions where an electron shifts from 4s to 3d.

Answer: (a) Be (2s²) has a completely filled stable 2s subshell, so more energy is needed to remove an electron. In B (2s² 2p¹) the electron to be removed is in the higher energy 2p orbital, which is shielded by the 2s electrons and is easier to remove. So IE of B < IE of Be.
(b) Fluorine is very small, so the added electron faces strong repulsion from the electrons already present in the compact 2p subshell, reducing the energy released. In Cl the 3p subshell is larger and the repulsion is less, so more energy is released and Cl has the more negative electron gain enthalpy.

Teacher's Note:
1. Link both answers to the stability of the electronic configuration and the size of the orbital.
2. Fluorine's small size is the reason for its unexpectedly low electron gain enthalpy.

Answer: 1. Energy order of molecular orbitals for O₂ (more than 14 electrons): σ1s < σ*1s < σ2s < σ*2s < σ2p_z < (π2p_x = π2p_y) < (π*2p_x = π*2p_y) < σ*2p_z. The 16 electrons are filled in this order.
2. Configuration: σ1s² σ*1s² σ2s² σ*2s² σ2p_z² (π2p_x² = π2p_y²) (π*2p_x¹ = π*2p_y¹).
3. Bond order = ½(N_b − N_a) = ½(10 − 6) = 2.
4. Two unpaired electrons in the π* orbitals make O₂ paramagnetic.

Teacher's Note:
1. On paper draw the diagram with the 1s, 2s and 2p atomic levels on the sides and the molecular levels in the centre, labelling every orbital.
2. Place one electron each in the two π* orbitals before pairing, as per Hund's rule.

Answer: 1. Consider n moles of an ideal gas expanding reversibly and isothermally at temperature T from volume V₁ to V₂. At every stage the external pressure differs from the gas pressure by an infinitesimal amount, so p_ex = p (the pressure of the gas).
2. Work for a small change in volume: dw = −p dV.
3. For an ideal gas p = nRT / V, so dw = −nRT dV / V.
4. Integrate from V₁ to V₂ at constant T: w = −nRT ∫ dV/V = −nRT ln(V₂/V₁).
5. Hence w = −2.303 nRT log (V₂/V₁). Since p₁V₁ = p₂V₂ at constant T, this can also be written w = −2.303 nRT log (p₁/p₂).
6. For an ideal gas ΔU = 0 in an isothermal process, so q = −w.

Teacher's Note:
1. Start from the definition dw = −p dV, as this earns the first marks.
2. Remember that p must be replaced by nRT/V because it changes during the reversible expansion.

Answer: (a) Wurtz reaction: 2 CH₃Br + 2 Na → CH₃–CH₃ + 2 NaBr (in dry ether).
(b) Dehydrohalogenation: CH₃CH₂Cl + alcoholic KOH → CH₂=CH₂ + KCl + H₂O (on heating).
(c) Friedel–Crafts acylation: C₆H₆ + CH₃COCl → C₆H₅COCH₃ (acetophenone) + HCl, in the presence of anhydrous AlCl₃.

Teacher's Note:
1. Always mention the reagent conditions such as dry ether, alcoholic KOH and anhydrous AlCl₃.
2. Dehydrohalogenation needs alcoholic KOH; aqueous KOH would give substitution to an alcohol instead.

Answer: Formal charge = (valence electrons) − (non-bonding electrons) − ½(bonding electrons).
(a) NO₂⁻: O=N–O⁻ with a lone pair on N. Formal charges: N = 5 − 2 − 3 = 0; double-bonded O = 6 − 4 − 2 = 0; single-bonded O = 6 − 6 − 1 = −1. Total = −1.
(b) H₃O⁺: O bonded to three H atoms with one lone pair on O. Formal charges: O = 6 − 2 − 3 = +1; each H = 1 − 0 − 1 = 0. Total = +1.
(c) O₃: O=O⁺–O⁻ (central O has one lone pair, the others have two and three lone pairs). Formal charges: terminal double-bonded O = 6 − 4 − 2 = 0; central O = 6 − 2 − 3 = +1; terminal single-bonded O = 6 − 6 − 1 = −1. Total = 0.

Teacher's Note:
1. Draw the skeleton, complete octets and then calculate formal charge for every atom using the formula.
2. The formal charges must add up to the overall charge on the species.

Answer: (a) Mesomeric (resonance) effect: the polarity produced in a molecule by the interaction of a lone pair or π electrons with a neighbouring π bond (conjugated system). Example: in CH₂=CH–CH=O the π electrons shift towards oxygen, giving CH₂⁺–CH=CH–O⁻ (–R effect); in aniline the lone pair of N enters the ring (+R effect).
(b) Hyperconjugation: the delocalisation of σ electrons of a C–H bond of an alkyl group into an adjacent empty p orbital or π orbital. Example: in propene CH₃–CH=CH₂ the σ electrons of C–H of CH₃ delocalise into the π bond; the more α-hydrogens, the more stable the carbocation or alkene. This is called 'no bond resonance'.
(c) Functional isomerism: compounds with the same molecular formula but different functional groups. Example: C₂H₆O is ethanol (CH₃CH₂OH) and dimethyl ether (CH₃OCH₃).
OR
(a) pH is the negative logarithm (base 10) of the hydronium ion concentration: pH = −log[H₃O⁺]. Here pH = −log(1.5 × 10⁻⁵) = 5 − log 1.5 = 5 − 0.176 = 4.824.
(b) Ionic product of water: K_w = [H₃O⁺][OH⁻] = 1 × 10⁻¹⁴ at 298 K. In pure water [H₃O⁺] = [OH⁻] = x, so x² = 10⁻¹⁴ and x = 10⁻⁷ M. Hence pH = −log 10⁻⁷ = 7.

Teacher's Note:
1. Give one clear example for each short note, as the question asks for it.
2. In the pH numerical, split log (1.5 × 10⁻⁵) into log 1.5 plus log 10⁻⁵ to avoid sign errors.

Answer: (a) Free radical substitution, for example CH₄ + Cl₂ → CH₃Cl + HCl in sunlight.
(b) CH₃CH₂Br + alc. KOH → CH₂=CH₂ + KBr + H₂O (dehydrohalogenation, an elimination reaction).
(c) CH₃CH₂OH + conc. H₂SO₄ (443 K) → CH₂=CH₂ + H₂O (dehydration, an elimination reaction).
(d) Electrophiles are electron-deficient species that accept an electron pair (electron-loving reagents). Examples: H⁺, NO₂⁺, carbocations (CH₃⁺), AlCl₃.

Teacher's Note:
1. Link the type of reaction to the bond-breaking pattern: halogenation of alkanes proceeds via free radicals.
2. Temperature matters in (c): around 443 K gives the alkene, whereas lower temperature gives an ether.

Answer: (a) Molarity (M) is the number of moles of solute dissolved in one litre of solution. Molality (m) is the number of moles of solute dissolved in one kilogram of solvent. Molarity depends on temperature, but molality does not.
(b) Take 100 g of solution: glucose = 10 g, water = 90 g.
Moles of glucose = 10 ÷ 180 = 0.0556 mol; moles of water = 90 ÷ 18 = 5 mol.
(i) Molality = 0.0556 ÷ 0.090 kg = 0.617 m.
(ii) Volume of solution = 100 ÷ 1.20 = 83.33 mL = 0.08333 L. Molarity = 0.0556 ÷ 0.08333 = 0.667 M.
(iii) Total moles = 5 + 0.0556 = 5.0556. Mole fraction of glucose = 0.0556 ÷ 5.0556 = 0.011; mole fraction of water = 5 ÷ 5.0556 = 0.989 (they add up to 1).

Teacher's Note:
1. Assume a convenient quantity such as 100 g of solution so that percentages convert directly into grams.
2. Use the mass of solvent for molality but the volume of the whole solution for molarity.

Answer: (a) Law of mass action: the rate of a chemical reaction at a given temperature is directly proportional to the product of the active masses (molar concentrations) of the reactants, each raised to a power equal to its stoichiometric coefficient.
(b) Use K_p = K_c(RT)^Δn.
(i) Δn = 2 − 3 = −1; T = 1000 + 273 = 1273 K.
K_c = K_p(RT) = 3.4 × 0.0831 × 1273 = 3.6 × 10² L mol⁻¹ (about 360).
(ii) Δn = 3 − 2 = +1; T = 1069 K.
K_p = K_c(RT) = 3.75 × 10⁻⁶ × 0.0831 × 1069 = 3.33 × 10⁻⁴ bar.

Teacher's Note:
1. First find Δn from gaseous products minus gaseous reactants, then place it as the power of RT.
2. Convert the temperature to kelvin and use R in bar-based units when pressure is in bar.

Answer: (a) C₆H₆ + HNO₃ → C₆H₅NO₂ (nitrobenzene) + H₂O (conc. H₂SO₄, 323–333 K); nitration.
(b) C₆H₆ + Cl₂ → C₆H₅Cl (chlorobenzene) + HCl (anhydrous AlCl₃); halogenation.
(c) C₆H₆ + H₂SO₄ → C₆H₅SO₃H (benzenesulphonic acid) + H₂O (fuming H₂SO₄, heat); sulphonation.
(d) C₆H₅OH + Zn → C₆H₆ (benzene) + ZnO (heat).
(e) C₆H₆ + CH₃Cl → C₆H₅CH₃ (toluene) + HCl (anhydrous AlCl₃); Friedel–Crafts alkylation.

Teacher's Note:
1. In (a), (b), (c) and (e) the electrophile replaces one hydrogen of the ring, which is electrophilic substitution.
2. Write the name of the organic product and any reaction condition, since both can earn marks.