📘 Chemistry Answer Key
Complete solutions · SSC / Board examination style
3.Match the following
Correct option: (c) (iv) (i) (ii)
(X) –CHO → Aldehyde (iv)
(Y) –O–R → Ether (i)
(Z) –OH → Hydroxyl (ii)
(Y) –O–R → Ether (i)
(Z) –OH → Hydroxyl (ii)
4.Aluminium manufacture
Correct option: (b) Statements (i) and (ii) are correct, but (iii) is not correct.
✔ (i) Ore treated with hot NaOH solution (Bayer’s process)
✔ (ii) Electricity used as reducing agent (Hall–Héroult)
✘ (iii) Aluminium obtained at cathode, not anode.
✔ (ii) Electricity used as reducing agent (Hall–Héroult)
✘ (iii) Aluminium obtained at cathode, not anode.
5.Thick white fumes
(a) Ammonium chloride (NH₄Cl)
(b) NH₃(g) + HCl(g) → NH₄Cl(s)
6.Displacement reactions
(a) Activity (ii) – Fe rod in CuSO₄ solution
(b) Iron (Fe) undergoes oxidation
7.Zinc blende (ZnS)
(a) Froth flotation
(b) Roasting
8.Homologous series (alkynes)
(a) Fifth member: C₆H₁₀ (CₙH₂ₙ₋₂)
(b) Position isomer example: CH₃–C≡C–CH₂–CH₂–CH₃ (hex-2-yne)
9.Manganese oxidation state
(a) Oxidation state of Mn in MnO₂ = +4
(b) Mn⁴⁺ electronic configuration: [Ar] 3d³
10.Option A (Methane combustion)
(a) Mass of O₂ required for 160 g CH₄ = 640 g
(b) Volume of CO₂ at STP from 80 g CH₄ = 112 L
🔁 OR (Option B)
(a) Mass of NH₃ (89.6 L at STP) = 68 g
(b) Number of molecules = 2.408 × 10²⁴ molecules
11.Option A (Polymerisation)
P: CH₂=CHCl (chloroethene / vinyl chloride)
Q: –[CH₂–CHCl]ₙ– (Polyvinyl chloride / PVC)
🔁 OR (Option B)
(a) A (from CO+2H₂) = CH₃OH (methanol)
(b) Methanol → ethanoic acid: CH₃OH + CO → CH₃COOH (catalytic carbonylation)
12.Periodic table (Option A)
(a) Atomic number = 12 (Mg)
(b) Oxide formula = MgO
(c) 4th period, same group (Ca): [Ar] 4s²
🔁 OR (Option B)
(a) Q (Period 3, Group 1): n = 3, l = 0
(b) P (Period 2, Group 16): outermost 2p → 3 orbitals
(c) Compound P + Q = Na₂O
13.Esters & IUPAC
(a) Ester: CH₃–CH₂–CH₂–COO–CH₂–CH₃ (ethyl butanoate)
(b) Acid = Butanoic acid | Alcohol = Ethanol
(c) CH₃CH₂CH₂COOH + CH₃CH₂OH ⇌ CH₃CH₂CH₂COOCH₂CH₃ + H₂O
14.Gas law & numerical
(a) Boyle’s Law
(b) New volume = 150 L
P₁V₁ = P₂V₂ → 2 × 450 = (2×3) × V₂ → 900 = 6V₂ → V₂ = 150 L
P₁V₁ = P₂V₂ → 2 × 450 = (2×3) × V₂ → 900 = 6V₂ → V₂ = 150 L
✅ Complete answer key — verified for all sections including optional OR choices
