CBSE Class 12 Chemistry The d and f Block Elements MCQs Set 15

Answer: (d) A–2, B–1, C–4, D–5

Explanation:
1. Lanthanoid oxides are used as phosphors in television screens (A–2); lanthanoid metals are used in the production of iron alloys/steel (B–1).
2. Misch metal is a lanthanoid metal + iron alloy (C–4); magnesium-based alloy with misch metal is used in bullets and shells (D–5).
3. Mixed oxides of lanthanoids are used as catalysts in petroleum cracking.

Teacher's Note:
1. Misch metal is about 95% lanthanoid metal and 5% iron with traces of S, C, Ca and Al.

Answer: (a) (i)–(d), (ii)–(a), (iii)–(b), (iv)–(e), (v)–(f)

Explanation:
1. Fe²⁺ salts are pale green, Ni²⁺ green, Mn²⁺ light pink and Co²⁺ pink.
2. Cu⁺ (3d¹⁰) has no unpaired electron, so Cu₂Cl₂ is colourless.

Teacher's Note:
1. Do not confuse Fe²⁺ (pale green) with Ni²⁺ (green), or Mn²⁺ (light pink) with Co²⁺ (pink).

Answer: (b) 4

Explanation:
1. μ = \( \sqrt{n(n+2)} \) = 4.90 BM ⇒ n(n + 2) = 24.
2. n = 4 unpaired electrons.

Teacher's Note:
1. Square the moment and solve n(n + 2) = μ²; the answer must be a whole number.

Answer: (d) Co³⁺

Explanation:
1. Co (Z = 27) = [Ar] 3d⁷ 4s², so Co³⁺ = [Ar] 3d⁶.
2. Ni³⁺ = 3d⁷, Mn³⁺ = 3d⁴, Fe³⁺ = 3d⁵.

Teacher's Note:
1. While forming cations, transition metals lose the 4s electrons first and then the 3d electrons.

Answer: (a) orange-red

Explanation:
1. The aqueous solution of CuSO₄ absorbs orange-red colour of light (The complementary colour, blue, is reflected).

Teacher's Note:
1. The colour we see is the complementary colour of the light absorbed, not the absorbed colour itself.

Answer: (b) They are chemically reactive

Explanation:
1. Interstitial compounds are chemically inert and not reactive. Hence, (b) is incorrect.

Teacher's Note:
1. Interstitial compounds (e.g. TiC, Fe₃H) are hard, high melting, conducting and chemically inert.

Answer: (a) Mn (Z = 25)

Explanation:
1. Mn (3d⁵ 4s²) shows all oxidation states from +2 to +7, the largest number in the first transition series.

Teacher's Note:
1. In the 3d series the number of oxidation states is largest in the middle, at Mn.

Answer: (b) (n − 1)d5 ns2

Explanation:
1. (n − 1)d5 ns2 (Mn) can use all 7 electrons in bonding, giving the highest oxidation state +7.

Teacher's Note:
1. Highest oxidation state = number of (n − 1)d + ns electrons, but only up to the middle of the series.

Answer: (c) 2.82 BM

Explanation:
1. In [NiCl₄]²⁻, Ni is present as Ni²⁺. Ni = [Ar]¹⁸ 3d⁸ 4s² ∴ Ni²⁺ = [Ar]¹⁸ 3d⁸
2. It has 2 unpaired electrons.
3. ∴ Magnetic moment = \( \sqrt{n(n+2)} \) BM = \( \sqrt{2 \times 4} \) = \( \sqrt{8} \) = 2.82 BM

Teacher's Note:
1. Cl⁻ is a weak field ligand, so the two unpaired electrons of Ni²⁺ (tetrahedral) stay unpaired.

Answer: (a) half-filled d-orbitals

Explanation:
1. Fe (Z = 26) = 3d⁶ 4s² ∴ Fe³⁺ = 3d⁵. Thus, Fe³⁺ ion is quite stable due to half-filled d-orbitals.

Teacher's Note:
1. Ferric = Fe³⁺, ferrous = Fe²⁺; do not mix up the two names.

Answer: (c) an oxidizing agent

Explanation:
1. In the highest oxidation state, it can undergo reduction and hence acts as oxidizing agent, e.g., in KMnO₄, Mn is in +7 oxidation state. It is one of the strongest oxidizing agent in solution.

Teacher's Note:
1. An element in its highest oxidation state can only gain electrons, so it can only be an oxidizing agent.

Answer: (c) Ni²⁺

Explanation:
1. Magnetic moment, μ = \( \sqrt{n(n+2)} \); μ = 2.84 BM when n = 2, i.e., there are 2 unpaired electrons.
2. Cr²⁺ = 3d⁴ (unpaired electrons = 4); Co²⁺ = 3d⁷ (unpaired electrons = 3); Ni²⁺ = 3d⁸ (unpaired electrons = 2); Ti³⁺ = 3d¹ (unpaired electrons = 1)

Teacher's Note:
1. Remember the values: 1 unpaired e⁻ = 1.73 BM, 2 = 2.83 BM, 3 = 3.87 BM, 4 = 4.90 BM, 5 = 5.92 BM.

Answer: (c) TiO > VO > CrO > FeO

Explanation:
1. Basic character of oxides of the same oxidation state decreases from left to right in a period, as the size of M²⁺ ion decreases. Hence TiO > VO > CrO > FeO.

Teacher's Note:
1. Same oxidation state, smaller cation → more covalent, less basic oxide.

Answer: (a) IO₃⁻

Explanation:
1. In alkaline (faintly alkaline) medium: 2MnO₄⁻ + I⁻ + H₂O → 2MnO₂ + IO₃⁻ + 2OH⁻

Teacher's Note:
1. In acidic medium iodide gives I₂, but in alkaline/neutral medium KMnO₄ oxidises it to iodate.

Answer: (b) MnO₂

Explanation:
1. Pyrolusite is the chief ore of manganese, MnO₂. It is the starting material for preparing KMnO₄.

Teacher's Note:
1. KMnO₄ is made by fusing pyrolusite (MnO₂) with KOH in air and then oxidising the manganate.

Answer: (a) Green Cr₂(SO₄)₃ is formed

Explanation:
1. K₂Cr₂O₇ + H₂SO₄ + 3SO₂ → K₂SO₄ + Cr₂(SO₄)₃ + H₂O
2. SO₂ is oxidised to sulphate and orange dichromate is reduced to green Cr³⁺.

Teacher's Note:
1. Orange → green is the test for SO₂ with acidified dichromate paper.

Answer: (b) Cr₂(SO₄)₃

Explanation:
1. Na₂SO₃ + H₂SO₄ gives SO₂, which reduces K₂Cr₂O₇ to green Cr₂(SO₄)₃ (Cr in +3 state).

Teacher's Note:
1. Green colour in dichromate reactions almost always means Cr³⁺ has been formed.

Answer: (d) M/6

Explanation:
1. Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
2. Each molecule gains 6 electrons (2 Cr from +6 to +3), so equivalent weight = M/6.

Teacher's Note:
1. Count the change in oxidation number for both Cr atoms: 2 × 3 = 6.

Answer: (b) 6 Cr–O bonds are equivalent

Explanation:
1. Cr₂O₇²⁻ has two tetrahedra sharing one corner (Cr–O–Cr bridge). The six terminal Cr–O bonds are equivalent; the two bridging Cr–O bonds are different (longer).

Teacher's Note:
1. Draw the two CrO₄ tetrahedra joined at one oxygen; 6 terminal + 2 bridging = 8 Cr–O bonds.

Answer: (c) 5

Explanation:
1. MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (Mn: +7 → +2)

Teacher's Note:
1. Electrons accepted: acidic 5 (Mn²⁺), neutral/weakly alkaline 3 (MnO₂), strongly alkaline 1 (MnO₄²⁻).

Answer: (a) 2

Explanation:
1. 2MnO₄⁻ + I⁻ + H₂O → 2MnO₂ + IO₃⁻ + 2OH⁻
2. I⁻ → IO₃⁻ loses 6 electrons; each MnO₄⁻ gains 3, so 2 moles of KMnO₄ are needed.

Teacher's Note:
1. Balance by electrons: 6 lost by one I⁻ = 2 × 3 gained by MnO₄⁻.

Answer: (b) SO₂

Explanation:
1. SO₂ readily decolourises acidified KMnO₄ solution:
2. 2KMnO₄ + 5SO₂ + 2H₂O → K₂SO₄ + 2MnSO₄ + 2H₂SO₄

Teacher's Note:
1. A gas that decolourises KMnO₄ must be a reducing agent; CO₂, NO₂ and P₂O₅ are not.

Answer: (d) +2, +3

Explanation:
1. Chromite ore is FeCr₂O₄. Ox. state of Fe = +2, Cr = +3.

Teacher's Note:
1. Check: (+2) + 2(+3) + 4(−2) = 0. Chromite can be written as FeO·Cr₂O₃.

Answer: (c) CrO₅

Explanation:
1. Blue colour is due to the formation of CrO₅ as follows:
2. Cr₂O₇²⁻ + 2H⁺ + 4H₂O₂ → 2CrO₅ + 5H₂O

Teacher's Note:
1. CrO₅ (chromium peroxide) has Cr in +6 state with two peroxo groups, so this is not a redox change for Cr.

Answer: (b) HgCl₂

Explanation:
1. HgCl₂ being covalent does not give chromyl chloride test.

Teacher's Note:
1. The chromyl chloride test needs ionic chloride; covalent chlorides such as HgCl₂ do not respond.

Answer: (d) (NH₄)₂Cr₂O₇

Explanation:
1. (NH₄)₂Cr₂O₇ →(Δ) Cr₂O₃ + N₂ + 4H₂O

Teacher's Note:
1. Ammonium dichromate gives N₂ (not O₂) because NH₄⁺ is oxidised by the dichromate within the same compound.

Answer: (b) K₂MnO₄, dark green

Explanation:
1. 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O (K₂MnO₄: Pot. manganate, green)

Teacher's Note:
1. Fusion gives green manganate (Mn +6); purple permanganate (Mn +7) is obtained only after further oxidation.

Answer: (a) disproportionates to MnO₂ and MnO₄⁻

Explanation:
1. MnO₄²⁻ disproportionates in acidic medium as follows:
2. 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O

Teacher's Note:
1. Mn(+6) goes to both +7 (MnO₄⁻) and +4 (MnO₂) – one species oxidised and reduced at once is disproportionation.

Answer: (d) Mn₂O₇

Explanation:
1. 2KMnO₄ + 2H₂SO₄ (conc.) → Mn₂O₇ + 2KHSO₄ + H₂O
2. Mn₂O₇ is formed which decomposes explosively on heating.

Teacher's Note:
1. This is why KMnO₄ is never dissolved in conc. H₂SO₄.

Answer: (d) Am

Explanation:
1. Am is an actinoid. Actinoids show more oxidation states than lanthanoids because 5f, 6d and 7s levels are close in energy; Am shows +2 to +6 (+3, +4, +5, +6 common).

Teacher's Note:
1. Lanthanoids mainly show +3 (with a few +2, +4); actinoids up to the middle (U to Am) show many states.

Answer: (c) Effective nuclear charge

Explanation:
1. 4f electrons shield each other poorly, so with each added proton the effective nuclear charge on the outer electrons increases and the size decreases steadily.

Teacher's Note:
1. The cause is poor (not effective) shielding by 4f electrons, which shows up as a steady rise in effective nuclear charge.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Ce⁴⁺ readily gains an electron to change to the common, more stable +3 state, so it is a good oxidising agent (E° Ce⁴⁺/Ce³⁺ = +1.74 V).

Teacher's Note:
1. The same idea explains why Eu²⁺ and Yb²⁺ are reducing agents: they also tend to reach +3.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. The order of complex formation follows the charge density of the ions, so A is true.
2. Actinides also form complexes with π-bonding ligands, which is true but does not explain the order.

Teacher's Note:
1. MO₂²⁺ has a higher effective charge on the metal than M³⁺, which is why it comes before M³⁺.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Both statements are true, but the correct explanation is: mercury has completely filled d-subshell.

Teacher's Note:
1. A transition element must have incompletely filled d orbitals in the atom or a common ion; being liquid has nothing to do with it.

Answer: (d) A is false, but R is true.

Explanation:
1. Correct A. Manganese shows a maximum oxidation state of +7 (as it has 3d⁵ 4s² configuration and the energy difference between 3d and 4s subshells is small).

Teacher's Note:
1. Count both 3d and 4s electrons for the maximum oxidation state: 5 + 2 = 7.

Answer: (c) A is true, but R is false.

Explanation:
1. Correct R. Ce⁴⁺ has the tendency to change to Ce³⁺ (as +3 oxidation state is more stable).

Teacher's Note:
1. An oxidizing agent is itself reduced: Ce⁴⁺ + e⁻ → Ce³⁺.

Answer: (d) A is false, but R is true.

Explanation:
1. Correct A. First ionization enthalpy of the 3d transition series increases with increase in atomic number.

Teacher's Note:
1. Ionisation enthalpy rises (irregularly) along the series as nuclear charge increases; it is the number of unpaired electrons that peaks in the middle.

Answer: (c) A is true, but R is false.

Explanation:
1. Correct R. The sum of the enthalpies of sublimation and ionization is not balanced by hydration enthalpy.

Teacher's Note:
1. Cu is the only 3d metal with positive E°(M²⁺/M), which is why it does not liberate H₂ from dilute acids.

Answer: (d) A is false, but R is true.

Explanation:
1. Correct A. The maximum oxidation state shown by osmium is +8.

Teacher's Note:
1. Os (and Ru) reach +8 in OsO₄, the highest oxidation state among transition elements.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Correct explanation. Cu²⁺ ion oxidizes iodide to iodine and CuI₂ itself is reduced to Cu₂I₂
2. (2Cu²⁺ + 4I⁻ → Cu₂I₂ + I₂)

Teacher's Note:
1. CuI₂ fails because of a redox reaction (I⁻ is a good reducing agent), not because of size.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. R is the correct explanation of A.
2. Cu²⁺ (3d⁹) and V⁴⁺ (3d¹) both have one unpaired electron.

Teacher's Note:
1. Ions with the same number of unpaired electrons often show similar colours and equal spin-only magnetic moments (1.73 BM here).