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

Answer: (b) Yb³⁺ < Pm³⁺ < Ce³⁺ < La³⁺

Explanation:
1. The overall decrease in atomic and ionic radii from La³⁺ to Lu³⁺ is called lanthanoid contraction. Hence, the correct order is
2. Yb³⁺ < Pm³⁺ < Ce³⁺ < La³⁺

Teacher's Note:
1. Place the four elements in order of atomic number first (La 57, Ce 58, Pm 61, Yb 70); the M³⁺ radius falls along that order.

Answer: (a) Cu²⁺ ion has 3d⁹ configuration while Zn²⁺ ion has 3d¹⁰ configuration

Explanation:
1. Cu²⁺ has 3d⁹ configuration i.e., one unpaired electron, hence, it is paramagnetic while Zn²⁺ has 3d¹⁰ configuration i.e., all orbitals are filled, hence it is diamagnetic in nature.

Teacher's Note:
1. Magnetism depends only on unpaired electrons; colour (option d) is a separate property and is not the reason.

Answer: (c) Fe³⁺ > Cr³⁺ > Ni²⁺ > Cu²⁺

Explanation:
1. Fe³⁺ – 3d⁵: No. of unpaired electrons = 5
2. Cr³⁺ – 3d³: No. of unpaired electrons = 3
3. Ni²⁺ – 3d⁸: No. of unpaired electrons = 2
4. Cu²⁺ – 3d⁹: No. of unpaired electrons = 1

Teacher's Note:
1. For 3dⁿ ions, unpaired electrons = n up to d⁵ and 10 − n after d⁵.

Answer: (a) MnO < Mn₃O₄ < Mn₂O₃ < MnO₂ < Mn₂O₇

Explanation:
1. Acidic strength of oxides of transition metals increases with increase in oxidation number.
2. Oxidation states of Mn: MnO (+2), Mn₃O₄ (+8/3), Mn₂O₃ (+3), MnO₂ (+4), Mn₂O₇ (+7)
3. Hence acidic strength is of the order of MnO (basic) < Mn₃O₄ < Mn₂O₃ (amphoteric) < MnO₂ < Mn₂O₇ (acidic)

Teacher's Note:
1. Work out the oxidation state of Mn in each oxide first; Mn₃O₄ has an average value of +8/3.

Answer: (d) Cr

Explanation:
1. ₂₄Cr → 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵; both 4s¹ and 3d⁵ are half-filled.

Teacher's Note:
1. Cu, Ag and Au have a half-filled ns subshell too, but their (n − 1)d subshell is completely filled (d¹⁰).

Answer: (c) Mn > Cr > Fe > Co

Explanation:
1. E° values for M²⁺/M with negative signs are
2. Cr = −0.91 V, Mn = −1.18 V, Fe = −0.44 V, Co = −0.28 V
3. Thus, the order is Mn > Cr > Fe > Co.

Teacher's Note:
1. Mn stands out because Mn²⁺ is the stable half-filled 3d⁵ ion, which makes Mn²⁺/Mn unusually negative.

Answer: (b) disproportionation

Explanation:
1. The reaction shows disproportionation since the oxidation state of Mn changes from +6 (MnO₄²⁻) to +7 (MnO₄⁻) and +4 (MnO₂).

Teacher's Note:
1. In disproportionation the same element in one species is both oxidised and reduced; check that one product is higher and one lower in oxidation state.

Answer: (d) Last element of third transition series – Cd

Explanation:
1. Hg is the last element of third transition series. (Cd is the last element of the second transition series.)

Teacher's Note:
1. Learn the end members: Zn (3d), Cd (4d) and Hg (5d) close the three transition series.

Answer: (a) 5.9 B.M

Explanation:
1. \( \mu = \sqrt{n(n+2)} \)
2. Electronic configuration of the element (25) = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s²; the divalent ion Mn²⁺ = [Ar]3d⁵
3. No. of unpaired electrons (n) = 5
4. \( \mu = \sqrt{5(5+2)} \) = √35 = 5.9 B.M.

Teacher's Note:
1. Remove the 4s electrons first when forming the M²⁺ ion; the 3d count stays at 5 for Mn²⁺.

Answer: (d) in covalent compounds, fluorine can form single bond only while oxygen forms double bond.

Explanation:
1. Oxygen can form multiple (double) bonds with the metal, as in Mn₂O₇ where each Mn is bonded to terminal O atoms by Mn=O bonds, whereas fluorine forms only single bonds.
2. So oxygen can stabilise a higher oxidation state than fluorine.

Teacher's Note:
1. The ability of oxygen to form pπ–dπ multiple bonds is the key reason oxides show higher oxidation states than fluorides.

Answer: (c) Zn²⁺

Explanation:
1. Zn²⁺ → 3d¹⁰ has no unpaired electrons to be excited. Hence, it is colourless.

Teacher's Note:
1. Ions with d⁰ or d¹⁰ configurations cannot show d–d transitions and are usually colourless.

Answer: (b) third ionisation enthalpy of Mn is much larger due to stable half filled d⁵ electronic configuration of Mn²⁺

Explanation:
1. The third ionisation energy of Mn required to change Mn²⁺ (d⁵) to Mn³⁺ (d⁴) is much larger due to stable half-filled d⁵ electronic configuration.

Teacher's Note:
1. A highly positive M³⁺/M²⁺ potential means the M³⁺ ion is readily reduced, i.e. M²⁺ is the more stable ion.

Answer: (c) They are chemically very reactive.

Explanation:
1. They are not chemically reactive. They are chemically inert.

Teacher's Note:
1. Remember the four properties of interstitial compounds: high m.p., very hard, metallic conductivity and chemical inertness.

Answer: (b) regular increase in ionisation enthalpy

Explanation:
1. Reactivity of transition elements decreases almost regularly from Sc to Cu because of regular increase in ionisation enthalpy.

Teacher's Note:
1. Lanthanoid contraction affects the 4d and 5d series, not the trend within the 3d series.

Answer: (a) Only Zn²⁺ is colourless and Ni²⁺, Cu²⁺ and Cr²⁺ are coloured.

Explanation:
1. Zn²⁺ (3d¹⁰) has zero unpaired electron (colourless).
2. Ni²⁺ (3d⁸) has 2 unpaired electrons (coloured).
3. Cu²⁺ (3d⁹) has 1 unpaired electron (coloured).
4. Cr²⁺ (3d⁴) has 4 unpaired electrons (coloured).

Teacher's Note:
1. Cu²⁺ is often wrongly called colourless; with 3d⁹ it has a vacancy for d–d transition and is blue in solution.

Answer: (d) Ionic radii of trivalent lanthanides steadily increases with increase in the atomic number.

Explanation:
1. Ionic radii of trivalent lanthanides decrease with increase in atomic number.

Teacher's Note:
1. In “false statement” questions, check each option; here (d) reverses the lanthanoid contraction trend.

Answer: (c) Magnetic moment of n unpaired electrons is given by \( \mu = \sqrt{n(n-2)} \) B.M.

Explanation:
1. The correct formula is \( \mu = \sqrt{n(n+2)} \) B.M.

Teacher's Note:
1. Check the spin-only formula with one electron: √(1 × 3) = 1.73 B.M.; the n(n − 2) form would give an impossible value.

Answer: (a) Cu(II) is more stable.

Explanation:
1. Though Cu(I) has 3d¹⁰ stable configuration while Cu(II) has 3d⁹ configuration, yet Cu(II) is more stable due to greater effective nuclear charge of Cu(II) (i.e., to hold 17 electrons instead of 18 in Cu(I)).

Teacher's Note:
1. In aqueous solution, the much more negative hydration enthalpy of Cu²⁺ more than compensates for the second ionisation enthalpy of Cu.

Answer: (b) V₂O₃ and CrO

Explanation:
1. In case of transition metal oxides, the oxides with metals in lower oxidation state are basic in nature.
2. O.S. of Mn in Mn₂O₇ = +7; V in V₂O₃ = +3; V in V₂O₅ = +5; Cr in CrO = +2; Cr in Cr₂O₃ = +3
3. Thus in V₂O₃, CrO and Cr₂O₃ transition metal ion is in lower oxidation state but Cr₂O₃ is amphoteric in nature. Hence V₂O₃ and CrO are basic in nature.

Teacher's Note:
1. Remember the two amphoteric exceptions often asked: Cr₂O₃ and V₂O₅.

Answer: (c) PtCl₂ catalyses the oxidation of ethyne to ethanal in the Wacker process.

Explanation:
1. In the Wacker process, the oxidation of ethyne to ethanal is catalysed by PdCl₂.

Teacher's Note:
1. Wacker process = PdCl₂ (palladium); do not confuse Pd with Pt in catalyst questions.

Answer: (a) La

Explanation:
1. Lanthanum is a d-block element which resembles lanthanides.

Teacher's Note:
1. La (5d¹ 6s², no 4f electron) is the prototype for the lanthanoids but its differentiating electron enters the 5d subshell.

Answer: (d) 1.732

Explanation:
1. The electronic configuration of Ce³⁺ is 4f¹
2. Hence, \( \mu = \sqrt{n(n+2)} \) = \( \sqrt{1(1+2)} \) = 1.732 B.M.

Teacher's Note:
1. One unpaired electron always gives √3 = 1.73 B.M. by the spin-only formula, whether it is a d or an f electron.

Answer: (a) Tl⁺, Tl³⁺

Explanation:
1. Tl⁺ is more stable than Tl³⁺ due to inert pair effect.
2. Cu²⁺ is more stable than Cu⁺. Cr³⁺ is more stable than Cr²⁺. V⁴⁺ in aqueous solution is more stable than V²⁺.

Teacher's Note:
1. Cr²⁺ and V²⁺ are strong reducing agents in water, which itself shows that their lower states are not the stable ones.

Answer: (b) CuF₂

Explanation:
1. Ag₂SO₄ → Ag⁺ (4d¹⁰) – colourless
2. CuF₂ → Cu²⁺ (3d⁹) – coloured
3. MgF₂ → Mg²⁺ (no d-electrons) – colourless
4. CuCl → Cu⁺ (3d¹⁰) – colourless

Teacher's Note:
1. Find the oxidation state of the metal first; only the ion with a partly filled d subshell (here Cu²⁺) is coloured.

Answer: (b) 2 only

Explanation:
1. Basic strength decreases from La(OH)₃ to Lu(OH)₃. Hence, (1) is incorrect.
2. Ce: [Xe] 4f¹ 5d¹ 6s²; Ce⁴⁺: [Xe] 4f⁰
3. Yb: [Xe] 4f¹⁴ 6s²; Yb²⁺: [Xe] 4f¹⁴
4. Lu: [Xe] 4f¹⁴ 5d¹ 6s²; Lu³⁺: [Xe] 4f¹⁴
5. The given ions contain no unpaired electrons and therefore, are diamagnetic.

Teacher's Note:
1. As the M³⁺ ion gets smaller (La to Lu), the M–OH bond becomes more covalent, so basic character falls.

Answer: (b) large value of second ionisation enthalpy of copper is compensated by much more negative hydration energy of Cu²⁺_(aq)

Explanation:
1. In aqueous solution, Cu⁺ disproportionates: 2Cu⁺ → Cu²⁺ + Cu.
2. This happens because the large second ionisation enthalpy of copper is more than compensated by the much more negative hydration enthalpy of Cu²⁺_(aq).

Teacher's Note:
1. Option (d) only describes what happens; the question asks for the reason, which is the energy balance in option (b).

Answer: (d) Sc(III), Ti(IV) are diamagnetic and Pd(II), Cu(II) are paramagnetic.

Explanation:
1. Sc³⁺ (3d⁰), Ti⁴⁺ (3d⁰) are diamagnetic while Pd²⁺ (4d⁸) and Cu²⁺ (3d⁹) are paramagnetic.

Teacher's Note:
1. d⁰ ions such as Sc³⁺ and Ti⁴⁺ have no unpaired electrons and are always diamagnetic.

Answer: (d) CrO₂Cl₂

Explanation:
1. In CrO₂Cl₂, O.S. of Cr = +6
2. MnO₄⁻, O.S. of Mn = +7
3. Cr(CN)₆³⁻, O.S. of Cr = +3
4. NiF₆²⁻, O.S. of Ni = +4

Teacher's Note:
1. For CrO₂Cl₂: x + 2(−2) + 2(−1) = 0 gives x = +6; always balance against the overall charge.

Answer: (a) Stability of complexes increases as the size of lanthanoid decreases.

Explanation:
1. As the size of the Ln³⁺ ion decreases, its charge density increases and it attracts ligands more strongly, so the stability of its complexes increases.

Teacher's Note:
1. Smaller ion with the same +3 charge means higher charge/size ratio, which is the general rule for stronger complex formation.

Answer: (b) configuration of Fe²⁺ is 3d⁶ while Fe³⁺ is 3d⁵

Explanation:
1. Fe²⁺ – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶
2. Fe³⁺ – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵
3. In Fe³⁺, orbital is half-filled hence provides extra stability.

Teacher's Note:
1. Higher charge alone does not give stability (Mn³⁺ is less stable than Mn²⁺); the half-filled d⁵ configuration is what matters.

Answer: (a) adopt multiple oxidation states and to form complexes

Explanation:
1. The transition metals and their compounds are known for their catalytic activity because of their ability to adopt multiple oxidation states and to form complexes.

Teacher's Note:
1. Variable oxidation states let the catalyst take part in electron transfer steps and then be regenerated.

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

Explanation:
1. Transition metals form substitutional alloys since they have nearly the same size, they can substitute one another in the crystal lattice.

Teacher's Note:
1. The reason for alloy formation is similar atomic size; the purpose of making alloys is a different, though true, statement.

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

Explanation:
1. Mn²⁺ = [Ar]3d⁵, Mn³⁺ = [Ar]3d⁴
2. Fe²⁺ = [Ar]3d⁶, Fe³⁺ = [Ar]3d⁵
3. Thus, Mn²⁺ has more stable configuration than Mn³⁺ while Fe³⁺ has more stable configuration than Fe²⁺.
4. Hence, reduction potential for Mn³⁺/Mn²⁺ couple is more positive than Fe³⁺/Fe²⁺.
5. As we move across the period, ionisation potential increases, thus, ionisation potential of Fe is more than that of Mn.

Teacher's Note:
1. The high E°(Mn³⁺/Mn²⁺) is explained by the stable half-filled d⁵ of Mn²⁺, not by the first ionisation enthalpy of the atoms.

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

Explanation:
1. A half-filled or fully-filled orbital is more stable than incompletely filled orbital.
2. Hence Mn²⁺ (3d⁵) is more stable than Mn³⁺ (3d⁴).

Teacher's Note:
1. This is the same reason Mn³⁺ is a strong oxidising agent: it readily gains an electron to reach d⁵.

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

Explanation:
1. Ti³⁺ has [Ar]3d¹ configuration. Thus, d–d transition is possible and thereby it shows colour.

Teacher's Note:
1. Colour depends on the presence of an unpaired d electron in that particular ion, not on the number of oxidation states the metal shows.

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

Explanation:
1. Chromium has maximum number of unpaired d electrons while Hg do not have any unpaired d electrons. Thus, Cr is hard but Hg is soft metal.

Teacher's Note:
1. Hardness and high melting point follow the strength of metallic bonding, which increases with the number of unpaired d electrons.

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

Explanation:
1. Even though Cu⁺ has completely filled 3d orbitals, [Ar]3d¹⁰ the nuclear charge in Cu is not enough to hold a core of 18 electrons in Cu⁺ and thus, Cu⁺ is unstable in comparison to Cu²⁺. Cu⁺ is diamagnetic in nature.

Teacher's Note:
1. A d¹⁰ ion has no unpaired electrons, so Cu⁺ must be diamagnetic.

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

Explanation:
1. Both Co and Ni have (IV) oxidation state. Ni (IV) has 3d⁶ electronic configuration.
2.

Teacher's Note:
1. Ni(IV) is rare but known, for example in K₂[NiF₆]; Ni: 3d⁸ 4s² minus four electrons gives 3d⁶.

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

Explanation:
1. Promethium does not occur naturally in appreciable amounts; it is radioactive and is obtained by artificial nuclear reactions, so it is called a man-made element.

Teacher's Note:
1. Promethium (Z = 61) is the only lanthanoid with no stable isotope.

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

Explanation:
1. Due to large surface area and variable valencies, transition metals form intermediate activated complexes easily, hence they are used as good catalysts.

Teacher's Note:
1. An example of a catalyst is not the reason for catalytic activity; the reason is variable oxidation states and large surface area.

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

Explanation:
1. The electronic configurations of europium (II) and cerium (II) are Eu²⁺: [Xe] 4f⁷ and Ce²⁺: [Xe] 4f¹ 5d¹
2. In Eu²⁺, f-subshell is half-filled thus, it is more stable.

Teacher's Note:
1. Questions comparing stability of lanthanoid ions are almost always decided by f⁰, f⁷ or f¹⁴ configurations.