JEE Chemistry The D And F Block Elements MCQs Set 20

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

Explanation:
1. Cu(I) has the stable d\(^{10}\) configuration, but the hydration enthalpy of Cu\(^{2+}\) is much higher than that of Cu\(^+\).
2. This large hydration energy more than compensates the second ionisation enthalpy, so Cu(II) is more stable in aqueous solution.

Teacher's Note:
1. Stability in water depends on hydration enthalpy as well as ionisation enthalpy.

Answer: (b) \( \ce{Cr2O7^{2-}} \) and \( \ce{H2O} \) are formed

Explanation:
1. Acidification shifts the equilibrium \( \ce{2CrO4^{2-} + 2H+ <=> Cr2O7^{2-} + H2O} \) to the right.
2. Chromate is converted to dichromate, and no change in oxidation state (+6) occurs.

Teacher's Note:
1. Yellow chromate turns orange in acid and the change is reversible.

Answer: (b) 3d\(^5\)

Explanation:
1. The number of unpaired electrons (high spin) is 3, 5, 2 and 2 respectively.
2. Magnetic moment \( \mu = \sqrt{n(n+2)} \) is greatest for n = 5, i.e. 3d\(^5\) (5.92 BM).

Teacher's Note:
1. Half-filled d\(^5\) has the maximum number of unpaired electrons.

Answer: (a) 2/5

Explanation:
1. \( \ce{MnO4^- + 8H+ + 5e^- -> Mn^{2+} + 4H2O} \) and \( \ce{S^{2-} -> S + 2e^-} \).
2. For 10 electrons: \( \ce{2MnO4^- + 5S^{2-}} \), so 1 mole of S\(^{2-}\) needs 2/5 mole of KMnO\(_4\).

Teacher's Note:
1. Balance electrons: 5 per permanganate, 2 per sulphide.

Answer: (a) [Xe] 4f\(^7\) 5d\(^1\) 6s\(^2\)

Explanation:
1. Gd is [Xe] 4f\(^7\) 5d\(^1\) 6s\(^2\).
2. The half-filled 4f\(^7\) subshell is extra stable, so one electron goes to 5d instead of making 4f\(^8\).

Teacher's Note:
1. Half-filled f\(^7\) is stable, like Eu and Gd.

Answer: (d) They are chemically very reactive

Explanation:
1. Interstitial compounds such as TiC, Fe\(_3\)H and Mn\(_4\)N have high melting points, are very hard and keep metallic conductivity.
2. They are chemically inert, so being very reactive is not a characteristic property.

Teacher's Note:
1. Interstitial compounds are hard, high-melting and chemically unreactive.

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

Explanation:
1. Oxygen can form multiple (double) bonds with Mn, which allows Mn to reach +7.
2. Fluorine forms only single bonds, which limits the highest oxidation state in fluorides.

Teacher's Note:
1. Multiple bonding ability of oxygen stabilises high oxidation states.

Answer: (c) both have similar atomic radius

Explanation:
1. Because of the lanthanoid contraction, the radius of Hf is almost the same as that of Zr.
2. This similar size gives them very similar properties.

Teacher's Note:
1. Lanthanoid contraction makes the second and third series elements alike.

Answer: (b) \( \ce{KMnO4} \) oxidises HCl into \( \ce{Cl2} \) which is also an oxidising agent

Explanation:
1. KMnO\(_4\) oxidises Cl\(^-\) to Cl\(_2\).
2. Some of the permanganate is used up in this side reaction, and the chlorine formed also acts as an oxidising agent, which spoils the titration.

Teacher's Note:
1. Dilute H\(_2\)SO\(_4\) is used to acidify permanganate solutions.

Answer: (c) 1 and 4

Explanation:
1. Co\(^{2+}\) is d\(^7\) (3 unpaired electrons) and Cr\(^{3+}\) is d\(^3\) (3 unpaired electrons).
2. Cr\(^{2+}\) has 4 and Mn\(^{2+}\) has 5 unpaired electrons.

Teacher's Note:
1. Compare the number of unpaired electrons.

Answer: (b) 2 and 4

Explanation:
1. Pu and Np reach +7, whereas U shows at most +6 and Am is mainly +3 to +6.
2. So the answer is Pu and Np.

Teacher's Note:
1. Actinoids in the middle of the series show the most oxidation states.

Answer: (b) \( \ce{Mn^{2+}} \)

Explanation:
1. The unpaired electrons are: Ti\(^{3+}\) (d\(^1\)) 1, Mn\(^{2+}\) (d\(^5\)) 5, Fe\(^{2+}\) (d\(^6\)) 4, Co\(^{3+}\) (d\(^6\)) 4.
2. Mn\(^{2+}\) has the most, so the highest \( \mu = \sqrt{35} = 5.92 \) BM.

Teacher's Note:
1. \( \mu = \sqrt{n(n+2)} \) BM.

Answer: (a) 1 and 2

Explanation:
1. Cr and Co both form stable trifluorides \( \ce{CrF3} \) and \( \ce{CoF3} \).
2. Cu and Ni form difluorides \( \ce{CuF2} \) and \( \ce{NiF2} \).

Teacher's Note:
1. Fluorine stabilises high oxidation states.

Answer: (a) \( \ce{V^{2+}} < \ce{Cr^{2+}} < \ce{Mn^{2+}} < \ce{Fe^{2+}} \): paramagnetic behaviour

Explanation:
1. The unpaired electrons are V\(^{2+}\) 3, Cr\(^{2+}\) 4, Mn\(^{2+}\) 5 and Fe\(^{2+}\) 4.
2. The paramagnetism increases to Mn\(^{2+}\) and then falls, so Fe\(^{2+}\) should come before Mn\(^{2+}\) and the given order is wrong.
3. The other three orders are correct.

Teacher's Note:
1. Paramagnetism peaks at d\(^5\).

Answer: (c) The complex is an outer orbital complex

Explanation:
1. Cr\(^{3+}\) (d\(^3\)) uses two inner 3d orbitals, giving d\(^2\)sp\(^3\) hybridisation, so it is an inner orbital complex.
2. It is paramagnetic with three unpaired electrons and gives AgCl with AgNO\(_3\) because Cl\(^-\) is outside the coordination sphere.
3. So (c) is wrong.

Teacher's Note:
1. d\(^2\)sp\(^3\) means inner orbital; sp\(^3\)d\(^2\) means outer orbital.

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

Explanation:
1. The values are Mn -1.18 V, Cr -0.91 V, Fe -0.44 V and Co -0.28 V.
2. The negative value decreases across the series except that Mn is more negative than Cr because of the stable d\(^5\) configuration of Mn\(^{2+}\).

Teacher's Note:
1. Mn\(^{2+}\) (d\(^5\)) and Zn\(^{2+}\) (d\(^{10}\)) are especially stable.

Answer: (b) In the highest oxidation states, the transition metal shows basic character and forms cationic complexes

Explanation:
1. In the highest oxidation states the metal oxides are acidic (for example \( \ce{CrO3} \), \( \ce{Mn2O7} \)) and form anionic species such as chromate and permanganate.
2. So the statement that they are basic and form cationic complexes is incorrect.

Teacher's Note:
1. Acidic character of oxides increases with oxidation state.

Answer: (b) a lesser energy difference between 5f and 6d than between 4f and 5d orbitals

Explanation:
1. The 5f, 6d and 7s orbitals of actinoids are close in energy, so electrons from all of them can take part in bonding.
2. For lanthanoids the 4f and 5d gap is larger, so fewer oxidation states are accessible.

Teacher's Note:
1. Closer 5f and 6d energies give more oxidation states.

Answer: (d) 4f and 5f orbitals are equally shielded

Explanation:
1. The 4f orbitals lie closer to the nucleus and are better shielded by inner electrons than the more diffuse 5f orbitals.
2. So the statement that they are equally shielded is incorrect.

Teacher's Note:
1. 5f orbitals extend further out than 4f orbitals.

Answer: (a) +3, +4

Explanation:
1. Cerium ([Xe] 4f\(^1\) 5d\(^1\) 6s\(^2\)) is mainly +3, and it also shows +4 because Ce\(^{4+}\) has the stable noble gas configuration.

Teacher's Note:
1. Ce(IV) is a good oxidising agent.