JEE Chemistry The D and F Block Elements MCQs Set 10

Answer: (c) V2+

Explanation:
1. \( \mu = \sqrt{n(n+2)} = 3.87 \) gives n = 3 unpaired electrons.
2. Cr2+ is d4 (4 unpaired), Mn2+ is d5 (5), V2+ is d3 (3) and Ti2+ is d2 (2).
3. So the ion is V2+.

Teacher's Note:
1. 3.87 B.M. corresponds to 3 unpaired electrons.
2. Cr3+ and V2+ are the common d3 ions.

Answer: (d) C and D only

Explanation:
1. Ce4+ (4f0) and Tb4+ (4f7) are strong oxidising agents because they readily gain an electron to reach the stable +3 state.
2. Sm2+ and Ce2+ are reducing agents as they tend to lose electrons to reach +3.

Teacher's Note:
1. The common lanthanoid oxidation state is +3.
2. Ce4+ is used as an oxidant in volumetric analysis.

Answer: (d) sulphur dioxide

Explanation:
1. Dichromate in acidic solution is orange and is reduced to green Cr3+ by a reducing agent.
2. SO2 is the reducing agent here: Cr2O7 2- + 3SO2 + 2H+ gives 2Cr3+ + 3SO4 2- + H2O.
3. CO2, SO3 and N2 cannot reduce dichromate.

Teacher's Note:
1. Orange to green shows reduction of Cr(VI) to Cr(III).
2. This is the test for SO2.

Answer: (b) 2

Explanation:
1. V3+ is d2 (2 unpaired), Cr3+ is d3 (3 unpaired), Fe2+ is d6 (4 unpaired) and Ni3+ is d7 (3 unpaired).
2. In the gaseous state there is no ligand field, so all are high spin.
3. Cr3+ and Ni3+ both have 3 unpaired electrons, so two ions have a similar magnetic moment.

Teacher's Note:
1. In the gaseous state use Hund's rule.
2. Count the unpaired electrons for each ion.

Answer: (b) +6 to +3

Explanation:
1. In acidic solution Cr2O7 2- + 14H+ + 6e- gives 2Cr3+ + 7H2O.
2. Chromium goes from +6 to +3.

Teacher's Note:
1. Six electrons are gained by one dichromate ion.
2. Cr3+ is green.

Answer: (c) V2+ and Cr2+

Explanation:
1. An ion M2+ can reduce H+ to H2 if its M3+/M2+ potential is below 0 V, because the oxidation M2+ to M3+ then has a positive potential.
2. This holds for V (-0.26 V) and Cr (-0.41 V).
3. Mn2+ and Co2+ have highly positive values and are stable.

Teacher's Note:
1. A negative E value means that the lower oxidation state is a reducing agent.
2. Cr2+ is a very strong reducing agent.

Answer: (b) (i), (ii) and (iv)

Explanation:
1. (i) is true because Mn2O7 has Mn in +7.
2. (ii) is true because RuO4 and OsO4 contain +8.
3. (iii) is false because Sc shows only +3, as its 3d and 4s electrons total 3.
4. (iv) is true because Cr(VI) as in dichromate and chromate is a strong oxidising agent.
5. So (i), (ii) and (iv) are correct.

Teacher's Note:
1. Scandium has only the +3 state.
2. The highest oxidation state is +8 for Ru and Os.

Answer: (d) blue

Explanation:
1. Acidified dichromate with H2O2 gives the blue peroxide CrO5, which is stabilised in ether or amyl alcohol.
2. CrO5 is blue in organic solvent.

Teacher's Note:
1. This blue colour is used as a test for chromate or dichromate.
2. CrO5 has one oxo group and two peroxo groups.

Answer: (c) green

Explanation:
1. The sulphite gives SO2 on warming with dilute H2SO4.
2. SO2 reduces orange Cr2O7 2- to green Cr3+ (as chromium(III) sulphate).
3. So the solution turns green.

Teacher's Note:
1. Orange to green is the confirmatory test for SO2.
2. SO2 is a reducing agent.

Answer: (b) 3

Explanation:
1. MnO4- + 4H+ + 3e- gives MnO2 + 2H2O.
2. Mn goes from +7 to +4, so 3 electrons are gained.

Teacher's Note:
1. In strongly acidic medium MnO4- goes to Mn2+ with 5 electrons.
2. To MnO2 it needs 3 electrons.

Answer: (b) 4f4

Explanation:
1. Nd (Z = 60) is [Xe] 4f4 6s2.
2. Nd2+ is formed by loss of the two 6s electrons, so it is [Xe] 4f4.

Teacher's Note:
1. The outer s electrons are lost first.
2. Lanthanoids generally prefer the +3 state, but Nd2+ is known.

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

Explanation:
1. The first ionisation enthalpies of 3d elements (about 630 to 760 kJ/mol) are higher than those of the group 2 metals such as Ca (590).
2. In the 3d series, electrons enter the inner (n-1)d subshell, which shields the nuclear charge poorly.
3. The effective nuclear charge increases, so ionisation enthalpy is higher, and R explains A.

Teacher's Note:
1. The nuclear charge rises while the added electrons go to an inner subshell.
2. Poor shielding by d electrons raises the ionisation enthalpy.

Answer: (a) B and D only

Explanation:
1. Eu is [Xe] 4f7 6s2 and Gd is [Xe] 4f7 5d1 6s2, so both have a half-filled 4f subshell.
2. Sm is 4f6, Pm is 4f5 and Tb is 4f9.

Teacher's Note:
1. Half-filled f subshell has 7 electrons.
2. Gd takes the 5d electron to keep 4f7.

Answer: (a) I2 and IO3-

Explanation:
1. In acidic solution 2MnO4- + 10I- + 16H+ gives 2Mn2+ + 5I2 + 8H2O, so I2 forms.
2. In neutral or faintly alkaline solution 2MnO4- + I- + H2O gives 2MnO2 + 2OH- + IO3-, so iodate forms.

Teacher's Note:
1. Alkaline medium gives a stronger oxidation of iodide to iodate.
2. Know both reactions of KMnO4 with iodide.

Answer: (a) A and C only

Explanation:
1. Mn2O7 has two MnO3 units joined through a bridging oxygen, O3Mn-O-MnO3.
2. Each Mn has four oxygen neighbours (three terminal and one bridging), so it is tetrahedral.
3. There is no Mn-Mn bond.
4. Correct: A and C.

Teacher's Note:
1. Mn2O7 is a covalent, explosive green oil.
2. Dichromate also has a bridge, Cr-O-Cr.

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

Explanation:
1. Cu+ (3d10) is stable in the gas phase, but in water Cu2+ is more stable.
2. The hydration enthalpy of Cu2+ (about -2121 kJ/mol) is far more negative than that of Cu+ (about -593 kJ/mol).
3. This large hydration energy more than compensates for the second ionisation enthalpy, so Cu2+ is the stable ion in water and R explains A.

Teacher's Note:
1. Cu+ disproportionates in water: 2Cu+ gives Cu2+ + Cu.
2. High charge density gives a high hydration enthalpy.