JEE Chemistry D and F Block Elements MCQs Set 01

Answer: (d) 4/5

Explanation:
1. \( \ce{MnO4-} + 8\ce{H+} + 5e^- \rightarrow \ce{Mn^2+} + 4\ce{H2O} \) and \( \ce{SO3^2-} \rightarrow \ce{SO4^2-} + 2e^- \).
2. Balanced: 2 MnO4- : 5 SO3 2-. Two moles of sulphite need \( 2\times 2/5 = 4/5 \) mol.

Teacher's Note:
1. Electron balance.
2. 2 moles x 2 e- = 4 e-.

Answer: (d) 86

Explanation:
1. Atomic number of Pt = 78. Oxidation state in \( \ce{[PtCl6]^2-} \) is +4, so 74 electrons remain.
2. Six ligands give 12 electrons: EAN = 74 + 12 = 86.

Teacher's Note:
1. EAN = Z - oxidation number + 2 x CN.
2. Equals radon.

Answer: (b) Cu

Explanation:
1. Gun metal is Cu, Sn, Zn; bronze is Cu and Sn; brass is Cu and Zn.
2. Copper is the common metal.

Teacher's Note:
1. Alloys of copper.
2. Cu is in all three.

Answer: (c) \( \ce{K2Cr2O7} \) solution becomes yellow on increasing pH beyond 7

Explanation:
1. The orange colour of dichromate is due to charge transfer, not d-d (Cr is d0), so A is false.
2. \( \ce{K2Cr2O7} \) is preferred as a primary standard since \( \ce{Na2Cr2O7} \) is hygroscopic, so B is false. In alkaline solution dichromate changes to yellow chromate, so C is true and D is false.

Teacher's Note:
1. Cr(VI) is d0.
2. Chromate-dichromate equilibrium.

Answer: (d) Both (A) and (B)

Explanation:
1. On heating NaCl and \( \ce{K2Cr2O7} \) with conc. \( \ce{H2SO4} \), chromyl chloride \( \ce{CrO2Cl2} \) forms as deep red vapours.
2. So both A and B are correct (chlorine is not formed).

Teacher's Note:
1. Chloride ion test.
2. Deep red vapours of CrO2Cl2.

Answer: (a) \( \ce{Ni^2+} \) and \( \ce{Ti^3+} \)

Explanation:
1. \( \ce{Ni^2+} \) is d8 and \( \ce{Ti^3+} \) is d1, so both have unpaired d electrons and are coloured.
2. \( \ce{Sc^3+} \) is d0 and \( \ce{Cu+} \) is d10, so they are colourless.

Teacher's Note:
1. d-d transition needs partly filled d orbitals.
2. d0 and d10 are colourless.

Answer: (d) \( \ce{Ce^4+} \) reacts very fast with \( \ce{H2O} \)

Explanation:
1. Zn (d10) is not a transition element, Pm is radioactive and \( \ce{La(OH)3} \) is more basic than \( \ce{Lu(OH)3} \) because of lanthanide contraction.
2. \( \ce{Ce^4+} \) is stable in aqueous solution and does not react quickly with water, so D is incorrect.

Teacher's Note:
1. Basic character falls across the series.
2. Check each.

Answer: (b) Sc

Explanation:
1. Sc has only the +3 oxidation state.
2. Mn and Cr show several oxidation states.

Teacher's Note:
1. Sc is 3d1 4s2.
2. Loses all 3 electrons.

Answer: (d) All of these

Explanation:
1. \( \ce{KMnO4} \) forms dark purple crystals; in acid it oxidises \( \ce{SO2} \) to \( \ce{SO4^2-} \); and its purple colour is due to charge transfer from O to Mn(VII).
2. So all the statements are correct.

Teacher's Note:
1. Mn(VII) is d0.
2. LMCT.

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

Explanation:
1. Gadolinium (Z = 64) has the configuration \( [\ce{Xe}]4f^7 5d^1 6s^2 \), because a half-filled 4f subshell is stable.
2. The Rn core options belong to actinides.

Teacher's Note:
1. Half-filled f7 stability.
2. Lanthanide series.

Answer: (b) 8, 4

Explanation:
1. Dichromate changes to chromate in alkaline medium, at about pH 8.
2. Chromate changes to dichromate in acidic medium, at about pH 4. So X = 8 and Y = 4.

Teacher's Note:
1. \( \ce{Cr2O7^2-} + 2OH^- \rightleftharpoons 2\ce{CrO4^2-} + \ce{H2O} \).
2. Acid shifts to dichromate.

Answer: (d) \( \ce{HNO3} \)

Explanation:
1. Concentrated \( \ce{HNO3} \) forms a protective oxide layer on iron and makes it passive.
2. The other reagents do not do this.

Teacher's Note:
1. Passivity.
2. Also seen in Al and Cr.

Answer: (c) \( \ce{MnO4^2-} \)

Explanation:
1. Chromate, dichromate and permanganate are d0 ions (Cr VI, Mn VII), so their colour is due to charge transfer.
2. \( \ce{MnO4^2-} \) (Mn VI) is d1, so its colour can arise from a d-d transition.

Teacher's Note:
1. d0 cannot show d-d transitions.
2. Mn(VI) is d1.

Answer: (d) Both (B) and (C)

Explanation:
1. Chromite ore fused with \( \ce{Na2CO3} \) in air gives sodium chromate, so A is \( \ce{CrO4^2-} \) (C is correct).
2. Acidification gives dichromate, so B is \( \ce{Cr2O7^2-} \) (B is correct). Hence both B and C.

Teacher's Note:
1. 4FeCr2O4 + 8Na2CO3 + 7O2 gives 8Na2CrO4.
2. Acid gives orange dichromate.