JEE Chemistry The D and F Block Elements MCQs Set 04

Answer: (b) 2/5

Explanation:
1. \( MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O \) (5 electrons per mole).
2. \( SO_3^{2-} \to SO_4^{2-} + 2e^- \) (2 electrons per mole).
3. Moles of permanganate = 2/5 per mole of sulphite.

Teacher's Note:
1. Use the n-factors: 5 for permanganate and 2 for sulphite.

Answer: (a) a catalytic promoter

Explanation:
1. Iron is the catalyst in the Haber process.
2. Molybdenum (or \( Al_2O_3 \)) increases the activity of the catalyst, so it is a promoter.

Teacher's Note:
1. Promoters enhance catalytic activity but are not catalysts themselves.

Answer: (d) NaOH and \( H_2SO_4 \)

Explanation:
1. \( Cr_2O_7^{2-} + 2OH^- \to 2CrO_4^{2-} + H_2O \): a base is needed.
2. \( 2CrO_4^{2-} + 2H^+ \to Cr_2O_7^{2-} + H_2O \): an acid. HI would reduce dichromate, so \( H_2SO_4 \) is used.

Teacher's Note:
1. Acid gives dichromate (orange) and alkali gives chromate (yellow).

Answer: (c) \( CrO_4^{2-} \)

Explanation:
1. \( CrO_3 \) is an acidic oxide: \( CrO_3 + 2OH^- \to CrO_4^{2-} + H_2O \).

Teacher's Note:
1. In alkaline solution the chromium(VI) species is the yellow chromate ion.

Answer: (b) green

Explanation:
1. In strongly alkaline medium \( MnO_4^- \) accepts one electron: \( MnO_4^- + e^- \to MnO_4^{2-} \).
2. Manganate ion \( MnO_4^{2-} \) is green.

Teacher's Note:
1. The product of permanganate depends on the medium: Mn2+ in acid, MnO2 in neutral, MnO42- in strong alkali.

Answer: (a) 18

Explanation:
1. Pure gold is 24 carat.
2. \( 0.75\times24 = 18 \) carat.

Teacher's Note:
1. Carat is parts of gold per 24 parts of alloy.

Answer: (d) charge transfer from O to Mn

Explanation:
1. Mn in \( MnO_4^- \) is +7 with a \( d^0 \) configuration, so d-d transitions are not possible.
2. The colour arises from ligand to metal charge transfer (O to Mn).

Teacher's Note:
1. The same applies to the colours of chromate and dichromate.

Answer: (c) \( KMnO_4 \)

Explanation:
1. Permanganate is deep purple and its reduction product Mn2+ is nearly colourless.
2. The first excess drop gives a permanent pink colour, so no separate indicator is needed.

Teacher's Note:
1. Titrations with permanganate are carried out in dilute sulphuric acid, never in HCl.

Answer: (b) HgS

Explanation:
1. Vermilion (cinnabar) is the red form of mercuric sulphide, HgS.

Teacher's Note:
1. Cinnabar is the chief ore of mercury.

Answer: (a) \( CrO_5 \) and \( H_2O \)

Explanation:
1. \( Cr_2O_7^{2-} + 4H_2O_2 + 2H^+ \to 2CrO_5 + 5H_2O \).
2. \( CrO_5 \) (a peroxo compound with Cr in +6) is blue.

Teacher's Note:
1. This is the test for chromate and dichromate or hydrogen peroxide.

Answer: (d) completion of (n-1)d orbitals

Explanation:
1. Zn is \( 3d^{10}4s^2 \). The d subshell is full and stable, so only the two 4s electrons are lost.
2. So it shows only +2.

Teacher's Note:
1. A filled d10 shell is not involved in bonding.

Answer: (c) \( Cr_2O_3 \)

Explanation:
1. \( (NH_4)_2Cr_2O_7 \to Cr_2O_3 + N_2 + 4H_2O \) (volcano experiment).

Teacher's Note:
1. The reaction is exothermic once started.

Answer: (b) Hg

Explanation:
1. Mercury is the liquid d-block metal; \( HgI_2 \) changes from red to yellow on heating.

Teacher's Note:
1. Mercury is a liquid because of its weak metallic bonding.

Answer: (a) a protective layer of \( Fe_3O_4 \)

Explanation:
1. Concentrated nitric acid forms a thin, firm layer of Fe3O4 on the surface.
2. This layer prevents further reaction.

Teacher's Note:
1. Al and Cr also become passive in concentrated nitric acid.

Answer: (d) cold 1% alkaline \( KMnO_4 \) solution

Explanation:
1. The reagent is cold dilute alkaline potassium permanganate.
2. The purple colour is decolourised and a brown precipitate of MnO2 forms.

Teacher's Note:
1. It converts alkenes to glycols by syn hydroxylation.

Answer: (c) \( NO_2 \)

Explanation:
1. \( Cu + 4HNO_3(conc) \to Cu(NO_3)_2 + 2NO_2 + 2H_2O \).

Teacher's Note:
1. With dilute nitric acid copper gives NO.

Answer: (b) \( Mn^{2+} \) acts as an autocatalyst

Explanation:
1. \( Mn^{2+} \) formed in the reaction catalyses the further oxidation of oxalate.

Teacher's Note:
1. An autocatalytic reaction speeds up as the product accumulates.

Answer: (a) \( K_3[Cu(CN)_4] \)

Explanation:
1. \( 2Cu^{2+} + 4CN^- \to 2CuCN + (CN)_2 \), then \( CuCN + 3CN^- \to [Cu(CN)_4]^{3-} \).

Teacher's Note:
1. Copper is reduced to Cu(I) and cyanogen gas is evolved.

Answer: (d) F and O

Explanation:
1. Fluorine and oxygen are small, highly electronegative atoms which can stabilise high oxidation states, for example \( MnO_4^- \), \( OsO_4 \), \( VF_5 \).

Teacher's Note:
1. Oxygen forms multiple bonds and fluorine has the highest oxidising ability.

Answer: (c) 1/5

Explanation:
1. Permanganate takes 5 electrons per mole and \( Fe^{2+} \) gives 1.
2. Moles of KMnO4 = 1/5.

Teacher's Note:
1. Moles of oxidant = n-factor of reductant divided by n-factor of oxidant.

Answer: (b) Zn and Fe

Explanation:
1. Insulin contains zinc. Haemoglobin contains iron (heme).

Teacher's Note:
1. Chlorophyll contains Mg and vitamin B12 contains Co.

Answer: (a) invar

Explanation:
1. Invar (Fe 64%, Ni 36%) has an extremely low thermal expansion, so lengths stay constant.

Teacher's Note:
1. Invar is also used in clocks and precision instruments.

Answer: (d) Au

Explanation:
1. Gold is attacked only by aqua regia. Ag, Pb and Hg react with nitric acid.

Teacher's Note:
1. Aqua regia is 3 parts conc. HCl to 1 part conc. HNO3.

Answer: (c) Au

Explanation:
1. Gold chloride, AuCl3, is used for toning photographs.

Teacher's Note:
1. Gold dissolves in aqua regia to give chloroauric acid.

Answer: (b) 4

Explanation:
1. Four water molecules are coordinated to \( Cu^{2+} \) in a square plane; the fifth is hydrogen bonded to sulphate.

Teacher's Note:
1. The formula is better written \( [Cu(H_2O)_4]SO_4\cdot H_2O \).