Download JEE MCQs for JEE Chemistry: The D and F Block Elements
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Chapter-wise Objective Questions: The D and F Block Elements
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Multiple Choice Questions
(a) Variable valency
(b) High ionisation energy
(c) Empty orbitals
(d) Free valency on the surface
Show Answer & Explanation
Answer: (b) High ionisation energy
Explanation:
1. Transition metals are good catalysts because of variable valency, vacant d orbitals and free valencies on the surface.
2. They do not need a high ionisation energy.
Teacher's Note:
1. Catalytic action needs adsorption and variable oxidation states.
(a) 3d series
(b) 4d series
(c) 5d series
(d) 6d series
Show Answer & Explanation
Answer: (b) 4d series
Explanation:
1. These are yttrium to cadmium, in which the 4d orbitals are filled.
Teacher's Note:
1. Z = 21-30 is 3d, 39-48 is 4d, 57 and 72-80 is 5d.
(a) Mn2O7, MnO2
(b) Mn2O, Mn2O3
(c) MnSO4, Mn2O3
(d) Mn2O3, MnO2
Show Answer & Explanation
Answer: (a) Mn2O7, MnO2
Explanation:
1. KMnO4 and cold conc. H2SO4 give the green oily Mn2O7.
2. It explodes on heating to MnO2, O2 (and ozone).
Teacher's Note:
1. Recall the electronic configuration of the ion.
2. Count unpaired electrons for magnetic questions.
(a) Mn3+, Co3+
(b) Fe3+, Cr3+
(c) Zn2+, Sc3+
(d) Ti2+, Cu2+
Show Answer & Explanation
Answer: (c) Zn2+, Sc3+
Explanation:
1. Zn2+ (3d10) and Sc3+ (3d0) have no unpaired d electrons, so there is no d-d transition.
Teacher's Note:
1. Colour needs a partly filled d subshell.
(a) Cr2O3
(b) CrO2
(c) Cr2O4
(d) CrO3
Show Answer & Explanation
Answer: (a) Cr2O3
Explanation:
1. (NH4)2Cr2O7 decomposes to N2, water and green Cr2O3.
Teacher's Note:
1. This is the volcano reaction.
(a) X = OH-, Y = H+
(b) X = H+, Y = OH-
(c) X = OH-, Y = H2O2
(d) X = H2O2, Y = OH-
Show Answer & Explanation
Answer: (a) X = OH-, Y = H+
Explanation:
1. Alkali converts dichromate (orange) to chromate (yellow); acid reverses it.
Teacher's Note:
1. Orange in acid, yellow in base.
(a) KMnO4
(b) Oxalic acid
(c) K2SO4
(d) MnSO4
Show Answer & Explanation
Answer: (a) KMnO4
Explanation:
1. The permanganate colour persists after the end point, so it acts as its own indicator.
Teacher's Note:
1. No external indicator is needed.
(a) F, O
(b) O, N
(c) O, Cl
(d) F, Cl
Show Answer & Explanation
Answer: (a) F, O
Explanation:
1. Only the most electronegative elements, fluorine and oxygen, can stabilise the highest oxidation states.
Teacher's Note:
1. Examples: MnO4-, OsF6.
(a) 4, 3
(b) 3, 5
(c) 4, 2
(d) 4, 1
Show Answer & Explanation
Answer: (c) 4, 2
Explanation:
1. \( \sqrt{n(n+2)}=\sqrt{24} \) gives 4 unpaired electrons.
2. Fe2+ (3d6) has 4 unpaired electrons, so n = 2.
Teacher's Note:
1. Spin-only formula.
(a) IO3-
(b) I2
(c) IO-
(d) IO4-
Show Answer & Explanation
Answer: (a) IO3-
Explanation:
1. In alkaline medium permanganate oxidises iodide to iodate (and is reduced to MnO2).
Teacher's Note:
1. Neutral or alkaline: MnO2 and IO3-.
(a) A layer of cupric oxide
(b) A layer of basic carbonate of copper
(c) A layer of cupric hydroxide
(d) (1) and (3) both
Show Answer & Explanation
Answer: (b) A layer of basic carbonate of copper
Explanation:
1. Copper reacts with CO2 and moisture to form green basic copper carbonate.
Teacher's Note:
1. Verdigris.
(a) CrO
(b) Cr2O3
(c) CrO3
(d) CrO5
Show Answer & Explanation
Answer: (b) Cr2O3
Explanation:
1. CrO is basic, CrO3 is acidic and Cr2O3 is amphoteric.
Teacher's Note:
1. Acidity rises with oxidation state.
(a) More stable
(b) Equally stable
(c) Less stable
(d) Stability depends on the copper salt
Show Answer & Explanation
Answer: (c) Less stable
Explanation:
1. The higher hydration enthalpy of Cu2+ outweighs the second ionisation energy, so Cu+ disproportionates in water.
Teacher's Note:
1. 2Cu+ gives Cu2+ and Cu.
(a) K2Cr2O7 reduces ferric sulphate to ferrous sulphate
(b) Iron does not form amalgam
(c) A permanent magnet is made of an alloy called Alnico
(d) In the lanthanides the ionic radius decreases from La3+ to Lu3+
Show Answer & Explanation
Answer: (a) K2Cr2O7 reduces ferric sulphate to ferrous sulphate
Explanation:
1. Dichromate is an oxidising agent: it oxidises ferrous to ferric, not the reverse.
Teacher's Note:
1. Cr(VI) is reduced to Cr(III).
(a) Acidic
(b) Basic
(c) Amphoteric
(d) Neutral
Show Answer & Explanation
Answer: (a) Acidic
Explanation:
1. Chromium in its highest oxidation state, +6, gives an acidic oxide.
Teacher's Note:
1. It forms chromic acid with water.
(a) KMnO4 + conc. HCl
(b) KCl + K2Cr2O7 + conc. H2SO4
(c) MnO2 + conc. HCl
(d) KCl + F2
Show Answer & Explanation
Answer: (b) KCl + K2Cr2O7 + conc. H2SO4
Explanation:
1. KCl, K2Cr2O7 and conc. H2SO4 give red chromyl chloride vapours.
2. The others liberate chlorine.
Teacher's Note:
1. Chromyl chloride test.
(a) d5
(b) d6
(c) d3
(d) d8
Show Answer & Explanation
Answer: (a) d5
Explanation:
1. d5 has five unpaired electrons.
Teacher's Note:
1. n(n+2) is largest for five unpaired electrons.
(a) MnO4 2- in acidic medium
(b) Cu+ in aqueous medium
(c) Cl2 in alkaline medium
(d) All of these
Show Answer & Explanation
Answer: (d) All of these
Explanation:
1. Manganate, Cu+ and Cl2 each undergo self oxidation and reduction in the stated media.
Teacher's Note:
1. Intermediate oxidation states disproportionate.
(a) TiO > FeO > VO > CrO
(b) TiO > VO > CrO > FeO
(c) VO > CrO > TiO > FeO
(d) CrO > VO > FeO > TiO
Show Answer & Explanation
Answer: (b) TiO > VO > CrO > FeO
Explanation:
1. Basic strength decreases as the metal-oxygen bond becomes more covalent, with the ionic radius decreasing from Ti to Fe.
Teacher's Note:
1. Radius falls across the series.
(a) 2Cu2+ + 4F- gives Cu2F2 + F2
(b) Cu2+ + 4NH3 gives [Cu(NH3)4]2+
(c) 2Cu2+ + 2CNS- + SO2 + 2H2O gives Cu2(CNS)2 + H2SO4 + 2H+
(d) 2Cu2+ + 4I- gives Cu2I2 + I2
Show Answer & Explanation
Answer: (d) 2Cu2+ + 4I- gives Cu2I2 + I2
Explanation:
1. The liberated iodine is titrated with hypo (thiosulphate).
Teacher's Note:
1. Iodometric estimation.
(a) The same crystal structure
(b) Almost the same electropositive character
(c) Transition metals contraction in a series
(d) The effect of lanthanide contraction
Show Answer & Explanation
Answer: (d) The effect of lanthanide contraction
Explanation:
1. The poor shielding by 4f electrons compensates the extra shell of gold.
Teacher's Note:
1. Lanthanide contraction.
(a) In d-block elements oxidation states differ by unity
(b) In p-block metals oxidation states differ by two units
(c) In a group of p-block lower oxidation states are favoured by the heavier members
(d) In a group of d-block higher oxidation states are favoured by the lighter members
Show Answer & Explanation
Answer: (d) In a group of d-block higher oxidation states are favoured by the lighter members
Explanation:
1. In the d-block, higher oxidation states are favoured by the heavier members.
Teacher's Note:
1. Opposite to the p-block trend.
(a) CrO3 > MoO3
(b) K2Cr2O7 > KMnO4
(c) Fe(CO)5 > Mn(CO)5
(d) V2O3 > V2O5
Show Answer & Explanation
Answer: (a) CrO3 > MoO3
Explanation:
1. Stability of the highest oxidation state increases down the group, so Cr(VI) is more oxidising than Mo(VI).
Teacher's Note:
1. Higher oxidation state: stronger oxidant for the lighter member.
(a) [Ar] 3d8 4s2
(b) [Ar] 3d10 4s1
(c) [Ar] 3d5 4s2
(d) [Ar] 3d7 4s2
Show Answer & Explanation
Answer: (c) [Ar] 3d5 4s2
Explanation:
1. Manganese (3d5 4s2) shows oxidation states +2 to +7.
Teacher's Note:
1. The middle of the series shows the most states.
(a) They are chemically inert
(b) They are soft and nonconductive
(c) They retain metallic conductivity
(d) They have high melting points
Show Answer & Explanation
Answer: (b) They are soft and nonconductive
Explanation:
1. Interstitial compounds are hard and conduct like metals.
Teacher's Note:
1. Carbides and nitrides of transition metals.
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Multiple Choice Questions (MCQs) for JEE Chemistry The D and F Block Elements
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