CBSE Class 12 Chemistry The d and f Block Elements MCQs Set 14

Answer: (c) La(OH)₃ is less basic than Lu(OH)₃.

Explanation:
1. La(OH)₃ is more basic than Lu(OH)₃. Due to lanthanoid contraction the size of lanthanoid ion decreases regularly with increase in atomic size.
2. Thus, ionic character between lanthanoid ion and OH⁻ decreases from La³⁺ to Lu³⁺.
3. Thus the basic character of hydroxides decreases from La(OH)₃ to Lu(OH)₃.

Teacher's Note:
1. Smaller cation means a more covalent M–OH bond and a weaker base, so basicity falls from La(OH)₃ to Lu(OH)₃.

Answer: (d) Poor shielding effect of 4f electrons

Explanation:
1. The lanthanoid contraction is due to poor shielding effect of 4f electrons. Thus, atomic radii decreases as the atomic number increases.

Teacher's Note:
1. The diffuse shape of 4f orbitals makes them shield poorly, so each added proton pulls the outer electrons in more strongly.

Answer: (b) Actinoids

Explanation:
1. Actinoids are elements of 5f block, so in actinoids, 5f orbitals are progressively filled.

Teacher's Note:
1. Lanthanoids fill 4f, actinoids fill 5f; do not mix up the two inner-transition series.

Answer: (d) Cu

Explanation:
1. The radii are almost the same, so the volume of the atoms changes very little, while the atomic mass increases from Fe to Cu.
2. Cu has the largest atomic mass (63.5 u), so it has the highest density.

Teacher's Note:
1. When atomic sizes are nearly equal, density is decided mainly by atomic mass.

Answer: (a) Copper liberates hydrogen from acids.

Explanation:
1. E°(Cu²⁺/Cu) = +0.34 V is positive, so copper cannot reduce H⁺ ions to H₂.
2. The other three statements are correct.

Teacher's Note:
1. Only metals with negative E° values liberate H₂ from dilute acids; Cu is the only 3d metal with a positive value.

Answer: (a) V₂O₅, Cr₂O₃

Explanation:
1. Mn₂O₇ and CrO₃ are acidic, CrO is basic, while V₂O₅ and Cr₂O₃ are amphoteric.

Teacher's Note:
1. Lowest oxidation state oxides are basic, highest are acidic and intermediate ones are usually amphoteric; V₂O₅ is an exception, being amphoteric though it is the highest oxide of vanadium.

Answer: (c) Ti < V < Cr < Mn, increasing melting point

Explanation:
1. Melting points rise up to Cr and then fall; Mn has a much lower melting point than Cr (and V).
2. So the order Ti < V < Cr < Mn for melting point is wrong.

Teacher's Note:
1. Mn is the odd one out in many 3d trends (melting point, enthalpy of atomisation) because of its stable 3d⁵ configuration.

Answer: (b) 3.87 B.M.

Explanation:
1. Cr³⁺ is 3d³ with n = 3 unpaired electrons.
2. \( \mu = \sqrt{n(n+2)} = \sqrt{15} \) = 3.87 B.M.

Teacher's Note:
1. Remember the values: 1.73, 2.83, 3.87, 4.90 and 5.92 B.M. for 1 to 5 unpaired electrons.

Answer: (a) Co²⁺ and Cr³⁺

Explanation:
1. Co²⁺ (3d⁷) and Cr³⁺ (3d³) both have 3 unpaired electrons, so both have μ = 3.87 B.M.
2. Cr²⁺ has 4 and Mn²⁺ has 5 unpaired electrons.

Teacher's Note:
1. A dⁿ and a d¹⁰⁻ⁿ ion have the same number of unpaired electrons in high spin state.

Answer: (b) CuF₂

Explanation:
1. In CuF₂, Cu²⁺ is 3d⁹ with one unpaired electron, so d–d transition gives colour.
2. Ag⁺, Zn²⁺ and Cu⁺ are d¹⁰ ions and are colourless.

Teacher's Note:
1. First find the oxidation state of the metal, then its d electron count; d⁰ and d¹⁰ ions are usually colourless.

Answer: (c) All ions are coloured.

Explanation:
1. Ions with d⁰ or d¹⁰ configuration, such as Sc³⁺, Ti⁴⁺, Cu⁺ and Zn²⁺, are colourless.

Teacher's Note:
1. Colour needs at least one electron that can undergo a d–d transition.

Answer: (a) Mn₂O₇

Explanation:
1. 2KMnO₄ + 2H₂SO₄ → Mn₂O₇ + 2KHSO₄ + H₂O
2. Mn₂O₇ is a green oily liquid which decomposes explosively.

Teacher's Note:
1. Mn₂O₇ is the acid anhydride of permanganic acid; Mn is in its highest (+7) oxidation state.

Answer: (d) Both (a) and (b)

Explanation:
1. KMnO₄ (Mn⁷⁺, d⁰) and Ce(SO₄)₂ (Ce⁴⁺, f⁰) are coloured due to charge transfer, not d–d transition.
2. TiCl₄ (Ti⁴⁺, d⁰) is colourless.

Teacher's Note:
1. The intense colour of MnO₄⁻ and CrO₄²⁻ comes from ligand-to-metal charge transfer, even though the metal has no d electrons.

Answer: (d) Mn²⁺ acts as autocatalyst.

Explanation:
1. Mn²⁺ ions formed in the reaction catalyse the reaction, so it speeds up once some Mn²⁺ is formed (autocatalysis).

Teacher's Note:
1. In oxalic acid titrations the solution is warmed to 60–70 °C to make the slow start faster.

Answer: (a) (i), (ii)

Explanation:
1. In (i) Cu(+1) goes to +2 and 0; in (ii) Mn(+6) goes to +7 and +4, so both are disproportionation.
2. In (iii) Mn is only reduced (oxygen is oxidised) and in (iv) Mn(+7) and Mn(+2) combine to give Mn(+4) (comproportionation).

Teacher's Note:
1. In disproportionation the same element in one oxidation state is both oxidised and reduced.

Answer: (a) \( \frac{2}{5} \)

Explanation:
1. 2MnO₄⁻ + 5S²⁻ + 16H⁺ → 2Mn²⁺ + 5S + 8H₂O
2. 5 mol S²⁻ need 2 mol KMnO₄, so 1 mol needs 2/5 mol.

Teacher's Note:
1. Balance electrons: each MnO₄⁻ takes 5 e⁻ in acid and each S²⁻ gives 2 e⁻.

Answer: (c) IO₃⁻

Explanation:
1. 2MnO₄⁻ + H₂O + I⁻ → 2MnO₂ + 2OH⁻ + IO₃⁻

Teacher's Note:
1. In acidic medium iodide gives I₂, but in neutral or alkaline medium it is oxidised to iodate.

Answer: (c) As a result of lanthanoid contraction, the properties of 4d series of the transition elements have no similarities with the 5d series of elements.

Explanation:
1. Because of lanthanoid contraction, 4d and 5d elements of the same group have almost equal radii and very similar properties (e.g., Zr and Hf).

Teacher's Note:
1. Zr/Hf, Nb/Ta and Mo/W pairs are the standard examples of the effect of lanthanoid contraction.

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

Explanation:
1. The 5f, 6d and 7s levels are close in energy, so a variable number of electrons can take part in bonding.

Teacher's Note:
1. 5f orbitals extend further out than 4f orbitals, which is another reason they take part in bonding more easily.

Answer: (c) both have similar atomic radius.

Explanation:
1. Zirconium (Zr) and hafnium (Hf) have similar atomic radius hence they show similar physical and chemical properties.

Teacher's Note:
1. The almost equal radii of Zr (160 pm) and Hf (159 pm) are a consequence of lanthanoid contraction: the poor shielding by 4f electrons cancels the expected increase in size.

Answer: (c) Tm

Explanation:
1. Tm (Thulium) is a lanthanoid.

Teacher's Note:
1. Actinoids run from Th (90) to Lr (103); U (92), Np (93) and Fm (100) are in this range, while Tm (69) is a 4f element.

Answer: (c) Sn⁴⁺

Explanation:
1. When acidified K₂Cr₂O₇ solution is added to Sn²⁺ salt, Sn²⁺ changes to Sn⁴⁺. The reaction is given here :
2. Cr₂O₇²⁻ + 14H⁺ + 3Sn²⁺ → 2Cr³⁺ + 3Sn⁴⁺ + 7H₂O
3. Sn²⁺ is oxidised to Sn⁴⁺ while dichromate is reduced to Cr³⁺.

Teacher's Note:
1. Dichromate is an oxidising agent, so the tin must go to a higher oxidation state; Sn³⁺ and Sn⁺ are not stable states of tin.

Answer: (c) Providing empty d-orbitals

Explanation:
1. Transition metal ions form complexes because they have small size, high charge/size ratio and vacant d-orbitals of suitable energy to accept lone pairs from ligands.
2. Hence, providing empty d-orbitals is the characteristic responsible.

Teacher's Note:
1. Remember it is a HIGH charge/size ratio, not a small one, that favours complex formation.

Answer: (a) 3d¹⁰4s¹

Explanation:
1. Cu (Z = 29): [Ar] 3d¹⁰4s¹.
2. A completely filled 3d¹⁰ subshell is extra stable, so one 4s electron shifts to 3d.

Teacher's Note:
1. Cu and Cr are the two exceptions in the 3d series: Cr is 3d⁵4s¹ and Cu is 3d¹⁰4s¹.

Answer: (c) 1/6 molecular weight

Explanation:
1. Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
2. Each Cr changes from +6 to +3, so 6 electrons are gained per formula unit. Equivalent weight = M/6.

Teacher's Note:
1. Equivalent weight = molar mass ÷ total change in oxidation number per formula unit; count both Cr atoms.

Answer: (b) KMnO₄

Explanation:
1. Oxidation states of Mn: K₂MnO₄ = +6, KMnO₄ = +7, MnO₂ = +4, Mn₃O₄ = +8/3.
2. Hence, the maximum is +7 in KMnO₄.

Teacher's Note:
1. +7 is the highest oxidation state of Mn, equal to its group number (7).

Answer: (c) M/3

Explanation:
1. In neutral medium: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻
2. Mn changes from +7 to +4 (3 electrons), so equivalent weight = M/3.

Teacher's Note:
1. Learn the three cases: acidic medium M/5 (Mn²⁺), neutral/weakly alkaline M/3 (MnO₂), strongly alkaline M/1 (MnO₄²⁻).

Answer: (a) CrO₄²⁻

Explanation:
1. CrO₃ is an acidic oxide: CrO₃ + 2NaOH → Na₂CrO₄ + H₂O
2. In alkaline medium chromium(VI) exists as the yellow chromate ion, CrO₄²⁻.

Teacher's Note:
1. Chromate (yellow) is stable in alkali and dichromate (orange) in acid; the oxidation state of Cr stays +6 in both.

Answer: (b) [Xe] 4f⁷ 5d¹ 6s²

Explanation:
1. Gd (Z = 64) has 64 − 54 = 10 electrons beyond Xe.
2. The half-filled 4f⁷ arrangement is extra stable, so the configuration is [Xe] 4f⁷ 5d¹ 6s².

Teacher's Note:
1. Count the electrons after [Xe] to check any option: 7 + 1 + 2 = 10 for Gd.

Answer: (b) Pu and Np

Explanation:
1. Np and Pu show oxidation states up to +7; U shows up to +6 and Am up to +6.

Teacher's Note:
1. The maximum oxidation state among actinoids rises from Th (+4) to Np/Pu (+7) and then falls again.

Answer: (b) Mn²⁺

Explanation:
1. Unpaired electrons: Ti³⁺ (3d¹) = 1, Mn²⁺ (3d⁵) = 5, Fe²⁺ (3d⁶) = 4, Co³⁺ (3d⁶) = 4.
2. Mn²⁺ has the most unpaired electrons, so μ = \( \sqrt{35} \) = 5.92 B.M. is the highest.

Teacher's Note:
1. Among 3d ions, d⁵ (Mn²⁺, Fe³⁺) has the maximum unpaired electrons and hence the highest spin-only moment.

Answer: (c) WO₃

Explanation:
1. In the heavier members of group 6 the higher oxidation states are more stable, so W(VI) in WO₃ is not easily reduced.
2. Cr(VI) in CrO₃ and CrO₄²⁻ and Mn(VII) in Mn₂O₇ are strong oxidising agents.

Teacher's Note:
1. Down a d-block group the higher oxidation states become more stable: Cr(VI) oxidises, but Mo(VI) and W(VI) do not.

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

Explanation:
1. Copper (Cu) does not liberate hydrogen from acids because of positive electrode potential. Only metal having negative reduction potential can liberate hydrogen as those are reactive than hydrogen.

Teacher's Note:
1. E°(Cu²⁺/Cu) = +0.34 V; Cu dissolves only in oxidising acids such as HNO₃ or hot conc. H₂SO₄, and then no H₂ is formed.

Answer: (d) A is false, but R is true.

Explanation:
1. Transition metals have high melting points because of the involvement of greater number of (n − 1)d and ns electrons in the interatomic metallic bonding.

Teacher's Note:
1. The reason actually predicts high melting points; melting points peak around the middle of each series (d⁵), with Mn and Tc as anomalies.

Answer: (d) A is false, but R is true.

Explanation:
1. Chromium is a transition element that belongs to 3d series as in chromium, 3d orbitals are filled whereas actinoids belong to f-block elements.

Teacher's Note:
1. Actinoids are Th to Lr (Z = 90–103) and fill 5f orbitals; Cr (Z = 24) is a 3d transition metal.

Answer: (c) A is true, but R is false.

Explanation:
1. Cerium exhibits +4 oxidation state because Ce⁴⁺ has 4f⁰ electronic configuration which is most stable.

Teacher's Note:
1. Ce is [Xe]4f¹5d¹6s²; losing four electrons gives the noble gas configuration 4f⁰. Ce⁴⁺ is still a good oxidising agent (E° = +1.74 V).

Answer: (c) A is true, but R is false.

Explanation:
1. Lanthanoids have poor tendency for making complexes due to their large size and low charge density (charge/size ratio).
2. Their low tendency to form complexes is not related to the density of the metals; lanthanoid metals actually have fairly high densities (about 6.8 to 9.8 g cm⁻³).

Teacher's Note:
1. Do not confuse density of the metal with charge density of the ion; it is the low charge density of the large Ln³⁺ ions that weakens complex formation.

Answer: (c) A is true, but R is false.

Explanation:
1. Zn, Cd and Hg have completely filled d¹⁰ orbitals in the ground state and in their common oxidation state, so they are not transition elements.
2. But they are d-block elements (group 12), so R is false.

Teacher's Note:
1. All transition elements are d-block elements, but all d-block elements are not transition elements.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Cu²⁺ is more stable in aqueous solution because its high hydration enthalpy more than compensates for the second ionisation enthalpy.
2. R is a general statement; the specific reason for A is the high hydration enthalpy of Cu²⁺.

Teacher's Note:
1. Cu⁺ (3d¹⁰) is stable in solids like Cu₂O and CuI, but disproportionates in water.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Complex formation is due to small size and high charge of the metal ions and the availability of vacant d orbitals.
2. Transition metals do have unpaired electrons, but this explains their paramagnetism and colour, not complex formation.

Teacher's Note:
1. Zn²⁺ (no unpaired electrons) also forms complexes, which shows unpaired electrons are not needed for complex formation.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. FeCl₃ hydrolyses in water on standing to give a brown precipitate of hydrated ferric oxide, Fe(OH)₃.
2. FeCl₃ is covalent and exists as Fe₂Cl₆ with chlorine bridges, but this does not explain the hydrolysis.

Teacher's Note:
1. The brown precipitate is due to hydrolysis of Fe³⁺; a little HCl is added to FeCl₃ solutions to prevent it.

Answer: (d) A is false, but R is true.

Explanation:
1. The most common oxidation state of actinoids is +3, so A is false.
2. All actinoids have the 7s² configuration, so R is true.

Teacher's Note:
1. Like lanthanoids, actinoids show +3 as the common oxidation state, but they also show many higher states (up to +7).