CBSE Class 12 Chemistry The d and f Block Elements MCQs Set 16

Answer: (a) +3

Explanation:
1. All lanthanoids show +3 oxidation state (removal of the two 6s and one 5d/4f electron); +2 and +4 occur only in a few.

Teacher's Note:
1. +2 is seen for Eu, Yb (and Sm); +4 for Ce, Tb – usually linked to 4f⁰, 4f⁷ or 4f¹⁴.

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

Explanation:
1. Due to lanthanoid contraction, size decreases from La³⁺ to Lu³⁺, covalent character of hydroxides increases and basic strength decreases. So La(OH)₃ is the most basic; statement (a) is wrong.

Teacher's Note:
1. Larger cation → more ionic M–OH bond → stronger base. La³⁺ is the largest.

Answer: (c) Zr⁴⁺, Hf⁴⁺

Explanation:
1. Zr⁴⁺ and Hf⁴⁺ have similar ionic radius due to lanthanoid contraction.

Teacher's Note:
1. Lanthanoid contraction cancels the expected size increase from the 4d to the 5d series, so Zr/Hf, Nb/Ta, Mo/W are near twins.

Answer: (c) The +4 oxidation state of cerium is not known in solutions.

Explanation:
1. Ce⁴⁺ exists in solution though it tends to change to more stable Ce³⁺. Hence statement (c) is incorrect.

Teacher's Note:
1. Ce⁴⁺ (4f⁰) solutions are used as an oxidant in volumetric analysis – so +4 certainly exists in solution.

Answer: (a) Ac

Explanation:
1. Ac (6d¹ 7s²) shows only +3 oxidation state.

Teacher's Note:
1. Th, Pa and U show +4, +5 and +6 respectively as well, because their 5f/6d electrons can also be used.

Answer: (a) Gd

Explanation:
1. Gadolinium (Gd) is a member of lanthanide series.

Teacher's Note:
1. At and Xe are p-block elements and Ac is an actinoid; only Gd is a lanthanoid.

Answer: (d) Terbium (Z = 65)

Explanation:
1. Actinoids are elements from Z = 89 to 103. Terbium (Tb, Z = 65) is not an actinoid but a lanthanoid.

Teacher's Note:
1. Lanthanoids: Z = 58–71 (Ce–Lu); actinoids: Z = 90–103 (Th–Lr), with La and Ac as parents.

Answer: (b) Np and Pu

Explanation:
1. Both Np and Pu show highest oxidation state of +7.

Teacher's Note:
1. Maximum oxidation states: Th +4, Pa +5, U +6, Np +7, Pu +7, Am +6.

Answer: (c) K Fe [Fe(CN)₆]

Explanation:
1. Prussian blue is K Fe^III[Fe^II(CN)₆].

Teacher's Note:
1. Prussian blue forms when Fe³⁺ meets [Fe(CN)₆]⁴⁻; it contains iron in both +3 and +2 states.

Answer: (c) 2KOH + Cl₂ → KCl + KOCl + H₂O

Explanation:
1. A disproportionation reaction is one in which the same substance undergoes oxidation as well as reduction. In (c),
2. Cl₂ (0) → KCl (−1) and also Cl₂ (0) → KOCl (+1).

Teacher's Note:
1. In disproportionation one element in one species ends up in two different oxidation states, one higher and one lower.

Answer: (b) Sn

Explanation:
1. Sn is a p-block element and not a transition element.

Teacher's Note:
1. Hg is a d-block element (though not a true transition element); Sn belongs to group 14.

Answer: (b) Mn²⁺

Explanation:
1. Fe²⁺ = 3d⁶ (four unpaired electrons); Mn²⁺ = 3d⁵ (five unpaired electrons); Cu⁺ = 3d¹⁰ (no unpaired electrons); Cr³⁺ = 3d³ (three unpaired electrons)
2. Thus, Mn²⁺ has the maximum number of unpaired electrons and hence has maximum magnetic moment.

Teacher's Note:
1. For 3d ions the maximum number of unpaired electrons (5) occurs at d⁵.

Answer: (b) V

Explanation:
1. Melting points: V 2183 K, Cr 2180 K, Fe 1811 K, Ni 1728 K; boiling points: V about 3680 K, Cr about 2944 K, Fe about 3134 K, Ni about 3186 K.
2. Vanadium has the highest melting point as well as the highest boiling point among these, because of strong metallic bonding involving its 3d³4s² electrons.

Teacher's Note:
1. Melting points rise to a maximum near the middle of the series (V/Cr), then fall; Mn is anomalously low.

Answer: (c) Cr > Mn > V > Ti

Explanation:
1. The second ionization enthalpy generally increases from Ti to Mn but Cr has higher value than Mn. This is because after the removal of first electron, Cr acquires a stable configuration of 3d⁵ and the removal of second electron is very difficult. Thus, the order for IE₂ is Cr > Mn > V > Ti.

Teacher's Note:
1. For IE₂ look at the configuration of M⁺: Cr⁺ is 3d⁵ (stable), while Mn⁺ is 3d⁵4s¹ and loses its 4s electron easily.

Answer: (d) all the three above

Explanation:
1. The large positive E° value for Mn³⁺/Mn²⁺ shows that Mn²⁺ is much more stable than Mn³⁺. This is due to the fact that Mn²⁺ has 3d⁵ configuration which is half-filled and hence, is very stable. Thus, the third ionization enthalpy of Mn will be very high. In fact, this is the reason that +3 state of Mn is of little importance.

Teacher's Note:
1. A large positive E°(M³⁺/M²⁺) means M³⁺ is easily reduced, i.e. M²⁺ is the stable state.

Answer: (a) Mn²⁺

Explanation:
1. μ_eff = \( \sqrt{n(n+2)} \) BM; 5.93 = \( \sqrt{n(n+2)} \)
2. or n(n + 2) = 35.16
3. or n = number of unpaired electrons = 5
4. ∴ Mn²⁺ ion (3d⁵) has magnetic moment of 5.93 B.M.

Teacher's Note:
1. Square the moment: 5.93² ≈ 35 = 5 × 7, so n = 5.

Answer: (c) Ti³⁺ (Z = 22)

Explanation:
1. Electronic configuration of Ti³⁺ is 3d¹. Thus Ti³⁺ contains an unpaired electron, so it will exhibit colour in aqueous solution due to d-d transition.

Teacher's Note:
1. Sc³⁺ and La³⁺ have d⁰/f⁰ and Lu³⁺ has f¹⁴ – no unpaired electrons, hence colourless.

Answer: (d) Pb⁺⁺

Explanation:
1. Pb⁺⁺, because it does not have vacant d-orbitals nor high nuclear charge and it does not belong to transition series.

Teacher's Note:
1. Ag⁺, Cu²⁺ and Cd²⁺ dissolve in excess NH₃ as ammine complexes; Pb²⁺ only gives Pb(OH)₂ precipitate.

Answer: (a) Ziegler-Natta catalyst

Explanation:
1. Ziegler-Natta catalyst [TiCl₄ + Al(CH₃)₃] is used for polymerization of olefin.

Teacher's Note:
1. Wilkinson's catalyst [(Ph₃P)₃RhCl] is for hydrogenation of alkenes; Raney Ni also for hydrogenation.

Answer: (a) TiO > VO > CrO > FeO

Explanation:
1. Basic character of oxides decreases from left to right in a period of periodic table.

Teacher's Note:
1. Along the period the M²⁺ ion gets smaller, so the oxide gets more covalent and less basic.

Answer: (c) Solution in test tube becomes green and lime water remains clear

Explanation:
1. Na₂SO₃ with dil. H₂SO₄ gives SO₂ gas which reacts with K₂Cr₂O₇ to give a green solution due to Cr₂(SO₄)₃. Since whole of the gas is consumed, the lime water will remain clear.
2. K₂Cr₂O₇ + 4H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 4H₂O + 3[O]
3. SO₂ + [O] + H₂O → H₂SO₄

Teacher's Note:
1. SO₂ could turn lime water milky, but here it is used up inside the test tube by the dichromate.

Answer: (b) N₂

Explanation:
1. (NH₄)₂Cr₂O₇ →(Δ) N₂ + 4H₂O + Cr₂O₃

Teacher's Note:
1. This "volcano" reaction is a convenient laboratory source of pure N₂.

Answer: (a) dilute H₂SO₄

Explanation:
1. KMnO₄ titrations are carried out only in presence of dilute H₂SO₄. HCl cannot be used as oxygen given out by KMnO₄ + HCl is partly used up to oxidize HCl to Cl₂. If HNO₃ is used, it itself acts as an oxidizing agent.

Teacher's Note:
1. The acid used must be neither a reducing agent (HCl) nor an oxidising agent (HNO₃).

Answer: (a) K₂MnO₄ and KMnO₄

Explanation:
1. 2MnO₂ + 4KOH + O₂ (Air) →(Fusion) 2K₂MnO₄ (Dark green, X) + 2H₂O
2. 3K₂MnO₄ + 4HCl →(Disproportionation) 2KMnO₄ (Intense purple, Y) + MnO₂ + 4KCl + 2H₂O

Teacher's Note:
1. Green = manganate (Mn +6), purple = permanganate (Mn +7).

Answer: (d) 5

Explanation:
1. 2KMnO₄ (2 moles) + 8H₂SO₄ + 10KBr →(Δ) 6K₂SO₄ + 2MnSO₄ + 5Br₂ (5 moles) + 8H₂O
2. Thus, 2 moles of KMnO₄ when heated with excess KBr in acidic medium produce 5 moles of Br₂.

Teacher's Note:
1. Electron count: 2 MnO₄⁻ accept 2 × 5 = 10 e⁻ = electrons from 10 Br⁻ → 5 Br₂.

Answer: (b) 16 and 2

Explanation:
1. In the balanced equation,
2. 2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 10CO₂ + 8H₂O
3. x = 16 and y = 2

Teacher's Note:
1. Balance electrons first (2 MnO₄⁻ take 10 e⁻ from 5 oxalate ions), then H⁺ and H₂O.

Answer: (d) Tm

Explanation:
1. Tm → Thulium → Atomic number 69.
2. It is a lanthanoid because lanthanoids are present in sixth period of periodic table from atomic number 57 (lanthanum) to atomic number 71 (lutetium).

Teacher's Note:
1. Tm (thulium) and Fm (fermium) look similar – learn which symbols end in -ium of the 5f series.

Answer: (d) 4f⁷ 5d¹ 6s²

Explanation:
1. The electronic configuration of ₆₄Gd = [Xe] 4f⁷ 5d¹ 6s²

Teacher's Note:
1. Half-filled 4f⁷ is extra stable, so Gd puts its eighth electron in 5d instead of 4f.

Answer: (b) +3, +4

Explanation:
1. Ce = [Xe] 4f² 5d⁰ 6s²; oxidation states shown are +3 and +4.

Teacher's Note:
1. Ce⁴⁺ has the noble gas configuration (4f⁰), which is why +4 is important for cerium.

Answer: (a) La³⁺

Explanation:
1. La³⁺ due to lanthanoid contraction has the largest ionic radius.

Teacher's Note:
1. Ionic radius falls steadily from La³⁺ (largest) to Lu³⁺ (smallest).

Answer: (a) negligible screening effect of f-orbitals

Explanation:
1. Lanthanoid contraction is the regular decrease in atomic and ionic radii of lanthanides with increase in atomic number. This is due to the imperfect shielding (or poor screening effect) of f-orbitals due to their diffused shapes which is unable to counterbalance the effect of the increased nuclear charge. Hence, the net result is a contraction in size of lanthanoids.

Teacher's Note:
1. The diffuse shape of 4f orbitals is the root cause; the rising nuclear charge acts only because it is not screened.

Answer: (b) In third transition series W : highest melting point :: Hg : lowest melting point

Explanation:
1. (b) is correct: tungsten has the highest melting point and mercury (a liquid) the lowest in the third (5d) series.
2. (a) is not correct because Cd is not considered as a transition element.
3. (c) is not correct because standard electrode potential of Cu (E°_Cu²⁺/Cu) = +ve and that of Cr (E°_Cr²⁺/Cr) = −ve.
4. (d) is not correct because Ti³⁺ (3d¹) is coloured while Ti⁴⁺ (3d⁰) is colourless.

Teacher's Note:
1. Zn, Cd and Hg (full d¹⁰ in atom and common ions) are all not regarded as transition metals.

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

Explanation:
1. R is the correct explanation of A.
2. MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻; equivalent mass = M/3 in both media.

Teacher's Note:
1. Only in strongly alkaline medium (with no reducing agent present) does MnO₄⁻ stop at MnO₄²⁻ (E = M).

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

Explanation:
1. R is the correct explanation of A.
2. 2Cu²⁺ + 4I⁻ → Cu₂I₂ + I₂

Teacher's Note:
1. This reaction is the basis of iodometric estimation of Cu²⁺.

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

Explanation:
1. Correct explanation. In lanthanoids, energy gap between 4f and 5d subshells is large while in case of actinoids, energy gap between 5f, 6d and 7s subshell is very small.

Teacher's Note:
1. When asked "why", always link to orbital energies: 5f, 6d and 7s are close, so more electrons can be used.

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

Explanation:
1. Higher enthalpies of atomisation are due to stronger metallic bonding on account of greater number of unpaired electrons present in them.
2. Thus R is the correct explanation of A.

Teacher's Note:
1. More unpaired d electrons → stronger interatomic (metallic) bonding → higher enthalpy of atomisation; the maximum is near the middle of each series.

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

Explanation:
1. Correct R. E° value for Cr³⁺/Cr²⁺ is negative (−0.41 V) whereas E° value for Mn³⁺/Mn²⁺ is positive (+1.57 V). Thus, Mn³⁺ can undergo reduction to form Mn²⁺ and hence is oxidizing.

Teacher's Note:
1. Cr²⁺ → Cr³⁺ gives stable t_2g³; Mn³⁺ → Mn²⁺ gives stable d⁵.

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

Explanation:
1. Correct explanation. Mercury has no partly or incompletely filled d-subshell in atomic state or common oxidation state of +2. Hence, it is not considered as a transition element.

Teacher's Note:
1. Hg (5d¹⁰ 6s²) and Hg²⁺ (5d¹⁰) both have a full d subshell.

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

Explanation:
1. Correct explanation. Mn²⁺ has 3d⁵ configuration which is half-filled and hence stable whereas Fe²⁺ has the configuration 3d⁶ and hence can lose one electron easily to form stable 3d⁵ configuration.

Teacher's Note:
1. Compare what each ion becomes on oxidation: Fe²⁺ → Fe³⁺ reaches d⁵, Mn²⁺ → Mn³⁺ leaves d⁵.

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

Explanation:
1. Correct A. Ti⁴⁺ is colourless (because it has 3d⁰ configuration, i.e., no unpaired 3d electron) whereas V⁴⁺ is coloured (because it has 3d¹ configuration, i.e., one unpaired electron is present in 3d).

Teacher's Note:
1. Same charge does not mean same colour: write the d configuration first (Ti⁴⁺ d⁰, V⁴⁺ d¹).

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

Explanation:
1. Correct explanation. Lanthanoids show limited number of oxidation states because energy gap between 4f and 5d subshells is large whereas actinoids show large number of oxidation states because of very small energy gap between 5f, 6d and 7s subshells.

Teacher's Note:
1. A statement can be a true fact and still not explain the assertion – check the cause, not just the truth.

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

Explanation:
1. Correct explanation. The most stable oxidation state of lanthanoids is +3. M²⁺ ions can lose electron to form M³⁺ and hence are reducing agents while M⁴⁺ ions can gain electron to form M³⁺ and hence are oxidizing agents.

Teacher's Note:
1. Rule for lanthanoids: anything other than +3 tends to return to +3 – +2 ions reduce, +4 ions oxidise.