CBSE Class 12 Chemistry The d and f Block Elements NEET JEE Questions Set 02

Single Correct Answer

Question 1. In which of the following ions, the colour is not due to d-d transition?
(a) Ti(H₂O)₆³⁺
(b) CoF₆³⁻
(c) CrO₄²⁻
(d) [Cu (NH₃)₄]²⁺
Answer:
Ans. (c)
In CrO₄²⁻, Cr is in +6 oxidation state. It has d⁰ configuration. The colour is due to charge transfer and not due to d–d transition.

Teacher's Note:
1. d⁰ oxo-anions (CrO₄²⁻, Cr₂O₇²⁻, MnO₄⁻) owe their intense colour to ligand-to-metal charge transfer.

Question 2. The bonds present in the structure of dichromate ion are (Kerala PET 2011)
(a) four equivalent Cr–O bonds only
(b) six equivalent Cr–O bonds and one O–O bond
(c) six equivalent Cr–O bonds and one Cr–Cr bond
(d) eight equivalent Cr–O bonds
(e) six equivalent Cr–O bonds and one Cr–O–Cr bond.
Answer:
Ans. (e)
Dichromate ion consists of two CrO₄ tetrahedra sharing one corner oxygen. Each Cr has three terminal oxygens, giving six equivalent terminal Cr–O bonds (163 pm), and the shared oxygen forms one Cr–O–Cr bridge (Cr–O = 179 pm, angle 126°).

Teacher's Note:
1. Draw it as two tetrahedra joined at a corner: 6 terminal O + 1 bridging O = 7 O atoms in Cr₂O₇²⁻.

Question 3. KMnO₄ can be prepared from K₂MnO₄ as per the reaction :
3MnO₄²⁻ + 2H₂O ⇄ 2MnO₄⁻ + MnO₂ + 4OH⁻
The reaction can go to completion by removing OH⁻ ions by adding (AIPMT 2013)
(a) HCl
(b) KOH
(c) CO₂
(d) SO₂
Answer:
Ans. (c)
The reaction can go to completion by removing OH⁻ ions by adding CO₂ as it reacts with OH⁻ ions to form CO₃²⁻ ions
4OH⁻ + 2CO₂ → 2CO₃²⁻ + 2H₂O

Teacher's Note:
1. HCl would be oxidised by KMnO₄ and SO₂ would reduce it; CO₂ removes OH⁻ without any redox side reaction.

Question 4. When manganous salt is fused with a mixture of KNO₃ and solid NaOH, the oxidation number of Mn changes from +2 to (West Bengal JEE 2010)
(a) +4
(b) +3
(c) +6
(d) +7
Answer:
Ans. (c)
Mn²⁺ + 2OH⁻ + (O) → MnO₂ + H₂O (alkaline oxidation of manganous salt)
MnO₂ + 2NaOH + (O) → Na₂MnO₄ + H₂O (O from KNO₃); Mn in Na₂MnO₄ is +6.

Teacher's Note:
1. Alkaline oxidative fusion of any manganese compound stops at manganate(VI), MnO₄²⁻ (green), not permanganate.

Question 5. The coloured spot of KMnO₄ on any article can be bleached by
(a) SO₂ + H⁺
(b) C₂O₄²⁻ + H⁺
(c) H₂O₂ + H⁺
(d) all of these
Answer:
Ans. (d)
KMnO₄ in presence of H⁺ ions can be reduced to colourless Mn²⁺ by all the given chemicals.

Teacher's Note:
1. Any reducing agent in acid medium decolourises permanganate, because Mn²⁺ (pale pink) is practically colourless in dilute solution.

Question 6. MnO₄⁻ reacts with Br⁻ in alkaline pH to give (Odisha JEE 2009)
(a) BrO₃⁻, MnO₂
(b) Br₂, MnO₄²⁻
(c) Br₂, MnO₂
(d) BrO⁻, MnO₄²⁻
Answer:
Ans. (a)
2KMnO₄ + KBr + H₂O → 2KOH + 2MnO₂ + KBrO₃

Teacher's Note:
1. Compare with iodide: in neutral/faintly alkaline medium permanganate oxidises I⁻ to IO₃⁻ and Br⁻ to BrO₃⁻, itself going to MnO₂.

Question 7. Amount of oxalic acid present in a solution can be determined by its titration with KMnO₄ solution in the presence of H₂SO₄. The titration gives unsatisfactory result when carried out the presence of HCl because HCl (AIEEE 2008)
(a) reduces permanganate to Mn²⁺
(b) oxidises oxalic acid to carbon dioxide and water
(c) gets oxidized by oxalic acid to chlorine
(d) furnishes H⁺ ions in addition to those from oxalic acid
Answer:
Ans. (a)
In presence of H₂SO₄, KMnO₄ oxidises oxalic acid to CO₂. In presence of HCl, KMnO₄ not only oxidises oxalic acid but also oxidises HCl to Cl₂ and itself it is reduced to Mn²⁺.

Teacher's Note:
1. Extra KMnO₄ is consumed by HCl, so the burette reading is too high; that is why only dilute H₂SO₄ is used for acidifying.

Question 8. KMnO₄ gets reduced to (AMU Engg. 2011)
(a) K₂MnO₄ in neutral medium
(b) MnO₂ in acidic medium
(c) Mn²⁺ in alkaline medium
(d) MnO₂ in neutral medium
Answer:
Ans. (d)
KMnO₄ is reduced to MnO₂ in neutral medium
2KMnO₄ + H₂O → 2MnO₂ + 2KOH + 3(O)
or MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻
In alkaline medium, it is reduced to K₂MnO₄ or MnO₂. In acidic medium, it is reduced to Mn²⁺.

Teacher's Note:
1. Remember the electrons gained: acidic 5 (Mn²⁺), neutral/weakly alkaline 3 (MnO₂), strongly alkaline 1 (MnO₄²⁻).

Question 9. The colour of KMnO₄ is due to (JEE Main 2015)
(a) L → M charge transfer transition
(b) σ → σ* transition
(c) M → L charge transfer transition
(d) d–d transition
Answer:
Ans. (a)
The deep purple colour of KMnO₄ is not due to d-d transition but due to charge transfer from O to Mn, i.e., from Ligand (L) to Metal (M) which reduces the oxidation state of Mn from +7 to +6 momentarily.

Teacher's Note:
1. Mn(VII) is d⁰, so a d–d transition is impossible; the electron moves from oxygen to the metal.

Question 10. In acidic medium, H₂O₂ changes Cr₂O₇²⁻ to CrO₅ which has two (–O–O–) bonds. Oxidation state of Cr in CrO₅ is (AIPMT 2014)
(a) +5
(b) +3
(c) +6
(d) −10
Answer:
Ans. (c)
CrO₅ has butterfly structure having two peroxo bonds (one Cr=O and two O–O groups bonded to Cr).
Oxidation state of peroxo oxygen = −1
∴ If oxidation state of Cr is x, then x + 4(−1) + 1(−2) = 0 or x = +6

Teacher's Note:
1. Never take −2 for every oxygen in peroxo compounds; doing so here gives the impossible value +10.

Question 11. The reaction of aqueous KMnO₄ with H₂O₂ in acidic conditions gives (AIPMT 2014)
(a) Mn⁴⁺ and O₂
(b) Mn²⁺ and O₂
(c) Mn²⁺ and O₃
(d) Mn⁴⁺ and MnO₂
Answer:
Ans. (b)
H₂O₂ is oxidized to H₂O and O₂ as under :
2KMnO₄ + 3H₂SO₄ + 5H₂O₂ → K₂SO₄ + 2MnSO₄ + 8H₂O + 5O₂
or 2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 8H₂O + 5O₂

Teacher's Note:
1. Against strong oxidants like MnO₄⁻, H₂O₂ acts as a reducing agent and its oxygen is released as O₂.

Question 12. The manganate and permanganate ions are tetrahedral due to (NEET 2019)
(a) the π-bonding involves overlap of d-orbitals of oxygen with d-orbitals of manganese
(b) the π-bonding involves overlap of p-orbitals of oxygen with d-orbitals of manganese
(c) there is no π-bonding
(d) the π-bonding involves overlap of p-orbitals of oxygen with p-orbitals of manganese
Answer:
Ans. (b)
MnO₄²⁻ (green manganate ion) and MnO₄⁻ (purple permanganate ion) are both tetrahedral, with one or two Mn–O⁻ and remaining Mn=O bonds.
In both the above ions, π-bonding takes place by overlap of p-orbitals of oxygen with d-orbitals of manganese.

Teacher's Note:
1. This pπ–dπ bonding is why all Mn–O bonds in each ion are equivalent (resonance) and the ions are tetrahedral.

Question 13. In order to oxidize a mixture of one mole of each of FeC₂O₄, Fe₂(C₂O₄)₃, FeSO₄ and Fe₂(SO₄)₃ in acidic medium, the number of moles of KMnO₄ required is (JEE Main 2019)
(a) 2
(b) 1.5
(c) 3
(d) 1
Answer:
Ans. (a)
Only oxidizable species are Fe²⁺ and C₂O₄²⁻ ions as Fe³⁺ and SO₄²⁻ ions are not oxidized.
Total Fe²⁺ = 2 moles (1 from FeC₂O₄ and 1 from FeSO₄)
Total C₂O₄²⁻ = 4 moles (1 from FeC₂O₄ and 3 from Fe₂(C₂O₄)₃)
2MnO₄⁻ + 16H⁺ + 10Fe²⁺ → 2Mn²⁺ + 8H₂O + 10Fe³⁺
∴ 2 moles Fe²⁺ will require KMnO₄ = \( \frac{2}{5} \) mole
2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 8H₂O + 10CO₂
∴ 4 moles of C₂O₄²⁻ will require KMnO₄ = \( \frac{2}{5}\times 4 = \frac{8}{5} \) moles
∴ Total moles of KMnO₄ required = \( \frac{2}{5}+\frac{8}{5} \) = 2 moles

Teacher's Note:
1. Equivalent method: electrons to be removed = 2 (Fe²⁺) + 4 × 2 (oxalate) = 10; each MnO₄⁻ takes 5, so 2 mol.

Question 14. On treating a compound with warm dil. H₂SO₄ gas X is evolved which turns acidified K₂Cr₂O₇ paper with dil. H₂SO₄ to a green compound Y. X and Y respectively are : (JEE Main 2021)
(a) X = SO₂, Y = Cr₂(SO₄)₃
(b) X = SO₂, Y = Cr₂O₃
(c) X = SO₃, Y = Cr₂O₃
(d) X = SO₃, Y = Cr₂(SO₄)₃
Answer:
Ans. (a)
3SO₂ (X) + K₂Cr₂O₇ + H₂SO₄ → Cr₂(SO₄)₃ (Y, green) + K₂SO₄ + H₂O

Teacher's Note:
1. SO₃ is not a reducing agent (S is already +6), so only SO₂ can turn orange dichromate paper green.

Question 15. The electronic configuration of Eu (Atomic No. 63), Gd (Atomic No. 64) and Tb (Atomic No. 65) are (NEET Phase I 2016)
(a) [Xe] 4f⁶5d¹6s², [Xe] 4f⁷5d¹6s² and [Xe] 4f⁸5d¹6s²
(b) [Xe] 4f⁷6s², [Xe] 4f⁷5d¹6s² and [Xe] 4f⁹6s²
(c) [Xe] 4f⁷6s², [Xe] 4f⁸6s² and [Xe] 4f⁸5d¹6s²
(d) [Xe] 4f⁶5d¹6s², [Xe] 4f⁷5d¹ and [Xe] 4f⁹6s²
Answer:
Ans. (b)
Eu (63) = [Xe] 4f⁷ 6s² (half-filled 4f); Gd (64) = [Xe] 4f⁷ 5d¹ 6s² (the extra electron goes to 5d to keep 4f⁷); Tb (65) = [Xe] 4f⁹ 6s².

Teacher's Note:
1. Only La, Ce, Gd and Lu carry a 5d¹ electron in the ground state; the rest are [Xe]4fⁿ6s².

Question 16. The atomic number of cerium (Ce) is 58. The correct electronic configuration of Ce³⁺ ion is (West Bengal JEE 2015)
(a) [Xe] 4f¹
(b) [Kr] 4f¹
(c) [Xe] 4f¹³
(d) [Kr] 4d¹
Answer:
Ans. (a)
Z = 54 is [Xe], Z = 56 is Ba = [Xe] 6s², Z = 57 (La) is [Xe] 5d¹ 6s² (electron enters 5d in place of expected 4f).
But in Z = 58 (Ce), electron enters 4f and 5d¹ also shifts to 4f. Hence, electronic configuration of Ce (58) is [Xe] 4f² 5d⁰ 6s².
∴ Ce³⁺ = [Xe] 4f¹.

Teacher's Note:
1. Lanthanoid ions lose the 6s (and 5d) electrons first; the core is always [Xe], never [Kr].

Question 17. The f-orbitals are half and completely filled respectively in lanthanide ions.
(Given : Atomic no. Eu 63, Sm 62, Tm 69, Tb 65, Yb 70, Dy 66) (JEE Main 2022)
(a) Eu²⁺ and Tm²⁺
(b) Sm²⁺ and Tm³⁺
(c) Tb⁴⁺ and Yb²⁺
(d) Dy³⁺ and Yb³⁺
Answer:
Ans. (c)
Tb (65) = [Xe] 4f⁹ 6s², so Tb⁴⁺ = [Xe] 4f⁷ (half-filled).
Yb (70) = [Xe] 4f¹⁴ 6s², so Yb²⁺ = [Xe] 4f¹⁴ (completely filled).
Eu²⁺ is 4f⁷ but Tm²⁺ is 4f¹³; Sm²⁺ is 4f⁶; Dy³⁺ is 4f⁹.

Teacher's Note:
1. Quick count: number of 4f electrons in Lnⁿ⁺ = Z − 54 − n (for these ions with no 5d electron).

Question 18. The electronic configuration of Gd²⁺ is (atomic no. of Gd is 64) (RE-AIPMT 2015, Karnataka CET 2015)
(a) [Xe] 4f⁷
(b) [Xe] 4f⁷ 5d¹
(c) [Xe] 4f⁸
(d) [Xe] 4f⁷ 5d¹ 6s²
Answer:
Ans. (b)
Gd (64) = [Xe] 4f⁷ 5d¹ 6s²
∴ Gd²⁺ = [Xe] 4f⁷ 5d¹

Teacher's Note:
1. The two 6s electrons are removed first; the stable half-filled 4f⁷ is left untouched.

Question 19. Because of lanthanoid contraction, which of the following pairs of elements have nearly same atomic radii? (Numbers in parenthesis are atomic numbers) (AIPMT 2015)
(a) Zr (40) and Hf (72)
(b) Zr (40) and Ta (73)
(c) Ti (22) and Zr (40)
(d) Zr (40) and Nb (41)
Answer:
Ans. (a)
Zr (40) and Hf (72) lie in the same group and Hf comes after La. Hence, Zr and Hf have nearly same size due to lanthanoid contraction.

Teacher's Note:
1. The pair must be in the same group, one in the 4d and one in the 5d series (Zr/Hf, Nb/Ta, Mo/W).

Question 20. The correct order of ionic radii of Y³⁺, La³⁺, Eu³⁺ and Lu³⁺ is (Kerala PMT 2007)
(a) Y³⁺ < La³⁺ < Eu³⁺ < Lu³⁺
(b) Lu³⁺ < Y³⁺ < Eu³⁺ < La³⁺
(c) Lu³⁺ < Eu³⁺ < La³⁺ < Y³⁺
(d) La³⁺ < Eu³⁺ < Lu³⁺ < Y³⁺
Answer:
Ans. (b)
Group 3: Sc³⁺ (75 pm), Y³⁺ (90 pm), La³⁺ (106 pm). Lanthanides: La³⁺ (106 pm), Eu³⁺ (95 pm), Lu³⁺ (86 pm).
Thus, Lu³⁺ < Y³⁺ < Eu³⁺ < La³⁺

Teacher's Note:
1. Because of lanthanoid contraction, Lu³⁺ ends up even smaller than Y³⁺ of the period above.

One or More Correct Answers

Question 21. Which of the following statements are correct with reference to the ferrous and ferric ions?
(a) Fe³⁺ gives brown colour with potassium ferricyanide
(b) Fe²⁺ gives blue precipitate with potassium ferricyanide
(c) Fe³⁺ gives red colour with potassium thiocyanate
(d) Fe²⁺ gives brown colour with ammonium thiocyanate.
Answer:
Ans. (a, b, c)
Fe³⁺ ions give a brown colouration (no precipitate) with K₃[Fe(CN)₆].
K₃[Fe(CN)₆] reacts with Fe²⁺ ions to give blue ppt. of Fe₃[Fe(CN)₆]₂ and potassium thiocyanate reacts with Fe³⁺ ions to give a red colour of Fe(CNS)₃.
Fe²⁺ gives no colour with thiocyanate, so (d) is wrong.

Teacher's Note:
1. Thiocyanate gives blood-red colour only with Fe³⁺; ferricyanide gives the deep blue precipitate only with Fe²⁺.

Question 22. Which of the following statements are correct when a mixture of NaCl and K₂Cr₂O₇ is gently warmed with conc. H₂SO₄?
(a) a deep red vapour is evolved
(b) The vapour when passed into NaOH solution gives yellow solution of Na₂CrO₄
(c) chlorine gas is evolved
(d) chromyl chloride is formed
Answer:
Ans. (a, b, d)
Deep red vapours of chromyl chloride are formed which on passing through NaOH solution give a yellow solution of Na₂CrO₄.
4NaCl + K₂Cr₂O₇ + 6H₂SO₄ → 2CrO₂Cl₂ + 2KHSO₄ + 4NaHSO₄ + 3H₂O; CrO₂Cl₂ + 4NaOH → Na₂CrO₄ + 2NaCl + 2H₂O

Teacher's Note:
1. Chlorine is not the product here; Cr stays at +6 in CrO₂Cl₂, so no redox occurs.

Question 23. Reduction of the metal centre in aqueous permanganate ion involves
(a) 3 electrons in neutral medium
(b) 5 electrons in neutral medium
(c) 3 electrons in alkaline medium
(d) 5 electrons in acidic medium
Answer:
Ans. (a, c, d)
In neutral medium: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻
In alkaline medium: MnO₄⁻ + e⁻ → MnO₄²⁻
But MnO₄²⁻ is further reduced to MnO₂ as shown below: MnO₄²⁻ + 2H₂O + 2e⁻ → MnO₂ + 4OH⁻
Complete reaction is MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻, which is the same as that for neutral medium.
In acidic medium: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Hence, the number of electrons involved in the reduction of metal centre in aqueous MnO₄⁻ ion in neutral, alkaline and acidic media are 3, 3 and 5 respectively.

Teacher's Note:
1. Count electrons from the change in oxidation number of Mn: +7 → +4 is 3 electrons, +7 → +2 is 5 electrons.

Passage-Based Questions

Teacher's Note:
1. The maximum number of oxidation states occurs in the middle of a series; in the 4d row Ru reaches +8 (RuO₄), like Os in the 5d row.

Teacher's Note:
1. d¹ and d⁹ ions have the same number of unpaired electrons (one electron or one ‘hole’), so they absorb similarly.

Question 26. Comprehension 2: Potassium permanganate is prepared from the mineral pyrolusite, MnO₂. Its crystals have deep purple colour. It acts as an oxidizing agent in the neutral, alkaline as well as acidic medium. In acidic medium, it is used in volumetric analysis for estimation of ferrous salts, oxalates etc. The titrations are carried out in presence of H₂SO₄. However, before using it as a titrant, it is first standardized with standard oxalic acid solution or Mohr salt solution. In one of the experiments on titration, 13.4 g of dry pure sodium oxalate (molar mass = 134 g mol⁻¹) was dissolved in 100 mL of distilled water and then 100 mL of 2M H₂SO₄ were added. The solution was cooled to 25-30°C. Now to this solution, 0.1 M KMnO₄ solution was added till a very faint pink colour persisted.
When pyrolusite is fused with KOH and KClO₃, we get
(a) KMnO₄
(b) K₂MnO₄
(c) Both KMnO₄ and K₂MnO₄
(d) None of these
Answer:
Ans. (b)
3MnO₂ + 6KOH + KClO₃ → 3K₂MnO₄ + KCl + 3H₂O

Teacher's Note:
1. Fusion gives only green manganate; conversion to permanganate needs a separate step (electrolytic oxidation, Cl₂, O₃ or CO₂ disproportionation).

Matrix Match Questions

Teacher's Note:
1. Cu⁺ disproportionates because the very negative hydration enthalpy of Cu²⁺ more than makes up for the second ionization enthalpy.

Teacher's Note:
1. Cu, Ag and Au (coinage metals) all have (n−1)d¹⁰ns¹; Zn, Cd and Hg have (n−1)d¹⁰ns².

Integer and Numerical Answer Questions

Question 29. The coefficient of H₂SO₄ on balancing the equation
K₂Cr₂O₇ + H₂SO₄ + KI → K₂SO₄ + Cr₂(SO₄)₃ + H₂O + I₂ is
Answer:
Ans. 7
The balanced equation is : K₂Cr₂O₇ + 7H₂SO₄ + 6KI → 4K₂SO₄ + Cr₂(SO₄)₃ + 3I₂ + 7H₂O.

Teacher's Note:
1. Balance electrons first (6 e⁻ for Cr₂O₇²⁻, so 6 KI), then K, then SO₄; H₂O balances H and O last.

Question 30. The oxidation number of Mn in the product of alkaline oxidative fusion of MnO₂ is (IIT 2009)
Answer:
Ans. 6
2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
Ox. state of Mn in K₂MnO₄ : 2 + x − 8 = 0 or x = +6

Teacher's Note:
1. The product of fusion is green potassium manganate, not permanganate.

Question 31. Consider the following list of reagents :
Acidified K₂Cr₂O₇, alkaline KMnO₄, CuSO₄, H₂O₂, Cl₂, O₃, FeCl₃, HNO₃ and Na₂S₂O₃
The total number of reagents that can oxidize aqueous iodide to iodine is (JEE Advanced 2014)
Answer:
Ans. 7
6KI + K₂Cr₂O₇ + 7H₂SO₄ → 4K₂SO₄ + Cr₂(SO₄)₃ + 3I₂ + 7H₂O
2CuSO₄ + 4KI → Cu₂I₂ + 2K₂SO₄ + I₂
H₂O₂ + 2KI → I₂ + 2KOH
Cl₂ + 2KI → 2KCl + I₂
O₃ + 2KI + H₂O → 2KOH + I₂ + O₂
2FeCl₃ + 2KI → 2FeCl₂ + 2KCl + I₂
4HNO₃ + 2KI → 2KNO₃ + 2NO₂ + I₂ + 2H₂O
Total = 7. (Alkaline KMnO₄ oxidises I⁻ to IO₃⁻, not I₂, and Na₂S₂O₃ is a reducing agent.)

Teacher's Note:
1. Read ‘to iodine’ carefully: alkaline permanganate takes iodide all the way to iodate.

Question 32. In neutral or faintly alkaline solution, 8 moles of permanganate anion quantitatively oxidize thiosulphate anions to produce X moles of sulphur containing product. The magnitude of X is (JEE Advanced 2016)
Answer:
Ans. 6
8MnO₄⁻ + 3S₂O₃²⁻ + H₂O → 8MnO₂ + 6SO₄²⁻ + 2OH⁻
∴ Moles of SO₄²⁻ formed from 8 moles of MnO₄⁻ = 6

Teacher's Note:
1. Electron check: 8 MnO₄⁻ × 3 = 24 e⁻; each S₂O₃²⁻ → 2SO₄²⁻ gives 8 e⁻, so 3 thiosulphate → 6 sulphate.

Question 33. Calculate the number of 4f electrons present in Gd²⁺ (Atomic number = 64). (JEE Main 2021)
Answer:
Ans. 7
Electronic configuration of Gd = [Xe] 4f⁷ 5d¹ 6s²
Electronic configuration of Gd²⁺ = [Xe] 4f⁷ 5d¹
∴ Number of 4f electrons in Gd²⁺ = 7

Teacher's Note:
1. Removing the two 6s electrons leaves the stable half-filled 4f⁷ (plus 5d¹) unchanged.

Assertion–Reason Questions

Question 34. Statement-1: CeO₂ can be used for oxidation of aldehydes and ketones.
Statement-2: Aqueous solution of EuSO₄ is a strong reducing agent. (JEE Main 2021)
(a) Both statement 1 and 2 are true.
(b) Both statements 1 and 2 are false.
(c) Statement 1 is true but statement 2 is false.
(d) Statement 1 is false but statement 2 is true.
Answer:
Ans. (a)
Ce(IV) is a strong oxidising agent (Ce⁴⁺ → Ce³⁺, E° = +1.74 V), so CeO₂ can oxidise aldehydes and ketones.
Eu²⁺ (4f⁷) readily changes to the common +3 state, so EuSO₄ solution is a strong reducing agent. Both statements are true.

Teacher's Note:
1. Lanthanoids in unusual states (+4 or +2) tend to return to +3: Ce⁴⁺ oxidises, Eu²⁺ and Yb²⁺ reduce.

Question 35. Statement-1: O₂, Cu²⁺ and Fe³⁺ are weakly attracted by magnetic field and are magnetised in the same direction as magnetic field.
Statement-2: NaCl and H₂O are weakly magnetised in opposite direction to magnetic field. (JEE Main 2022)
(a) Both statement 1 and 2 are true.
(b) Both statements 1 and 2 are false.
(c) Statement 1 is true but statement 2 is false.
(d) Statement 1 is false but statement 2 is true.
Answer:
Ans. (a)
O₂, Cu²⁺ and Fe³⁺ have 2, 1 and 5 unpaired electrons respectively. So they are paramagnetic species. Hence, they are attracted by the magnetic field.
NaCl and H₂O are diamagnetic species. So they are repelled by the magnetic field.

Teacher's Note:
1. Paramagnetic substances are magnetised along the field; diamagnetic ones are magnetised opposite to it and are weakly repelled.

Question 36. Assertion: The second ionization energies of ₂₃V, ₂₄Cr and ₂₅Mn are in the order V < Cr < Mn.
Reason: Ionization energies show a regular increase along a period with increase of atomic number.
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (d)
Correct A. The correct order of IE₂ is ₂₃V < ₂₄Cr > ₂₅Mn.
Correct R. Cr after losing one electron has 3d⁵ stable configuration from which removal of second electron is difficult; ionization energies do not increase regularly in the transition series.

Teacher's Note:
1. Irregularities in ionization enthalpy across the 3d series come from extra stability of d⁵ and d¹⁰ configurations.

Question 37. Assertion: Mn atom loses ns electrons first during ionization as compared to (n−1) d electrons.
Reason: The effective nuclear charge experienced by (n−1) d electrons is greater than that by ns electrons.
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
R is the correct explanation of A. Once the 3d orbitals are occupied, the 4s electrons are held less firmly and are removed first.

Teacher's Note:
1. Fill 4s before 3d, but remove 4s before 3d: Mn²⁺ is 3d⁵, not 3d³4s².

Question 38. Assertion: E° for Mn³⁺/Mn²⁺ is more positive than for Cr³⁺/Cr²⁺.
Reason: The third ionization energy of Mn is larger than that of Cr. (AIIMS 2006)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
R is the correct explanation of A. Removing the third electron from Mn breaks the stable 3d⁵ configuration of Mn²⁺, so Mn³⁺ is readily reduced to Mn²⁺ (E° = +1.57 V), while Cr³⁺ (t_2g³) is stable (E° = −0.41 V).

Teacher's Note:
1. A large third ionization enthalpy means the +3 state is hard to form and easily reduced, i.e. a more positive E°_{M³⁺/M²⁺}.

Question 39. Assertion: Mercury vapour have shining silvery appearance.
Reason: Mercury is a metal with shining silvery appearance. (AIIMS 2012)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
(e) If assertion is false, but reason is true.
Answer:
Ans. (e)
Correct A. Mercury vapours are invisible and no metallic bonding is possible in vapour state.
R is true: mercury is a liquid metal with shining silvery appearance.

Teacher's Note:
1. Lustre is a property of the bulk metal (free electrons); isolated Hg atoms in the vapour show no metallic appearance.