CBSE Class 12 Chemistry The d and f Block Elements NEET JEE Questions Set 01

Single Correct Answer

Question 1. Which of the following transition metals of 3d series has the lowest melting point? (Kerala PMT 2015)
(a) Ti (Z = 22)
(b) V (Z = 23)
(c) Cr (Z = 24)
(d) Mn (Z = 25)
(e) Fe (Z = 26)
Answer:
Ans. (d)
In the first transition series, Mn has abnormally low melting point. This is due to the fact that Mn has exactly half-filled d-orbitals. As a result, electronic configuration is very stable, i.e., electrons are held tightly by the nucleus so that the delocalization is less and the metallic bond is much weaker than the elements on its either side.

Teacher's Note:
1. Melting point rises to a maximum near the middle of the series (Cr, V) and then falls; Mn is the well-known dip in this trend.

Question 2. Which one of the following ions is the most stable in aqueous solution? (At. No. Ti = 22, V = 23, Cr = 24, Mn = 25) (AIPMT 2007)
(a) Mn²⁺
(b) Cr³⁺
(c) V³⁺
(d) Ti³⁺
Answer:
Ans. (b)
Out of the given species, Cr³⁺ has the highest negative reduction potential. Hence, it cannot be reduced to Cr²⁺ and, therefore, is the most stable ion in aqueous solution.
Alternatively, Mn³⁺ = [Ar] 3d⁴, Cr³⁺ = [Ar] 3d³, V³⁺ = [Ar] 3d², Ti³⁺ = [Ar] 3d¹.
In Cr³⁺, all three d electrons enter into lower energy t_2g orbitals. The lowering of energy is maximum and hence stability is maximum.

Teacher's Note:
1. A half-filled t_2g level (t_2g³) gives Cr³⁺ the largest crystal field stabilisation, which is why chromium(III) is so common.

Question 3. Four successive members of the first row transition elements are listed below with atomic numbers. Which one of them is expected to have the highest E°_{M³⁺/M²⁺} value? (JEE Main 2013)
(a) Co (Z = 27)
(b) Cr (Z = 24)
(c) Mn (Z = 25)
(d) Fe (Z = 26)
Answer:
Ans. (a)
Highest E°_{M³⁺/M²⁺} value means highest reduction potential for M³⁺ + e⁻ → M²⁺, i.e., the case in which M³⁺ is reduced most easily to M²⁺, i.e., M³⁺ ion is less stable than M²⁺ ion.
Generally, standard reduction potential increases from left to right along the period in d-block but there are some exceptions. Cr³⁺ is more stable than Cr²⁺. Fe³⁺ (3d⁵) is more stable than Fe²⁺. Mn²⁺ (3d⁵) is more stable than Mn³⁺ but Co²⁺ ions are still more stable than Co³⁺ in aqueous solution as Co³⁺ are easily reduced to Co²⁺ by H₂O. Hence Co³⁺/Co²⁺ has highest reduction potential.
(E°_{Cr³⁺/Cr²⁺} = −0.41 V, E°_{Mn³⁺/Mn²⁺} = +1.57 V, E°_{Fe³⁺/Fe²⁺} = +0.77 V, E°_{Co³⁺/Co²⁺} = +1.97 V)

Teacher's Note:
1. Learn the four M³⁺/M²⁺ values in order: Cr (−0.41) < Fe (+0.77) < Mn (+1.57) < Co (+1.97) V.

Question 4. Four successive members of the first series of the transition metals are listed below. For which one of them, the standard potential (E°_M²⁺/M) value has a positive sign? (AIPMT Main 2012)
(a) Cu (Z = 29)
(b) Fe (Z = 26)
(c) Co (Z = 27)
(d) Ni (Z = 28)
Answer:
Ans. (a)
Copper does not liberate H₂ gas from acids and thus shows positive value of E° (E° for Cu²⁺/Cu is +0.34 V). For other metals, E° value is negative.

Teacher's Note:
1. Cu is the only 3d metal with a positive E°_M²⁺/M; its high atomisation enthalpy and low hydration enthalpy are not balanced.

Question 5. In context with the transition elements, which of the following statements is incorrect? (AIEEE 2009)
(a) In the highest oxidation states, the transition metals show basic character and form cationic complexes.
(b) In the highest oxidation states of the first five transition elements (Sc to Mn), all the 4s and 3d electrons are used for bonding.
(c) Once the d⁵ configuration is exceeded, the tendency to involve all the 3d electrons in bonding decreases.
(d) In addition to the normal oxidation states, the zero oxidation state is also shown by these elements in complexes.
Answer:
Ans. (a)
In the highest oxidation state, transition metals show acidic character and form anionic complexes.

Teacher's Note:
1. High oxidation state means high covalent/acidic character: Mn₂O₇ and CrO₃ are acidic and form anions like MnO₄⁻, CrO₄²⁻.

Question 6. For the four successive transition elements (Cr, Mn, Fe and Co), the stability of +2 oxidation state will be there in which of the following order? (AIPMT Prelim 2011)
(a) Cr > Mn > Co > Fe
(b) Mn > Fe > Cr > Co
(c) Fe > Mn > Co > Cr
(d) Co > Mn > Fe > Cr
Answer:
Ans. (b)
Mn²⁺ (d⁵) gets highest stabilization due to half filled configuration. In case of Fe²⁺ (d⁶), addition of one extra electron in the subshell destabilizes it. Addition of two electrons in case of Co²⁺ (d⁷) destabilizes it more. Cr²⁺ (d⁴) has one vacant subshell. Fe²⁺ gets more stabilization as compared to Cr²⁺ through exchange energy. So, the correct order is Mn > Fe > Cr > Co.

Teacher's Note:
1. Cross-check with E°_{M³⁺/M²⁺}: Cr²⁺ is a strong reducing agent, so its +2 state is the least stable among Cr, Mn and Fe.

Question 7. The titanium (atomic number 22) compound that does not exist is (Kerala PMT 2011)
(a) TiO
(b) TiO₂
(c) K₂TiF₆
(d) TiCl₃
(e) K₂TiO₄
Answer:
Ans. (e)
Oxidation states will be: TiO (+2), TiO₂ (+4), K₂TiF₆ (+4), TiCl₃ (+3), K₂TiO₄ (+6).
Ti shows oxidation states of +2, +3 and +4 only. So K₂TiO₄ does not exist.

Teacher's Note:
1. The maximum oxidation state of an early transition metal equals the total number of (n−1)d + ns electrons; for Ti (3d²4s²) that is +4.

Question 8. Out of TiF₆²⁻, CoF₆³⁻, Cu₂Cl₂ and NiCl₄²⁻ (Z of Ti = 22, Co = 27, Cu = 29, Ni = 28), the colourless species are (AIPMT 2009)
(a) Cu₂Cl₂ and NiCl₄²⁻
(b) TiF₆²⁻ and Cu₂Cl₂
(c) CoF₆³⁻ and NiCl₄²⁻
(d) TiF₆²⁻ and CoF₆³⁻
Answer:
Ans. (b)
Cu₂Cl₂ (Cu⁺ = 3d¹⁰), TiF₆²⁻ (Ti⁴⁺ = 3d⁰). With no unpaired d electron, no d–d transition is possible, so both are colourless.

Teacher's Note:
1. First find the oxidation state, then the d count: d⁰ and d¹⁰ ions are colourless (unless charge transfer occurs).

Question 9. The catalytic activity of transition metals and their compounds is ascribed mainly to (AIPMT Main 2012)
(a) their ability to adopt variable oxidation states
(b) their chemical reactivity
(c) their magnetic behaviour
(d) their unfilled d-orbitals
Answer:
Ans. (a)
The catalytic activity of transition metals and their compounds is ascribed mainly to their ability to adopt variable oxidation states due to which they easily absorb and re-emit wide range of energies to provide the necessary activation energy.
Also, because of the variable oxidation states, they easily combine with one of the reactants to form an intermediate which reacts with the second reactant to form the final products.

Teacher's Note:
1. Classic example: Fe³⁺/Fe²⁺ catalysing the reaction between I⁻ and S₂O₈²⁻ by switching oxidation state.

Teacher's Note:
1. K₂Cr₂O₇ is coloured but diamagnetic (d⁰); its colour comes from charge transfer, so ‘coloured’ alone does not imply paramagnetic.

Question 11. The value of ‘spin only’ magnetic moment for one of the following configurations is 2.84 BM. The correct one is
(a) d⁴ (in strong ligand field)
(b) d⁴ (in weak ligand field)
(c) d³ (in weak as well as strong fields)
(d) d⁵ (in strong ligand field)
Answer:
Ans. (a)
\( \mu = \sqrt{n(n+2)} \) = 2.84 BM means n = 2. Only d⁴ in strong field can have two unpaired electrons (t_2g⁴: ↑↓ ↑ ↑).

Teacher's Note:
1. Memorise: 1 unpaired = 1.73, 2 = 2.83, 3 = 3.87, 4 = 4.90, 5 = 5.92 BM.

Question 12. Which of the following statement(s) is/are correctly applicable to all the transition elements with atomic numbers 21-29?
1. The lowest oxidation state is +1.
2. The 4s orbital is completely filled in the ground state.
3. The 3d orbital is incompletely filled in the ground state.
4. The ion in the +2 oxidation state is paramagnetic. (IAS Prelim 2010)
(a) 1, 2 and 3
(b) 3 and 4 only
(c) 4 only
(d) 3 only
Answer:
Ans. (c)
Lowest oxidation state is +1 only in case of Cu. Hence, (1) is wrong.
4s orbital is not completely filled in Cu and Cr. Hence, (2) is wrong.
3d orbital is not incompletely filled in case of Cu. Hence, (3) is wrong.
Ions in +2 oxidation state of all elements (21-29) have unpaired electrons and, therefore, paramagnetic. Hence, (4) is correct.

Teacher's Note:
1. ‘All elements’ statements are best tested against the exceptions Cr (3d⁵4s¹) and Cu (3d¹⁰4s¹).

Question 13. Which of the following arrangements does not represent the correct order of the property stated against it? (JEE Main 2013)
(a) Sc < Ti < Cr < Mn : number of oxidation states
(b) V²⁺ < Cr²⁺ < Mn²⁺ < Fe²⁺ : paramagnetic behaviour
(c) Ni²⁺ < Co²⁺ < Fe²⁺ < Mn²⁺ : ionic size
(d) Co³⁺ < Fe³⁺ < Cr³⁺ < Sc³⁺ : stability in aqueous solution
Answer:
Ans. (b)
(a) Number of oxidation states increases along a period. Hence, (a) is correct.
(b) V = 3d³ 4s², V²⁺ = 3d³ = 3 unpaired e⁻; Cr = 3d⁵ 4s¹, Cr²⁺ = 3d⁴ = 4 unpaired e⁻; Mn = 3d⁵ 4s², Mn²⁺ = 3d⁵ = 5 unpaired e⁻; Fe = 3d⁶ 4s², Fe²⁺ = 3d⁶ = 4 unpaired e⁻.
(c) Ionic radii (M²⁺) decrease along the period, so the size order Ni²⁺ < Co²⁺ < Fe²⁺ < Mn²⁺ is correct.
(c) Ionic radii (M²⁺) decrease along the period. Hence, (c) is correct.
(d) Stability of M³⁺ ion in aqueous solution increases as E°_{M³⁺/M²⁺} values decrease. Hence, (d) is correct.

Teacher's Note:
1. Mn²⁺ (d⁵) always has the maximum number of unpaired electrons (5) among the 3d M²⁺ ions; Fe²⁺ drops back to 4.

Question 14. The highest value of the spin only magnetic moment (in BM) among all the transition metal complexes is : (JEE Main 2019)
(a) 5.92
(b) 3.87
(c) 6.93
(d) 4.90
Answer:
Ans. (a)
\( \mu = \sqrt{n(n+2)} \) BM, n = number of unpaired electrons.
Maximum number of unpaired electrons = 5 (e.g. in Mn²⁺)
∴ Maximum value of \( \mu = \sqrt{5(5+2)} = \sqrt{35} \) = 5.92 BM

Teacher's Note:
1. A d subshell can hold at most 5 unpaired electrons (d⁵ high spin), so no spin-only value can exceed 5.92 BM.

Question 15. Basic properties of TiO₂, ZrO₂ and HfO₂ are in the order :
(a) TiO₂ < ZrO₂ < HfO₂
(b) ZrO₂ < HfO₂ < TiO₂
(c) HfO₂ < TiO₂ < ZrO₂
(d) TiO₂ < HfO₂ < ZrO₂
Answer:
Ans. (a)
Ti, Zr and Hf lie in Group 4. The metal oxides are basic and their basic character increases down the group.

Teacher's Note:
1. As in main groups, basic character of oxides of the same oxidation state increases down a group with increasing metallic character.

Question 16. Among the oxides, Mn₂O₇ (I), V₂O₃ (II), V₂O₅ (III), CrO (IV) and Cr₂O₃ (V), the basic oxides are (Kerala PMT 2010)
(a) I and II
(b) II and III
(c) III and IV
(d) II and IV
(e) III and V
Answer:
Ans. (d)
Generally, the oxides in the lower oxidation states of metals are basic. V₂O₃ (+3) and CrO (+2) are basic; Mn₂O₇ is acidic, V₂O₅ and Cr₂O₃ are amphoteric.

Teacher's Note:
1. Rule of thumb: low oxidation state oxide → basic, intermediate → amphoteric, highest → acidic (e.g. CrO, Cr₂O₃, CrO₃).

Question 17. Nature of the oxides of V₂O₃ and CrO respectively is (JEE Main 2021)
(a) Amphoteric and Basic
(b) Basic and Basic
(c) Basic and amphoteric
(d) Acidic and acidic
Answer:
Ans. (b)
They are in the lowest oxidation state (+3 and +2) and hence basic.

Teacher's Note:
1. For vanadium the change is basic V₂O₃ → less basic V₂O₄ → amphoteric V₂O₅.

Question 18. K₂Cr₂O₇ reacts with NH₄Cl in presence of H₂SO₄. The product formed is
(a) chromyl chlorate with green vapour
(b) chromous chloride with white vapour
(c) chromous chloride with blue vapour
(d) chromyl chloride with deep red colour
Answer:
Ans. (d)
A chloride heated with K₂Cr₂O₇ and conc. H₂SO₄ gives deep red vapours of chromyl chloride (chromyl chloride test):
K₂Cr₂O₇ + 4NH₄Cl + 3H₂SO₄ → 2CrO₂Cl₂ + K₂SO₄ + 2(NH₄)₂SO₄ + 3H₂O

Teacher's Note:
1. Chromyl chloride (CrO₂Cl₂) is the basis of the test for chloride ion; the red vapours give yellow Na₂CrO₄ with NaOH.

Question 19. When sulphur dioxide is passed through an acidified K₂Cr₂O₇ solution, the oxidation state of sulphur is changed from (Karnataka CET 2008)
(a) +4 to +6
(b) +6 to +4
(c) +4 to 0
(d) +4 to +2
Answer:
Ans. (a)
K₂Cr₂O₇ + 4H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 4H₂O + 3(O)
[SO₂ + (O) + H₂O → H₂SO₄] × 3
K₂Cr₂O₇ + H₂SO₄ + 3SO₂ → K₂SO₄ + Cr₂(SO₄)₃ + H₂O
In SO₂, ox. state of S = +4. In SO₄²⁻, ox. state of S = +6.

Teacher's Note:
1. Dichromate is the oxidant, so SO₂ must be oxidised; this rules out every option in which the oxidation state of S falls.

Question 20. How many moles of iodine are liberated when 1 mole of potassium dichromate reacts with potassium iodide? (AFMC 2010)
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
Ans. (c)
K₂Cr₂O₇ + 4H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 4H₂O + 3(O)
[2KI + H₂SO₄ + (O) → K₂SO₄ + I₂ + H₂O] × 3
K₂Cr₂O₇ + 7H₂SO₄ + 6KI → 4K₂SO₄ + Cr₂(SO₄)₃ + 7H₂O + 3I₂

Teacher's Note:
1. Quick electron count: Cr₂O₇²⁻ gains 6 electrons (2Cr: +6 → +3), and each I₂ needs 2 electrons removed, so 6/2 = 3 I₂.

One or More Correct Answers

Question 21. Transition elements have greater tendency to form complexes because
(a) They have vacant d orbitals
(b) They have large size
(c) They have large charge/size ratio
(d) They have two electrons in their outermost shells.
Answer:
Ans. (a, c)
Transition metal ions are small and highly charged (large charge/size ratio) and have vacant d-orbitals of suitable energy to accept lone pairs from ligands. Hence (a) and (c) are correct.
Their size is small, not large, and having two outer s electrons is not the reason for complex formation.

Teacher's Note:
1. Complex-forming ability needs both a strongly polarising cation and empty orbitals to receive ligand electron pairs.

Question 22. Which of the following show oxidation state of +4?
(a) Ce
(b) Ac
(c) Th
(d) U
Answer:
Ans. (a, c, d)
Ce shows +4 (Ce⁴⁺, 4f⁰), Th shows +4 as its common oxidation state and U shows +3, +4, +5 and +6.
Ac shows only +3 oxidation state.

Teacher's Note:
1. Actinium, like lanthanum, shows only the +3 state; the wide range of oxidation states begins with Th.

Question 23. Potassium manganate (K₂MnO₄) is formed when
(a) Chlorine is passed through aqueous KMnO₄ solution.
(b) Manganese dioxide is fused with potassium hydroxide in air.
(c) Formaldehyde reacts with potassium permanganate in presence of strong alkali.
(d) Potassium permanganate reacts with H₂SO₄.
Answer:
Ans. (b, c)
(b) 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O (fusion in air).
(c) KMnO₄ in presence of KOH forms K₂MnO₄ and oxygen produced oxidizes HCHO to HCOOH and then to CO₂ and H₂O.

Teacher's Note:
1. Cl₂ converts manganate to permanganate (the reverse change), and with conc. H₂SO₄ KMnO₄ gives Mn₂O₇.

Passage-Based Questions

Teacher's Note:
1. Sc starts the 3d series (largest in that row), while La and Hg belong to the 5d series; so the answer must be the 3d element at the right.

Teacher's Note:
1. Second ionization of Cr breaks into the stable 3d⁵ of Cr⁺, so Cr shows a peak in IE₂.

Matrix Match Questions

Teacher's Note:
1. The equivalent weight of an oxidant depends on the medium, because the number of electrons it accepts changes with the product formed.

Teacher's Note:
1. Here the free (high-spin) ions are meant; with the value written as √n(n+2), just read off n from the number under the root.

Integer and Numerical Answer Questions

Question 28. The highest oxidation state shown by any transition element is
Answer:
Ans. 8
The highest oxidation state shown by the transition element is +8 viz. Ru in 4d series and Os in 5d series (RuO₄, OsO₄).

Teacher's Note:
1. In the 3d series the maximum is only +7 (Mn in MnO₄⁻); +8 needs the heavier Ru and Os.

Question 29. Oxidation state of Cr in CrO₅ is
Answer:
Ans. 6
CrO₅ is a peroxo compound, CrO(O₂)₂, with one oxo oxygen (−2) and two peroxo groups (four O, each −1) attached to Cr.
i.e., x + 4(−1) + (−2) = 0 or x = +6.

Teacher's Note:
1. The blue CrO₅ formed in the chromate–H₂O₂ test still has Cr in +6; no element in group 6 can exceed +6.

Question 30. The number of metals collectively called ‘platinum metals’ in the transition elements is
Answer:
Ans. 6
The elements of 2nd and 3rd triads of Groups 8, 9 and 10 (i.e., Ru, Rh, Pd; Os, Ir, Pt) are collectively called platinum metals: 6.

Teacher's Note:
1. Fe, Co and Ni (the first triad) are excluded; the platinum group is the 4d and 5d members of groups 8–10.

Question 31. The number of electrons present in the 4f-subshell of Terbium (atomic no. = 65) is
Answer:
Ans. 9
Electronic configuration of Terbium (Z = 65) is [Xe]⁵⁴ 4f⁹ 5d⁰ 6s², so 4f has 9 electrons.

Teacher's Note:
1. Tb does not keep a 5d¹ electron like Gd; it is 4f⁹6s².

Assertion–Reason Questions

Question 32. Statement-1: Ti(III) salts are coloured whereas Ti(IV) salts are white.
Statement-2: Ti(III) is less stable than Ti(IV).
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation of Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation of Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (b)
Correct explanation. Ti(III) salts have one unpaired electron in d-subshell (3d¹) whereas Ti(IV) salts have no electron in the d-subshell (3d⁰).

Teacher's Note:
1. Colour depends on the presence of d electrons for d–d transitions, not on the relative stability of the oxidation states.

Question 33. Statement-1: The metals of 4d and 5d transition series have greater enthalpies of atomisation than the corresponding elements of the 3d series.
Statement-2: The metal-metal bond in 4d and 5d series are stronger than those in the 3d series.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation of Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation of Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (a)
Statement-2 is the correct explanation of Statement-1. The larger 4d and 5d orbitals overlap better, giving stronger metal–metal bonding.

Teacher's Note:
1. This is why heavy transition metals (W, Os, Re) have very high melting points and form metal–metal bonded compounds more readily.

Question 34. Statement-1: MnO is basic whereas Mn₂O₇ is acidic.
Statement-2: Higher the oxidation state of a transition metal in its oxide, greater is the acidic character.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation of Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation of Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (a)
Statement-2 is the correct explanation of Statement-1. In MnO, Mn is +2 (ionic, basic oxide); in Mn₂O₇, Mn is +7 (covalent, acidic oxide).

Teacher's Note:
1. Mn₂O₇ dissolves in water to give permanganic acid, HMnO₄, showing its acidic nature.

Question 35. Statement-1: Lanthanoids show a limited number of oxidation states whereas actinoids show a large number of oxidation states.
Statement-2: Energy gap between 4f, 5d and 6s subshells is small whereas that between 5f, 6d and 7s subshells is large.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation of Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation of Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (c)
Statement-2 is False because energy gap between 4f, 5d and 6s is large whereas that between 5f, 6d and 7s is small.

Teacher's Note:
1. Statement-2 is the reverse of the truth; small energy gaps (as in actinoids) allow more electrons to be used in bonding.

Question 36. Statement-1: To a solution of potassium chromate, if a strong acid is added, it changes its colour from yellow to orange.
Statement-2: The colour change is due to the change in the oxidation state of potassium chromate. (Bihar Board 2012)
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation of Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation of Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (c)
On adding acid, yellow potassium chromate changes to orange coloured potassium dichromate: 2CrO₄²⁻ + 2H⁺ → Cr₂O₇²⁻ + H₂O.
Statement-2 is false because Cr is in +6 oxidation state in both CrO₄²⁻ and Cr₂O₇²⁻; there is no change in oxidation state.

Teacher's Note:
1. Chromate–dichromate interconversion is a pH-controlled equilibrium, not a redox reaction.

Question 37. Statement-1: Potassium permanganate on heating at 573 K forms potassium manganate.
Statement-2: Both potassium permanganate and potassium manganate are tetrahedral and paramagnetic in nature. (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. (c)
2KMnO₄ —573 K→ K₂MnO₄ + MnO₂ + O₂, so Statement-1 is true.
Correct Statement-2. KMnO₄ (Mn +7, d⁰) is diamagnetic but K₂MnO₄ (Mn +6, d¹) is paramagnetic.

Teacher's Note:
1. Both ions are tetrahedral, but only manganate has an unpaired electron.