Match List-I with List-II :
| List-I | List-II |
| (A) Mn2+ | (I) Pyrolusite ore |
| (B) Spin only Magnetic Moment | (II) An alloy of 4f metal, iron and traces of S, C, Al and Ca |
| (C) MnO2 | (III) \( \mu_s=\sqrt{n(n+2)} \) BM |
| (D) Misch metal | (IV) Highest oxidation states |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (IV), (B) - (III), (C) - (I), (D) - (II)
The items are all from the chemistry of manganese and of the d- and f-block elements.
(A) \( \ce{Mn^{2+}} \) — This matches (IV) Highest oxidation states, by elimination. Manganese shows the highest oxidation state of the first transition series, +7, because Mn (Z = 25, [Ar]3d54s2) can use all seven 3d and 4s electrons. The ion \( \ce{Mn^2+} \) itself, though, is manganese in its lowest common state (3d5), not its highest.
(B) Spin only Magnetic Moment — This matches (III) μs = √n(n+2) BM, where n is the number of unpaired electrons; for \( \ce{Mn^2+} \) (five unpaired electrons) it gives √35 = 5.92 BM.
(C) \( \ce{MnO2} \) — This matches (I) Pyrolusite ore. Pyrolusite is the chief ore of manganese; it is fused with KOH in air to make potassium manganate on the way to \( \ce{KMnO4} \).
\( \displaystyle \ce{2MnO2 + 4KOH + O2 -> 2K2MnO4 + 2H2O} \)
(D) Misch metal — This matches (II) An alloy of 4f metal, iron and traces of S, C, Al and Ca. It is about 95% lanthanoid metal and 5% iron, with traces of S, C, Ca and Al; it is used to make bullets, shells and lighter flints.
Note: It seems that there is a mistake in the original NTA paper, which prints Mn2+ in item (A); Mn2+ is manganese in its lowest common oxidation state, while (IV) Highest oxidation states fits manganese (up to +7).