Match List I with List II
| List I | List II |
| (A) Transition metal that doesn't show variable oxidation state | (I) \(Mn\) |
| (B) Metal of 3d-series that exhibits maximum number of oxidation states | (II) \(Zn\) |
| (C) Transition metal with positive \(E^\circ_{M^{2+}/M}\) value | (III) \(Sc\) |
| (D) Element of \(3d\)-series which is colourless in its compounds | (IV) Cu |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-III, B-I, C-IV, D-II
(A) Transition metal that doesn’t show variable oxidation state — This matches (III) Sc. Sc (Z = 21) is [Ar] 3d1 4s2; losing all three outer electrons gives \( \ce{Sc^3+} \), 3d0, its only state (+3).
(B) Metal of 3d-series that exhibits maximum number of oxidation states — This matches (I) Mn. Mn (Z = 25) is [Ar] 3d5 4s2; it shows every state from +2 to +7, where all seven outer electrons are used.
(C) Transition metal with positive E°(M2+/M) value — This matches (IV) Cu. Cu (Z = 29) is [Ar] 3d10 4s1; it is the only 3d metal with a positive value (+0.34 V), so it does not liberate \( \ce{H2} \) from acids.
(D) Element of 3d-series which is colourless in its compounds — This matches (II) Zn. Zn (Z = 30) is [Ar] 3d10 4s2; \( \ce{Zn^2+} \) is 3d10, so no d–d transition is possible and its compounds are white. (\( \ce{Sc^3+} \), 3d0, is also colourless, but Sc is already matched with A.)