| List-I (Species) | List-II (Electronic distribution) |
|---|---|
| (A) Cr⁺² | (I) 3d⁸ |
| (B) Mn^+ | (II) 3d⁵4s¹ |
| (C) Ni⁺² | (III) 3d⁴ |
| (D) V^+ | (IV) 3d³4s¹ |
Solution
Core Logic
Let's determine the electronic configuration for each species by first writing the neutral atom's configuration, and then removing electrons starting from the outermost 4s orbital.
(A) Cr (Z=24): [Ar] 3d⁵ 4s¹ arrow Cr²⁺: [Ar] 3d⁴ (B) Mn (Z=25): [Ar] 3d⁵ 4s² arrow Mn^+: [Ar] 3d⁵ 4s¹ (C) Ni (Z=28): [Ar] 3d⁸ 4s² arrow Ni²⁺: [Ar] 3d⁸ (D) V (Z=23): [Ar] 3d³ 4s² arrow V^+: [Ar] 3d³ 4s¹
Step 1: Match execution
A arrow III B arrow II C arrow I D arrow IV
Chapter Mix
Class 12 Chemistry: The d- and f-Block Elements