Match the LIST-I with LIST-II.LIST-IComplex ionLIST-IInumber of unpai…

The correct answer is Option (2) → A-III, B-IV, C-I, D-II

In an octahedral field the d orbitals split into a lower t_2g set and a higher e_g set. A strong field ligand such as \( \ce{CN^{-}} \) gives a large splitting (Δ_o > P), so electrons pair up in t_2g (low spin). A weak field ligand such as \( \ce{Cl^{-}} \) or \( \ce{F^{-}} \) gives a small splitting (Δ_o < P), so electrons enter e_g before pairing (high spin).

Each ligand carries a charge of −1, so in all four complexes the oxidation state x of the metal is:

    \( \displaystyle x+6(-1)=-3\quad\Rightarrow\quad x=+3 \)

Mn (Z = 25) is [Ar]3d⁵4s²; losing the two 4s electrons and one 3d electron gives \( \ce{Mn^{3+}} \): 3d⁴. Fe (Z = 26) is [Ar]3d⁶4s²; losing the two 4s electrons and one 3d electron gives \( \ce{Fe^{3+}} \): 3d⁵.

• (A) \( \ce{[Mn(CN)6]^{3-}} \) — This matches (III) n = 2. With the strong field \( \ce{CN^{-}} \), the four d electrons of \( \ce{Mn^{3+}} \) all go into t_2g: t_2g⁴e_g⁰, which leaves two unpaired electrons.

• (B) \( \ce{[MnCl6]^{3-}} \) — This matches (IV) n = 4. With the weak field \( \ce{Cl^{-}} \), the fourth electron goes up to e_g: t_2g³e_g¹, with four unpaired electrons.

• (C) \( \ce{[Fe(CN)6]^{3-}} \) — This matches (I) n = 1. With the strong field \( \ce{CN^{-}} \), the five d electrons of \( \ce{Fe^{3+}} \) all go into t_2g: t_2g⁵e_g⁰, which leaves one unpaired electron.

• (D) \( \ce{[FeF6]^{3-}} \) — This matches (II) n = 5. With the weak field \( \ce{F^{-}} \), the electrons spread over both sets: t_2g³e_g², with all five electrons unpaired.

Shortcut: The pair (C)–(I) appears only in Option (2).

Note: The same/similar question has been asked in different years by NTA as per the details given below: