Match the complexes given in List-I with the number of unpaired elect…

The correct answer is Option (1) → (A)-(I), (B)-(III), (C)-(IV), (D)-(II)

List-I Complexes	List-II Number of unpaired electrons
(A) $[Ti(H_2O)_6]^{3+}$	(I) 1
(B) $[NiF_6]^{4-}$	(III) 2
(C) $[Fe(CN)_6]^{4-}$	(IV) 0
(D) $[MnF_6]^{4-}$	(II) 5

(A) $[Ti(H_2O)_6]^{3+}$

• Oxidation State: Titanium is in the $+3$ state ($Ti^{3+}$).
• Electronic Configuration: Neutral $Ti$ is $[Ar] 3d^2 4s^2$. Therefore, $Ti^{3+}$ is $[Ar] 3d^1$.
• Unpaired Electrons: There is 1 unpaired electron in the $t_{2g}$ orbital.
• Match: (A) — (I)

(B) $[NiF_6]^{4-}$

• Oxidation State: Nickel is in the $+2$ state ($Ni^{2+}$).
• Electronic Configuration: Neutral $Ni$ is $[Ar] 3d^8 4s^2$. Therefore, $Ni^{2+}$ is $[Ar] 3d^8$.
• Ligand Field: $F^-$ is a weak field ligand, so electrons fill according to Hund's rule.
• Unpaired Electrons: In an octahedral field, $d^8$ fills as $(t_{2g})^6 (e_g)^2$. This results in 2 unpaired electrons.
• Match: (B) — (III)

(C) $[Fe(CN)_6]^{4-}$

• Oxidation State: Iron is in the $+2$ state ($Fe^{2+}$).
• Electronic Configuration: Neutral $Fe$ is $[Ar] 3d^6 4s^2$. Therefore, $Fe^{2+}$ is $[Ar] 3d^6$.
• Ligand Field: $CN^-$ is a strong field ligand, causing all electrons to pair up in the lower $t_{2g}$ orbitals.
• Unpaired Electrons: Configuration is $(t_{2g})^6 (e_g)^0$. This results in 0 unpaired electrons.
• Match: (C) — (IV)

(D) $[MnF_6]^{4-}$

• Oxidation State: Manganese is in the $+2$ state ($Mn^{2+}$).
• Electronic Configuration: Neutral $Mn$ is $[Ar] 3d^5 4s^2$. Therefore, $Mn^{2+}$ is $[Ar] 3d^5$.
• Ligand Field: $F^-$ is a weak field ligand, resulting in a high-spin complex.
• Unpaired Electrons: Configuration is $(t_{2g})^3 (e_g)^2$. This results in 5 unpaired electrons.
• Match: (D) — (II)