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