The correct answer is Option (1) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
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List-I Complex
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List-II Property
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(A) $[CoF_6]^{3-}$
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(II) $d^6$, high spin
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(B) $[Fe(CN)_6]^{4-}$
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(IV) $d^6$, low spin
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(C) $[FeF_6]^{3-}$
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(I) $d^5$, high spin
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(D) $[Mn(CO)_6]^{2+}$
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(III) $d^5$, low spin
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(A) $[CoF_6]^{3-}$
- Oxidation State: Cobalt is in the $+3$ state $(Co^{3+})$.
- Electronic Configuration: Neutral Cobalt is $[Ar]3d^7 4s^2$. $Co^{3+}$ is $d^6$.
- Spin State: Fluoride $(F^-)$ is a weak field ligand. It does not cause electron pairing, leading to a high spin complex.
- Match: $(A) \rightarrow (II) \ d^6$, high spin.
(B) $[Fe(CN)_6]^{4-}$
- Oxidation State: Iron is in the $+2$ state $(Fe^{2+})$.
- Electronic Configuration: Neutral Iron is $[Ar]3d^6 4s^2$. $Fe^{2+}$ is $d^6$.
- Spin State: Cyanide $(CN^-)$ is a strong field ligand. It forces the electrons to pair up, resulting in a low spin complex.
- Match: $(B) \rightarrow (IV) \ d^6$, low spin.
(C) $[FeF_6]^{3-}$
- Oxidation State: Iron is in the $+3$ state $(Fe^{3+})$.
- Electronic Configuration: $Fe^{3+}$ has a $d^5$ configuration.
- Spin State: Fluoride $(F^-)$ is a weak field ligand, resulting in a high spin complex where all $5$ electrons remain unpaired.
- Match: $(C) \rightarrow (I) \ d^5$, high spin.
(D) $[Mn(CO)_6]^{2+}$
- Oxidation State: Manganese is in the $+2$ state $(Mn^{2+})$.
- Electronic Configuration: Neutral Manganese is $[Ar]3d^5 4s^2$. $Mn^{2+}$ is $d^5$.
- Spin State: Carbon Monoxide $(CO)$ is one of the strongest field ligands. It causes maximum electron pairing, leading to a low spin complex.
- Match: $(D) \rightarrow (III) \ d^5$, low spin.
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