Match the complex given in List-I with its property given in List-II
|
List-I Complex |
List-II Property |
|
(A) \( [CoF_6]^{3-} \) |
(I) \( d^5 \), high spin |
|
(B) \( [Fe(CN)_6]^{4-} \) |
(II) \( d^6 \), high spin |
|
(C) \( [FeF_6]^{3-} \) |
(III) \( d^5 \), low spin |
|
(D) \( [Mn(CO)_6]^{2+} \) |
(IV) \( d^6 \), low spin |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
In an octahedral complex the d orbitals split into the lower \( t_{2g} \) and the upper \( e_g \) set. Whether electrons pair up in \( t_{2g} \) or go up to \( e_g \) depends on the ligand. A weak field ligand (\( \ce{F-} \)) gives a small splitting, \( \Delta_o \lt P \) (pairing energy), so the electrons stay unpaired: high spin. A strong field ligand (\( \ce{CN-} \), CO) gives \( \Delta_o \gt P \), so the electrons pair in \( t_{2g} \): low spin. The d count comes from the oxidation state of the metal.
(A) \( \ce{[CoF6]^3-} \) — This matches (II) \(d^6\), high spin. \( x + 6(-1) = -3 \Rightarrow x = +3 \). Co (Z = 27) is \( \mathrm{[Ar]\,3d^{7}\,4s^{2}} \); losing the two 4s electrons and one 3d electron gives \( \ce{Co^3+} \): \( \mathrm{3d^{6}} \). With weak field \( \ce{F-} \): \( t_{2g}^{4}\,e_g^{2} \), high spin.
(B) \( \ce{[Fe(CN)6]^4-} \) — This matches (IV) \(d^6\), low spin. \( x + 6(-1) = -4 \Rightarrow x = +2 \). Fe (Z = 26) is \( \mathrm{[Ar]\,3d^{6}\,4s^{2}} \); losing the two 4s electrons gives \( \ce{Fe^2+} \): \( \mathrm{3d^{6}} \). With strong field \( \ce{CN-} \): \( t_{2g}^{6}\,e_g^{0} \), low spin.
(C) \( \ce{[FeF6]^3-} \) — This matches (I) \(d^5\), high spin. \( x + 6(-1) = -3 \Rightarrow x = +3 \). Losing the two 4s electrons and one 3d electron gives \( \ce{Fe^3+} \): \( \mathrm{3d^{5}} \). With weak field \( \ce{F-} \): \( t_{2g}^{3}\,e_g^{2} \), high spin.
(D) \( \ce{[Mn(CO)6]^2+} \) — This matches (III) \(d^5\), low spin. CO is neutral, so Mn is +2. Mn (Z = 25) is \( \mathrm{[Ar]\,3d^{5}\,4s^{2}} \); losing the two 4s electrons gives \( \ce{Mn^2+} \): \( \mathrm{3d^{5}} \). With strong field CO: \( t_{2g}^{5}\,e_g^{0} \), low spin.
Shortcut: (A)-(II) appears in Options (1) and (2), and (B)-(IV) then leaves only Option (1).