Match the complexes given in List-I with their properties given in List-II
|
List-I Complex |
List-II Property |
|
(A) \( [FeF_6]^{3-} \) |
(I) \( d^6 \), high spin |
|
(B) \( [CoF_6]^{3-} \) |
(II) \( d^5 \), high spin |
|
(C) \( [Mn(CN)_6]^{3-} \) |
(III) \( d^6 \), low spin |
|
(D) \( [Co(CN)_6]^{3-} \) |
(IV) \( d^4 \), low spin |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
In an octahedral complex the d orbitals split into a lower \( t_{2g} \) set and an upper \( e_g \) set. A weak field ligand such as \( \ce{F-} \) gives a small splitting, so the electrons occupy all five orbitals singly before pairing: a high spin complex. A strong field ligand such as \( \ce{CN-} \) gives a large splitting, so the electrons pair up in the \( t_{2g} \) set: a low spin complex. In all four complexes x + 6(−1) = −3, so the metal is +3.
(A) \([FeF_6]^{3-}\) — This matches (II) \(d^5\), high spin. Fe (Z = 26) is \( \mathrm{[Ar]\,3d^{6}\,4s^{2}} \); \( \ce{Fe^3+} \) is \( \mathrm{3d^{5}} \). With \( \ce{F-} \) it is high spin, \( t_{2g}^{3}e_g^{2} \), five unpaired electrons.
(B) \([CoF_6]^{3-}\) — This matches (I) \(d^6\), high spin. Co (Z = 27) is \( \mathrm{[Ar]\,3d^{7}\,4s^{2}} \); \( \ce{Co^3+} \) is \( \mathrm{3d^{6}} \). With \( \ce{F-} \) it is high spin, \( t_{2g}^{4}e_g^{2} \), four unpaired electrons (paramagnetic, outer orbital \( \mathrm{sp^{3}d^{2}} \)).
(C) \([Mn(CN)_6]^{3-}\) — This matches (IV) \(d^4\), low spin. Mn (Z = 25) is \( \mathrm{[Ar]\,3d^{5}\,4s^{2}} \); \( \ce{Mn^3+} \) is \( \mathrm{3d^{4}} \). With \( \ce{CN-} \) it is low spin, \( t_{2g}^{4} \), two unpaired electrons.
(D) \([Co(CN)_6]^{3-}\) — This matches (III) \(d^6\), low spin. \( \ce{Co^3+} \) is \( \mathrm{3d^{6}} \); with \( \ce{CN-} \) it is low spin, \( t_{2g}^{6} \), with no unpaired electron (diamagnetic, inner orbital \( \mathrm{d^{2}sp^{3}} \)).
Shortcut: Only iron(III) has five d electrons, so (A)-(II); this leaves Options (3) and (4). Fluoride is a weak field ligand, so \( \ce{[CoF6]^3-} \) is high spin, (B)-(I), which gives Option (3).