Coordination Compounds - Result Question 51
####57. Of the following complex ions, which is diamagnetic in nature?
[2011]
(a) $[NiCl_4]^{2-}$
(b) $[Ni(CN)_4]^{2-}$
(c) $[CuCl_4]^{2-}$
(d) $[CoF_6]^{3-}$
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Solution:
- (b) $Ni^{2+}=3 d^{8} 4 s^{\circ}$
Since, the coordination number of $Ni$ in this complex is 4 , the configuration of $Ni^{2+}$ at first sight shows that the complex is paramagnetic with two unpaired electron. However, experiments show that the complex is diamagnetic. This is possible when the $3 d$ electrons rearrange against the Hund’s rule as shown below. This is in
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accordance with the fact that the ligand involved here is strong i.e., $CN^{-}$ion.
$Ni^{2+}$ (after rearrangement)
Hence, now $d s p^{2}$ hybridization involving one $3 d$, one $4 s$ and two $4 p$ orbitals, takes place leading to four $d s p^{2}$ hybrid orbitals, each of which accepts four electron pairs from $CN^{-}$ion forming $[Ni(CN)_4]^{2-}$ ion.
$[Ni(CN)_4]^{2-}$
Thus, the complex is diamagnetic as it has no unpaired electron.
58 .
(c) $Cr^{2+}, d^{4}$
4 unpaired $e^{-} s$
$Mn^{2+}, d^{5}$
5 unpaired $e^{-} s$
4 unpaired $e^{-} s$
3 unpaired $e^{-} s$
Minimum paramagnetic behaviour $=[Co(H_2 O)_6]^{2+}$
