Coordination Compounds - Result Question 44
####47. Which of these statements about $[Co(CN)_6]^{3-}$ is true?
[2015]
(a) $[Co(CN)_6]^{3-}$ has four unpaired electrons and will be in a low-spin configuration.
(b) $[Co(CN)_6]^{3-}$ has four unpaired electrons and will be in a high spin configuration. (c) $[Co(CN)_6]^{3-}$ has no unpaired electrons and will be in a high-spin configuration.
(d) $[Co(CN)_6]^{3-}$ has no unpaired electrons and will be in a low-spin configuration.
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Solution:
- (d) $In[Co(CN)_6]^{-3}$, O.N. of Co is +3
$\therefore Co^{+3}=3 d^{6} 4 s^{0}$
$CN^{-}$is a strong field ligand
$\therefore \quad$ Pairing of electrons occurs so in this complex no unpaired electron is present and it is low spin complex.
48 .
(b) Due to $d^{5}$ configuration CFSE is zero.
(b) cis-platin
(d) $[Cu(NH_3)_4]^{2+}$ hybridisation $d s p^{2}$
$Cu^{+2}-3 d^{9}$ has one unpaired $e^{-}$
So, magnetic moment
$ \begin{aligned} \mu & =\sqrt{n(n+2)}=\sqrt{1(1+2)} \\ & =\sqrt{3}=1.73 \end{aligned} $
