Neet Solved Paper 2013 Question 14
Question: A magnetic moment of 1.73 BM will be shown by one among the following
Options:
A) $ {{[\text{ Cu(N}{H_3}{{)}_4}]}^{\text{2+}}} $
B) $ {{[\text{ (NiCN}{{)}_4}]}^{\text{2-}}} $
C) $ TiC{l_4} $
D) $ {{[\text{ CoC}{l_6}]}^{4-}} $
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Answer:
Correct Answer: A
Solution:
Magnetic moment, p. is related with number of unpaired electrons as
$ \mu =\sqrt{n(n+2)}BM $ $ {{(1.73)}^{2}}=n(n+2) $ On solving n = 1.
Thus, the complex compound having one unpaired electron exhibit a magnetic moment of 1.73 BM.
In $ \ln ,{{[Cu{{(NH _3)}_4}]}^{2+}} $ $ C{u^{2+}}=[Ar]3d^{9} $
(Although in the presence of strong field ligand $ NH _3 $ , the unpaired electron gets excited to higher energy level but it still remains unpaired).
ln $ {{[Nl{{(CN)}_4}]}^{2-}} $ $ N{i^{2+}}=[Ar],3d^{8} $
But $ C{N^{-}} $ being strong field ligand pair up the unpaired electrons and hence in this complex, number of unpaired electrons = 0.
In $ [TiCl _4] $ $ T{i^{4+}}=[Ar] $ No unpaired electron.
ln $ {{[CoCl _6]}^{4-}} $ $ C{o^{2+}}=[Ar]3d^{7} $
It contains three unpaired electrons. Thus, $ {{[Co{{(NH _3)}_4}]}^{2+}} $ is the complex that exhibits a magnetic moment 1.73 BM.
