A) \[{{\left[ Cu{{\left( N{{H}_{3}} \right)}_{4}} \right]}^{2+}}\]
B) \[{{\left[ Ni{{\left( CN \right)}_{4}} \right]}^{2-}}\]
C) \[TiC{{l}_{4}}\]
D) \[{{\left[ CoC{{l}_{6}} \right]}^{4-}}\]
Correct Answer: A
Solution :
Magnetic moment, \[\mu \]is related with number of unpaired electrons as: |
\[\mu =\sqrt{n\left( n+2 \right)}BM\] |
\[{{\left( 1.73 \right)}^{2}}=n(n+2)\] |
On solving, \[n=1\] |
Thus, the complex/compound having one unpaired electron exhibit a magnetic moment of\[1.73\text{ }BM\]. |
[a] In \[{{\left[ Cu{{\left( N{{H}_{3}} \right)}_{4}} \right]}^{2+}}\] the electronic configuration of \[C{{u}^{2+}}=\left[ Ar \right]3{{d}^{9}}\] |
(Although in the presence of strong field ligand\[N{{H}_{3}}\], the unpaired electrons gets excited to higher energy level, but it still remains unpaired.) |
[b] In \[{{\left[ Ni{{\left( CN \right)}_{4}} \right]}^{2-}}\] the electronic configuration of \[N{{i}^{2+}}=\left[ Ar \right]3{{d}^{4}}\] |
But \[C{{N}^{-}}\] being strong field ligand pair up the unpaired electrons and hence, in this complex, number of unpaired electrons = 0 |
[c] In \[\left[ TiC{{l}_{4}} \right]\]the electronic configuration of\[T{{i}^{2+}}=\left[ Ar \right]\], no unpaired electron. |
[d] In \[{{\left[ CoC{{l}_{6}} \right]}^{4-}}C{{o}^{2+}}=\left[ Ar \right]3{{d}^{7}}\] |
It contains three unpaired electrons. [Thus, \[{{\left[ Cu{{\left( N{{H}_{3}} \right)}_{4}} \right]}^{2+}}\]is the complex that exhibits a magnetic moment of 1.73 BM.] |
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