question_answer

A magnetic moment of 1.73 BM will be shown by one among the following.

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.]