Question

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

• A. ${\left[Cu\right(N{H}_3{)}_4]}^{2+}$
• B. ${\left[Ni\right(CN{)}_4]}^{2-}$
• C. $TiC{l}_4$
• D. ${\left[CoC{l}_6\right]}^{4-}$

Answer 1

The correct option is A ${\left[Cu\right(N{H}_3{)}_4]}^{2+}$

A magnetic moment of 1.73 BM will be shown by ${\left[Cu\right(N{H}_3{)}_4]}^{2+}$.
Magnetic moment is given by the expression $\left(\mu \right)=\sqrt{n(n+2)}$
$1.73=\sqrt{n(n+2)}$
$n=1$
So, compound must contain one unpaired electron. The compound is $\left[Cu\right(N{H}_3{)}_4{]}^{2+}$ as $C{u}^{2+}$ has $3d^9$ electronic configuration.