Question

A magnetic moment of 1.73 BM will be shown by:

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

Answer 1

Answer: (A)

$\left[Cu\right(N{H}_3{)}_4{]}^{2+}$

Magnetic moment$\left(\mu \right)$ is related with the no. of unpaired electron

$\mu =\sqrt{n(n+2)}\to 1$ Bohr Magneton (BM)-Units

The given magnetic moment in the question is $1.73BM$.

Putting that value $\left(\mu \right)$ in equation $1$; we get:-

${\left(1.73\right)}^2=n(n+2)$

On solving,

$\Rightarrow n=1$

Thus, the complex/compound having one unpaired electrons exhibit a magnetic moment of $1.73BM$

a. In ${\left[Cu{\left(N{H}_3\right)}_4\right]}^{2+}$

${Cu}^{2+}=\left[Ar\right]3d^9$, $n=1$

b. In ${\left[Ni{\left(CN\right)}_4\right]}^{2-}$

${Ni}^{2+}=\left[Ar\right]3d^8$ , $n=2$

c. In $\left[Ti{Cl}_4\right]$

${Ti}^{4+}=\left[Ar\right]3d^0$, $n=0$

d. In ${\left[Co{Cl}_6\right]}^{4-}$

${Co}^{2+}=\left[Ar\right]3d^7$ , $n=3$

In above all configuration only ${\left[Cu{\left(N{H}_3\right)}_4\right]}^{2+}$ has one unpaired electron.

$\therefore$Correct answer :-

Option A ${\left[Cu{\left(N{H}_3\right)}_4\right]}^{2+}$ has a dipole moment of $1.73BM$