Question

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

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

Answer 1

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

Magnetic moment 1.73 BM
$\mu =\sqrt{n(n+2)}B.M$
n = no. of unpaired $e^{-}$
$\mu =1.73$
$1.73=\sqrt{n(n+2)}B.M$
$n=1$

* $[CoC{l}_6{]}^{4-}\to C{o}^{2+};d^7$
$C{l}^{-}$ (weak field ligand) - $t_{2g}^5{e}_g^2$ unpaired $e^{-}=3$

* $\left[Cu\right(N{H}_3{)}_4{]}^{2+}\to C{u}^{2+};d^9$
$N{H}_3$ is strong field ligand, hybridisation is $ds{p}^2$
Unpaired $e^{-}$ = 1

* $Ni(CN{)}_4{]}^{2-}\to N{i}^{+2}-d^8$
$C{N}^{-}$ Strong field ligand, $ds{p}^2$ hybridisation
unpaired $e^{-}=0$


* $TiC{l}_4\to T{i}^{+4};d^0$
unpaired $e^{-}$= 0.

Hence, option (b) is correct.