Question

A magnetic moment of $1.73\text{BM}$ will be shown by :

• A. $H{g}_{2}C{l}_2$
• B. $TiC{l}_4$
• C. $[CoC{l}_6{]}^{4-}$
• D. $\left[Cu\right(N{H}_3{)}_4{]}^{2+}$

Answer 1

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

$\left[Cu\right(N{H}_3{)}_4{]}^{2+}$ hybridisation $ds{p}^2$
$C{u}^{+2}-3d^9$ has one unpaired $e^{-}$
so magnetic moment.
$\mu =\sqrt{n(n+2)}=\sqrt{1(1+2)}=\sqrt{3}=1.73BM$
In $TiC{l}_4$ & $H{g}_{2}C{l}_2$ = no unpaired electrons
In $[CoC{l}_6{]}^{4-}$ = 3 unpaired electrons.
$\mu =\sqrt{n(n+2)}=\sqrt{3(3+2)}=\sqrt{15}=3.87BM$