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

Oxidation state of Cu in ${\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]}^{2+}$is +2

It has one unpaired electron (n = 1).

$\mathrm{\mu }=\sqrt{\mathrm{n}\left(\mathrm{n}+2\right)}\mathrm{BM}$

$\mathrm{\mu }=\sqrt{1\left(1+2\right)}=\sqrt{3}=1.73\mathrm{BM}$