1.73 Bohr magneton implies one unpaired electron according to the expression $\sqrt{n\left(m+2\right)}PB$. The species  $\mathrm{Cu},\mathrm{Ni}$$\mathrm{Ti}$ and Co in the complexes have $(3\mathrm{d}{)}^9,(3\mathrm{d}{)}^8,(3\mathrm{d}{)}^0$  and $(3\mathrm{d}{)}^7$ electronic configurations, respectively. This involve  ${\mathrm{sp}}^2\mathrm{d},{\mathrm{dsp}}^2,{\mathrm{sp}}^3$ and ${\mathrm{d}}^{2}s{\mathrm{p}}^3$ hybridization respectively. Thus the complex  ${\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]}^{2+}$ has one unpaired electron.