$1.73$ Bohr magneton implies one unpaired electron according to the expression $\sqrt{n(n+2)} \mu_{\mathrm{B}}$. The species $\mathrm{Cu}, \mathrm{Ni}$, 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 $\mathrm{d}^{2} \mathrm{sp}^{3}$ hybridization, respectively. Thus, the complex $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}ight)_{4}ight]^{2+}$ has one unpaired electron.