The 'Spin only’ Magnetic moment for $\left[\mathrm{Ni}\right({\mathrm{NH}}_3{)}_6]{ }^{2+}$ is______$\times {10}^{–1}$ BM. (given = Atomic number of $\mathrm{Ni} : 28$)

Round off your answer to the nearest integer.

In the presence of ligand ${\mathrm{NH}}_3$, pairing does not occur in   ${\mathrm{Ni}}^{2+}$ because pairing of electrons cannot create two vacant d-orbitals. Hence, nickel ion undergoes ${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridisation and it has two unpaired electrons.

${\mathrm{Ni}}^{2+}=3{\mathrm{d}}^8$

No. of unpaired electron $=2$

The spin only magnetic moment 

$$\begin{gathered} \mathrm{\mu }=\sqrt{\mathrm{n}\left(\mathrm{n}+2\right)} \\ =\sqrt{8} \\ =2.82 \mathrm{BM} \end{gathered}$$

$=28.2\times {10}^{-1} \mathrm{BM}$

$\mathrm{x}=28$