1. Find the electronic configuration of $\mathrm{Ni}$

   Atomic number of $\mathrm{Ni}=28$.

   $\mathrm{Ni}: [Ar] \, 3d^8 4s^2$

2. Find the oxidation state in $[\mathrm{Ni}(\mathrm{NH}_3)_6]^{2+}$

   Since $\mathrm{NH}_3$ is a neutral ligand,

   $\text{Oxidation state of Ni} = +2$

   So,

   $\mathrm{Ni}^{2+}: [Ar] \, 3d^8$

3. Determine the number of unpaired electrons

   $\mathrm{NH}_3$ is not strong enough here to force pairing in a $d^8$ octahedral $\mathrm{Ni}^{2+}$ ion beyond the usual arrangement.

   For octahedral $d^8$:

   $t_{2g}^6 e_g^2$

   This gives:

• $t_{2g}^6$ : all paired

• $e_g^2$ : two unpaired electrons

  Hence, number of unpaired electrons,

  $n=2$

4. Use spin-only magnetic moment formula

   $\mu = \sqrt{n(n+2)}\, \text{BM}$

   Substituting $n=2$:

   $\mu = \sqrt{2(2+2)} = \sqrt{8} = 2\sqrt{2} \approx 2.828\, \text{BM}$

5. Match with the form asked in the question

   The question asks for

   $\underline{\hspace{2cm}} \times 10^{-1}\, \text{BM}$

   Since

   $2.828\, \text{BM} = 28.28 \times 10^{-1}\, \text{BM}$

   As an integer answer, this is taken as:

   $\boxed{28}$

Comparison with stored answer

Stored correct answer = $28$

Our derived answer = $28$

So, the answer agrees.