The 'Spin only' Magnetic moment for $$[Ni(NH_3)_6]^{2+}$$ is ______ $$\times 10^{-1}$$ BM. (given = Atomic number of Ni : 28) Round off your answer to the nearest integer.

$$[Ni(NH_3)_6]^{2+}$$: Ni²⁺ is d⁸. NH₃ is weak field in octahedral... actually NH₃ is moderate/strong field. For octahedral Ni²⁺ with 8 d-electrons: t₂g⁶ eg² gives 2 unpaired electrons.

$$\mu = \sqrt{n(n+2)} = \sqrt{2(4)} = \sqrt{8} = 2\sqrt{2} \approx 2.83$$ BM.

In units of $$10^{-1}$$ BM: 28.3 ≈ 28.

The answer is $$\boxed{28}$$.