Question

The 'Spin only’ Magnetic moment for \([Ni (NH_3)_6]^{2+}\) is ____ \(\times 10 ^{-1}\) BM.

(given = Atomic number of Ni : 28)

Answer 1

Correct answer is : 28

\(NH_3\) act as WFL with \(Ni^{2+}\)

\(Ni^{2+}= 3d^8\)

No. of unpaired electron = 2

\(\mu = \sqrt {n(n+2)}= \sqrt {8} = 2.82 \,BM\)

\(= 28.2 \times 10^{-1}BM\)

x = 28