Ncert Exercises & Solutions

8.8: Calculate the ‘spin only’ magnetic moment of M $^{2 +}$ $_{(a q)}$ (aq) ion (Z = 27)

Z = 27

[Ar]

3 d^{7} 4 s^{2}

∴ M^{2 +} = [A r] 3 d^{7}

3 d^{7} =

i.e., 3 unpaired electrons

∴ n = 3

\Rightarrow \sqrt{n (n + 2)} = μ

\Rightarrow \sqrt{3 (3 + 2)} = μ

\Rightarrow \sqrt{15} = μ

μ \approx 4 B M