The calculated magnetic moment (in Bohr magnetons of $\mathrm{Cu}^{2+}$ ion is)

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Aicharya · Accepted Answer

Electronic configuration of $\mathrm{Cu}^{2+}$ is $1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 3 d^9$ (because $2 e^{-}$goes from $4 s$ orbital) Number of unpaired electrons $=1$ $ \begin{aligned} 	ext { Magnetic moment } & =\sqrt{n(n+2)} \ & =\sqrt{1(1+2)} \ & =\sqrt{3}=1.732 \end{aligned} $

The calculated magnetic moment (in Bohr magnetons of $\mathrm{Cu}^{2+}$ ion is)

Solution