The calculated magnetic moment (in Bohr magne | vedclass

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(A) The atomic number of $Cu$ is $29$. The electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$.For $Cu^{2+}$ ion,two electrons are removed (one fr

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$1.73$

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(A) The atomic number of $Cu$ is $29$. The electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$.For $Cu^{2+}$ ion,two electrons are removed (one from $4s$ and one from $3d$),resulting in the configuration $[Ar] 3d^9$.In the $3d^9$ configuration,there is $1$ unpaired electron $(n = 1)$.The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} 	ext{ BM}$.$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 	ext{ BM}$.

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

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