What is the numerical value of spin only magn | vedclass

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(B) The atomic number of copper $(Cu)$ is $29$. The electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$. In the $+2$ oxidation state, $Cu^{2+}$ lo

Vedclass · Accepted Answer

$1.73$

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(B) The atomic number of copper $(Cu)$ is $29$. The electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$. In the $+2$ oxidation state, $Cu^{2+}$ loses two electrons, resulting in the configuration $[Ar] 3d^9$. In the $3d^9$ configuration, there is $n = 1$ unpaired electron. The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$. Substituting $n = 1$, we get $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.

What is the numerical value of spin only magnetic moment of copper in $+2$ state?

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