For a t_{2g}^6 e_g^2 system,the value of the | vedclass

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(A) The magnetic moment is calculated using the formula $\mu = \sqrt{n(n + 2)}$,where $n$ is the number of unpaired electrons.In the $t_{2g}^6 e_g^2$

Vedclass · Accepted Answer

$2.83$

Vedclass · Answer

(A) The magnetic moment is calculated using the formula $\mu = \sqrt{n(n + 2)}$,where $n$ is the number of unpaired electrons.In the $t_{2g}^6 e_g^2$ configuration,the $t_{2g}$ orbitals are fully filled ($6$ electrons),and the $e_g$ orbitals contain $2$ electrons.According to Hund's rule,the $2$ electrons in the $e_g$ orbitals will remain unpaired.Therefore,the number of unpaired electrons $n = 2$.Substituting the value of $n$ into the formula: $\mu = \sqrt{2(2 + 2)} = \sqrt{8} \approx 2.83 \ B.M.$

For a $t_{2g}^6 e_g^2$ system,the value of the magnetic moment $(\mu)$ is equal to ..... $B.M.$

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