Although Cr^{3+} and Co^{2+} ions have the sa | vedclass

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(B) Both $Cr^{3+}$ $(3d^3)$ and $Co^{2+}$ $(3d^7)$ have $3$ unpaired electrons.According to the spin-only formula,$\mu = \sqrt{n(n+2)} \ B.M.$,both sh

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Due to orbital contribution in $Co^{2+}$ ion.

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(B) Both $Cr^{3+}$ $(3d^3)$ and $Co^{2+}$ $(3d^7)$ have $3$ unpaired electrons.According to the spin-only formula,$\mu = \sqrt{n(n+2)} \ B.M.$,both should have a magnetic moment of $\sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ B.M.$However,$Cr^{3+}$ has a symmetrical $t_{2g}^3$ configuration,resulting in no orbital contribution to the magnetic moment.In contrast,$Co^{2+}$ has an unsymmetrical configuration,which leads to a significant orbital contribution,increasing the observed magnetic moment to approximately $4.87 \ B.M.$

Although $Cr^{3+}$ and $Co^{2+}$ ions have the same number of unpaired electrons,why is the magnetic moment of $Cr^{3+}$ $3.87 \ B.M.$ and that of $Co^{2+}$ $4.87 \ B.M.?$

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