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

(N/A) The magnetic moment $(\mu)$ is calculated using the spin-only formula: $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For $Cr^{3+}$ $(Z=24)$: The electronic configuration is $[Ar] 3d^3$. Here,$n = 3$. Thus,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
For $Co^{2+}$ $(Z=27)$: The electronic configuration is $[Ar] 3d^7$. This results in $n = 3$ unpaired electrons. However,for $Co^{2+}$,there is a significant orbital contribution to the magnetic moment due to the nature of the $d$-orbitals,which increases the observed value to approximately $4.87 \ BM$.