Question

$\left[Cr{\left(H_{2}O\right)}_6\right]C{l}_3$ ( atomic number of $Cr=24$ ) has a magnetic moment of 3.83 B.M. The correct distribution
of 3d electrons in the chromium present in the complex is :

• A. $3d^1{}_{xy},3d^1{}_{yz},3d^1{}_{zx}$
• B. $3d^1{}_{xy},3d^1{}_{yz},3d^1{}_{z^2}$
• C. $3d_{(x^2-y^2)}^1,3d_{z^2}^1,3d_{xz}^1$
• D. $3d_{xy}^1,3d_{(x^2-y^2)}^1,3d_{yz}^1$

Answer 1

The correct option is A $3d^1{}_{xy},3d^1{}_{yz},3d^1{}_{zx}$

As, magnetic moment $\mu =\sqrt{n(n+2)}$, where n is no. of unpaired electron.
Here $\mu =3.83$ so n$=3$. So Cr has $d^3$ configuration and these electron will be present in $t_{2g}$ orbitals.