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$[Cr(H_2O)_6]Cl_3$ (at. no. of Cr = 24) has a magnetic moment of 3.83 B.M. The correct distribution of 3d electrons in the chromium of the complex is :-
A
$3d^1_{x^2-y^2}$, $3d^1_{z^2}$, $3d^1_{xz}$
B
$3d^1_{xy}$, $3d^1_{x^2-y^2}$, $3d^1_{yz}$
C
$3d^1_{xy}$, $3d^1_{yz}$, $3d^1_{xz}$
D
$3d^1_{xy}$, $3d^1_{yz}$, $3d^1_{z^2}$
Detailed Solution
In $[Cr(H_2O)_6]Cl_3$ chromium is in the +3 oxidation state. Cr is $[Ar]3d^5 4s^1$, so $Cr^{3+}$ is $3d^3$.
Magnetic moment $\mu = \sqrt{n(n + 2)}$ BM. For $n = 3$: $\mu = \sqrt{15} = 3.87$ BM, which agrees with the observed value of 3.83 BM. So there are three unpaired electrons.
The complex is octahedral. In an octahedral field the five d orbitals split into a lower energy $t_{2g}$ set ($d_{xy}$, $d_{yz}$, $d_{xz}$) and a higher energy $e_g$ set ($d_{x^2-y^2}$, $d_{z^2}$).
The $t_{2g}$ orbitals point between the ligands, so they are lower in energy and are filled first.
The three electrons occupy the three $t_{2g}$ orbitals singly (Hund's rule): $t_{2g}^3\,e_g^0$.
So the distribution is $3d^1_{xy}$, $3d^1_{yz}$, $3d^1_{xz}$.
Magnetic moment $\mu = \sqrt{n(n + 2)}$ BM. For $n = 3$: $\mu = \sqrt{15} = 3.87$ BM, which agrees with the observed value of 3.83 BM. So there are three unpaired electrons.
The complex is octahedral. In an octahedral field the five d orbitals split into a lower energy $t_{2g}$ set ($d_{xy}$, $d_{yz}$, $d_{xz}$) and a higher energy $e_g$ set ($d_{x^2-y^2}$, $d_{z^2}$).
The $t_{2g}$ orbitals point between the ligands, so they are lower in energy and are filled first.
The three electrons occupy the three $t_{2g}$ orbitals singly (Hund's rule): $t_{2g}^3\,e_g^0$.
So the distribution is $3d^1_{xy}$, $3d^1_{yz}$, $3d^1_{xz}$.