$\left[Cr{\left(H_{2}O\right)}_6\right]C{l}_3$ (at no. of Cr = 24) has a magnetic moment of $3.83B.M$. The correct distribution of $3d$ electrons the chromium of the complex.

Magnetic moment $\left(\mu \right)=\sqrt{n\left(n+2\right)}$ BM
$\text{or }3.83=\sqrt{n\left(n+2\right)}$
$\text{or }3.83\times 3.83=n^2+2n$
$14.6689=n^2+2n$
$n=3$
Hence, number of unpaired electrons in d-subshell of chromium (Cr= 24) = 3.
So. the configuration of chromium ion is
$C{r}^{3+}=1s^2,2s^2,2p^6,3s^{2}3p^{6}3d^3$
In $\left[Cr{\left(H_{2}O\right)}_6\right]C{l}_2,$ oxidation state of Cr is + 3.
Hence, in $3d^3$ the distribution of electrons
$3d_{xy}^1,3d_{yz}^1,3d_{zx}^1,3d_{x^{2-y^2}}^0,3d_{z^2}^0$