$\left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]\mathrm{C}{l}_3$ (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

$$\begin{gathered} \mathrm{\mu } = \sqrt{\mathrm{n}(\mathrm{n}+2)}\mathrm{BM} \\ 3.83 = \sqrt{\mathrm{n}(\mathrm{n}+2)} \\ 3.83 \times 3.83 = {\mathrm{n}}^2 + 2\mathrm{n} \\ 14.6689 = {\mathrm{n}}^2 + 2\mathrm{n} \end{gathered}$$

On solving this we get n = 3
As a result, there are a number of unpaired electrons in the d- subshell of the penultimate shell of chromium (Cr = 24).

So, the chromium ion configuration is Cr3+= 1s²2s²2p⁶3s²3p⁶3d³.
Cr's oxidation state in [Cr(H₂O)₆]Cl₂ is +3.

Hence the correct option ( A) ($3{\mathrm{d}}_{\mathrm{xy}}^1,3{\mathrm{d}}_{\mathrm{yz}}^1,3{\mathrm{d}}_{\mathrm{xz}}^1$)