`[Cr(H_(2)O)_(6)]Cl_(3)` has a spin only magnetic moment of `3.83BM`. The correct distribution of `3d` electrons in the chromium of this complex is

Options

A. `3d_(xy)^1, 3d^(xz)^1, 3d_(xz)^1`
B. `3d_(xy)^1,3d_(yz)^1, 3d_(z^2)^1`
C. `3d_((x^2 - y^2))^1 , 3d_(x^2)^1, 3d_(xz)^1`
D. `3d_(xy)^1, 3d_((x^2-y^2))^1 , 3d_(yz)^1`

Answer

A

Explanation

Megnetic moment `= sqrt(n (n + 2)) B.M. = 3.83 B.M.` (Given). Hence,n = 3, ie, there are three unpaired electrons. Thus, we have
In `d^2sp^3` hybridisation,the orbitals taking part are `d_(x^2-y^2) and d_(x^2)` . Hence, unpaired electrons are present in `3d_(xy), 3d_(yz), 3d_(xz)`.