`[Cr(H_2 O)_6]Cl_3` (at no. of Cr = 24) has a magnetic moment of `3.83 B.M`. The correct distribution of `3d` electrons the chromium of the complex.

Options

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

Answer

A

Explanation

Magnetic moment =`sqrt(n(n+2))BM = 3.83 BM.` (Given). Hence = 3 i.e. there are three unpaired electrons . Thus we have

In `d^(2)sp^(3)` hybridisation, the orbital taking part are `d_(x)^(2)-y^(2) and dz^(2)` . Hence unpaired electrons are prsesnt in `3dxy, 3dyz, 3dxz`