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

Answer

A

Explanation

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

In `d^(2)sp^(3)` hybridisation, the orbitals taking part are `d_(x^(2)-y^(2)) and d_(z^(2))`. Hence, unpaired electrons are present in `3d_(xy), 3d_(yz), 3d_(xz)`.