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

Answer

D

Explanation

Since magnetic moment of `3.83` BM, which lies between 3 and 4, there are three unpaired electrons. Altranatively
`mu = sqrt(n (n + 2))` BM
`3.83 BM = sqrt(n(n + 2))` BM
`3.83xx3.83 = n^(2) + 2n`
`14.6689 = n^(2) + 2n`
`n = 3`
or, in complex `[Cr(H_(2)O)_(6)C1_(3)]` , the oxidation numbe4r of Cr is `+3` .
`Cr(3d^(5) 4s^(1)) implies Cr^(3+) (3d^(3))`
Since `H_(2)O` is a weak field ligand, all the 3 electrons will be placed singly in the `t_(2g)` set of d - orbitals which are `d_(xy)` , `d_(yz)` and `d_(xz)` . Thus, the correct distribution of 3d electrons in the chromimum of the complex is `3d_(xy)^(1)` , `3d^(1)` and `3d_(z^(2))^(1)` .