`[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_(z2)^(1)`
B. `3d_(zx^(-2) - y^(2))^(1) , 3d_(z)^(1/2) , 3d_(xz)^(1)`
C. `3d_(xy)^(1) , 3d_(x^(2) - y^(2))^(1) , 3d_(z)^(1)`
D. `3d_(xy)^(1) , 3d_(yz)^(1) , 3d_(xz)^(1)`

Answer

D

Explanation

Magnetic moment = `sqrt n(n + 2) `
3.83 = `sqrt n(n + 2)`
14.6689 (`approx` 15 ) = n(n + 2)
15 = `n^(2)` + 2n
On calculation n `approx` 3
`therefore` `Cr^(3+)` ion has the configuration [Ar]`3d^(3)` or `3d_(sy)^1`, `3d_(yz)^(1)`, `3d_(xz)^(1)`