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

Answer

D

Explanation

Magnetic moment `(mu)=sqrt(n(n+2))` BM
`"or "3.83=sqrt(n(n+2))`
`"or "3.83xx3.83=n^(2)+2n`
`14.6689=n^(2)+2n`
`n=3`
Hence, number of unpaired electrons in d-subshell of chromium (Cr= 24) = 3.
So. the configuration of chromium ion is
`Cr^(3+)=1s^(2),2s^(2),2p^(6),3s^(2)3p^(6)3d^(3)`
In `[Cr(H_(2)O)_(6)]Cl_(2),` oxidation state of Cr is + 3.
Hence, in `3d^(3)` the distribution of electrons
`3d_(xy)^(1),3d_(yz)^(1),3d_(zx)^(1),3d_(x^(2-y^(2)))^(0),3d_(z^(2))^(0)`