Question
$$\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]C{l_3}$$ ( at. no. of $$Cr=24$$ ) has a magnetic moment of $$3.83\,BM,$$ the correct distribution of $$3d$$ electrons in the chromium of the complex is
A.
$$3d_{xy}^1,3d_{yz}^1,3d_{{z^2}}^1$$
B.
$$3{d_{\left( {{x^2} - {y^2}} \right)}},3d_{{z^2}}^1,3d_{xz}^1$$
C.
$$3{d_{xy}},3{d_{\left( {{x^2} - {y^2}} \right)}},3d_{yz}^1$$
D.
$$3d_{xy}^1,3d_{yz}^1,3d_{zx}^1$$
Answer :
$$3d_{xy}^1,3d_{yz}^1,3d_{zx}^1$$
Solution :
Magnetic moment $$\left( \mu \right) = \sqrt {n\left( {n + 2} \right)} \,\,BM$$
$$\eqalign{
& {\text{or}}\,\,\,3.83 = \sqrt {n\left( {n + 2} \right)} \cr
& {\text{or}}\,\,\,3.83 \times 3.83 = {n^2} + 2n \cr
& 14.6689 = {n^2} + 2n \cr
& n \simeq 3 \cr} $$
Hence, number of unpaired electrons in $$d$$ - subshell of chromium $$(Cr= 24) =3.$$
So, the configuration of chromium ion is $$C{r^{3 + }} = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^3}$$
In $$\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]C{l_2},$$ oxidation state of $$Cr$$ is + 3.
Hence, in $$3{d^3}$$ the distribution of electrons $$3d_{xy}^1,3d_{yz}^1,3d_{zx}^1,3d_{{x^2} - {y^2}}^0,3d_{{z^2}}^0$$