Question

$\left[Cr{\left(H_{2}O\right)}_6\right]C{l}_3$ (at. No. of $Cr=24$) has a magnetic moment of $3.83B.M$. The correct distribution of $3d$ electrons in the Chromium of the complex is:

• A. ${(3d{x}^2-y^2)}^1,3d{z}^{2^1},3dx{z}^1$
• B. $3dx{y}^1,{(3d{x}^2-y^2)}^1;3dy{z}^1$
• C. $3dx{y}^1,3dy{z}^1,3dx{z}^1$
• D. $3dx{y}^1,3dy{z}^1,3d{z}^{2^1}$

Answer 1

Answer: (C)

$3dx{y}^1,3dy{z}^1,3dx{z}^1$

$u=\sqrt{n(n+2)}$

where, $n=$ unpaired electrons.

$⟹3.83=\sqrt{n(n+2)}$

$⟹n=3$

$\left[Cr\right(H_{2}O{)}_6]C{l}_3\Rightarrow C{r}^{+3}$

Crystal field splitting is shown above.

$3dx{y}^1,3dy{z}^1,3dx{z}^1$