Question

$[Cr(H_{2}O{)}_6]C{l}_3$ has a magnetic moment of $3.83B.M$. The correct distribution of $3d$ electrons in the chromium of the complex is

• A. $3dx{y}^1,3dy{z}^1,3d{z}^2$
• B. $3d^{1}xy,3d^{1}yz,3d^{1}zx$
• C. $3d_{{\left(x^2-Y^2\right)}^{1,}}3d^1{z}^2,3d^{1}xz$
• D. $3dx{y}^1,3d^1{x}^2-y^2,3d^1{y}^2$

Answer 1

Answer: (B)

$3.83=\sqrt{n\left(n+2\right)}so,n=3$
So, these are three unpaired $e^{-}s$. Based on crystal field splitting the orbital of $t_{2g}$ set have one $e^{-}$ each.