Question

In the complex ion $M{L}_6^{n+},M^{n+}$ has four d-electrons and L is a strong field neutral ligand. According to crystal field theory, the magnetic properties of the complex ion correspond to how many unpaired electrons?

• A. 4
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is C 2

According to the quesiton, $M^{n+}$ have 4 d- $e^{-},$ So we can write for $M^{4+}$
$\underset{(n-1)d}{\begin{array}{ccccc}\uparrow & \uparrow & \uparrow & \uparrow & \end{array}}\underset{\left(ns\right)}{\begin{array}{c}\end{array}}\underset{\left(np\right)}{\begin{array}{ccc}& & \end{array}}$
In the presence of 6 strong field ligand, pairing occurs.
According to CFT , for strong field ligand:
$t_{2g}^4{e}_g^0$

$\underset{e_g}{\begin{array}{cc}& \end{array}}$

$\underset{t_{2g}}{\begin{array}{ccc}\uparrow \downarrow & \uparrow & \uparrow \end{array}}$

Number of unpaired $e^{-}=2$
Therefore, the correct option is (c).