Question

The complex, ${\left[Ni{\left(CN\right)}_4\right]}^{2-}$ has square planar configuration with zero magnetic moment. What would be the magnetic moment if it was tetrahedral?

• A. $2.83$ BM
• B. $1.73$ BM
• C. $4.84$ BM
• D. $5.9$ BM

Answer 1

The correct option is A $2.83$ BM

The oxidation state of $Ni$ in the given complex is $+2$. It has $3d^8$ electronic configuration. The number of electrons in the outermost orbit is eight.

We have to calculate the magnetic moment in a tetrahedral geometry.

The electronic configuration will be $d_{x^2-y^2}^2$, $d_{z^2}^2$, $d_{xy}^2$, $d_{yz}^1$, $d_{xz}^1$.

Hence, the number of unpaired electrons ($n$) is $2$ (according to the $d$ orbital splitting in the tetrahedral complex as shown in the figure).

We substitute this value in the formula of magnetic moment which is given as follows:

$\mu =\sqrt{n(n+2)}$ $=\sqrt{2(2+2)}=2.83$

Therefore, $\mu$ will be $2.83$ B.M.