Question

Magnetic moment of $[MnC{l}_4{]}^{2-}is5.92B.M$. Explain giving reason.

Answer 1

For the given complex $[MnC{l}_4{]}^{2-}$, the oxidation state of $Mn$ will be;

$x+4(-1)=-2$

$x=-2+4=+2$

Thus, oxidation state of $Mnis+2$.

Electronic configuration of
$M{n}^{2+}=\left[Ar\right]3d^{5}4s^0$

In the given complex , $C{l}^{-}$ is a weak field ligand thus it does not cause pairing of electrons.


Thus, $[MnC{l}_4{]}^{2-}has5$ unpaired electrons.

Now, magnetic moment is calculated by the formula:

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

where n = number of unpaired electrons

𝜇 = √(5(5+2)) = √(35 ) = 5.92 𝐵.𝑀.

So, magnetic moment of $[MnC{l}_4{]}^{2-}$ is $5.92BM$. due to the presence of 5 unpaired electrons.