Question

The species that has a spin only magnetic moment of $5.9BM,$ is

• A. $Ni(CO{)}_4\left(T_d\right)$
• B. ${\left[MnB{r}_4\right]}^{2-}\left(T_d\right)$
• C. ${\left[NiC{l}_4\right]}^{2-\left(T_d\right)}$
• D. ${\left[Ni(CN{)}_4\right]}^{2-}$ (square planar)

Answer 1

Answer: (B)

1. $\left[Ni(CO{)}_4\right]$

$Ni\to [Ar]3d^{8}4s^2$

due to presence of strong field ligand back pairing of electrons will take place, so

$Ni\to [Ar]3d^{10}4s^0$



Hybridisation $=s{p}^3,$ tetrahedral

Number of unpaired $e^{-}=0$

Hence ${\mu }_m=0$

2. ${\left[MnB{r}_4\right]}^{2-}$

$M{n}^{+2}\to [Ar]3d^{5}4s^0$



Since $B{r}^{-}$ is weak field ligand, so

hybridisation $=s{p}^3$, tetrahedral

number of unpaired $e^{-}=5$

${\mu }_m=\sqrt{n(n+2)}=\sqrt{35}BM=5.91BM$

3. ${\left[NiC{l}_4\right]}^{2-}$

$N{i}^{+2}\to [Ar]3d^{8}4s^0$



No. of unparied electron $=2$

${\mu }_m=\sqrt{8}BM=2.83BM$

4. ${\left[Ni(CN{)}_4\right]}^{2-}$

$N{i}^{+2}\to [Ar]3d^{8}4s^0$

$CN$ is strong field ligand so rearrangement of electrons takes place here.



number of unpaired electron $=0$

${\mu }_m=0$