Question

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

• A. $Ni ( CO )_{4}\left( T _{ d }\right)$
• B. $\left[ MnBr _{4}\right]^{2-}\left( T _{ d }\right)$
• C. $\left[ NiCl _{4}\right]^{2-\left( T _{ d }\right)}$
• D. $\left[ Ni ( CN )_{4}\right]^{2-}$ (square planar)

Answer 1

Answer: (B)

1. $\left[ Ni ( CO )_{4}\right]$

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

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

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



Hybridisation $= sp ^{3},$ tetrahedral

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

Hence $\mu_{ m }=0$

2. $\left[ MnBr _{4}\right]^{2-}$

$ Mn ^{+2} \rightarrow[ Ar ] 3 d ^{5} 4 s ^{0} $



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

hybridisation $= sp ^{3}$, tetrahedral

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

$\mu_{ m }=\sqrt{ n ( n +2)}=\sqrt{35} BM =5.91 BM$

3. $\left[ NiCl _{4}\right]^{2-}$

$Ni ^{+2} \rightarrow[ Ar ] 3 d ^{8} 4 s ^{0}$



No. of unparied electron $=2$

$\mu_{ m }=\sqrt{8} BM =2.83 BM$

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

$Ni ^{+2} \rightarrow[ Ar ] 3 d ^{8} 4 s ^{0}$

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



number of unpaired electron $=0$

$\mu_{ m }=0$