Question

'Spin only' magnetic moment of $Ni$ in $\left[Ni\right(DMG{)}_2]$ is same as that found in:

• A. $Ni$ in $\left[NiC{l}_2\right(PP{h}_3{)}_2]$
• B. $Mn$ in $[Mn{O}_4{]}^{-}$
• C. $Co$ in $[CoB{r}_4{]}^{2-}$
• D. None

Answer 1

The correct option is B $Mn$ in $[Mn{O}_4{]}^{-}$

DMG is Dimethylglyoximato ligand having net charge of -1.
In $Ni(dmg{)}_2$ complex dmg acts as a strong field ligand and compells the $3d^8$ electron of $N{i}^{2+}$ to be paired. So, it is square planar and diamagnetic.
(A) In $\left[NiC{l}_2\right(PP{h}_3{)}_2]$, the $N{i}^{2+}$ ion is present having electronic configuration of $\left[Ar\right]3d^8$.
But here both the ligands act as weak field ligand. So, the electrons don't paired up and is paramagnetic.
(B) $Mn$ in $[Mn{O}_4{]}^{-}$ is in + 7 oxidation state having ${}_{18}\left[Ar\right]3d^0$ configuration. Complex ion is tetrahedral and diamagnetic.
(C) In $[CoB{r}_4{]}^{2-}$ Co is on +2 oxidation state having ${}_{18}\left[Ar\right]3d^7$. Here the $B{r}^{-}$ ligand is weak field ligand and elctrons don't paired up. The central metal ion $C{o}^{2+}$ has 3 unpaired electrons. So, the complex is paramagnetic.

Hence option B is correct.