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 $[Mn{O}_4{]}^{2-}$
• C. Co in $[CoB{r}_4{]}^{2-}$
• D. Pt in $\left[Pt\right(H_{2}O{)}_{2}B{r}_2]$

Answer 1

The correct option is D Pt in $\left[Pt\right(H_{2}O{)}_{2}B{r}_2]$

Ni $\left[Ni\right(dmg{)}_2]$ complex is square planar and dimagnetic. Pt (II) have $5d^8$ configuration. Complex is square planar and therefore, dimagnetic. So, ${\mu }_{B.M}=0$.