Question

The correct order of magnetic moments(spin only values in B.M.) among the following is:

• A. ${\left[Mn{Cl}_4\right]}^{2-}>{\left[Co{Cl}_4\right]}^{2-}>{\left[Fe{\left(CN\right)}_6\right]}^{4-}$
• B. ${\left[Fe{\left(CN\right)}_6\right]}^{4-}>{\left[Co{Cl}_4\right]}^{2-}>{\left[Mn{Cl}_4\right]}^{2-}$
• C. ${\left[Fe{\left(CN\right)}_6\right]}^{4-}>{\left[Mn{Cl}_4\right]}^{2-}>{\left[Co{Cl}_4\right]}^{2-}$
• D. ${\left[Mn{Cl}_4\right]}^{2-}>{\left[Fe\left(CN\right)6\right]}^{4-}>{\left[Co{Cl}_4\right]}^{2-}$

Answer 1

Answer: (A)

${\left[Mn{Cl}_4\right]}^{2-}>{\left[Co{Cl}_4\right]}^{2-}>{\left[Fe{\left(CN\right)}_6\right]}^{4-}$

Because $CN$ ligand being strong it causes pairing of the $e^{-}s$ in the 3d-orbital of $Fe$ thereby making it diamagnetic. hence $Fe=0$

And,

In $[MnC{l}_4{]}^{2-}$ and $[CoC{l}_4{]}^{2-}$; $Cl$ being weak ligand, the pairing does not occurs thereby having $M{n}^{2+}$ having more number of unpaired $e^{-}s$ i.e; $5$ has more value of magnetic moment where $C{o}^{2+}$ has only $3$ unpaired $e^{-}s$.