Question

Which of the following has highest magnetic moment?

• A. ${\left[Fe{F}_6\right]}^{3-}$
• B. ${\left[Co{\left({NH}_3\right)}_6\right]}^{3+}$
• C. ${\left[Fe{\left(CN\right)}_6\right]}^{4-}$
• D. ${\left[Mn{\left(CN\right)}_6\right]}^{4-}$

Answer 1

Answer: (A)

${\left[Fe{F}_6\right]}^{3-}$

$[Fe{F}_6{]}^{3-}$

Oxidation state of $Fe$; $x=6-3$

$F^{3+}=4s^{0}3d^5$ ($5$ unpaired $e^{-}$)

$\left[Co\right(N{H}_3{)}_6{]}^{3+}$

Oxidation state of $Co;x=+3$

$C{o}^{3+}=4s^{0}3d^6$ ($4$ unpaired $e^{-}$)

$\left[Fe\right(CN{)}_6{]}^{4-}$

Oxidation state of $Fe;x=6-4=+2$

$F^{2+}=4s^{0}3d^6$ ($4$ unpaired $e^{-}$)

$\left[Mn\right(CN{)}_6{]}^{4-}$

Oxidation state of $Mn;x=6-4=+2$

$M{n}^{2+}=4s^{0}3d^5$ ($5$ unpaired $e^{-}$)

It is known, more is the number of unpaired $e^{-}$ more is the value of magnetic moment. Here $[Fe{F}_6{]}^{3-}$ and $\left[Mn\right(CN{)}_6{]}^{4-}$ have $5$ unpaired $e^{-}s$ but since $C{N}^{-}$ is strong ligand than $F^{-}$ thereby in $\left[Mn\right(Cn{)}_6{]}^{4-}$ pairing of $e^{-}$ occurs thereby making its magnetic moment less, whereas in $[Fe{F}_6{]}^{3-}$ as $F^{-}$ is weak ligand the unpaired $e^{-}s$ remains as it is thereby making its magnetic moment high.