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  4. The complex having the highest spin only magnetic moment is

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Q. The complex having the highest spin only magnetic moment is

KVPYKVPY 2018Coordination Compounds

Solution:

The spin only magnetic moment for a complex can be calculated using formula
$\mu=\sqrt{n(n+2)}$
(a) $\left[ Fe ( CN )_{6}\right]^{3-}$
The oxidation number $Fe$ in $\left[ Fe ( CN )_{6}\right]^{3-}$ is $+3$.
The electronic configuration of $Fe ^{3+}$ is $-( Ar ) 3 d^{5} 4 s^{0}$
image
As $CN ^{-}$is a strong ligand pairing will occur.
$\left[ Fe ( CN )_{6}\right]^{3-}$
image
$\therefore n=1 \Rightarrow \mu=\sqrt{1(1+2)}=\sqrt{3}$
(b) $\left[ Fe \left( H _{2} O \right)_{6}\right]^{2+}$
The oxidation number of $Fe$ in $\left[ Fe \left( H _{2} O \right)_{6}\right]^{2+}$ is $+2$.
The electronic configuration of $Fe ^{2+}$ is $-[ Ar ] 3 d^{6} 4 s^{0}$
As $H _{2} O$ is weak ligand, so pairing will not occur.
$\left[ Fe \left( H _{2} O \right)_{6}\right]^{2-}$
image
$\therefore n=4 $
$ \mu=\sqrt{4(4+2)}=\sqrt{24}$
(c) $\left[ MnF _{6}\right]^{4-}$
The oxidation number of $Mn$ is $+2$.
It's configuration is $-[ Ar ] 3 d^{5} 4 s^{0}$
As $F ^{-}$is a weak field ligand, so pairing will not occur.
$\left[ MnF _{6}\right]^{4-}$
image
$\therefore n=5 $
$\mu=\sqrt{5(5+2)}=\sqrt{35}$
(d) $\left[ NiCl _{4}\right]^{2-}$
In $NiCl _{4}^{2-}$, the oxidation number of $Ni$ will be $+2$.
It's configuration is $-[ Ar ] 3 d^{8} 4 s^{0}$
As $Cl ^{-}$is a weak field ligand, so pairing will not occur.
$\left[ NiCl _{4}\right]^{2-}$
image
$n=2 $
$\mu=\sqrt{2(2+2)}=\sqrt{8}$
Thus, $\left[ MnF _{6}\right]^{4-}$ will have the highest spin only magnetic moment.