Question

The complex having the highest spin only magnetic moment is

• A. ${\left[Fe{\left(CN\right)}_6\right]}^{3-}$
• B. ${\left[Fe{\left(H_{2}O\right)}_6\right]}^{2+}$
• C. ${\left[Mn{F}_6\right]}^{4-}$
• D. ${\left[NiC{l}_4\right]}^{2-}$

Answer 1

Answer: (C)

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

As $C{N}^{-}$is a strong ligand pairing will occur.
${\left[Fe(CN{)}_6\right]}^{3-}$

$\therefore n=1\Rightarrow \mu =\sqrt{1(1+2)}=\sqrt{3}$
(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 $F{e}^{2+}$ is $-[Ar]3d^{6}4s^0$
As $H_{2}O$ is weak ligand, so pairing will not occur.
${\left[Fe{\left(H_{2}O\right)}_6\right]}^{2-}$

$\therefore n=4$
$\mu =\sqrt{4(4+2)}=\sqrt{24}$
(c) ${\left[Mn{F}_6\right]}^{4-}$
The oxidation number of $Mn$ is $+2$.
It's configuration is $-[Ar]3d^{5}4s^0$
As $F^{-}$is a weak field ligand, so pairing will not occur.
${\left[Mn{F}_6\right]}^{4-}$

$\therefore n=5$
$\mu =\sqrt{5(5+2)}=\sqrt{35}$
(d) ${\left[NiC{l}_4\right]}^{2-}$
In $NiC{l}_4^{2-}$, the oxidation number of $Ni$ will be $+2$.
It's configuration is $-[Ar]3d^{8}4s^0$
As $C{l}^{-}$is a weak field ligand, so pairing will not occur.
${\left[NiC{l}_4\right]}^{2-}$

$n=2$
$\mu =\sqrt{2(2+2)}=\sqrt{8}$
Thus, ${\left[Mn{F}_6\right]}^{4-}$ will have the highest spin only magnetic moment.