Question

Amongst the following the lowest degree of paramagnetism per mol of the compound at $298{\text{ K}}$ will be shown by:
A) ${\text{FeS}}{{\text{O}}_{\text{4}}}{\text{.6}}{{\text{H}}_{\text{2}}}{\text{O}}$
B) ${\text{NiS}}{{\text{O}}_{\text{4}}}{\text{.6}}{{\text{H}}_{\text{2}}}{\text{O}}$
C) ${\text{MnS}}{{\text{O}}_{\text{4}}}{\text{.4}}{{\text{H}}_{\text{2}}}{\text{O}}$
D) ${\text{CuS}}{{\text{O}}_{\text{4}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}$

Answer 1

We know that the complexes having unpaired electrons are paramagnetic in nature and those having no unpaired electrons are diamagnetic in nature. The crystal field theory helps to describe if the compound is paramagnetic or diamagnetic.

Complete step by step answer:
The ligand bonds to the central metal atom and donates a pair of electrons to the central metal atom and thus, form a coordination complex.
We are given four complexes, ${\text{FeS}}{{\text{O}}_{\text{4}}}{\text{.6}}{{\text{H}}_{\text{2}}}{\text{O}}$, ${\text{NiS}}{{\text{O}}_{\text{4}}}{\text{.6}}{{\text{H}}_{\text{2}}}{\text{O}}$, ${\text{MnS}}{{\text{O}}_{\text{4}}}{\text{.4}}{{\text{H}}_{\text{2}}}{\text{O}}$ and ${\text{CuS}}{{\text{O}}_{\text{4}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}$. In the given complexes, the ligand is the water molecule.
Water is a weak field ligand. Thus, it will not cause the pairing of the electrons of the metal atom and high spin complexes are formed.
The oxidation state of ${\text{Fe}}$ in ${\text{FeS}}{{\text{O}}_{\text{4}}}{\text{.6}}{{\text{H}}_{\text{2}}}{\text{O}}$ is +2. Thus, the outer shell electronic configuration of ${\text{F}}{{\text{e}}^{2 + }}$ is $3{d^6}4{s^0}$. Thus,

The oxidation state of ${\text{Ni}}$ in ${\text{NiS}}{{\text{O}}_{\text{4}}}{\text{.6}}{{\text{H}}_{\text{2}}}{\text{O}}$ is +2. Thus, the outer shell electronic configuration of ${\text{N}}{{\text{i}}^{2 + }}$ is $3{d^8}4{s^0}$. Thus,

The oxidation state of ${\text{Mn}}$ in ${\text{MnS}}{{\text{O}}_{\text{4}}}{\text{.4}}{{\text{H}}_{\text{2}}}{\text{O}}$ is +2. Thus, the outer shell electronic configuration of ${\text{M}}{{\text{n}}^{2 + }}$ is $3{d^5}4{s^0}$. Thus,

The oxidation state of ${\text{Cu}}$ in ${\text{CuS}}{{\text{O}}_{\text{4}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}$ is +2. Thus, the outer shell electronic configuration of ${\text{C}}{{\text{u}}^{2 + }}$is $3{d^9}4{s^0}$. Thus,

From the electronic configurations, we can see that the ${\text{M}}{{\text{n}}^{2 + }}$ has five unpaired electrons. Thus, ${\text{MnS}}{{\text{O}}_{\text{4}}}{\text{.4}}{{\text{H}}_{\text{2}}}{\text{O}}$ shows highest degree of paramagnetism. ${\text{F}}{{\text{e}}^{2 + }}$ has four unpaired electrons. Thus, the degree of paramagnetism of ${\text{FeS}}{{\text{O}}_{\text{4}}}{\text{.6}}{{\text{H}}_{\text{2}}}{\text{O}}$ is lower than that of ${\text{MnS}}{{\text{O}}_{\text{4}}}{\text{.4}}{{\text{H}}_{\text{2}}}{\text{O}}$. ${\text{N}}{{\text{i}}^{2 + }}$ has two unpaired electrons. Thus, the degree of paramagnetism of ${\text{NiS}}{{\text{O}}_{\text{4}}}{\text{.6}}{{\text{H}}_{\text{2}}}{\text{O}}$ is lower than that of ${\text{FeS}}{{\text{O}}_{\text{4}}}{\text{.6}}{{\text{H}}_{\text{2}}}{\text{O}}$. ${\text{C}}{{\text{u}}^{2 + }}$ has one unpaired electron. Thus, ${\text{CuS}}{{\text{O}}_{\text{4}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}$ shows lowest degree of paramagnetism.

Thus, the correct option is (D) ${\text{CuS}}{{\text{O}}_{\text{4}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}$.

Note: Strong field ligands form low spin complexes. The examples of strong field ligands are ${\text{C}}{{\text{O}}^ - }$, ${\text{C}}{{\text{N}}^ - }$. Weak field ligands form high spin complexes. The examples of weak field ligands are ${{\text{F}}^ - }$, ${\text{C}}{{\text{l}}^ - }$, ${{\text{H}}_{\text{2}}}{\text{O}}$.