Question

Consider the following complex ions P, Q and R.
P = ${{[Fe{{F}_{6}}]}^{3-}}$ , Q = ${{[V{{({{H}_{2}}O)}_{6}}]}^{2+}}$ and R = ${{[Fe{{({{H}_{2}}O)}_{6}}]}^{2+}}$
The correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.) is _____.
a.) R < Q < P
b.) Q < R < P
c.) R < P < Q
d.) Q < P < R

Answer 1

Hint: The spin only magnetic moment is the moment which is caused due to the spin of the particle. It can be found by writing the correct electronic configuration.

Complete Solution :
Here in compound P, iron is present in +3 oxidation state
Atomic number of iron= 26
Electronic configuration of iron= $[Ar]3{{d}^{6}}4{{s}^{2}}$
Electronic configuration of iron in +3 oxidation state = $[Ar]3{{d}^{5}}$
Number of unpaired electron=5
Ligand in the complex is a weak ligand hence the number of unpaired electrons remains same
in compound Q, vanadium is present in +2 oxidation state
Atomic number of vanadium= 23
Number of unpaired electron= 3
Ligand in the complex is a weak ligand hence the number of unpaired electrons remains same
in compound R, iron is present in +2 oxidation state
Atomic number of iron = 26
Electronic configuration of iron= $[Ar]3{{d}^{6}}4{{s}^{2}}$
Electronic configuration of iron in +2 oxidation state= $[Ar]3{{d}^{6}}$
Number of unpaired electron=4
Ligand in the complex is a weak ligand hence the number of unpaired electrons remains same
Spin only magnetic moment is calculated by using the formulae:
\[\mu =\sqrt{n(n+2)}B.M.\]
Where n is the number of unpaired electrons

Spin only magnetic moment of $F{{e}^{+3}}$
Number of unpaired electrons = 5
So, \[\mu =\sqrt{n(n+2)}B.M.\] = $\sqrt{5(5+2)}$= 5.91 B.M.

Spin only magnetic moment of ${{V}^{+2}}$
Number of unpaired electrons = 3
So, \[\mu =\sqrt{n(n+2)}B.M.\] = $\sqrt{3(3+2)}$= 3.87 B.M.

Spin only magnetic moment of $F{{e}^{+2}}$
Number of unpaired electrons = 4
So, \[\mu =\sqrt{n(n+2)}B.M.\] = $\sqrt{4(4+2)}$= 4.69 B.M.
So, the correct answer is “Option B”.

Note: While writing the electronic configuration we should take care of the charge present on the element. If there is a positive charge on the element remove that number of electrons while writing the electronic configuration similarly if there is a negative charge on the element then pass that number of electron while writing the electronic configuration.