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Question: Consider the following complex ions P, Q and R. P = \({{[Fe{{F}_{6}}]}^{3-}}\) , Q = \({{[V{{({{H}...

Consider the following complex ions P, Q and R.
P = [FeF6]3−{{[Fe{{F}_{6}}]}^{3-}} , Q = [V(H2O)6]2+{{[V{{({{H}_{2}}O)}_{6}}]}^{2+}} and R = [Fe(H2O)6]2+{{[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

Explanation

Solution

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]3d64s2[Ar]3{{d}^{6}}4{{s}^{2}}
Electronic configuration of iron in +3 oxidation state = [Ar]3d5[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]3d64s2[Ar]3{{d}^{6}}4{{s}^{2}}
Electronic configuration of iron in +2 oxidation state= [Ar]3d6[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:
μ=n(n+2)B.M.\mu =\sqrt{n(n+2)}B.M.
Where n is the number of unpaired electrons

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

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

Spin only magnetic moment of Fe+2F{{e}^{+2}}
Number of unpaired electrons = 4
So, μ=n(n+2)B.M.\mu =\sqrt{n(n+2)}B.M. = 4(4+2)\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.