Correct representation between the pairing energy (P) and the C.F.S.E (${{\Delta }_{0}}$) in complex ion ${{[Ir{{({{H}_{2}}O)}_{6}}]}^{3+}}$ is
(A) ${{\Delta }_{0}}$ $<$ $P$
(B) ${{\Delta }_{0}}>P$
(C) ${{\Delta }_{0}}=P$
(D) Cannot comment

Hint: According to crystal field theory there is a difference of 10${{D}_{q}}$ between the energy states i.e. ${{t}_{2g}}and\text{ }{{e}_{g}}$. When an electron goes into the ${{t}_{2g}}$ state which is the lower energy state it stabilizes the system by an amount of -4${{D}_{q}}$ and when an electron goes into the ${{e}_{g}}$ it destabilizes the system by +6${{D}_{q}}$.

Complete step by step solution:
The stability which results due to the placing of a transition metal into a field that is caused by the set of the ligands which surrounds it is called the crystal field stabilization energy.
Pairing energy is the energy required to place two electrons in the same orbital. If the value of crystal field splitting is small because of the weak bond ligand then the value of the pairing energy will be larger similarly if the value of the crystal field splitting energy is larger because of the strong field ligands then the value of the pairing energy will be smaller.
In the given compound i.e. ${{[Ir{{({{H}_{2}}O)}_{6}}]}^{3+}}$ water acts as a weak ligand, the crystal field energy of water is smaller than the pairing energy.

Hence the correct answer is option (A) i.e. Representation between the pairing energy (P) and the C.F.S.E (${{\Delta }_{0}}$) in complex ion ${{[Ir{{({{H}_{2}}O)}_{6}}]}^{3+}}$is ${{\Delta }_{0}}$ $<$ $P$.

Note: Complexes which have a higher number of unpaired electrons are known as high spin complexes and the complexes which have a low number of unpaired electrons are known as low spin complexes. Most of the time, the high spin complexes have weak field ligands and the low spin complexes have strong field ligands.