Question

The heat of hydration of $ C{{r}^{2+}} $ ion is $ 460\text{ }Kcal/mole. $ For $ {{\left[ Cr{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}} $ $ {{\Delta }_{0}}=13900\text{ }c{{m}^{-1}}. $ What heat of hydration would be, if there were no crystal field stabilization energy?
(A) $ -436Kcal/mole. $
(B) $ -245Kcal/mole. $
(C) $ -420Kcal/mole. $
(D) $ +436 Kcal/mole. $

Answer 1

Hint: We know that the octahedral complexes are formed in the element in which there is the presence of d-orbital or we can say that in octahedral complexes, there is the involvement of d-orbitals.

Complete answer:
As we know that the crystal field stabilization energy so that the crystal field stabilization energy is the energy of electronic configuration in the ligand field - Electronic configuration in the isotropic field. Complexes which have a higher number of unpaired electrons are called high spin complexes and the ones which have low number of unpaired electrons are called low spin complexes. Most of the time, high spin complexes have weak field ligands and hence their splitting energy has lower value.
Conversely, low spin complexes have strong field ligands and hence have a higher value of splitting energy. Also, it must be noted that when the ligand is high spin complex then the pairing of the electron will occur and when the ligand is low spin complex then first the pairing of an electron in a set will occur then the electron will move into the orbital. The possibility to make a mistake is that you may use the crystal field splitting energy formula for tetrahedral, not octahedral. But all the compounds are octahedral so do not confuse between them and for $ {{\left[ Cr{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}=-0.6{{\Delta }_{0}} $ thus, the experimental value $ - $ calculated value is (i.e., $ -24Kcal/mole $ )
So, the Heat of hydration is $ 460-24=436 Kcal/mol. $
Therefore, the correct answer is option D.

Note:
Remember that the splitting pattern in each complex compound is octahedral. If a strong field ligand is present, then back pairing will take place and splitting energy can never be zero, since no electron can go into orbitals.