Question

Identify the correct trend given below: (atomic number of Ti = 22, Cr = 24, Mo = 42)
(A) ${{\Delta }_{\circ }}\text{ of }{{\left[ Cr{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}>{{\left[ Mo{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}\text{ and }{{\Delta }_{\circ }}\text{ of }{{\left[ Ti{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{3+}}>{{\left[ Ti{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}$
(B) ${{\Delta }_{\circ }}\text{ of }{{\left[ Cr{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}<{{\left[ Mo{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}\text{ and }{{\Delta }_{\circ }}\text{ of }{{\left[ Ti{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{3+}}<{{\left[ Ti{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}$
(C) ${{\Delta }_{\circ }}\text{ of }{{\left[ Cr{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}<{{\left[ Mo{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}\text{ and }{{\Delta }_{\circ }}\text{ of }{{\left[ Ti{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{3+}}>{{\left[ Ti{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}$
(D) ${{\Delta }_{\circ }}\text{ of }{{\left[ Cr{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}>{{\left[ Mo{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}\text{ and }{{\Delta }_{\circ }}\text{ of }{{\left[ Ti{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{3+}}>{{\left[ Ti{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}$

Answer 1

Hint: The crystal field splitting depends upon certain factors. Higher is the oxidation state, higher will be the crystal field splitting. Also, remember that $4d$ orbital is more diffuse than $3d$ and thus splitting is higher for $4d$-orbitals.

Complete step by step solution:
To answer this question, firstly we have to know the meaning of ${{\Delta }_{\circ }}$.
Under a spherical electrostatic field like in s-orbitals, the degeneracy of the orbital is maintained but under non-spherical electrostatic field or unsymmetrical field like in p, d- orbitals, splitting of orbitals occur which give rise to two non-degenerate energy levels and this is known as crystal field splitting. ${{\Delta }_{\circ }}$ is the crystal field splitting energy of a complex.
This energy depends on certain factors. To answer this question, we have to discuss them and then compare the ${{\Delta }_{\circ }}$ of the given complexes.
The first factor is geometry as it changes the d-orbital splitting pattern.
The next factor is the nature of the metal.
- With the increase in oxidation state, the extent of metal-ligand electrostatic interaction increases leading to more splitting.
- More the $d$-orbital is diffused higher will be the tendency of the ligand to interact and thus the splitting will increase.
And the last factor is the nature of the ligand depending upon the spectro-chemical series.
Now, let us see the complexes given to us.
Here, the ligand is the same for all the complexes i.e. ${{H}_{2}}O$. So let us compare the other factors i.e. the nature of the metal.
Here, we have titanium in +3 and +2 oxidation states. The +3 oxidation state will have a higher splitting as we have already discussed above.
Then we have chromium and molybdenum. The charge is the same for both so we will see the nature of $d$-orbitals.
Molybdenum has $4d$-orbitals and chromium have $3d$-orbitals. $4d$ is more diffused than $3d$ thus has higher splitting.

Therefore, the correct answer is an option- (C) ${{\Delta }_{\circ }}\text{ of }{{\left[ Cr{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}<{{\left[ Mo{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}\text{ and }{{\Delta }_{\circ }}\text{ of }{{\left[ Ti{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{3+}}>{{\left[ Ti{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}$.

Note: We can arrange the ligands according to their strengths in form of a series. This series is known as the spectro-chemical series- ${{I}^{-}}In this series, ligands towards the left (iodide) are weak field ligands and ligands towards the right ($CO$) are stronger field ligands. Strong field ligands have a higher splitting and result in higher crystal field stabilisation energy whereas weak field ligands have a lower crystal field splitting. Generally, weak field ligands are known to form high spin complexes and strong field ligands form low spin complexes.