Identify the correct trend given below:
(Atomic No. = Ti : 22, Cr : 24 and Mo : 42)

We have to compare the octahedral crystal-field splitting parameter, written as $$\Delta_o$$ (usually shown simply as $$\Delta$$), for three pairs of complexes that carry the same ligand $$H_2O$$. Remember that $$H_2O$$ is a weak-field ligand and therefore all the differences we are going to discuss come only from the metal centre and its oxidation state.

First we recall two general rules of crystal-field theory.

Rule 1 (same period, different oxidation state) — For a given metal ion and a given ligand, a higher oxidation state produces a larger splitting. Symbolically

$$\Delta_o \propto \text{oxidation state of the metal}. \quad -(1)$$

This happens because the higher positive charge pulls the d-electrons closer, allowing the ligand field to interact more strongly with them.

Rule 2 (same group, going down the group) — For the same oxidation state and the same ligand, $$4d$$ and $$5d$$ metals experience a larger splitting than their $$3d$$ congeners:

$$\Delta_o(4d) \gt \Delta_o(3d), \qquad \Delta_o(5d) \gt \Delta_o(4d). \quad -(2)$$

The underlying reason is that the more diffuse $$4d$$ and $$5d$$ orbitals overlap better with the ligand orbitals, strengthening the ligand field.

Now we apply these rules to the complexes given in the question.

Comparison 1: $$[Cr(H_2O)_6]^{2+}$$ versus $$[Mo(H_2O)_6]^{2+}$$

Both ions are in the +2 oxidation state, so Rule 1 does not favour either. Chromium is a $$3d$$ element (period 4) whereas molybdenum is a $$4d$$ element (period 5). By Rule 2 we therefore have

$$\Delta_o\bigl([Mo(H_2O)_6]^{2+}\bigr) \gt \Delta_o\bigl([Cr(H_2O)_6]^{2+}\bigr).$$

So the inequality is

$$\Delta_o\,[Cr(H_2O)_6]^{2+} \lt \Delta_o\,[Mo(H_2O)_6]^{2+}. \quad -(3)$$

Comparison 2: $$[Ti(H_2O)_6]^{3+}$$ versus $$[Ti(H_2O)_6]^{2+}$$

Here the metal is the same (titanium, a $$3d$$ element), and the ligand is the same, so Rule 2 is irrelevant. Instead we invoke Rule 1. The +3 ion has a larger positive charge than the +2 ion, so we expect a larger splitting for the +3 species:

$$\Delta_o\bigl([Ti(H_2O)_6]^{3+}\bigr) \gt \Delta_o\bigl([Ti(H_2O)_6]^{2+}\bigr). \quad -(4)$$

Putting both pieces together

From (3) we need the symbol < between the chromium(II) and molybdenum(II) complexes, and from (4) we need the symbol > between the titanium(III) and titanium(II) complexes. Scanning the four options, only Option C contains both these inequalities in exactly this order.

Hence, the correct answer is Option 3.