1. Identify the comparison

A metal $M$ forms two chlorides:

• $MCl_2$ where metal is in $+2$ oxidation state
• $MCl_4$ where metal is in $+4$ oxidation state

We must compare their properties.

2. Use Fajans' rule / polarization concept

A cation with:

• higher positive charge,
• smaller size,

has greater polarising power and hence produces more covalent character in its compounds.

So, compared to $M^{2+}$, the ion $M^{4+}$ has much higher polarising power.

Therefore: $MCl_4 \text{ is more covalent, while } MCl_2 \text{ is more ionic.}$

So option A is correct.

3. Check the remaining options

Option B: $MCl_2$ is more easily hydrolysed than $MCl_4$

More covalent chlorides are generally more easily hydrolysed.

Since $MCl_4$ is more covalent than $MCl_2$, we expect: $MCl_4 \text{ is more easily hydrolysed than } MCl_2$ Hence B is false.

Option C: $MCl_2$ is more volatile than $MCl_4$

More covalent compounds are generally more volatile, while ionic compounds are less volatile.

Since $MCl_4$ is more covalent, it is expected to be more volatile. Thus C is false.

Option D: $MCl_2$ is more soluble in anhydrous ethanol than $MCl_4$

Covalent compounds tend to dissolve better in organic solvents such as anhydrous ethanol, while ionic compounds prefer polar solvents like water.

Since $MCl_4$ is more covalent, it should be more soluble in anhydrous ethanol. Thus D is false.

4. Final conclusion

The correct statement is: $\boxed{\text{A: } MCl_2 \text{ is more ionic than } MCl_4}$

5. Comparison with stored correct answer

Stored correct answer: A

Derived answer: A

They match.