A metal, $$M$$ forms chlorides in its $$+2$$ and $$+4$$ oxidation states. Which of the following statements about these chlorides is correct?

According to Fajans’ rules, the covalent character of a compound increases when the cation possesses

• a high positive charge (large polarising power), and
• a small ionic radius (greater charge density).

Comparing the two cations here:

$$M^{2+} \quad \text{vs} \quad M^{4+}$$

• $$M^{4+}$$ carries twice the charge of $$M^{2+}$$ but has practically the same size; therefore its charge density is far larger.
• Consequently, $$M^{4+}$$ polarises the electron cloud of the chloride ion much more strongly than $$M^{2+}$$.

Greater polarisation leads to greater sharing of electron density between $$M^{4+}$$ and $$Cl^-$$, i.e. a higher covalent character. Hence

$$MCl_4$$ (with $$M^{4+}$$) is appreciably more covalent, whereas $$MCl_2$$ (with $$M^{2+}$$) is comparatively more ionic.

This reasoning directly validates statement C.

For completeness, examine the other options:

A more covalent compound is usually more volatile because intermolecular forces are weaker. Since $$MCl_4$$ is more covalent, it will be more volatile, not $$MCl_2$$. Therefore Option A is wrong.

Covalent chlorides dissolve better in organic solvents such as anhydrous ethanol. Hence $$MCl_4$$ (covalent) is more soluble than $$MCl_2$$ (ionic). Option B is wrong.

Covalent chlorides are more susceptible to hydrolysis by water. Thus $$MCl_4$$ is hydrolysed more readily than $$MCl_2$$, making Option D wrong.

Only Option C is consistent with Fajans’ rules and the observed trends in volatility, solubility, and hydrolysis.

Option C which is: $$MCl_2$$ is more ionic than $$MCl_4$$