JEE Main · 2006 · 30 Apr 2006 · Offline · Chemical Bonding and Molecular Structure

A metal,

M

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

(A) $MCl_{2}$ is more volatile than $MCl_{4}$
(B) $MCl_{2}$ is more soluble in anhydrous ethanol than $MCl_{4}$
(C) $MCl_{2}$ is more ionic than $MCl_{4}$ ✓ Correct
(D) $MCl_{2}$ is more easily hydrolysed than $MCl_{4}$

Answer: (C)

Solution

Fajans' rules say that a cation of higher charge and smaller size polarises an anion more strongly and so builds in greater covalent character. The metal in

MC l_{4}

is in the

+ 4

state, smaller and more highly charged than the

+ 2

metal in

MC l_{2}

, so

MC l_{4}

is the more covalent chloride; being more covalent, it is also the more volatile, the more soluble in a solvent like anhydrous ethanol and the more readily hydrolysed. The remaining chloride,

MC l_{2}

, is thus the more ionic of the two.

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option

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JEE Main · 2006 · 30 Apr 2006 · Offline · Chemical Bonding and Molecular Structure

Solution

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