Question

Covalent character of the following compounds $LiCl$, $BaCl_{2}$, $AlCl_{3}$, $CCl_{4}$ is in the order

• A. $LiCl <\, BaCl_{2} <\, CCl_{4} < \,AlCl_{3}$
• B. $CCl_{4} <\, AlCl_{3} <\, LiCl <\, BaCl_{2}$
• C. $BaCl_{2} < \,LiCl < AlCl_{3} <\, CCl_{4}$
• D. $AlCl_{3} <\, LiCl <\, BaCl_{2} < \,CCl_{4}$

Answer 1

Answer: (C)

Crystal radii $\begin{matrix}Li \to1.52\,\mathring{A}\\ Ba \rightarrow2.22\, \mathring{A}\\ Al \rightarrow1.43\,\mathring{A}\\ C \to0.77\,\mathring{A}\end{matrix}$
Carbon is smallest in size. So, it has highest polarising power and so most covalent. So, the order is $BaCl_{2}<\, LiCl <\, AlCl_{3}<\, CCl_{4}$