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The correct sequence of increasing covalent character is represented by -
A
$BeCl_2 \lt NaCl \lt LiCl$
B
$NaCl \lt LiCl \lt BeCl_2$
C
$BeCl_2 \lt LiCl \lt NaCl$
D
$LiCl \lt NaCl \lt BeCl_2$
Detailed Solution
According to Fajans' rules, the covalent character of an ionic compound increases with the polarising power of the cation, which is high for a small cation with a high charge.
All three compounds have the same anion, $Cl^-$, so the cation decides the order.
$Na^+$ is the largest and has a single charge, so it has the least polarising power; NaCl is the most ionic.
$Li^+$ is much smaller than $Na^+$, so LiCl has more covalent character than NaCl.
$Be^{2+}$ is very small and doubly charged, so it has the highest polarising power; $BeCl_2$ is essentially covalent.
So the increasing order of covalent character is $NaCl \lt LiCl \lt BeCl_2$.
All three compounds have the same anion, $Cl^-$, so the cation decides the order.
$Na^+$ is the largest and has a single charge, so it has the least polarising power; NaCl is the most ionic.
$Li^+$ is much smaller than $Na^+$, so LiCl has more covalent character than NaCl.
$Be^{2+}$ is very small and doubly charged, so it has the highest polarising power; $BeCl_2$ is essentially covalent.
So the increasing order of covalent character is $NaCl \lt LiCl \lt BeCl_2$.