Question

Which of the following is correct order of increasing conductance?

• A. ${\wedge }_{m}HCl$> ${\wedge }_{m}KCl$ > ${\wedge }_{m}NaCl$ >${\wedge }_{m}LiCl$
• B. ${\wedge }_{m}LiCl$> ${\wedge }_{m}NaCl$ > ${\wedge }_{m}KCl$ >${\wedge }_{m}HCl$
• C. ${\wedge }_{m}LiCl$> ${\wedge }_{m}NaCl$ > ${\wedge }_{m}HCl$ >${\wedge }_{m}KCl$
• D. ${\wedge }_{m}LiCl>{\wedge }_{m}HCl>{\wedge }_{m}NaCl>{\wedge }_{m}KCl$

Answer 1

The correct option is A ${\wedge }_{m}HCl$> ${\wedge }_{m}KCl$ > ${\wedge }_{m}NaCl$ >${\wedge }_{m}LiCl$

First check the anion here.
In all the compounds, anion is same, i.e., $C{l}^{-}$.
Now, we know the conductance of an ion in the solution is inversely propositional to its size in the aqueous solution.
So, it will only depend on size of cation in the aqueous solution.
We have to note two things here.
Size of $H^{\oplus }andO{H}^{-}$ ions are extremely small yet they have a very high conductance value in the solution (as opposed to what you might have expected). Infact, $H^{\oplus }$ ions have the highest ${\wedge }_m$ at any temperature followed by $O{H}^{-}$ ions. This is due to the mechanism by which they travel in the aqueous solution.
$H^{\oplus }$ moves in the solution as shown:

$L{i}^{\oplus }$ ion has a very high charge-to-size ratio (charge density) due to which it gets hydrated to a large extent, when dissolved in water.
Thus, Size of $L{i}^{\oplus }$ (aq) >> size of $L{i}^{\oplus }$ (g).
Note:
$r_K{\oplus }_{\left(g\right)}$>$r_{Na}{\oplus }_{\left(g\right)}$>$r_{Li}{\oplus }_{\left(g\right)}but{r}_K{\oplus }_{\left(aq\right)}$<$r_{Na}{\oplus }_{\left(aq\right)}\Rightarrow {\wedge }_m{K}^{\oplus }\left(aq\right)$>${\wedge }_{m}N{a}^{\oplus }\left(aq\right)$>${\wedge }_{m}L{i}^{\oplus }\left(aq\right)$
A heavily hydrated ion drags water molecules along with it when it moves in a solution which makes it less mobile (or less conducting).
Now,
$Since,{\wedge }_{m{H}^{+}}$>${\wedge }_{m{K}^{+}}$>${\wedge }_{mN{a}^{+}}$>${\wedge }_{mL{i}^{+}}$
Hence,
${\wedge }_{m}HCl$> ${\wedge }_{m}KCl$ > ${\wedge }_{m}NaCl$ >${\wedge }_{m}LiCl$