Periodicity: Question 5

Worked solution

Step 1: Recall how Period 3 chlorides behave in water

• $\text{NaCl}$ is an ionic solid. It simply dissolves in water, separating into hydrated $\text{Na}^+$ and $\text{Cl}^-$ ions. No bonds are broken or new bonds formed with the water itself, so the solution stays neutral (pH $\approx 7$).
• $\text{MgCl}_2$ is also predominantly ionic, but $\text{Mg}^{2+}$ has a higher charge density than $\text{Na}^+$. It polarises the water molecules coordinated around it enough to release a very small amount of $\text{H}^+$, giving a solution that is only very slightly acidic (pH just below $7$).
• $\text{AlCl}_3$ has significant covalent character, and $\text{Al}^{3+}$ has a much higher charge density still; the hydrated $\text{Al}^{3+}$ ion polarises surrounding water molecules strongly enough to release a noticeable concentration of $\text{H}^+$, giving a solution that is clearly acidic (pH around $3$).
• $\text{SiCl}_4$ is covalent and simple molecular. It hydrolyses vigorously (it even fumes in moist air), reacting completely with water to give silicon dioxide (or silicic acid) and hydrogen chloride, so the resulting solution is strongly acidic (pH around $1$–$2$).

Step 2: Identify the closest-to-neutral chloride

Only $\text{NaCl}$ simply dissolves without hydrolysing or otherwise reacting with water, so it is the only chloride here that gives a solution with a pH close to $7$.

Why the other options are wrong

• B: $\text{MgCl}_2$ does hydrolyse slightly (its solution is a little acidic), so its pH is somewhat below $7$, though closer to neutral than C or D.
• C: $\text{AlCl}_3$ hydrolyses significantly, giving a distinctly acidic solution (pH around $3$).
• D: $\text{SiCl}_4$ hydrolyses completely and vigorously in water, giving the most acidic solution of the four (pH around $1$–$2$).

Final answer

• The chloride that gives a solution closest to pH $7$ is sodium chloride, $\text{NaCl}$, option A.