Q42 / 180 · AIIMS 2008 · Chemistry

What is the correct relationship between the \(\mathrm{pH}\) of isomolar solutions of sodium oxide \(\left(\mathrm{pH}_1\right)\), sodium sulphide \(\left(\mathrm{pH}_2\right)\), sodium selenide \(\left(\mathrm{pH}_3\right)\) and sodium telluride \(\left(\mathrm{pH}_4\right)\) ?

  • A
    \(\mathrm{pH}_1>\mathrm{pH}_2=\mathrm{pH}_3=\mathrm{pH}_4\)
  • B
    \(\mathrm{pH}_1 < \mathrm{pH}_2 < \mathrm{pH}_3 < \mathrm{pH}_4\)
  • C
    \(\mathrm{pH}_1 < \mathrm{pH}_2>\mathrm{pH}_3 < \mathrm{pH}_4\)
  • Correct answer
    D
    \(\mathrm{pH}_1>\mathrm{pH}_2>\mathrm{pH}_3>\mathrm{pH}_4\).

Solution

$\begin{array}{ll}\text { I } & \mathrm{Na}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaOH} \\ \text { II } & \mathrm{Na}_2 \mathrm{~S}+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{~S} \\ \text { III } & \mathrm{Na}_2 \mathrm{Se}+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{Se} \\ \text { IV } & \mathrm{Na}_2 \mathrm{Te}+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{Te}\end{array}$ I will have the highest $\mathrm{pH}$. Among $\mathrm{H}_2 \mathrm{~S}, \mathrm{H}_2 \mathrm{Se}$, $\mathrm{H}_2$ Te acidity goes on increasing on going down the group as bond length increases on increasing size of central atom. So, $\mathrm{pH}_4 < \mathrm{pH}_3 < \mathrm{pH}_2$ Overall order is $\mathrm{pH}_1>\mathrm{pH}_2>\mathrm{pH}_3>\mathrm{pH}_4$

Hence, the correct answer is option: 'D'.

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