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Step-by-Step Explanation
Step 1: Identify the Species Involved
We are comparing the pH of isomolar solutions of sodium oxide ($\text{Na}_2\text{O}$), sodium sulphide ($\text{Na}_2\text{S}$), sodium selenide ($\text{Na}_2\text{Se}$), and sodium telluride ($\text{Na}_2\text{Te}$).
Step 2: Recognize the Basic Nature of These Compounds
All these salts produce basic solutions in water because they form corresponding anions ($\text{O}^{2-}$, $\text{S}^{2-}$, $\text{Se}^{2-}$, $\text{Te}^{2-}$) which are strong or moderately strong bases. However, the extent of basicity differs down the group.
Step 3: Analyze Basic Strength Trend Down the Group
In the periodic table (Group 16), as we move from oxygen to tellurium, the size of the anion increases. Larger anions tend to be less capable of holding onto protons ($\text{H}^+$) effectively, which leads to a decrease in their basic strength:
$\text{O}^{2-}$ (most basic) > $\text{S}^{2-}$ > $\text{Se}^{2-}$ > $\text{Te}^{2-}$
Step 4: Relate Basic Strength to pH
Since a more basic anion will generate a higher concentration of $\text{OH}^{-}$ ions in solution, it will produce a higher pH. Therefore, the pH of an isomolar solution of these salts follows the same trend as their basicity.
Step 5: Conclude the Order of pH
Thus, the order of the pH of the solutions is:
$ \text{pH}_1 > \text{pH}_2 > \text{pH}_3 > \text{pH}_4 $,
where pH1 corresponds to $\text{Na}_2\text{O}$, pH2 to $\text{Na}_2\text{S}$, pH3 to $\text{Na}_2\text{Se}$, and pH4 to $\text{Na}_2\text{Te}$.
