What is the correct relationship between the pHs of isomolars...

It is given that the isomolar solutions are sodium salt of Group 16 elements are present in equal molar concentration. The salts on hydrolysis with water, it produces an acid and base, as follows: I. $\mathrm{Na}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{NaOH}$ II. $\mathrm{Na}_2 \mathrm{~S}+\mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{~S}$ III. $\mathrm{Na}_2 \mathrm{Se}+\mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{NaOH}$ $+\mathrm{H}_2 \mathrm{Se}$ IV. $\mathrm{Na}_2 \mathrm{Te}+\mathrm{H}_2 \mathrm{O}>2 \mathrm{NaOH}$ $+\mathrm{H}_2 \mathrm{Te}$ The acidic strength of the hydrides is in the order $\mathrm{H}_2 \mathrm{O} < \mathrm{H}_2 \mathrm{~S} < \mathrm{H}_2 \mathrm{Se} <$ $\mathrm{H}_2 \mathrm{Te}$, due to increase in the atomic size. As the acidic character of the hydride increases down the group, the basic character of the salt decreases down the group. Therefore, the order of basic character of the salts will be: $\mathrm{Na}_2 \mathrm{O}>\mathrm{Na}_2 \mathrm{~S}>\mathrm{Na}_2 \mathrm{Se}>\mathrm{Na}_2 \mathrm{Te}$ Therefore, the order of $\mathrm{pH}$ of the given aqueous solution is $\mathrm{pH}_1>\mathrm{pH}_2>$ $\mathrm{pH}_3>\mathrm{pH}_4$ Related Theory The $\mathrm{pH}$ of the isomolar solution, can be determined with respect to the basic character of the sodium salts of the Group 16 elements. These salts undergo hydrolysis, forming hydrides, whose acidic character can account for the basic character of the salt, formed through neutralisation. A Caution In the reverse reaction of neutralisation, from the acidic strength of the hydrides, we can obtain the order of ease of neutralisation of the sodium hydroxide with the hydrides. As more the acidity of the hydride, more will be the neutralisation to form the salt.