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

The correct order of pH of isomolar solution in sodium oxide $\left(\mathrm{pH}_1\right)$, sodium sulphide \left(\mathrm{pH}_2\right)$, sodium selenide \left(\mathrm{pH}_3\right) and sodium telluride \left(\mathrm{pH}_4\right)$ is $\mathrm{pH}_1\gt\mathrm{pH}_2\gt\mathrm{pH}_3\gt\mathrm{pH}_4$ because in aqueous solution, they are hydrolysed as follows. $\mathrm{Na}_2 \mathrm{O}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\text{ Base }}{2 \mathrm{NaOH}}+{\mathrm{H}_2 \mathrm{O}}{ } \mathrm{Na}_2 \mathrm{~S}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\text{ Strong base }}{2 \mathrm{NaOH}+} \underset{\text{ Weak acid }}{\mathrm{H}_2 \mathrm{~S}} \mathrm{Na}_2 \mathrm{Se}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\text{ Strong base }}{2 \mathrm{NaOH}+} \begin{gathered}\mathrm{H}_2 \mathrm{Se} \\ \text{ Weak acid }\end{gathered} \mathrm{Na}_2 \mathrm{Te}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\text{ Strong base }}{2 \mathrm{NaOH}+} \begin{gathered}\mathrm{H}_2 \mathrm{Te} \\ \text{ Weak acid }\end{gathered}$ Order of acidic strength $\mathrm{H}_2 \mathrm{Te}\gt\mathrm{H}_2 \mathrm{Se}\gt\mathrm{H}_2 \mathrm{~S}\gt\mathrm{H}_2 \mathrm{O}$ Order of neutralisation of NaOH $\mathrm{H}_2 \mathrm{Te}\gt\mathrm{H}_2 \mathrm{Se}\gt\mathrm{H}_2 \mathrm{~S}\gt\mathrm{H}_2 \mathrm{O}$ Hence, their aqueous solutions have the following order of basic character due to neutralisation of NaOH with $\mathrm{H}_2 \mathrm{O}, \mathrm{H}_2 \mathrm{~S} . \mathrm{H}_2 \mathrm{Se} and \mathrm{H}_2 \mathrm{Te} \mathrm{Na}_2 \mathrm{O}\gt\mathrm{Na}_2 \mathrm{~S}\gt\mathrm{Na}_2 \mathrm{Se}\gt\mathrm{Na}_2 \mathrm{Te} (\therefore \mathrm{pH}$ of basic solution is higher than acidic or least basic solution.)