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

The correct order of pH of isomolar solution of sodium oxide $\left(p{H}_1\right)$, sodium sulphide $\left(p{H}_2\right)$, sodium selenide $\left(p{H}_3\right)$ and sodium telluride $\left(p{H}_4\right)$ is $p{H}_1>p{H}_2>p{H}_3>p{H}_4$ because in aqueous solution, they are hydrolysed as follows.
$N{a}_{2}O+2H_{2}O\to \underset{\text{Base}}{2NaOH}+H_{2}O$
$N{a}_{2}S+2H_{2}O\to \underset{\text{Strong base}}{2NaOH}+\underset{\text{Weak acid}}{H_{2}Se}$
$N{a}_{2}Te+2H_{2}O\to \underset{\text{strong base}}{2NaOH}+\underset{\text{Weak acid}}{H_{2}Te}$
On moving down the group acidic character of oxides increases.
Order of acidic strength
Order of acidic strength
$H_{2}Te>H_{2}Se>H_{2}S>H_{2}O$
Order of neutralisation of NaOH
$H_{2}Te>H_{2}Se>H_{2}S>H_{2}O$
Hence, their aqueous solution have the following order of basic character due to neutralisation of NaOH with $H_{2}O,H_{2}S,H_{2}Se$ and $H_{2}Te$.
($\because pH$ of basic solution is higher than acidic or least basic solution)