What is the correct relationship between the `pH` of isomolar solutions of sodium oxide `(pH_(1))`, sodium sulphide `(pH_(2))`, sodium selenide `(pH_(3))` , and sodium telluride `(pH_(4)) ` ?

Options

A. `pH_(1)gt pH_(2)~~pH_(3)gt pH_(4)`
B. `pH_(1)lt pH_(2)lt pH_(3) lt pH_(4)`
C. `pH_(1)lt pH_(2)lt pH_(3)~~ pH_(4)`
D. `pH_(1)gt pH_(2)gt pH_(3)gtpH_(4)`

Answer

D

Explanation

The correct order of pH of isomolar solution of sodium oxide `(pH_(1))`, sodium sulphide `(pH_(2))`, sodium selenide `(pH_(3))` and sodium telluride `(pH_(4))` is `pH_(1)gt pH_(2)gt pH_(3)gt pH_(4)` because in aqueous solution, they are hydrolysed as follows.
`Na_(2)O+2H_(2)O rarr underset("Base")(2NaOH)+H_(2)O`
`Na_(2)S+2H_(2)O rarr underset("Strong base")(2NaOH)+underset("Weak acid")(H_(2)Se)`
`Na_(2)Te+2H_(2)O rarr underset("strong base")(2NaOH)+underset("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 gt H_(2)Se gt H_(2)S gt H_(2)O`
Order of neutralisation of NaOH
`H_(2)Te gt H_(2)Se gt H_(2)S gt 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)