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Hydrides of Group 16

What is the c...

Question

What is the correct relationship between the pHs of isomolar solutions of sodium oxide $(p H_{1})$ , sodium sulphide $(p H_{2})$ , sodium selenide $(p H_{3})$ and sodium telluride $(p H_{4})$ ?

p H_{1} > p H_{2} = p H_{3} > p H_{4}

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p H_{1} < p H_{2} < p H_{3} < p H_{4}

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p H_{1} < p H_{2} < p H_{3} = p H_{4}

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p H_{1} > p H_{2} > p H_{3} > p H_{4}

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Solution

The correct option is D $p H_{1} > p H_{2} > p H_{3} > p H_{4}$
When the above compounds are dissolved in water, each form their respective bases.
As the basicity of these oxides decreases, when placed in water, the basicity of solution also decreases.

${N a}_{2} O + 2 H_{2} O ⟶ 2 N a O H + H_{2} O$
${N a}_{2} S + 2 H_{2} O ⟶ 2 N a O H + H_{2} S$
${N a}_{2} S e + 2 H_{2} O ⟶ 2 N a O H + H_{2} S e$
${N a}_{2} T e + 2 H_{2} O ⟶ 2 N a O H + H_{2} T e$

Order of neutralisation of NaOH is
$H_{2} T e > H_{2} S e > H_{2} S > H_{2} O$

Hence their aqueous solutions have the following order of basic character due to neutralisation of NaOH with $H_{2} O, H_{2} S, H_{2} S e, H_{2} T e$ .
$N a_{2} O > N a_{2} S > N a_{2} S e > N a_{2} T e$ .

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