What is the correct relationship between the $p H$ of isomer solutions of sodium oxide, $N a_{2} O (p H_{1}), sodium sulphide, N a_{2} S (p H_{2})$ $sodium selenide, N a_{2} S e (p H_{3})$ $and sodium telluride N a_{2} T e (p H_{4}) ?$

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 A $p H_{1} > p H_{2} > p H_{3} > p H_{4}$
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.

$N a_{2} O + 2 H_{2} O \to 2 N a O H + H_{2} O$
$N a_{2} S + 2 H_{2} O \to 2 N a O H + H_{2} S$

$N a_{2} S e + 2 H_{2} O \to 2 N a O H + H_{2} S e$

$N a_{2} T e + 2 H_{2} O \to 2 N a O H + H_{2} T e$

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 $N a O H with H_{2} O, H_{2} S, H_{2} S e, H_{2} T e$ .
$N a_{2} O > N a_{2} S > N a_{2} S e > N a_{2} T e$ .

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(p H)_{1} + (p H)_{2} = 9.5 u n i t s

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