On hydrolysis of $P_{4} O_{10}$ with water we obtain:

H_{3} P O_{3} a n d P H_{3}

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H_{3} P O_{4} a n d H_{3} P O_{3}

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H_{3} P O_{4}

only

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H_{4} P_{2} O_{7} a n d H_{2}

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Solution

The correct option is C $H_{3} P O_{4}$ only
$P_{4} O_{10} + 6 H_{2} O ⟶ 4 H_{3} P O_{4}$

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