${PH}_{3}$ (Phosphine) when passed in aqueous solution. of ${CuSO}_{4}$ it produces:

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dark blue solution of $[Cu {({PH}_{3})}_{4}] {SO}_{4}$

blue precipitate of $Cu {(OH)}_{2}$

colourless solution of ${[Cu {(H_{2} O)}_{4}]}^{+}$

black precipitate of ${Cu}_{3} P_{2}$

answer is C.

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Detailed Solution

Water, sulfuric acid, and copper (II) phosphide are the byproducts. A variety of compounds are created when phosphine is added to an aqueous solution of copper sulphate. Using the reaction's stoichiometry:

1 mole of copper (II) phosphide, 3 moles of sulfuric acid, and 15 moles of water are produced when 3 moles of copper sulphate and 2 moles of phosphine react.

As a result, water, sulfuric acid, and copper (II) phosphide are produced.

The equation for the reaction between the two follows:

$3 {CuSO}_{4} \cdot 5 H_{2} O (a q .) + 2 {PH}_{3} \to C_{3} P_{2} + 3 H_{2} {SO}_{4} + 15 H_{2} O$ .