Question

Calcium phosphide reacts with water to form $Ca ( OH )_{2}$ and $X .$ When $X$ is passed into $CuSO _{4}$ solution. $Y$ and $H _{2} SO _{4}$ are formed. What is $Y ?$

• A. $\left[ Cu \left( PH _{3}\right)_{4}\right]^{2+}$
• B. $\left[ Cu \left( PH _{3}\right)_{6}\right]^{2+}$
• C. $Cu _{3} P _{2}$
• D. $CuHPO _{4}$

Answer 1

Answer: (C)

Calcium phosphide reacts with water to form calcium hydroxide and phosphine.

$\underset{\text{Calcium Phosphide}}{Ca _{3} P _{2}}+ \underset{\text{Water}}{6 H _{2} O} \longrightarrow \underset{\text{Calcium hydroxide}}{3 Ca ( OH )_{2} a q}+ \underset{\text{Phosphine (Y)}}{2 PH _{3}(g)}$

Phosphine when passed into $CuSO _{4}$ solution, it reacts to produce black precipitate of cupric phosphide and $H _{2} SO _{4}$.

$\underset{\text{Phosphine}}{3 PH _{3}}+ \underset{\text{Copper sulphate}}{3 CuSO _{4}} \longrightarrow \underset{\text{Cupric phosphide}}{Cu _{3} P _{2}} +\underset{\text{Sulphuric acid}}{3 H _{2} SO _{4}}$