Question
The hydride ion $${H^ - }$$ is stronger base than its hydroxide ion $$O{H^ - }.$$ Which of the following reactions will occur if sodium hydride $$\left( {NaH} \right)$$ is dissolved in water?
A.
$$2{H^ - }\left( {aq} \right) + {H_2}O\left( l \right) \to {H_2}O + {H_2} + 2{e^ - }$$
B.
$${H^ - }\left( {aq} \right) + {H_2}O\left( l \right) \to O{H^ - } + {H_2}$$
C.
$${H^ - } + {H_2}O\left( l \right) \to {\text{No reaction}}$$
D.
$${\text{None of the above}}$$
Answer :
$${H^ - }\left( {aq} \right) + {H_2}O\left( l \right) \to O{H^ - } + {H_2}$$
Solution :
Sodium hydride dissolved in water as
$$\eqalign{
& NaH + {H_2}O \to NaOH + {H_2} \cr
& {\text{or}}\,\,\,{H^ - }\left( {aq} \right) + {H_2}O\left( l \right) \to O{H^ - } + {H_2} \uparrow \cr} $$
In the above reaction hydride ion take proton from water molecule and hydrogen gas is evolved.