The hydride ion, H^- is a stronger base than hydroxide ion, OH^-, which of the following reactions will occur if sodium hydride is dissolved in H₂O?

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$H^{-} + H_{2} O \to_{}^{} \underset{(a q)}{H_{3} O^{+}} + e^{-}$

$H^{-} + H_{2} O \to_{}^{} O H_{(a q)}^{-} + \underset{(g)}{H_{2}}$

$H^{+} + H^{-} \to_{}^{} \underset{(l)}{H_{2} O}$

$H^{-} + H_{2} O \to_{}^{} H_{3} O_{(a q)}^{-}$

answer is B.

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Detailed Solution

$N a H + H_{2} O \to_{}^{} \underset{(a q)}{N a O H} + \underset{(g)}{H_{2}} ↑$ . The reason for this reaction is hydride ion is a very strong base and abstracts hydrogen ion from water and converts itself into hydrogen.

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