The hydride ion, $H^{-}$ is a stronger base than its hydroxide ion $O H^{-}$ . Which of the following reactions will occur, if sodium hydride $(N a H)$ is dissolved in water?

H^{-} (a q) + H_{2} O \to H_{2} O

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H^{-} (a q) + H_{2} O (l) \to O H^{-} + H_{2}

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H^{-} + H_{2} O \to N o r e a c t i o n

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N o n e o f t h e a b o v e

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Solution

The correct option is B $H^{-} (a q) + H_{2} O (l) \to O H^{-} + H_{2}$
Sodium hydride $(N a H)$ react with water to form hydrogen gas and the sodium hydroxide $(N a O H)$ .
$N a H + H_{2} O \to H_{2} + N a O H$
i.e., $N a^{+}, H^{-}$ and $H^{+}, O H^{-}$ give rise to $H_{2}$ and $N a O H$ . So, it is similar to option B.

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