Hydride ion $(H^{-})$ is a stronger base than hydroxide ion $(O H^{-})$ . In the light of the above fact, which of the following reactions will take place if $N a H$ is dissolved in water?

H_{(a q)}^{-} + H_{2} O \to H_{3} O_{(a q)}^{+}

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

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H_{(a q)}^{-} + H_{2} O \to H_{3} O^{-}

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

No reaction

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Solution

The correct option is B $H_{(a q)}^{-} + H_{2} O \to O H^{-} + H_{2}$
It is given that hydride ion is a stronger base than $O H^{-}$ . Now, if $N a H$ is dissolved in water, the following reaction occurs.

As $H^{-}$ will act as stronger base than ${O H}^{-}$ ion, it will accept a proton from water to give $H_{2}$ and ${O H}^{-}$ .

$H^{-} + H^{+} {O H}^{-} \to H_{2} + {O H}^{-}$

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