Arrange the following in order of increasing reducing property.

$N a H$ , $M g H_{2}$ and $H_{2} O$

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Solution

$N a H$ is an ionic hydride and can easily donate its electrons. It is the most reducing.

$M g H_{2}$ and $H_{2} O$ are covalent hydrides. $H_{2} O$ has lower reducing power than $M g H_{2}$ as it's bond dissociation energy is higher.

Hence, the increasing order of the reducing property is $H_{2} O < M g H_{2} < N a H$ .

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