Out of the following metals which will give $H_{2}$ reaction with $N a O H$ ?
$I : Z n$
$I I : M g$
$I I I : A l$
$I V : B e$

I, I I, I I I, I V

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I, I I I, I V

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I I, I V

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I, I I I

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Solution

The correct option is A $I, I I I, I V$
The reactions of following metals with $N a O H$ can be given as:-
$Z n + 2 N a O H ⟶ N a_{2} Z n O_{2} + H_{2}$
$M g + 2 N a O H ⟶ 2 N a + M g (O H)_{2}$
$2 A l + 2 N a O H + 2 H_{2} O ⟶ 2 N a A l O_{2} + 3 H_{2}$
$B e + N a O H + H_{2} O ⟶ N a_{2} [B e (O H)_{4}] + H_{2}$
From the above reactions, it is clear that $Z n, A l$ and $B e$ react with $N a O H$ to give $H_{2}$ while $M g$ does not.

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