Which of the following metal(s) on treatment with NaOH will liberate $H_{2}$ gas?

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Solution

The correct options are
A Zn
B Sn
C Al
Amphoteric metals react with boiling alkali, in this case NaOH to liberate $H_{2}$ gas.
$Z n + 2 N a O H \to N a_{2} Z n O_{2} + H_{2}$
$S n + 2 N a O H + H_{2} O \to N a_{2} S n O_{3} + 2 H_{2}$
$2 A l + 2 N a O H + 2 H_{2} O \to 2 N a A l O_{2} + 3 H_{2}$
and $M g + N a O H \to M g O + N a H$

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