$Z n^{+ 2}$ dissolves in excess of NaOH due to the formation of:

Z n (O H)_{2}

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Z n O

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N a_{2} Z n O_{2}

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all the above

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Solution

The correct option is C $N a_{2} Z n O_{2}$
Sodium hydroxide ( $N a O H$ ) reacts with zinc to form soluble sodium zincate $(N a_{2} Z n O_{2})$ along with the liberation of inflammable hydrogen gas.

$Z n + 2 N a O H \to N a_{2} Z n O_{2} ↓ + H_{2}$

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