Metal carbides on reaction with $H_{2} O$ forms $C H_{4}$ .

Carbide can be:

C a C_{2}

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M g_{3} C_{2}

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B e_{2} C

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S i C

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Solution

The correct option is C $B e_{2} C$
Different metal carbides react with $H_{2} O$ , give the following compounds.

$B e_{2} C (s) + 4 H_{2} O (l) \to 2 B e (O H)_{2} (s) + C H_{4} (g)$

$C a C_{2} + 2 H_{2} O \to C_{2} H_{2} + C a (O H)_{2}$

$M g_{3} C_{2} + 4 H_{2} O \to 2 M g (O H)_{2} + C_{3} H_{4}$

Here, beryllium carbide is a metal carbide that forms methane.

The correct option is ( C )

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