$C H_{4}$ can be prepared by the reaction of $H_{2} O$ with:

M g_{2} C_{3}

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C a C_{2}

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B e_{2} C

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A l_{4} C_{3}

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Solution

The correct options are
C $A l_{4} C_{3}$
D $B e_{2} C$
Carbides of Be and Al contain methanide ion $(C^{4 -}$ and thus give $C H_{4}$ gas (due to the diagonal relationship of Be with Al).
$B e_{2} C \to 2 C e^{2 +} + C^{4 -}, A l_{4} C_{3} \to 4 A l^{3 +} + C^{4 -}$
$B e_{2} C + 4 H_{2} O \to C H_{4} + 3 B e (O H)_{2}$
$A l_{4} C_{3} + 12 H_{2} O \to 3 C H_{4} 4 A l (O H)_{3}$
Whereas the carbides of Mg, i.e. $M g_{2} C_{3},$ contain propynide ion $(C_{3}^{4 -}$ ) and hence give propyne.

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Which is not obtained when metal carbides react with H₂O?

B e_{2} C, C a C_{2}, M g_{2} C_{3}, A l_{4} C_{3}, C H_{4},

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X

Y

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