`CH_(14)` can be prepared by the reaction of `H_(2)O` with:

Options

A. (a) `Mg_(2)C_(3)`
B. (b) `CaC_(2)`
C. (c) `Be_(2)C`
D. (d) `A1_(4)C_(3)`

Answer

C, D

Explanation

Carbides of Be and A1 conrtain methinide ion `(C^(4-)` and thus give `CH_(4)` gas (due to the diagonal relationship of Be with A1).
`Be_(2)C rarr 2Be^(2) + C^(4-)` , `A1_(4)C_(3) rarr 4A1^(3+) + C^(4-)`
`Be_(2)C + 4H_(2)O rarr CH_(4) + 2Be(OH)_(2)`
`A1_(4)C_(3) + 12H_(2)O rarr 3CH_(4) + 4A1(OH)_(3)`
Where the carbides of Mg, i.e., `Mg_(2)C_(3)` , contain propynide ion `(C_(3)^(4-))` and hence give propyne
`(Me - -= - H)` gas.
`Mg_(2)C_(3) + 4H_(2)O rarr CH_(3) - C -= CH + 2Mg (OH)_(2)`
The carbides of rest of the second group, i.e., Ca. Sr, and Ba `(CaC_(2), SrC_(2)`, and `BaC_(2))` give ethyne (acetylene) gas because they contain accetylide or ethynide ion `(C_(2)^(2))` .
`CaC_(2) rarr Ca^(2) + C_(2)^(2-)`