Question

$C{H_4}$ can be prepared by the reaction of ${H_2}O$ with:
(This question has multiple correct options)
A.$M{g_2}{C_3}$
B.$Ca{C_2}$
C.$B{e_2}C$
D.$A{l_4}{C_3}$

Answer 1

Hint: Firstly in the given question we have to observe that options possess a pattern of diagonal relationship among them in the periodic table. Which is an important prospect as it leads to the sharing of many important chemical properties. Now we have to take into account that the carbides of the $Be$ and $Al$ shows the generation of the methanide ion when reacted with the carbides.

Complete step by step answer: In the given question firstly we have to characterise out the elements for the characteristic equations shown in the given conditions. Then we can start observing the trends and the methanide ion generation.
We know that the carbides of the $Be$ and $Al$ contain the Methanide ion ( the ${C^{4 - }}$ ion which reacts to the methane ).
This is because of the diagonal relationship between both the elements ( very important property in inorganic chemistry ).
Therefore the reactions they both will give on reacting with ${H_2}O$ would be as follows:
For the $Be$ the reactions are :-
$
  B{e_2}C \to 2B{e^{2 + }} + {C^{4 - }} \\
  B{e_2}C + 4{H_2}O \to C{H_4} + 3Be{(OH)_2} \\
 $
For the $Al$ the reactions are :-
$
  A{l_4}{C_3} \to 4A{l^{3 + }} + {C^{4 - }} \\
  A{l_4}{C_3} + 12{H_2}O \to 3C{H_4} + 4Al{(OH)_3} \\
 $
On the other hand the carbides of $Mg$ will yield propynide $({C_3}^{4 - })$ , therefore it will give propyne on the reaction.
Hence, the correct options are C and D.

Note:The reaction of calcium carbide with water, producing acetylene and calcium hydroxide, was discovered by Friedrich Wöhler in \[1862\] . This reaction was the basis of the industrial manufacture of acetylene, and is the major industrial use of calcium carbide.