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${\text{X}}\,{\text{ + }}\,{{\text{H}}_{\text{2}}}{\text{O}}\, \to {\text{Y}}\,{\text{ + }}\,{\text{C}}{{\text{H}}_4}$
In the above reaction an unknown component X on hydrolysis gives Y and methane gas. Identify X and Y.
A. ${\text{Ca}}{{\text{C}}_{\text{2}}}{\text{,}}\,{\text{Al}}{\left( {{\text{OH}}} \right)_{\text{3}}}$
B. ${\text{A}}{{\text{l}}_4}{{\text{C}}_3}{\text{,}}\,{\text{Al}}{\left( {{\text{OH}}} \right)_{\text{3}}}$
C. ${\text{Mg}}{{\text{C}}_{\text{2}}}{\text{,}}\,{\text{Mg}}{\left( {{\text{OH}}} \right)_{\text{3}}}$
D. None of the above

Answer
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Hint: Methanides on hydrolysis gives methane gas. In the process of hydrolysis water is added to a substance. The carbides that contain ${{\text{C}}^{4 - }}$ are known as methanides.

Complete step by step answer:
Determine the type of anion in ${\text{Ca}}{{\text{C}}_{\text{2}}}$ as follows:
${\text{Ca}}{{\text{C}}_{\text{2}}}\, \to \,{\text{C}}{{\text{a}}^{2 + }}\, + \,{\left[ {{{\text{C}}_2}} \right]^{2 - }}$
${\text{Ca}}{{\text{C}}_{\text{2}}}$on dissociation gives ${\left[ {{{\text{C}}_2}} \right]^{2 - }}$ ion so, it is not methanide.
The hydrolysis of ${\text{Ca}}{{\text{C}}_{\text{2}}}$ is as follows:
Calcium carbide on hydrolysis gives calcium hydroxide and acetylene gas.
\[{\text{Ca}}{{\text{C}}_{\text{2}}}\, + \,{{\text{H}}_2}{\text{O}}\, \to {\text{Ca}}{\left( {{\text{OH}}} \right)_2}{\text{ + }}\,{{\text{C}}_{\text{2}}}{{\text{H}}_2}\,\]
The hydrolysis of ${\text{Ca}}{{\text{C}}_{\text{2}}}$ does not give ${\text{Al}}{\left( {{\text{OH}}} \right)_{\text{3}}}$ and methane. So, ${\text{Ca}}{{\text{C}}_{\text{2}}}{\text{,}}\,{\text{Al}}{\left( {{\text{OH}}} \right)_{\text{3}}}$ are not X and Y so, option (A) is incorrect.
Determine the type of anion in ${\text{A}}{{\text{l}}_4}{{\text{C}}_3}$ as follows:
${\text{A}}{{\text{l}}_4}{{\text{C}}_3}\, \to \,4\,{\text{A}}{{\text{l}}^{3 + }}\, + 3\,\,{{\text{C}}^{4 - }}$
${\text{A}}{{\text{l}}_4}{{\text{C}}_3}$ on dissociation gives $\,{{\text{C}}^{4 - }}$ ion so, it is a methanide.
The hydrolysis of ${\text{A}}{{\text{l}}_4}{{\text{C}}_3}$ is as follows:
Aluminium carbide on hydrolysis gives aluminium hydroxide and methane gas.
\[{\text{A}}{{\text{l}}_4}{{\text{C}}_3}\, + \,12\,{{\text{H}}_2}{\text{O}}\, \to 4\,{\text{Al}}{\left( {{\text{OH}}} \right)_3}{\text{ + }}\,3\,{\text{C}}{{\text{H}}_4}\,\]
The hydrolysis of ${\text{A}}{{\text{l}}_4}{{\text{C}}_3}$ gives ${\text{Al}}{\left( {{\text{OH}}} \right)_{\text{3}}}$ and methane. So, ${\text{A}}{{\text{l}}_4}{{\text{C}}_3}{\text{,}}\,{\text{Al}}{\left( {{\text{OH}}} \right)_{\text{3}}}$ are not X and Y so, option (B) is correct.
Determine the type of anion in ${\text{Mg}}{{\text{C}}_2}$ as follows:
${\text{Mg}}{{\text{C}}_2}\, \to \,{\text{M}}{{\text{g}}^{2 + }}\, + \,{\left[ {{{\text{C}}_2}} \right]^{2 - }}$
${\text{Mg}}{{\text{C}}_2}$ on dissociation gives ${\left[ {{{\text{C}}_2}} \right]^{2 - }}$ ion so, it is not methanide.
The hydrolysis of ${\text{Mg}}{{\text{C}}_2}$ is as follows:
Magnesium carbide on hydrolysis gives magnesium hydroxide and acetylene gas.
\[{\text{Mg}}{{\text{C}}_{\text{2}}}\, + \,{{\text{H}}_2}{\text{O}}\, \to {\text{Mg}}{\left( {{\text{OH}}} \right)_2}{\text{ + }}\,{{\text{C}}_{\text{2}}}{{\text{H}}_2}\,\]
The hydrolysis of ${\text{Mg}}{{\text{C}}_{\text{2}}}$ does not give methane. So, ${\text{Mg}}{{\text{C}}_{\text{2}}}{\text{,}}\,{\text{Mg}}{\left( {{\text{OH}}} \right)_{\text{3}}}$ are not X and Y so, option (C) is incorrect.

Therefore, option (B) ${\text{A}}{{\text{l}}_4}{{\text{C}}_3}{\text{,}}\,{\text{Al}}{\left( {{\text{OH}}} \right)_{\text{3}}}$ is correct.

Note: Carbides are complex composed of metals and carbon. The carbides that contain ${\left[ {{{\text{C}}_2}} \right]^{2 - }}$ are known as acetylides. ${\text{Mg}}{{\text{C}}_2}$ is an acetylides whereas the ${\text{M}}{{\text{g}}_2}{\text{C}}$ is a methanide. Acetylides on hydrolysis gives acetylene. The carbides that contain ${\left[ {{{\text{C}}_3}} \right]^{4 - }}$ are known as allylide.