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Question: Which of the following is a formula of methanides: A. \(B{e_2}C\) B. \(Ca{C_2}\) C. \(M{g_2...

Which of the following is a formula of methanides:
A. Be2CB{e_2}C
B. CaC2Ca{C_2}
C. Mg2C3M{g_2}{C_3}
D. Al4C3A{l_4}{C_3}

Explanation

Solution

Methanides are carbon containing compounds ( ionic carbides ) which can be considered to be derived from methane ( CH4C{H_4} ) . Beryllium carbide upon reacting with water gives beryllium hydroxide along with methane gas and hence they are considered as carbides. Aluminium carbide gives Aluminium Hydroxide along with methane gas.

Complete step by step answer:
Methanides upon reacting with water form methane CH4C{H_4} , since they contain carbon atoms having a valency of 4{4^ - } . Out of these options , the ones that are methanides are option A and option D. This is because these compounds have carbide ions as anions and release them upon reacting with water , and thus they form methane on combining with free hydrogen ions in the reaction.
The other two options B and C do not form methane upon reacting with water. Calcium Carbide forms acetylene on reacting with water and hence called as being derived from acetylides. Similarly, Magnesium carbide Mg2C3M{g_2}{C_3} upon reacting with water gives propyne gas and magnesium hydroxide. This shows that they do not have C4{C^{4 - }} ions and hence do not form methane. Thus they are not methanides.
Beryllium carbide upon reacting with water gives beryllium hydroxide along with methane gas and hence they are considered as carbides. Calcium carbide gives acetylene gas ( having Carbon-Carbon triple bond) along with calcium hydroxide. Aluminium carbide gives Aluminium Hydroxide along with methane gas and thus is one of the best-characterized forms of methanides along with Beryllium Carbide.
The following chemical equations give the understanding of the above concepts of methanides:
Be2C+4H2O2Be(OH)2+CH4B{e_2}C + 4{H_2}O \to 2Be{(OH)_2} + C{H_4}
CaC2+2H2OCa(OH)2+C2H2Ca{C_2} + 2{H_2}O \to Ca{(OH)_2} + {C_2}{H_2}
Mg2C3+4H2O2Mg(OH)2+C3H4M{g_2}{C_3} + 4{H_2}O \to 2Mg{(OH)_2} + {C_3}{H_4}
Al4C3+12H2O4Al(OH)3+3CH4A{l_4}{C_3} + 12{H_2}O \to 4Al{(OH)_3} + 3C{H_4}

So, the correct answer is Option A,D.

Note: Do not consider the compounds that give products other than methane such as ethane, propane, etc. as methanides. The best examples of methanides have already been given in the solution above.