Question

Methanides are:
[A] $M{{g}_{2}}{{C}_{3}},\text{ B}{{\text{e}}_{2}}C,\text{ A}{{\text{l}}_{4}}{{S}_{3}}\text{ and Ca}{{\text{C}}_{2}}$
[B] $M{{g}_{2}}{{C}_{3}},\text{ B}{{\text{e}}_{2}}C\text{ and A}{{\text{l}}_{4}}{{C}_{3}}\text{ }$
[C] $\text{ B}{{\text{e}}_{2}}C,\text{ A}{{\text{l}}_{4}}{{C}_{3}}\text{ and Ca}{{\text{C}}_{2}}$
[D] $\text{ B}{{\text{e}}_{2}}C\text{ and A}{{\text{l}}_{4}}{{C}_{3}}$

Answer 1

To answer this question, you should know that carbides are compounds containing carbon along with other less electronegative metals. A methanide is a carbide with a ${{C}^{4-}}$ carbon centre. Find out the particular carbon centre type in the given options to find out the correct answer.

Complete step by step solution:
To answer this question, firstly we have to know what methanides are. We also know methanides as carbides. They are basically compounds containing carbon and other elements with lower electronegativity. They are formed at a very high temperature, above 1500 degree Celsius from different metals or metal oxides. The metal atom is combined with the carbon atom and forms carbide. They are stable and have high melting points. We can divide carbides into different types depending upon their nature of chemical bonding. These types of carbides are – covalent carbides, interstitial carbides and salt-like carbides. Now, let us try to answer the question. The carbides containing ${{C}^{4-}}$ type of carbon atoms are methanides. Let us go through the options and figure out the methanides among them.
Firstly we have $M{{g}_{2}}{{C}_{3}},\text{ B}{{\text{e}}_{2}}C,\text{ A}{{\text{l}}_{4}}{{S}_{3}}\text{ and Ca}{{\text{C}}_{2}}$. Here, $M{{g}_{2}}{{C}_{3}}$ is a poly-atomic carbide but it is not a methanide. This type of carbide is known as allylenide. $\text{A}{{\text{l}}_{4}}{{S}_{3}}\text{ }$ is clearly not a carbide as it has no carbon atom. $\text{Ca}{{\text{C}}_{2}}$ is not ${{C}^{4-}}$ type but of ${{C}_{2}}^{2-}$ and is known as acetylide so it is not a carbide either. However, $\text{B}{{\text{e}}_{2}}C$ is a methanide but all the others are not so this option is not correct. Then we have $M{{g}_{2}}{{C}_{3}},\text{ B}{{\text{e}}_{2}}C\text{ and A}{{\text{l}}_{4}}{{C}_{3}}\text{ }$. Here also $M{{g}_{2}}{{C}_{3}}$ is not a methanide. So, this option is incorrect. Next we have $\text{ B}{{\text{e}}_{2}}C,\text{ A}{{\text{l}}_{4}}{{C}_{3}}\text{ and Ca}{{\text{C}}_{2}}$. Here, $\text{Ca}{{\text{C}}_{2}}$ is not a methanide. So, this option is incorrect. And lastly, we have $\text{ B}{{\text{e}}_{2}}C\text{ and A}{{\text{l}}_{4}}{{C}_{3}}$. Both aluminium carbide and beryllium carbides are examples of methanides as they have ${{C}^{4-}}$ carbon centres. So, this option is correct.

Therefore, we can say that the correct answer is option [D] $\text{ B}{{\text{e}}_{2}}C\text{ and A}{{\text{l}}_{4}}{{C}_{3}}$

Note: Here, all the carbides given to us are ionic carbides and are also known as salt – like or saline carbides. Aluminium is the only group 14 element that forms carbide. Salt – like carbides include methanide, acetylides and allylides (also known as sesqui carbides). They have isolated carbon centres as ${{C}^{4-}}$, ${{C}_{2}}^{2-}$ and ${{C}_{3}}^{4-}$ respectively.