Question

Consider the following carbides:
\begin{matrix}
{Ca{C_2}} &{B{e_2}C} &{Mg{C_2}} &{Sr{C_2}} \\
 I& II& III&IV
\end{matrix}
Select the carbide which gives different products on hydrolysis than other carbides.
A.$I$
B.$II$
C.$III$
D.$IV$

Answer 1

Hint:We have to know that carbides are binary compounds that are made up of element carbon and lower (or) about equal electronegativity. Based on chemical bonding, we can classify carbides as salt-like carbides, interstitial (or) alloy like carbides, and covalent carbides. At high temperatures when we heat an element (or) its oxide with hydrocarbon (or) carbon, carbides are obtained.

Complete answer:
We must remember that salts like carbides are formed by metals present in groups IA, IIA and IIIA other than boron, coinage metals, zinc, cadmium, and certain lanthanides. These carbides on hydrolysis (or) dilute hydrochloric acids give different hydrocarbons.

We have to remember that the calcium carbide $\left( {Ca{C_2}} \right)$ is a salt like carbide. It yields acetylene on hydrolysis. We know that calcium is an alkaline earth metal and it forms $M{C_2}$ kind carbide. We can write the hydrolysis reaction of calcium carbide as,
$Ca{C_2} + 2{H_2}O \to {C_2}{H_2} + Ca{\left( {OH} \right)_2}$

We have to know that the Beryllium carbide $\left( {B{e_2}C} \right)$ is a salt like carbide. It yields methane on hydrolysis. We know that beryllium is an alkaline earth metal and the derivatives of methane by hydrolysis of beryllium carbide are called methanides. We can write the hydrolysis reaction of beryllium carbide as,
$B{e_2}C + 4{H_2}O \to C{H_4} + 2Be{\left( {OH} \right)_2}$

We also remember that the Magnesium carbide $\left( {Mg{C_2}} \right)$ is a salt like carbide. It yields acetylene on hydrolysis. We know that magnesium is an alkaline earth metal and it forms $M{C_2}$ kind carbide. The hydrolysis reaction of magnesium carbide is written as,
$Mg{C_2} + 2{H_2}O \to {C_2}{H_2} + Mg{\left( {OH} \right)_2}$

Strontium carbide $\left( {Sr{C_2}} \right)$ is a salt like carbide. It yields acetylene on hydrolysis. We know that strontium is an alkaline earth metal and it forms $M{C_2}$ kind carbide. The hydrolysis reaction of strontium carbide is written as,
$Sr{C_2} + 2{H_2}O \to {C_2}{H_2} + Sr{\left( {OH} \right)_2}$

We can see that beryllium carbide on hydrolysis gives methanides, whereas the rest all gives acetylides. So, Beryllium carbide is the carbide that gives different products on hydrolysis.
Therefore, the option (B) is correct.

Note:
Some other compounds that give methanides are $A{l_4}{C_3}$, $B{e_2}C$, $M{n_3}C$ etc. We also have to know that mixed carbides on hydrolysis gives a mixture of hydrocarbons. Some of the examples of mixed carbides are carbides of iron, $U{C_2}$ and $Th{C_2}$. Allylides are formed by hydrolysis of methyl acetylene. An example of this type is $M{g_2}{C_3}$.