Question

Which of the following will not be stable?

• A. $C{H}_3$
• B. $C_{2}H$
• C. $C{H}_4$
• D. $C_2{H}_2$

Answer 1

Answer: (A), (B)

The correct options are
A

$C{H}_3$

B

$C_{2}H$

Let's roll back a little bit. What is the electronic configuration of carbon?

So, carbon has 4 electrons in its valence shell.

Therefore, it can make four covalent bonds to achieve the stable octet configuration.

$\to$Now, In case of $C{H}_3$

Only 3 electrons will covalently bond with Hydrogen, so one electron will be left and the octet configuration will not be attained.

$\to$ In case of $C_{2}H$

We know, carbon can covalently bond with carbon.

But, carbon can make maximum three covalent bonds with carbon. [Why?]

Because it doesn't have enough atomic orbitals pointing in the right direction

[* Try to make quadruple bond between two carbon using the shapes of orbitals]

Therefore,

is not possible

$C\equiv C-H$ is possible

because octet configuration of one carbon atom is not achieved. So, not stable.

$\to C{H}_4$

This is stable because 4 electrons of carbon will be covalently bonded with four Hydrogens.

$\to C_2{H}_2$

This is stable because carbon will make three covalent bonds with the other carbon and one bond each with two Hydrogens.

$H-C\equiv C-H$