Question

Draw the electron dot diagram of $C{H_4}$ and justify your answer.

Answer 1

The ground state electronic configuration of carbon atom is $1{s^2}2{s^2}2{p^2}$ and ground state electronic configuration of a hydrogen atom is $1{s^1}$. We show only the valence electron in the electron dot representation.

Complete step by step solution:
We will draw the electron dot diagram for $C{H_4}$.
-Let’s write the electronic configurations of C and H atoms in order to determine how many electrons there will be in their valence shell of electrons.
Electronic configuration of C: $1{s^2}2{s^2}2{p^2}$
Electronic configuration of H: $1{s^1}$
Now, we know that C-H bonds in $C{H_4}$ are covalent bonds. So, they are formed by sharing one electron each from both the atoms.
-Now, carbon’s valence orbitals undergo hybridization and so the electronic configuration of the excited state of carbon atom will be $1{s^2}2{s^1}2{p^3}$
-So, we can see that there are four electrons in the valence orbital of the carbon atom. Valence orbital is the outermost orbital of the atom.
-We can see from the electronic configuration of the hydrogen atom that there is only one electron in the valence shell of the hydrogen atom.
So, we can form the electron dot diagram for $C{H_4}$ as follows:
Here, the valence electrons of the carbon atom are represented by dot and valence electron of H-atom is represented by *. So, we can clearly see that one electron of carbon and one electron of hydrogen atom combine to form a covalent bond. There are four such covalent bonds present in $C{H_4}$.

Note: Remember that the carbon atom in $C{H_4}$ is $s{p^3}$ hybridized. All the $s{p^3}$ hybridized atoms have a tetrahedral shape. Methane is also tetrahedral in shape. As carbon has four electrons in its valence shell, it has more tendency to share its electrons rather than losing or gaining electrons to form ionic bonds.