Question

The bond angle in methane
A) ${{109}^{\circ }}31'$
B) ${{107}^{\circ }}48'$
C) ${{180}^{\circ }}$
D) ${{198}^{\circ }}{{28}^{'}}$

Answer 1

Hint: In methane there is one carbon atom and four hydrogen atoms present in the molecule.
The methane molecule is having the $s{{p}^{3}}$hybridization.

Complete step by step answer:
So here in the question we are asked to say about the bond angle of methane molecules.
We know that before passing a comment on the bond angle of the methane molecule, we should know about the number of atoms in methane, the hybridization of the molecule and shape of the molecule. This gives the idea about the bond angle of the molecule.
So now let’s solve the question.
The molecular formula of methane is $C{{H}_{4}}$.
From the molecular formulae it is clear that it contains one carbon atom and four hydrogen atoms in the molecule. Methane is the first molecule of the alkane homologous series having the formulae ${{C}_{n}}{{H}_{2n+2}}$,
n is the number of carbon atoms in the molecule.
The atomic number of carbon is 6 and hence the electronic configuration is,
$Electronic\,configuration\,of\,C=1{{s}^{2}}2{{s}^{2}}2{{p}^{2}}$
The atomic number of H is 1 and it has one electron and the electronic configuration is as,
$Electronic\,configuration\,of\,H=1{{s}^{1}}$
In a carbon atom there are four valence electrons and H is having a valency of +1.
Now let’s see the hybridization, C is the central atom with four valence electrons to form a molecule, each atom should have complete octet configuration. So the carbon atom needs four more electrons to complete the octet configuration and hydrogen needs one more electron to form stable configuration.
The fours hydrogen shares its 4 electrons with the carbon so that the C and H gets the stable configuration.
We know that the methane is having $s{{p}^{3}}$ hybridization, where $2s$ and $2p$ orbitals of the carbons undergoes the hybridization and forms the four $s{{p}^{3}}$ hybrid orbitals with same shape and energy. And the four orbitals are filled by the sharing of four electrons from the C atom and the four electrons from the H atoms. By sharing the electrons C forms four covalent bonds with H atoms.
So there are four bond pairs of electrons and the geometry corresponding to the $s{{p}^{3}}$ hybridization is tetrahedral.
The methane molecule is arranged in a tetrahedral geometry to reduce the bond pair-bond pair (bp-bp) repulsion.
The standard value of bond angle for tetrahedral geometry is${{109}^{\circ }}31'$.

Hence the correct answer for the question is option (A).

Note: Here the bond length of C-H bond is $1.09{{A}^{\circ }}$, all the C-H bonds are having the same bond length in methane molecules. The methane molecule is nonpolar in nature as the dipole value is zero, since the substituents attached to the C atom are all the same ones so the dipole is in the same direction i.e. toward the C atom and the dipoles cancel out.