Question

Considering x-axis as the internuclear axis, which one out of the following will not form a sigma bond?
 (A) 1s and 1s
 (B) $\text {1s and}$ $2{p_x}$
 (C) $2{p_y}$ and $2{p_y}$
 (D) 1s and 2s

Answer 1

Hint: The sigma bonds are formed if the atomic orbitals overlap head to head along the internuclear axis and pi bonds are formed if they overlap sideways or lateral in a direction perpendicular to the internuclear axis. Sigma bonds are stronger due to large overlapping areas as compared to pi bonds.

Complete step by step answer:
-First we will see what sigma and pi bond and how they are formed.
We all know that covalent bonds are formed by the overlapping of different atomic orbitals. There are 2 types of covalent bonds based on the type of overlapping of atomic orbitals: sigma bond and pi bond. The head to head overlapping of the atomic orbitals leads to formation of sigma bonds while lateral overlapping leads to formation of pi bonds.

-Sigma (σ) bonds: This bond will be formed if the overlapping orbitals overlap head on positive or the orbitals lie in the same phase along the internuclear axis. Due to this direct overlapping of the orbitals, sigma bonds are the strongest type of covalent bond and the involved electrons are known as σ electrons. Usually all the single bonds are sigma bonds and can be formed via following combinations:

(1) s-s overlapping: This occurs when the involved s orbitals are half filled and one s orbital from each of the participating atoms undergo head to head overlapping along the internuclear axis. For example: in ${H_2}$.

(2) s-p overlapping: In this one half filled s orbital will overlap with another half filled p orbital along the internuclear axis. This can be formed by overlapping s orbital with ${p_x}$, ${p_y}$ and ${p_z}$ orbitals in ammonia ($N{H_3}$) molecule.

(3) p-p overlapping: It is formed when one half filled p orbital from each of the participating atoms undergoes head to head overlapping along the internuclear axis. For example: the $3{p_z} - 3{p_z}$ overlapping of the 2 chlorine atoms in $C{l_2}$.

-Pi (π) bonds: If two atomic orbitals overlap laterally or sidewise along a direction which is perpendicular to the internuclear axis, leads to the formation of pi bonds. For this the two orbitals overlapping should be parallel to each other but perpendicular to the principal axis. These bonds are much weaker than sigma bonds due to low degree of overlapping.

Hence we can say that on taking x-axis as internuclear axis sigma bonds can be formed by: 1s-1s, 1s-$2{p_x}$ and 1s-2s overlappings because they overlap head on head along the x-axis. But overlapping of $2{p_y}$ and $2{p_y}$ occurs laterally and in a direction perpendicular to the x-axis, so it will form a pi bond.
Hence the correct answer is: $2{p_y}$ and $2{p_y}$.
So, the correct answer is “Option C”.

Note: Sigma bonds can exist independently and are stronger while the pi bonds always exist along with the sigma bonds and are weaker, but a combination of sigma and pi bonds is stronger than a single sigma bond. Also sigma bonds have the ability to determine the shape of the molecule while the shape is not affected by the pi bonds.