The outer orbitals of C in ethene molecule can be considered to be hybridized to give three equivalent sp² orbitals. The total number of sigma (s) and pi (p) bonds in ethene molecule is

The correct option is 5 sigma (s) and 1 pi (p) bonds.

To understand why this is correct, let us analyze the structure and bonding of the ethene molecule step-by-step:

1. Structure of Ethene:
Ethene is a hydrocarbon with the molecular formula:

$$C_2{H}_4$$

It consists of two carbon atoms bonded to each other, with each carbon atom also bonded to two hydrogen atoms. Structurally, it is represented as:

$$H_{2}C=C{H}_2$$

2. Hybridization of Carbon:
Each carbon atom in ethene is hybridized as:

$$s{p}^2$$

This hybridization is achieved by mixing one 2s orbital and two 2p orbitals, resulting in three equivalent hybrid orbitals aligned in a planar trigonal geometry. One unhybridized 2p orbital remains perpendicular to this plane on each carbon atom.

3. Counting Sigma (σ) Bonds:
Sigma bonds are formed by the head-on overlap of atomic or hybridized orbitals. In ethene, these are formed as follows:
• Carbon-Carbon (C-C) overlap: One hybrid orbital from the first carbon atom overlaps head-on with a hybrid orbital from the second carbon atom to form a single C-C sigma bond.
• Carbon-Hydrogen (C-H) overlap: The remaining two hybrid orbitals on each carbon atom overlap with the 1s orbitals of hydrogen atoms. This forms 4 C-H sigma bonds in total (2 per carbon atom).
Summing these up:

$$1\text{ (C-C sigma bond) }+4\text{ (C-H sigma bonds) }=5\text{ sigma bonds}$$

4. Counting Pi (π) Bonds:
Pi bonds are formed by the lateral (sideways) overlap of unhybridized p-orbitals.
• The single unhybridized 2p orbital on each of the two carbon atoms overlap sideways with one another to form 1 pi bond.

Conclusion:
Therefore, the double bond between the two carbon atoms consists of one sigma bond and one pi bond, while the four carbon-hydrogen single bonds are all sigma bonds. This gives a total of 5 sigma (s) bonds and 1 pi (p) bonds.