Q.How many σ and π bonds are present in each of the following molecules?

Sigma Pi Bond Counting: From Intuition to Precision

Imagine you're building a molecular model with sticks and balls. Every single bond you see — a single line between two atoms — is made of one sigma bond. That's the backbone. A double bond? That's one sigma plus one pi bond. A triple bond? One sigma plus two pi bonds.

This is the core idea: sigma bonds are the first bond formed between any two atoms; any additional bonds are pi bonds.

Why sigma comes first

When two atoms approach each other, their orbitals overlap end-to-end along the line joining the nuclei. That head-on overlap creates a sigma bond — strong, cylindrically symmetric, and free to rotate. If the atoms need to share more electrons (to satisfy octets, for example), they can't form another sigma bond because the orbitals are already used up in that direction. Instead, they use sideways overlap of p-orbitals above and below the internuclear axis. That sideways overlap is a pi bond — weaker, and it locks the molecule into a plane (no free rotation).

So the rule is simple: between any two bonded atoms, exactly one bond is sigma; the rest are pi.

The precise counting method

For any molecule, you can count sigma and pi bonds systematically:

1. Count sigma bonds: Every single bond is one sigma. Every double bond contributes one sigma (and one pi). Every triple bond contributes one sigma (and two pi). Also, every bond to hydrogen is sigma.

2. Count pi bonds: For each multiple bond, subtract 1 from the bond order. That remainder is the number of pi bonds.

So:

• Single bond ($n=1$): 1 sigma, 0 pi
• Double bond ($n=2$): 1 sigma, 1 pi
• Triple bond ($n=3$): 1 sigma, 2 pi

A worked example: ethene ($\ce{C2H4}$)

Draw the structure: each carbon is double-bonded to the other, and each carbon has two single bonds to hydrogen.

• The C=C double bond: 1 sigma + 1 pi
• Each C–H single bond: 1 sigma (4 such bonds)
• Total: 5 sigma bonds, 1 pi bond

Check: The molecule has 5 sigma bonds holding the skeleton together, and 1 pi bond in the double bond region.

A trickier case: benzene ($\ce{C6H6}$)

Benzene has six C–C bonds that are all equivalent — each is 1.5 bonds (resonance hybrid). But for counting purposes, treat each ring bond as a single bond (sigma) plus a delocalised pi system.

• 6 C–H bonds: all sigma
• 6 C–C ring bonds: each is sigma
• The pi system: 3 pi bonds (delocalised over the ring)

Total: 12 sigma bonds, 3 pi bonds.

Why this matters

Sigma-pi counting is not just a classification exercise. It explains:

• Rotation barriers: Single bonds (pure sigma) rotate freely; double bonds (sigma + pi) do not.
• Reactivity: Pi bonds are weaker and more exposed — they're where addition reactions happen (e.g., $\ce{Br2}$ adding across a double bond).
• Hybridisation: The number of sigma bonds around an atom determines its hybridisation ($sp^3$ for 4 sigma bonds, $sp^2$ for 3, $sp$ for 2).

The one-sentence summary

Every bond has exactly one sigma bond; any additional bond order comes from pi bonds.

This topic is commonly searched as "Sigma Pi Bond Counting 11 chemistry important questions" or "Sigma Pi Bond Counting formula and examples", and it maps cleanly onto the Class 11 Chemistry portion of the NCERT/CBSE syllabus. Because sigma pi bond counting shows up repeatedly in JEE Main, NEET and state CET Chemistry papers, mastering the underlying idea (not just the formula) is genuinely worth the extra time.