Q.What is the hybridisation of each carbon in H2C=C=CH2?

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. …