Q.What is meant by hybridisation? Compound CH2=C=CH2 contains sp or sp2 hybridised carbon atoms? Will it be a planar molecule?

Concept: Orbital Hybridization Theory – mixing of atomic orbitals to form equivalent hybrid orbitals for bonding.

Reasoning:

1. In $\mathrm{CH_2=C=CH_2}$ (allene), the central carbon is bonded to two other carbons via two double bonds. It uses two $sp$ hybrid orbitals for sigma bonds and two unhybridised $p$ orbitals for the two perpendicular pi bonds.
2. Each terminal carbon is bonded to two hydrogens and one carbon via a double bond. Each uses three $sp^2$ hybrid orbitals (for three sigma bonds) and one unhybridised $p$ orbital for the pi bond. …

The Concept: Why Hybridisation Exists

Hybridisation is not a real physical event — it is a mathematical model that explains observed molecular geometry. Carbon in its ground state has the configuration $1s^2 2s^2 2p^2$, which would suggest only two unpaired electrons and thus only two bonds. But carbon almost always forms four bonds. The theory says: one $2s$ orbital and three $2p$ orbitals mix (hybridise) to form four equivalent $sp^3$ orbitals, each with one electron, pointing to the corners of a tetrahedron. The number and type of hybrid orbitals depend on how many $p$ orbitals are mixed in — $sp$ uses one $p$, $sp^2$ uses two, $sp^3$ uses all three.

The key insight: the geometry of the hybrid orbitals determines the shape of the molecule. $sp$ gives linear (180°), $sp^2$ gives trigonal planar (120°), $sp^3$ gives tetrahedral (109.5°). You can identify the hybridisation of an atom by counting the number of atoms bonded to it plus the number of lone pairs — this is the steric number.

Step-by-Step Analysis of $\mathrm{CH_2=C=CH_2}$ (Allene)

1. Draw the structure. Allene has three carbon atoms in a chain: $\mathrm{H_2C=C=CH_2}$. The central carbon is doubly bonded to each terminal carbon. Each terminal carbon is also bonded to two hydrogen atoms.

2. Find hybridisation of the central carbon. The central carbon forms two double bonds — that means it is bonded to two atoms (the two terminal carbons). It has no lone pairs. Steric number = 2. This requires $sp$ hybridisation. The two $sp$ hybrid orbitals lie 180° apart, giving a linear $\mathrm{C=C=C}$ backbone.

3. Find hybridisation of each terminal carbon. Each terminal carbon is bonded to the central carbon (one double bond) and to two hydrogen atoms (two single bonds). That is three atoms bonded, no lone pairs. Steric number = 3. This requires $sp^2$ hybridisation. The three $sp^2$ hybrid orbitals lie in a plane at 120° to each other.

4. What about the remaining $p$ orbitals? In $sp$ hybridisation, two $p$ orbitals remain unhybridised (pure $p$). In $sp^2$ hybridisation, one $p$ orbital remains unhybridised. These unhybridised $p$ orbitals form the $\pi$ bonds. The central carbon has two pure $p$ orbitals, perpendicular to each other. One of these $p$ orbitals overlaps with a $p$ orbital from the left terminal carbon to form one $\pi$ bond; the other $p$ orbital overlaps with a $p$ orbital from the right terminal carbon to form the second $\pi$ bond. …