Assertion (A): All the carbon atoms in H2C=C=CH2 are sp2 hybridised. Reason (R): In this molecule all the carb

NCERT Exemplar · Q58

Q. Assertion (A): All the carbon atoms in H2C=C=CH2 are sp2 hybridised. Reason (R): In this molecule all the carbon atoms are attached to each other by double bonds.

(i) Both A and R are correct and R is the correct explanation of A.

(ii) Both A and R are correct but R is not the correct explanation of A.

(iii) Both A and R are not correct.

(iv) A is not correct but R is correct.

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Concept understanding — Hybrid Orbital Overlap

The Intuition: Why Atoms Don't Just Use Their "Natural" Orbitals

Imagine you're trying to build a stable molecule. Carbon has four valence electrons — two in the 2s orbital and two in the 2p orbitals. If carbon used its pure s and p orbitals to form bonds, you'd expect two identical bonds (from the two p electrons) and two different, weaker bonds (from the s electrons). But experiment shows methane ( $\mathrm{CH_4}$ ) has four identical bonds, all at $109.5^\circ$ angles.

Nature has a trick: before bonding, the atom's orbitals mix — like blending primary colours to get new shades. This mixing is called hybridisation, and the resulting orbitals are hybrid orbitals.

Note

Hybridisation is a mathematical model, not a physical event. The atom doesn't "decide" to hybridise — we use hybrid orbitals to explain the observed geometry and bond equivalence.

The Precise Statement

Hybrid orbital overlap is the process where two atoms form a covalent bond by overlapping their hybrid orbitals along the internuclear axis. The strength of the bond depends on how well the orbitals overlap — more overlap means a stronger bond.

The key idea: hybrid orbitals are directional and concentrated in specific regions of space, which allows them to overlap more effectively than pure s or p orbitals.

How Hybrid Orbitals Are Constructed

Take $sp^3$ hybridisation (as in methane):

One 2s orbital + three 2p orbitals → four equivalent $sp^3$ hybrid orbitals
Each $sp^3$ orbital has 25% s-character and 75% p-character
They point toward the corners of a tetrahedron ( $109.5^\circ$ apart)

The mathematical form for an $sp^3$ hybrid orbital is:

\psi_{sp^3} = \frac{1}{2}\psi_{2s} + \frac{1}{2}\psi_{2p_x} + \frac{1}{2}\psi_{2p_y} + \frac{1}{2}\psi_{2p_z}

sp^n

Overlap in Action: Methane

When a hydrogen 1s orbital approaches a carbon $sp^3$ hybrid orbital along the line joining the nuclei:

The $sp^3$ lobe points directly at the hydrogen — maximum overlap
The electron density concentrates between the nuclei
A sigma ( $\sigma$ ) bond forms — cylindrical symmetry about the bond axis

Compare this to using a pure carbon 2p orbital: the p orbital has a node at the nucleus and lobes pointing in two opposite directions. Overlap with hydrogen would be weaker and less directional.

Watch out

Hybrid orbitals do not exist in isolated atoms. They are a mathematical convenience for bonded atoms. An isolated carbon atom has pure s and p orbitals — hybridisation only makes sense in the context of bonding.

Why Hybridisation Matters for Overlap

Property	Pure p orbital	$sp^3$ hybrid
Shape	Dumbbell (two lobes)	One large lobe, one small lobe
Directionality	Two opposite directions	One concentrated direction
Overlap with H 1s	Moderate (sideways)	Maximum (head-on)
Bond strength	Weaker	Stronger

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The central carbon in allene ( $H_2C=C=CH_2$ ) is sp-hybridised, not sp², so Assertion (A) is false. Reason (R) is true but irrelevant. The correct choice is (iv).

The question tests your understanding of hybridisation in cumulated dienes — molecules with consecutive double bonds. The key is to recognise that the central carbon in allene forms two sigma bonds (one to each adjacent carbon) and has two perpendicular pi bonds, which forces sp hybridisation, not sp².

Let’s break it down step by step.

Identify the bonding pattern in allene.

The molecule is $H_2C=C=CH_2$ . The two terminal carbons each have two C–H sigma bonds and one C=C sigma bond, plus one pi bond to the central carbon. The central carbon has two C=C sigma bonds (one to each terminal carbon) and two pi bonds — one to each terminal carbon, but these pi bonds lie in perpendicular planes.
Determine hybridisation of the terminal carbons.

Each terminal carbon is bonded to three atoms (two H and one C) and has no lone pairs. Three sigma bonds means three hybrid orbitals are needed, so the terminal carbons are sp² hybridised. This is correct.
Determine hybridisation of the central carbon.

The central carbon is bonded to only two atoms (the two terminal carbons) via sigma bonds. It has no lone pairs. Two sigma bonds require two hybrid orbitals, so the central carbon is sp hybridised. The two remaining p orbitals on the central carbon (pure p) form the two pi bonds — one with each terminal carbon’s p orbital.

Watch out

A common mistake is to assume that because all bonds are double bonds, every carbon must be sp². But the central carbon in a cumulated diene uses sp hybridisation — it has two sigma bonds and two pi bonds, not three sigma bonds.

Evaluate Assertion (A). …

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Q.Assertion (A): All the carbon atoms in H2C=C=CH2 are sp2 hybridised. Reason (R): In this molecule all the carbon atoms are attached to each other by double bonds.

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Q.Assertion (A): All the carbon atoms in H2C=C=CH2 are sp2 hybridised.
Reason (R): In this molecule all the carbon atoms are attached to each other by double bonds.

Unlock everything free for 14 days