Why Carbon Is Tetravalent
Carbon sits at the head of Group 14 with the electronic configuration $\ce{1s^2 2s^2 2p^2}$. Four of its six electrons lie in the valence shell, two short of the stable octet. Carbon cannot reasonably gain four electrons to form $\ce{C^{4-}}$ nor lose four to form $\ce{C^{4+}}$, because both demand prohibitively large amounts of energy. The only practical route to an octet is to share its four valence electrons, forming four covalent bonds. This is the meaning of tetravalence: in essentially every stable organic compound, each carbon atom is joined to four other atoms or groups by four bonds.
The NIOS lesson uses this rule constantly while building structures. When it instructs you to "attach H-atoms to the C-atoms of the main chain to satisfy the tetravalency of carbon," it is invoking precisely this principle: a carbon skeleton is only complete once every carbon shows four bonds. A double bond counts as two of those four, and a triple bond as three, so the total of four is preserved whether the bonds are single, double or triple.
Tetravalence is about number of bonds, not number of atoms
A carbon with one double bond and two single bonds (as in $\ce{H2C=CH2}$) is bonded to only three atoms, yet it is still tetravalent because the double bond contributes two bonds. Students who count attached atoms instead of bonds mis-assign hybridisation and shape.
Count bonds: single = 1, double = 2, triple = 3. They must always sum to four on carbon.
Hybridisation: The Bridge to Shape
Tetravalency explains how many bonds carbon forms; it does not by itself explain the angles between them. To predict shape we use hybridisation — the mixing of the one 2s and the three 2p orbitals of carbon into a set of equivalent hybrid orbitals that point in fixed directions. The number of hybrid orbitals equals the number of atomic orbitals mixed, and their orientation minimises electron-pair repulsion.
Carbon shows three hybridisation states that matter for the basic molecules: sp³ (one s + three p), sp² (one s + two p) and sp (one s + one p). The geometry of the hybrid set — tetrahedral, trigonal planar or linear — is what we see as the shape of methane, ethene and ethyne respectively. The detailed orbital picture and the energetics of orbital mixing belong to the Chemical Bonding and Molecular Structure chapter; here we apply the result to carbon compounds.
Source note. The NIOS Chapter 23 text confirms tetravalency and the σ/π bond composition of multiple bonds, but does not lay out the numerical bond angles and orbital mixing. The hybridisation geometries and angles (109.5°, 120°, 180°) below are stated from the standard NEET-syllabus treatment of bonding and are cross-checked against the NEET PYQ bank, which repeatedly assumes them.
sp³ & the Tetrahedral Methane
In methane, $\ce{CH4}$, carbon mixes its 2s orbital with all three 2p orbitals to give four equivalent sp³ hybrid orbitals. These four orbitals point towards the corners of a regular tetrahedron, the arrangement that keeps the four bonding pairs as far apart as possible. Each sp³ orbital overlaps head-on with a hydrogen 1s orbital to form a C–H σ bond, so methane has four identical C–H bonds and the symmetric tetrahedral shape.
Every carbon in a saturated compound — every alkane, and every $\ce{-CH3}$, $\ce{-CH2-}$ or $\ce{>CH-}$ unit — is sp³ hybridised and locally tetrahedral. This is why the carbon skeleton of organic molecules zig-zags rather than lying in a straight line.
sp² & the Planar Ethene
When carbon forms a double bond, as in ethene $\ce{C2H4}$, it mixes its 2s orbital with only two of its 2p orbitals, giving three sp² hybrid orbitals in a plane, separated by $120^\circ$. The remaining unhybridised 2p orbital stands perpendicular to that plane. Each carbon uses its three sp² orbitals to form three σ bonds — two to hydrogen and one to the other carbon — and the leftover p orbitals overlap sideways to form the π bond.
Because all three sp² directions lie in a single plane and both carbons share that plane, ethene is a planar molecule with all bond angles close to $120^\circ$. The same sp² geometry appears at any doubly bonded carbon — in alkenes, in carbonyl groups, and in the carbon atoms of an aromatic ring.
Once you can assign hybridisation you can read a skeletal formula at a glance — see Structural Representations of Organic Compounds.
sp & the Linear Ethyne
In ethyne (acetylene), $\ce{C2H2}$, each carbon forms a triple bond. The carbon mixes its 2s orbital with just one 2p orbital, producing two sp hybrid orbitals oriented at $180^\circ$ — a straight line. The two unhybridised 2p orbitals remain perpendicular both to the molecular axis and to each other. Each carbon uses its two sp orbitals to make two σ bonds (one to H, one to the other carbon); the two pairs of p orbitals then overlap sideways to give two π bonds.
Sigma & Pi Bonds in C–C
The NIOS lesson states the bond composition of multiple bonds directly: the carbon–carbon double bond of an alkene "contains two types of bonds; one σ (sigma) bond and another π (pi) bond," while in alkynes "out of the three carbon–carbon bonds, one is σ (sigma) bond and the other two are π (pi) bonds." A σ bond arises from head-on (axial) overlap and concentrates electron density along the internuclear axis; a π bond arises from sideways overlap of unhybridised p orbitals, placing density above and below that axis.
Crucially, the text adds that "the π (pi) bond is weaker than the σ (sigma) bond and breaks easily." This single sentence explains the entire reactivity pattern of unsaturated hydrocarbons: because the π bond is the weak, exposed part of a multiple bond, alkenes and alkynes readily undergo addition reactions — for instance the decolourisation of bromine water, $\ce{CH2=CH2 + Br2 -> BrCH2-CH2Br}$, or catalytic hydrogenation, $\ce{CH3-CH=CH2 + H2 ->[Ni/Pt] CH3-CH2-CH3}$.
Count the σ and π bonds in propyne, $\ce{CH3-C#CH}$.
The $\ce{CH3}$ carbon has four single bonds (3 C–H + 1 C–C): 4 σ. The triple bond contributes 1 σ + 2 π. The terminal $\equiv$C–H is 1 σ. Totals: σ bonds $= 3 + 1 + 1 + 1 = 6$; π bonds $= 2$. The two triple-bond carbons are sp hybridised; the methyl carbon is sp³.
This same molecule underlies a real NEET item: in the propynyl carbanion $\ce{CH3-C#C^-}$, the negative carbon is sp hybridised because it belongs to the triple bond, so its lone pair sits in an sp orbital.
s-Character, Angle & Bond Strength
The percentage of s-character in a hybrid orbital is the fraction contributed by the s orbital: sp³ is $\tfrac{1}{4}=25\%$, sp² is $\tfrac{1}{3}\approx 33\%$ and sp is $\tfrac{1}{2}=50\%$. An s orbital is closer to the nucleus than a p orbital, so the more s-character a hybrid carries, the more tightly it holds its bonding electrons. Three measurable trends follow as we go sp³ → sp² → sp.
| Property | sp³ (e.g. $\ce{CH4}$) | sp² (e.g. $\ce{C2H4}$) | sp (e.g. $\ce{C2H2}$) |
|---|---|---|---|
| s-character | 25% | ≈ 33% | 50% |
| Bond angle | $109.5^\circ$ | $120^\circ$ | $180^\circ$ |
| C–H bond length | longest | intermediate | shortest |
| C–H bond strength / dissociation energy | lowest | intermediate | highest |
| Electronegativity of carbon | lowest (≈ 2.5) | intermediate (≈ 2.75) | highest (≈ 3.25) |
The electronegativity values in the last row are exactly those the NEET 2017 acidity question relies on: because an sp carbon is the most electronegative, it stabilises a negative charge best, making terminal alkynes the most acidic of the simple hydrocarbons, $\ce{HC#CH > CH3-C#CH > CH2=CH2 > CH3-CH3}$. The bond-strength row, in turn, is the basis of the NEET 2025 bond-dissociation-energy question: the C–H bond on an sp carbon has the highest dissociation energy of the three.
Higher s-character is acidic but its C–H bond is also stronger
These two facts feel contradictory but are not. Acidity is about how well the anion (after the proton leaves) is stabilised — sp carbon does this best. C–H bond dissociation energy is about breaking the bond homolytically — that bond is strongest at sp carbon. NEET tests both directions in different years.
More s-character → stronger, shorter C–H bond and a more stable conjugate base.
Hybridisation–Geometry Table
The following table is the single most exam-relevant summary of this subtopic. If you can reproduce it from memory you can answer almost any direct hybridisation, geometry or bond-angle question in the paper.
| Hybridisation | Orbitals mixed | Geometry | Bond angle | σ / π on each C | Example |
|---|---|---|---|---|---|
| sp³ | 1 s + 3 p | Tetrahedral | $109.5^\circ$ | 4 σ, 0 π | Methane, $\ce{CH4}$ |
| sp² | 1 s + 2 p | Trigonal planar | $120^\circ$ | 3 σ, 1 π | Ethene, $\ce{C2H4}$ |
| sp | 1 s + 1 p | Linear | $180^\circ$ | 2 σ, 2 π | Ethyne, $\ce{C2H2}$ |
Five things to lock in
- Carbon is tetravalent: configuration $\ce{1s^2 2s^2 2p^2}$ → shares four electrons → four bonds, always summing to four (single = 1, double = 2, triple = 3).
- sp³ → tetrahedral, $109.5^\circ$ ($\ce{CH4}$); sp² → planar, $120^\circ$ ($\ce{C2H4}$); sp → linear, $180^\circ$ ($\ce{C2H2}$).
- Single bond = 1 σ; double bond = 1 σ + 1 π; triple bond = 1 σ + 2 π. The π bond is weaker and breaks first, driving addition reactions.
- s-character rises 25% → 33% → 50%; with it the bond angle widens, the C–H bond shortens and strengthens, and carbon's electronegativity rises.
- Higher s-character means both a stronger C–H bond (high dissociation energy) and a more stable conjugate base (greater acidity of terminal alkynes).