Q.Select the correct one: The hybridization of IF7 is:
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Orbital Hybridization Theory
Orbital Hybridization Theory – From Intuition to Precision
The Problem That Started It All
Imagine you are looking at a methane molecule, CH4. Carbon has four valence electrons: two in the 2s orbital and two in the 2p orbitals. If carbon used its pure atomic orbitals to bond, you would expect two bonds from the 2s (identical, but one direction) and two from the 2p (at 90∘ to each other). That would give you three different bond types and bond angles of 90∘ and something else.
But experiment says methane is perfectly tetrahedral: all four bonds are identical in length, strength, and energy, and the bond angle is 109.5∘, not 90∘. Something is fundamentally wrong with the "pure orbital" picture.
This is the puzzle that hybridization theory solves.
The Core Intuition
Think of atomic orbitals as shapes that an electron can occupy. The s orbital is a sphere. The p orbitals are dumbbells along the x, y, and z axes. When an atom forms bonds, it wants to mix these shapes together to create new, hybrid shapes that point in directions that maximise bond strength and minimise repulsion.
It is like mixing primary colours to get new colours. You don't have to use red, blue, and yellow separately — you can blend them to get green, orange, or purple. Similarly, an atom can blend its s and p orbitals to get new hybrid orbitals that are better suited for bonding.
The key insight: hybridization is a mathematical mixing of atomic orbitals on the same atom to produce an equal number of new, equivalent hybrid orbitals. The number of hybrid orbitals formed always equals the number of atomic orbitals mixed.
The Precise Statement
Orbital Hybridization Theory: When an atom forms covalent bonds, its valence atomic orbitals (one s and up to three p orbitals) can linearly combine to form an equal number of new, equivalent hybrid orbitals. These hybrid orbitals have specific directional properties that match the observed molecular geometry.
The theory rests on three pillars:
- Conservation of orbitals: Mixing n atomic orbitals gives exactly n hybrid orbitals. No orbitals are created or destroyed.
- Energy averaging: The hybrid orbitals have energies that are intermediate between the original s and p energies.
- Directionality: Hybrid orbitals point in specific directions to minimise electron pair repulsion, which directly determines molecular shape.
The Three Common Hybridizations
| Hybridization | Orbitals Mixed | Number of Hybrids | Geometry | Bond Angle | Example |
|---|---|---|---|---|---|
| sp | one s + one p | 2 | Linear | 180∘ | BeCl2 |
| sp2 | one s + two p | 3 | Trigonal planar | 120∘ | BF3 |
| sp3 | one s + three p | 4 | Tetrahedral | 109.5∘ | CH4 |
The superscript in sp2 or sp3 tells you how many p orbitals were mixed. sp3 means one s and three p orbitals were blended. It does not mean there are three s orbitals — there is only one s orbital per shell.
How It Works: The Methane Example
Carbon in its ground state has the configuration 1s22s22px12py1. Only two unpaired electrons — it should form only two bonds. But we know carbon forms four bonds.
Step 1: Promotion. One electron from the 2s orbital is promoted (excited) to the empty 2pz orbital. This costs a small amount of energy, but it is more than compensated by the energy released when four strong bonds form instead of two.
Step 2: Hybridization. The one 2s orbital and three 2p orbitals mix to form four equivalent sp3 hybrid orbitals. Each hybrid has 25% s character and 75% p character.
Step 3: Bonding. Each sp3 hybrid overlaps with the 1s orbital of a hydrogen atom, forming four identical σ bonds. The hybrids point to the corners of a tetrahedron, giving the 109.5∘ angle. …
The hybridization of the central atom is found by counting its total bond pairs plus lone pairs (steric number). …
Iodine in IF7 forms 7 sigma bonds to fluorine with zero lone pairs, so its steric number is 7, requiring sp3d3 hybridization (pentagonal bipyramidal shape).
Iodine has 7 valence electrons. In IF7, all 7 electrons are used to form 7 single bonds to 7 fluorine atoms, leaving no lone pair on iodine.
Steric number = number of bond pairs + number of lone pairs = 7 + 0 = 7.
…
Showing the 12 most recent of 30 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.The hybridisation of 'N' in NH3 is(a) sp3d(b) sp3d2(c) sp3(d) sp2
›Reveal solutionSolution
Nitrogen in NH3 has 5 valence electrons: 3 are used to form 3 N-H sigma bonds, and 2 remain as a lone pair, giving 4 electron domains total -> sp3 hybridisation.
Nitrogen (Group 15) has 5 valence electrons. In NH3, three of these electrons pair up with the 3 hydrogen atoms to form 3 N-H sigma bonds, and the remaining 2 electrons stay on nitrogen as a lone pair.
Total electron domains around N = 3 bond pairs + 1 lone pair = 4 domains.
…
- CBSE 2026Set ANNUAL1 markMCQQ.Example of Sp3d2 hybridization is -(a) CH4(b) PCl5(c) NH3(d) SF6
›Reveal solutionSolution
SF6 is the example of sp3d2 hybridisation among the given options; it has 6 bond pairs and 0 lone pairs around sulphur, giving octahedral geometry.
Check each option by counting the electron domains (sigma bonds + lone pairs) on the central atom:
- CH4: carbon has 4 bond pairs, 0 lone pairs → sp3 hybridisation (tetrahedral). Not this.
- PCl5: phosphorus has 5 bond pairs, 0 lone pairs → sp3d hybridisation (trigonal bipyramidal). Not this. …
- CBSE 2026Set ANNUAL1 markQ.Draw the molecular orbital diagram of Ethane.
›Reveal solutionSolution
Figure — The stem asks to draw a labelled diagram of the Daniell cell, and the catalog figure is exactly the Daniell ce Ethane's bonding is described not by a diatomic-style MO energy diagram, but by sp3 orbital-overlap (hybridisation): a C-C sigma bond plus six C-H sigma bonds.
Worth noting up front: the formal molecular orbital (MO) theory energy-level diagrams taught at this level (showing sigma, sigma*, pi, pi* levels and bond order calculations) are constructed for homonuclear diatomic molecules such as H2, He2, Li2, N2, O2, F2 — not for polyatomic organic molecules like ethane. For a molecule like ethane (C2H6), NCERT instead explains the bonding using valence bond theory with orbital hybridisation.
…
- CBSE 2026Set ANNUAL1 markMCQQ.Which of the following is not true for hybridisation?(a) The orbitals present in the valence shell of the atom are hybridised.(b) The orbital undergoing hybridisation should have almost equal energy.(c) Promotion of electron is essential condition prior to hybridisation.(d) It is not necessary that only half filled orbitals participate in hybridisation. In some case, filled and even empty orbitals of valence shell take part in hybridisation.
›Reveal solutionSolution
Promotion of an electron is common but not an essential precondition for hybridisation — that is the false statement.
Checking each option:
- (a) True — only valence-shell orbitals of the atom undergo hybridisation.
- (b) True — the orbitals mixing must have nearly the same energy for effective hybridisation.
- (c) Not true — promotion of an electron to an empty orbital is a common accompanying feature (e.g. in carbon, 2s→2p promotion before sp3 mixing) but is not an essential condition; hybridisation can also occur among orbitals that are already singly/doubly occupied without any promotion step. …
- CBSE 2026Set ANNUAL1 markQ.Write ethane, ethene and ethyne in order of decreasing electronegativity of carbon in them.
›Reveal solutionSolution
More s-character means the hybrid orbital's electrons are held closer to the nucleus, making that carbon more electronegative; ethyne's sp carbon has the most s-character.
The carbon atoms in these hydrocarbons are hybridised differently:
- Ethane (C2H6): sp3 hybridised carbon, 25% s-character.
- Ethene (C2H4): sp2 hybridised carbon, 33% s-character.
- Ethyne (C2H2): sp hybridised carbon, 50% s-character. …
- CBSE 2026Set ANNUAL1 markMCQQ.Hybridization state in CH3Cl is:(a) sp3, sp2(b) sp3, sp(c) sp2(d) sp3
›Reveal solutionSolution
The carbon in CH3Cl is bonded to four different atoms (3 H + 1 Cl) by four single bonds, so it is sp3 hybridized with tetrahedral geometry.
To determine hybridization, count the number of sigma bonds and lone pairs on the central atom. In CH3Cl (chloromethane), the carbon atom forms:
- 3 sigma bonds to hydrogen atoms
- 1 sigma bond to the chlorine atom
- 0 lone pairs …
- CBSE 2026Set ANNUAL1 markMCQQ.The hybridization of carbon in CO2 is(a) sp(b) sp^2(c) sp^3(d) sp^3 d
›Reveal solutionSolution
Carbon in CO2 is sp hybridised.
CO2 is O=C=O. The central carbon forms two sigma bonds (to the two oxygens) and has no lone pairs, so it needs two hybrid orbitals → sp hybridisation. This gives a linear shape with a 180° …
- CBSE 2025Set ANNUAL1 markMCQQ.Which of the following central atoms contains sp3-hybridisation?(a) PCl3(b) SO3(c) BF3(d) NO3-
›Reveal solutionSolution
PCl3's central phosphorus atom is sp3 hybridised.
Phosphorus in PCl3 has 3 bonding pairs (to the 3 Cl atoms) plus 1 lone pair, totalling 4 electron domains, which corresponds to sp3 hybridisation and a pyramidal shape. SO3 and BF3 each have 3 electron domains around …
- CBSE 2025Set ANNUAL1 markMCQQ.The correct order regarding the electronegativity of hybrid orbitals of carbon is(a) sp < sp2 < sp3(b) sp < sp2 > sp3(c) sp > sp2 < sp3(d) sp > sp2 > sp3
›Reveal solutionSolution
Electronegativity of carbon's hybrid orbitals follows sp > sp2 > sp3.
The percentage of s-character is 50% in sp, 33% in sp2, and 25% in sp3 hybrid orbitals. Since s-orbitals hold electrons closer to (and more tightly bound to) the nucleus than p-orbitals, a hybrid orbital with more s-character …
- CBSE 2025Set ANNUAL1 markMCQQ.The hybridisation of P in PCl5 is(a) dsp2(b) sp3d(c) sp2d2(d) spd3
›Reveal solutionSolution
PCl5 has phosphorus bonded to 5 chlorine atoms with no lone pairs, needing 5 hybrid orbitals — one s, three p, and one d orbital combine to form sp3d hybrids.
Step 1 — Count electron domains on P: Phosphorus in PCl5 forms 5 sigma bonds to 5 chlorine atoms and has no lone pairs remaining (P uses all 5 valence electrons, one per bond, made possible because P can expand its octet using an empty 3d orbital — it's in period 3).
Step 2 — Determine hybridisation from domain count: 5 electron domains require 5 equivalent hybrid orbitals. These are built by mixing:
1 s orbital + 3 p orbitals + 1 d orbital = 5 hybrid orbitals -> sp3d hybridisation
…
- CBSE 2025Set ANNUAL1 markMCQQ.State of hybridisation of carbon in HCHO is:(a) sp(b) sp2(c) sp3(d) dsp2
›Reveal solutionSolution
Carbon in formaldehyde forms 3 sigma bonds (2 C-H, 1 C-O) plus one pi bond to O, which is the classic sp2 + unhybridised p-orbital picture.
Formaldehyde, HCHO (methanal), has the structure:
H−∥CO−H
The central carbon atom forms:
- 2 sigma (σ) bonds to the two H atoms,
- 1 sigma (σ) bond to O, plus 1 pi (π) bond to O (the C=O double bond). …
- CBSE 2025Set sz1 markMCQQ.Select the correct one: Carbon in ethylene involves the hybridisation:(a) sp3(b) sp2(c) sp(d) None of these
›Reveal solutionSolution
Each carbon atom in ethylene (C2H4) is sp2 hybridised.
In ethylene, each carbon is bonded to two hydrogen atoms and one other carbon atom by sigma bonds (three sigma bonds total per carbon), arranged in a trigonal planar geometry with about 120 degree bond angles. This requires one s orbital and two p orbitals on each carbon to mix and form three equivalent sp2 hybrid orbitals. The one remaining unhybridised p orbital on each carbon, oriented perpendicular to the plane of the molecule, overlaps sideways with the corresponding p orbital on the other carbon to form a pi bond — together with …
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.