Which of the following molecular orbitals has maximum number of nodal planes? (i) σ*1s (ii) σ*2pz (iii) π2pₓ (

NCERT Exemplar · Q32

Q. Which of the following molecular orbitals has maximum number of nodal planes?

(i) σ*1s

(ii) σ*2p_z

(iii) π2p_x

(iv) π*2p_y

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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, $\text{CH}_4$ . 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 $90^\circ$ to each other). That would give you three different bond types and bond angles of $90^\circ$ 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 $109.5^\circ$ , not $90^\circ$ . 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

s

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	$180^\circ$	$\text{BeCl}_2$
$sp^2$	one $s$ + two $p$	3	Trigonal planar	$120^\circ$	$\text{BF}_3$
$sp^3$	one $s$ + three $p$	4	Tetrahedral	$109.5^\circ$	$\text{CH}_4$

Note

The superscript in $sp^2$ or $sp^3$ tells you how many $p$ orbitals were mixed. $sp^3$ 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 $1s^2 2s^2 2p_x^1 2p_y^1$ . 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 $2p_z$ 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 $sp^3$ hybrid orbitals. Each hybrid has 25% $s$ character and 75% $p$ character.

Step 3: Bonding. Each $sp^3$ hybrid overlaps with the $1s$ orbital of a hydrogen atom, forming four identical $\sigma$ bonds. The hybrids point to the corners of a tetrahedron, giving the $109.5^\circ$ angle. …

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Rank the candidates by nodal planes: $\sigma^*1s$ has one, $\pi 2p_x$ has one, $\pi^*2p_y$ has two, and $\sigma^*2p_z$ has the most — the two nodal planes carried by its parent $2p_z$ orbitals (perpendicular to the bond axis, through each nucleus) plus the extra internuclear node created by the antibonding combination. The answer is (ii) $\sigma^*2p_z$ .

Counting the nodes

$\sigma^*1s$ : the 1s orbitals are nodeless; subtraction adds exactly one nodal plane, midway between the nuclei. Total: 1.
$\pi 2p_x$ (bonding): inherits the single nodal plane of the p orbitals — the plane that contains the internuclear axis. Total: 1. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2026Set 2026-M1 markMCQ
$\mathrm{NO}_2{ }^{+}, \mathrm{NO}_3{ }^{-}$
›Reveal solutionSolution
The hybridisation of the central nitrogen atom in each species is determined by its steric number (number of sigma bonds + lone pairs). For $\mathrm{NO}_2^{+}$ , steric number 2 gives $sp$ ; for $\mathrm{NO}_3^{-}$ , steric number 3 gives $sp^2$ ; for $\mathrm{NH}_4^{+}$ , steric number 4 gives $sp^3$ . The correct option is (A).

The key idea is that hybridisation is a property of the central atom, and it depends only on how many regions of electron density (sigma bonds and lone pairs) surround that atom. The number of pi bonds or the overall charge doesn't directly change the hybridisation — it only affects the electron count used to find lone pairs.

Let’s work through each ion step by step.

$\mathrm{NO}_2^{+}$ (nitronium ion)

Nitrogen is the central atom. Count its valence electrons: nitrogen has 5. The positive charge means we remove 1 electron, so total valence electrons to distribute = $5 - 1 = 4$ .

Each oxygen atom (as a terminal atom) typically forms a double bond with nitrogen in this ion. Two double bonds mean two sigma bonds and two pi bonds. There are no lone pairs left on nitrogen because all 4 electrons are used in bonding (each sigma bond uses 1 electron from N, and the remaining 2 electrons go into pi bonds).

Steric number = number of sigma bonds + number of lone pairs = $2 + 0 = 2$ .

Steric number 2 corresponds to $sp$ hybridisation (linear geometry).

$\mathrm{NO}_3^{-}$ (nitrate ion)

Nitrogen has 5 valence electrons. The negative charge adds 1 electron, so total = $5 + 1 = 6$ .

The nitrate ion has three oxygen atoms bonded to nitrogen. One of the N–O bonds is a double bond (one sigma, one pi), and the other two are single bonds (each sigma) with a negative charge on those oxygens. So nitrogen forms three sigma bonds (one to each oxygen) and has no lone pair — the remaining 2 electrons after forming three sigma bonds (using 3 electrons) go into the pi bond.

Steric number = $3 + 0 = 3$ .

Steric number 3 gives $sp^2$ hybridisation (trigonal planar). …
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KCET 2026Set D31 markMCQ
$NO_2^+$
›Reveal solutionSolution

Determine the hybridisation of the central nitrogen in each ion from its geometry, which in turn follows from the number of sigma-bonded groups (and lone pairs, if any) around it.

Step 1 — $NO_2^+$ (nitronium ion)

Nitrogen is doubly bonded to two oxygen atoms with no lone pair on nitrogen, giving a linear geometry (O=N=O, bond angle 180°). Two regions of electron density around nitrogen correspond to sp hybridisation.

Step 2 — $NO_3^-$ (nitrate ion)

Nitrogen is bonded to three oxygen atoms (with delocalised double-bond character shared among them by resonance) and has no lone pair, giving a trigonal planar geometry (bond angle 120°). Three regions of electron density correspond to $sp^2$ hybridisation.

Step 3 — $NH_4^+$ (ammonium ion) …
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COMEDK 2024Set 2024-E1 markMCQ
$\mathrm{NH}_3$
›Reveal solutionSolution
Statement C is wrong: methyl carbocation is $sp^2$ (trigonal planar), not $sp^3$ .

Evaluate each:

A — Isoelectronic species (e.g. $N_2$ , $CO$ , $CN^-$ , $NO^+$ ) share the same bond order. Correct.

B — In $NH_3$ the lone-pair and bond dipoles add; in $NF_3$ they oppose, so $\mu(NH_3)>\mu(NF_3)$ . Correct. …
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KCET 2021Set B-21 markMCQ
$Q. The number of six membered and five membered rings in Buckminster Fullerene respectively is (A) 20, 12 (B) 12, 20 (C) 14, 18 (D) 14, 11$
›Reveal solutionSolution
$C_{60}$ is a truncated icosahedron — 20 hexagonal and 12 pentagonal faces — verifiable from Euler's formula $V - E + F = 2$ .

Step 1 — What Buckminsterfullerene is

Discovered in 1985 and named after the architect Buckminster Fuller (whose geodesic domes it resembles), $C_{60}$ is an allotrope of carbon shaped like a football (soccer ball) — a closed cage of 60 carbon atoms. Every carbon is $sp^2$ hybridised, bonded to three others, with the remaining $p$ -electron delocalised over the cage.

Step 2 — The ring count

The cage is a truncated icosahedron, containing:

20 six-membered rings (hexagons)

12 five-membered rings (pentagons)

A key structural rule: no two pentagons are adjacent (the isolated-pentagon rule) — each pentagon is surrounded entirely by hexagons, which is what makes the cage stable and closes it into a sphere.

Step 3 — Verify with Euler's polyhedron formula

For any convex polyhedron: $V - E + F = 2$ .

Vertices $V = 60$ (one carbon each).

Edges: each carbon forms 3 bonds, and every bond is shared by 2 carbons:

$E = \frac{60\times 3}{2} = 90$

Faces: $F = 2 - V + E = 2 - 60 + 90 = 32$

Now let $h$ = number of hexagons and $p$ = number of pentagons:

$h + p = 32 \qquad \text{(total faces)}$ …
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KCET 2019Set A-11 markMCQ
$\text{SO}_2$
›Reveal solutionSolution
The net dipole moment depends on both bond polarity and molecular geometry. Only SO₂ has a bent shape that prevents the bond dipoles from cancelling, giving it a net dipole moment.

The key idea is simple: a molecule has a net dipole moment only if two conditions are met — the bonds themselves must be polar (different electronegativities), and the molecular geometry must be such that those bond dipoles do not cancel each other out. Symmetry is the enemy of a net dipole.

Let’s examine each molecule one by one.

SO₂ (Sulfur dioxide)

Sulfur is bonded to two oxygen atoms. The S–O bond is polar because oxygen is more electronegative than sulfur. But the real question is geometry. SO₂ has a bent shape (V-shaped) because sulfur has a lone pair of electrons that pushes the two oxygens down. The bond dipoles point from S to each O, and because the molecule is bent, these two vectors add up — they do not cancel. The result is a net dipole moment pointing roughly upward, away from the lone pair.

So SO₂ does possess a net dipole moment.

BeCl₂ (Beryllium chloride)

Beryllium forms two bonds with chlorine. The Be–Cl bond is polar (Cl is more electronegative). But BeCl₂ is linear — the two chlorines are exactly opposite each other at 180°. The bond dipoles are equal in magnitude and point in exactly opposite directions. They cancel perfectly.

Net dipole moment = zero.

BF₃ (Boron trifluoride)

Boron is bonded to three fluorine atoms. The B–F bond is highly polar (F is the most electronegative element). However, BF₃ is trigonal planar with 120° bond angles. The three bond dipoles are symmetrically arranged. When you add three equal vectors at 120° to each other, the resultant is zero.

No net dipole moment.

CO₂ (Carbon dioxide) …
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Q.Which of the following molecular orbitals has maximum number of nodal planes?

Counting the nodes

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