JEE Mains Chemistry · Chemical Bonding and Molecular Structure

Molecular Orbital Theory

Atomic orbitals of the same symmetry combine into bonding and antibonding molecular orbitals; filling them in order gives the bond order, ½(Nb − Na), and the number of unpaired electrons.

All Chemical Bonding and Molecular Structure notes JEE Mains Chemistry strategy

Why this matters

Thirty-six PYQs, sixteen of them numerical, and three from 2026 — the chapter's largest page. Eight test how atomic orbitals combine: the conditions, symmetry about the axis and the shapes of bonding and antibonding orbitals; fifteen find or rank bond orders; thirteen count unpaired electrons or sort species into paramagnetic and diamagnetic. Three ideas cover the page.

Concept 1 of 3: Combining atomic orbitals (LCAO)

Two atomic orbitals that overlap make two molecular orbitals. Adding the waves in phase piles electron density between the nuclei and gives a bonding orbital of lower energy. Subtracting them leaves a node between the nuclei and gives an antibonding orbital of higher energy.

Definition

Conditions: the atomic orbitals must have comparable energy, the same symmetry about the molecular axis, and overlap as much as possible.
Bonding: $\psi_{MO} = \psi_A + \psi_B$ . Antibonding: $\psi_{MO} = \psi_A - \psi_B$ , with a node between the nuclei.
$n$ atomic orbitals give $n$ molecular orbitals, half bonding and half antibonding. The 2s and 2p orbitals of two atoms (8 in all) give 8 molecular orbitals, 4 of them antibonding.
Symmetry about the z axis: s, $p_z$ and $d_{z^2}$ are σ type; $p_x$ , $p_y$ , $d_{xz}$ and $d_{yz}$ are π type; $d_{xy}$ and $d_{x^2-y^2}$ are δ type.
A π bonding orbital has its density above and below the axis and a nodal plane containing the axis; π* has, in addition, a node between the nuclei.

Pair of orbitals (axis z)	Symmetry of each	Do they combine?
1s and 1s	σ and σ	Yes: σ1s and σ*1s
$2p_z$ and $2p_z$	σ and σ	Yes: σ2p and σ*2p (head-on)
$2p_x$ and $2p_x$	π and π	Yes: π2p and π*2p (sideways)
2s and $2p_z$	σ and σ	Yes, if their energies are close
2s and $2p_y$	σ and π	No: zero net overlap
$2p_x$ and $2p_y$	π, but at right angles	No: they are orthogonal
$3d_{xz}$ and $2p_x$	π and π	Yes: a π overlap
$3d_{xy}$ and $3d_{x^2-y^2}$	δ and δ, but rotated 45°	No: orthogonal to each other Both are δ type, yet they cancel; same symmetry label is not enough when the lobes are turned 45°.

Same symmetry about the axis and a matching orientation are both needed for a net overlap.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 24 Jan 2025 · Q30Moderate

Example 1 · Chemical Bonding and Molecular Structure · Molecular Orbital Theory

Which of the following linear combination of atomic orbitals will lead to formation of molecular orbitals in homonuclear diatomic molecules [internuclear axis in z-direction] ? (A)

2p_{z}

and

2p_{x}

(B) 2 s and

2p_{x}

(C)

3d_{xy}

and

3d_{x^{2}-y^{2}}

(D) 2 s and

2p_{z}

(E)

2p_{z}

and

3d_{x^{2}-y^{2}}

Bonding π density is not low above the axis

A π bonding orbital puts its density above and below the internuclear axis, with none on the axis itself. A statement that it has lower density above and below the axis is false.

Maximum overlap, not minimum

The three LCAO conditions are comparable energy, same symmetry and maximum overlap. 'Minimum overlap' or 'different symmetry' in a list of conditions is always a wrong option.

Drill 7 more on combining atomic orbitals (lcao)

Concept 2 of 3: Bond order from the MO diagram

Electrons in bonding orbitals hold the atoms together and electrons in antibonding orbitals push them apart. Bond order is the net count of shared pairs. A higher bond order means a shorter, stronger bond; a bond order of zero means the molecule does not exist.

Definition

Bond order $= \tfrac{1}{2}(N_b - N_a)$ : $N_b$ electrons in bonding orbitals, $N_a$ in antibonding orbitals.
Up to 14 electrons ( $\mathrm{B_2}$ , $\mathrm{C_2}$ , $\mathrm{N_2}$ ): $\sigma1s < \sigma^*1s < \sigma2s < \sigma^*2s < \pi2p_x = \pi2p_y < \sigma2p_z < \pi^* < \sigma^*$ .
From 15 electrons ( $\mathrm{O_2}$ , $\mathrm{F_2}$ ): $\sigma2p_z$ drops below the two π2p orbitals.
Values: $\mathrm{B_2}$ 1, $\mathrm{C_2}$ 2, $\mathrm{N_2}$ 3, $\mathrm{O_2}$ 2, $\mathrm{F_2}$ 1; $\mathrm{He_2}$ , $\mathrm{Be_2}$ , $\mathrm{Ne_2}$ 0 (they do not exist); $\mathrm{He_2^+}$ 0.5.
Isoelectronic species share a bond order: 14 electrons ( $\mathrm{N_2}$ , CO, $\mathrm{NO^+}$ , $\mathrm{CN^-}$ , $\mathrm{C_2^{2-}}$ , $\mathrm{O_2^{2+}}$ ) all give 3.
Removing an electron from an antibonding orbital strengthens the bond ( $\mathrm{O_2 \to O_2^+}$ , $\mathrm{NO \to NO^+}$ ); removing one from a bonding orbital weakens it ( $\mathrm{N_2}$ , $\mathrm{C_2}$ , $\mathrm{B_2}$ ).

Bond order

\text{bond order} = \tfrac{1}{2}(N_b - N_a)

Worked example

Find the bond orders of

\mathrm{B_2}

\mathrm{C_2}

and

\mathrm{N_2}

and rank their bond strengths.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 1 · Q28Moderate

Example 2 · Chemical Bonding and Molecular Structure · Molecular Orbital Theory

Given below are two statements : Statement (I) : The correct sequence of bond lengths in the following species is :

O_{2}^{+} < O_{2} < O_{2}^{-} < O_{2}^{2 -}

Statement (II) : The correct sequence of number of unpaired electrons in the following species is :

O_{2} > O_{2}^{+} > O_{2}^{-} > O_{2}^{2 -}

In the light of the above statements, choose the correct answer from the options given below :

Count every electron, core included, or none

Either count all electrons (σ1s and σ*1s cancel) or only the valence ones; the bond order is the same. Mixing the two, for example counting σ1s as bonding but skipping σ*1s, adds one to the answer.

A bond order of zero means no molecule

\mathrm{Be_2}

and

\mathrm{He_2}

have as many antibonding as bonding electrons, so they do not exist.

\mathrm{He_2^+}

\mathrm{He_2^-}

and

\mathrm{O_2^{2-}}

have positive bond orders and do.

Drill 14 more on bond order from the mo diagram

Concept 3 of 3: Unpaired electrons and magnetism

A species with any unpaired electron is drawn into a magnetic field: it is paramagnetic. If every electron is paired, it is diamagnetic. The unpaired electrons sit in the highest filled orbitals, usually the π or π* pair, and Hund's rule spreads them out singly.

Definition

Paramagnetic: at least one unpaired electron. Diamagnetic: none.
$\mathrm{O_2}$ has two unpaired electrons in π*, which the Lewis structure cannot show; $\mathrm{B_2}$ has two in π2p.
Any species with an odd electron count is paramagnetic: NO, $\mathrm{NO_2}$ , $\mathrm{ClO_2}$ , $\mathrm{KO_2}$ (the superoxide ion $\mathrm{O_2^-}$ ).
Spin-only magnetic moment: $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons.
$\mathrm{S_2}$ , like $\mathrm{O_2}$ , is paramagnetic; $\mathrm{N_2}$ , $\mathrm{F_2}$ and $\mathrm{Cl_2}$ are diamagnetic.

Species	Electrons	Bond order	Unpaired electrons	Magnetism
$\mathrm{H_2^+}$ , $\mathrm{He_2^+}$	1, 3	0.5	1	Paramagnetic
$\mathrm{Li_2}$	6	1	0	Diamagnetic
$\mathrm{B_2}$	10	1	2	Paramagnetic
$\mathrm{C_2}$	12	2	0	Diamagnetic
$\mathrm{C_2^-}$ , $\mathrm{N_2^+}$	13	2.5	1	Paramagnetic
$\mathrm{N_2}$ , CO, $\mathrm{CN^-}$ , $\mathrm{NO^+}$	14	3	0	Diamagnetic
$\mathrm{N_2^-}$ , $\mathrm{O_2^+}$ , NO	15	2.5	1	Paramagnetic
$\mathrm{O_2}$ , $\mathrm{N_2^{2-}}$	16	2	2	Paramagnetic
$\mathrm{O_2^-}$	17	1.5	1	Paramagnetic
$\mathrm{O_2^{2-}}$ , $\mathrm{F_2}$	18	1	0	Diamagnetic O₂²⁻ has 10 electrons in bonding orbitals and 8 in antibonding ones.

Species with the same electron count have the same bond order and the same number of unpaired electrons.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 2 · Q31Moderate

Example 3 · Chemical Bonding and Molecular Structure · Molecular Orbital Theory

Pair of species among the following having same bond order as well as paramagnetic character will be-

O₂⁺ and O₂⁻ have the same number of unpaired electrons

\mathrm{O_2^+}

has one π* electron and

\mathrm{O_2^-}

has three, one of them unpaired. Both have exactly one unpaired electron, so any order that puts one above the other is false.

N₂²⁻ looks like N₂ but behaves like O₂

Adding two electrons to

\mathrm{N_2}

gives 16, the count of

\mathrm{O_2}

. So

\mathrm{N_2^{2-}}

has bond order 2 and two unpaired electrons: it is paramagnetic.

Drill 12 more on unpaired electrons and magnetism

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Bond order from the MO diagram
Bond order
$\text{bond order} = \tfrac{1}{2}(N_b - N_a)$

Reference tables (2)

Combining atomic orbitals (LCAO)8 rows

Pair of orbitals (axis z)	Symmetry of each	Do they combine?
1s and 1s	σ and σ	Yes: σ1s and σ*1s
$2p_z$ and $2p_z$	σ and σ	Yes: σ2p and σ*2p (head-on)
$2p_x$ and $2p_x$	π and π	Yes: π2p and π*2p (sideways)
2s and $2p_z$	σ and σ	Yes, if their energies are close
2s and $2p_y$	σ and π	No: zero net overlap
$2p_x$ and $2p_y$	π, but at right angles	No: they are orthogonal
$3d_{xz}$ and $2p_x$	π and π	Yes: a π overlap
$3d_{xy}$ and $3d_{x^2-y^2}$	δ and δ, but rotated 45°	No: orthogonal to each other Both are δ type, yet they cancel; same symmetry label is not enough when the lobes are turned 45°.

Same symmetry about the axis and a matching orientation are both needed for a net overlap.

Unpaired electrons and magnetism10 rows

Species	Electrons	Bond order	Unpaired electrons	Magnetism
$\mathrm{H_2^+}$ , $\mathrm{He_2^+}$	1, 3	0.5	1	Paramagnetic
$\mathrm{Li_2}$	6	1	0	Diamagnetic
$\mathrm{B_2}$	10	1	2	Paramagnetic
$\mathrm{C_2}$	12	2	0	Diamagnetic
$\mathrm{C_2^-}$ , $\mathrm{N_2^+}$	13	2.5	1	Paramagnetic
$\mathrm{N_2}$ , CO, $\mathrm{CN^-}$ , $\mathrm{NO^+}$	14	3	0	Diamagnetic
$\mathrm{N_2^-}$ , $\mathrm{O_2^+}$ , NO	15	2.5	1	Paramagnetic
$\mathrm{O_2}$ , $\mathrm{N_2^{2-}}$	16	2	2	Paramagnetic
$\mathrm{O_2^-}$	17	1.5	1	Paramagnetic
$\mathrm{O_2^{2-}}$ , $\mathrm{F_2}$	18	1	0	Diamagnetic O₂²⁻ has 10 electrons in bonding orbitals and 8 in antibonding ones.

Species with the same electron count have the same bond order and the same number of unpaired electrons.

Watch out for (6)

Bonding π density is not low above the axis
→ Combining atomic orbitals (LCAO)
Maximum overlap, not minimum
→ Combining atomic orbitals (LCAO)
Count every electron, core included, or none
→ Bond order from the MO diagram
A bond order of zero means no molecule
→ Bond order from the MO diagram
O₂⁺ and O₂⁻ have the same number of unpaired electrons
→ Unpaired electrons and magnetism
N₂²⁻ looks like N₂ but behaves like O₂
→ Unpaired electrons and magnetism

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