What is the bond order in superoxide O2 minus?

O2 minus has 17 electrons, with one extra electron in an antibonding pi-star 2p MO. Bond order equals (10 minus 7) divided by 2, giving 1.5. Superoxide is paramagnetic with one unpaired electron.

Why do B2, C2, N2 use a different MO order than O2, F2, Ne2?

In lighter diatomics the 2s-2p energy gap is small, so s-p mixing raises sigma 2p above pi 2p. In heavier diatomics the gap is large and no mixing occurs, so sigma 2p stays below pi 2p.

Compare stability of N2, N2 plus, and N2 minus.

N2 has bond order 3 (most stable, diamagnetic). N2 plus has bond order 2.5 (electron removed from bonding sigma 2p). N2 minus has bond order 2.5 (electron added to antibonding pi star). N2 plus is slightly more stable than N2 minus.

Why is NO plus more stable than NO?

NO has 15 electrons with the last one in an antibonding pi-star MO. Removing it gives NO plus (14 electrons, bond order 3) with a stronger, shorter bond than neutral NO (bond order 2.5).

Molecular Orbital Theory

Q: What is a molecular orbital and how does it differ from an atomic orbital?

A molecular orbital extends over multiple atoms (polycentric), while an atomic orbital is centred on one nucleus (monocentric). MOs form by linear combination of atomic orbitals (LCAO) and follow the same filling rules as atomic orbitals.

Q: What is the difference between a bonding and an antibonding molecular orbital?

A bonding MO forms by constructive interference of atomic orbitals, has lower energy, and increases electron density between nuclei. An antibonding MO forms by destructive interference, has higher energy, and has a node between the nuclei with zero electron density there.

Q: Why does He2 not exist?

He2 would have configuration (sigma1s)2 (sigma*1s)2 giving bond order zero. With no net bonding, the two atoms cannot be held together. However, He2+ with 3 electrons has bond order 0.5 and exists transiently.

Q: Why is O2 paramagnetic?

O2 has 16 electrons. The two highest-occupied MOs are the degenerate pi-star 2p orbitals with one electron each (unpaired, following Hund's rule). Two unpaired electrons make O2 paramagnetic - MOT's most famous prediction.

Q: Which species is isoelectronic with N2 and has the same bond order?

CO, CN minus, and NO plus all have 14 electrons like N2 and share bond order 3.

Q: How does MOT explain fractional bond orders?

Fractional bond orders arise naturally when odd numbers of electrons distribute between bonding and antibonding MOs. H2 plus has bond order 0.5; O2 minus has 1.5. Classical Lewis theory cannot describe these but MOT handles them straightforwardly.

ChemistryChemical Bonding And Molecular StructureFor JEE aspirants

Molecular Orbital Theory (MOT), developed by F. Hund and R. S. Mulliken in 1932, treats a molecule as a whole entity where atomic orbitals combine to form new molecular orbitals that extend over the entire molecule. Electrons in the molecule occupy these molecular orbitals following the same Aufbau, Pauli, and Hund's rules as in atoms. Every pair of combining atomic orbitals gives a lower-energy bonding MO and a higher-energy antibonding MO. The stability of the molecule is measured by the bond order, $\frac{1}{2} (N_{b} - N_{a})$ : a positive bond order means a stable molecule, zero means it does not exist. MOT beats Valence Bond Theory at explaining paramagnetism of $O_{2}$ , existence of $H_{2}^{+}$ and $H e_{2}^{+}$ , non-existence of $H e_{2}$ and $B e_{2}$ , and fractional bond orders in ions like $O_{2}^{+}$ and $O_{2}^{-}$ .

Key Ideas - Quick Reference

Bond order $= \frac{1}{2} (N_{b} - N_{a})$ , where $N_{b}$ = electrons in bonding MOs, $N_{a}$ = electrons in antibonding MOs.
Positive bond order $\Rightarrow$ stable molecule; zero $\Rightarrow$ molecule does not exist.
Higher bond order $\Rightarrow$ shorter bond length and larger bond energy.
Any unpaired electrons $\Rightarrow$ paramagnetic; all paired $\Rightarrow$ diamagnetic.
Standard MO order (for $O_{2}$ , $F_{2}$ , $N e_{2}$ ): $σ 1 s < σ^{*} 1 s < σ 2 s < σ^{*} 2 s < σ 2 p_{z} < (π 2 p_{x} = π 2 p_{y}) < (π^{*} 2 p_{x} = π^{*} 2 p_{y}) < σ^{*} 2 p_{z}$ .
Modified order (for $B_{2}$ , $C_{2}$ , $N_{2}$ , due to $2 s$ - $2 p$ mixing): $(π 2 p_{x} = π 2 p_{y}) < σ 2 p_{z}$ (the $π$ MOs come below the $σ$ ).

1. The MOT Approach

In Valence Bond Theory, electrons stay localised in specific atomic orbitals and form bonds by pairing. In Molecular Orbital Theory, atoms lose their individual identities within a molecule. All the electrons of the molecule are distributed among a new set of molecular orbitals that extend over two or more nuclei.

A molecular orbital is a region of space around two or more nuclei where the probability of finding an electron is high. It is polycentric (unlike an atomic orbital, which is monocentric) and is filled with electrons following the Aufbau principle, Pauli exclusion principle, and Hund's rule of maximum multiplicity.

Salient features of MOT

Electrons in a molecule occupy molecular orbitals just as electrons in an atom occupy atomic orbitals.
Molecular orbitals are formed by linear combination of atomic orbitals (LCAO) of comparable energy and proper symmetry.
The number of molecular orbitals formed equals the number of atomic orbitals combined. Combining two AOs gives two MOs: one bonding, one antibonding.
A bonding molecular orbital has lower energy than the parent atomic orbitals and greater electron density between the nuclei.
An antibonding molecular orbital has higher energy than the parent atomic orbitals, with a nodal plane between the nuclei (zero electron density between them).
Electrons fill MOs in order of increasing energy.

2. Formation of Molecular Orbitals - LCAO Method

According to LCAO, molecular orbitals are built by adding or subtracting the wave functions of the combining atomic orbitals. For two atoms A and B with atomic orbital wave functions $ψ_{A}$ and $ψ_{B}$ :

$ψ_{bonding} = ψ_{A} + ψ_{B}$ (constructive interference)

$ψ_{antibonding}^{*} = ψ_{A} - ψ_{B}$ (destructive interference)

In the bonding MO, the two electron waves reinforce each other. Electron density between the nuclei increases, shielding the two nuclei from each other's repulsion and holding them together. In the antibonding MO, the waves cancel in the internuclear region, creating a node where electron density is zero. This actually pushes the nuclei apart.

Figure 1: Two atomic orbitals

ψ_{A}

and

ψ_{B}

on adjacent atoms combine to form two molecular orbitals. The in-phase sum (

ψ_{A} + ψ_{B}

) gives the bonding MO

σ

at lower energy than the atomic orbitals; the out-of-phase difference (

ψ_{A} - ψ_{B}

) gives the antibonding MO

σ^{*}

at higher energy. Electrons fill the bonding MO first.

Conditions for combining atomic orbitals

Similar energy: The two atomic orbitals must have similar (nearly equal) energies. A $1 s$ orbital cannot combine with a $2 s$ orbital.
Same symmetry about the molecular axis: Only orbitals of matching symmetry can combine. The $2 p_{z}$ of one atom can combine with the $2 p_{z}$ of another, but not with the $2 p_{x}$ or $2 p_{y}$ because of different symmetries.
Significant overlap: Atomic orbitals must overlap to an appreciable extent for MO formation to be effective.

3. Types of Molecular Orbitals

Molecular orbitals are classified based on the symmetry of the electron density around the internuclear axis.

Sigma ( $σ$ ) MOs

Symmetric about the internuclear axis. Formed by head-on overlap of s-s, s-p, or p-p orbitals along the axis. Examples: $σ 1 s$ , $σ^{*} 1 s$ , $σ 2 s$ , $σ^{*} 2 s$ , $σ 2 p_{z}$ , $σ^{*} 2 p_{z}$ (taking the z axis as the internuclear axis).

Pi ( $π$ ) MOs

Not symmetric about the internuclear axis. Formed by lateral overlap of parallel $2 p_{x}$ or $2 p_{y}$ orbitals. Examples: $π 2 p_{x}$ , $π 2 p_{y}$ , $π^{*} 2 p_{x}$ , $π^{*} 2 p_{y}$ . Since $2 p_{x}$ and $2 p_{y}$ are degenerate atomic orbitals, the resulting $π$ MOs are also degenerate (equal energy).

4. Energy Level Diagrams

The order of molecular orbital energies has been determined experimentally from spectroscopic data. There are two standard orderings, which apply to different homonuclear diatomic molecules.

Standard order (for $O_{2}$ , $F_{2}$ , $N e_{2}$ )

$σ 1 s < σ^{*} 1 s < σ 2 s < σ^{*} 2 s < σ 2 p_{z} < (π 2 p_{x} = π 2 p_{y}) < (π^{*} 2 p_{x} = π^{*} 2 p_{y}) < σ^{*} 2 p_{z}$

Modified order (for $B_{2}$ , $C_{2}$ , $N_{2}$ )

$σ 1 s < σ^{*} 1 s < σ 2 s < σ^{*} 2 s < (π 2 p_{x} = π 2 p_{y}) < σ 2 p_{z} < (π^{*} 2 p_{x} = π^{*} 2 p_{y}) < σ^{*} 2 p_{z}$

Why two orders? In lighter diatomics (

B_{2}

C_{2}

N_{2}

), the energy gap between

2 s

and

2 p

is small, so

2 s

and

2 p_{z}

mix (called

s

p

mixing). This pushes

σ 2 p_{z}

above the

π 2 p

level. In heavier diatomics (

O_{2}

onwards), the

2 s

2 p

gap is large enough that mixing is negligible, and

σ 2 p_{z}

stays below

π 2 p

Figure 2: Formation of an antibonding

σ^{*}

molecular orbital. Two

s

atomic orbitals of opposite phase (positive and negative) undergo destructive interference. Their amplitudes cancel between the nuclei, producing a nodal plane where the electron probability is zero. Electrons in the

σ^{*}

MO are pushed outward instead of being held between the nuclei, which destabilises the molecule.

Figure 3: MO energy diagram for the lighter homonuclear diatomics

B_{2}

C_{2}

, and

N_{2}

. Because the

2 s

and

2 p_{z}

atomic orbitals are close in energy, they mix (

s

p

mixing), pushing the

σ 2 p_{z}

MO above the

π 2 p

pair. Fill order (bottom to top): $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \pi 2p_x = \pi 2p_y < \sigma 2p_z < \pi^* 2p_x = \pi^* 2p_y < \sigma^* 2p_z$.

Figure 4: MO energy diagram for

O_{2}

F_{2}

N e_{2}

. The larger

2 s

2 p

energy gap in these elements suppresses

s

p

mixing, so the natural ordering $\sigma 2p_z < \pi 2p$ is restored. Fill order (bottom to top): $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < \pi 2p_x = \pi 2p_y < \pi^* 2p_x = \pi^* 2p_y < \sigma^* 2p_z$. This ordering is essential for $\mathrm{O_2}$: filling the two degenerate $\pi^* 2p$ MOs with one electron each gives the two unpaired electrons that make oxygen paramagnetic.

5. Bond Order, Bond Length, Bond Energy, and Magnetic Behaviour

Bond order is defined as half the difference between the number of electrons in bonding MOs and antibonding MOs:

$Bond order = \frac{N _{b} - N _{a}}{2}$

Interpretation of bond order:

Positive integer (1, 2, 3): single, double, triple bond respectively.
Fractional (0.5, 1.5, 2.5): stable species with delocalised or partial multiple bonds.
Zero: no net bond, molecule does not exist.
Higher bond order $\Rightarrow$ shorter bond length and larger bond energy.

Magnetic behaviour: If all electrons in MOs are paired, the species is diamagnetic (weakly repelled by a magnetic field). If one or more electrons are unpaired, it is paramagnetic (attracted to a magnetic field).

6. Application to Homonuclear Diatomic Molecules

Hydrogen ( $H_{2}$ ) - 2 electrons

Configuration: $(σ 1 s)^{2}$ . Bond order $= \frac{1}{2} (2 - 0) = 1$ . Single bond, diamagnetic. Bond dissociation energy $\approx 436 kJ mo l^{- 1}$ ; bond length 74 pm.

Hydrogen molecule ion ( $H_{2}^{+}$ ) - 1 electron

Configuration: $(σ 1 s)^{1}$ . Bond order $= \frac{1}{2} (1 - 0) = 0.5$ . Paramagnetic. This shows that a half-bond can exist, something classical Lewis theory cannot describe.

Helium molecule ( $H e_{2}$ ) - 4 electrons

Configuration: $(σ 1 s)^{2} (σ^{*} 1 s)^{2}$ . Bond order $= \frac{1}{2} (2 - 2) = 0$ . So $H e_{2}$ does not exist. MOT explains this without invoking any "noble gas rule."

Helium molecule ion ( $H e_{2}^{+}$ ) - 3 electrons

Configuration: $(σ 1 s)^{2} (σ^{*} 1 s)^{1}$ . Bond order $= \frac{1}{2} (2 - 1) = 0.5$ . Paramagnetic. Detected in discharge tubes.

Lithium ( $L i_{2}$ ) - 6 electrons

Configuration: $K K (σ 2 s)^{2}$ where $K K$ represents the filled inner shell $(σ 1 s)^{2} (σ^{*} 1 s)^{2}$ . Bond order $= \frac{1}{2} (2 - 0) = 1$ . Diamagnetic. Detected in Li vapour.

Beryllium ( $B e_{2}$ ) - 8 electrons

Configuration: $K K (σ 2 s)^{2} (σ^{*} 2 s)^{2}$ . Bond order $= \frac{1}{2} (4 - 4) = 0$ . Molecule does not exist.

Boron ( $B_{2}$ ) - 10 electrons

Uses the modified order. Configuration: $K K (σ 2 s)^{2} (σ^{*} 2 s)^{2} (π 2 p_{x})^{1} (π 2 p_{y})^{1}$ . Bond order $= \frac{1}{2} (6 - 4) = 1$ . Two unpaired electrons in degenerate $π$ MOs $\Rightarrow$ paramagnetic. This paramagnetism is the direct evidence for the modified MO order in light diatomics.

Carbon ( $C_{2}$ ) - 12 electrons

Configuration: $K K (σ 2 s)^{2} (σ^{*} 2 s)^{2} (π 2 p_{x})^{2} (π 2 p_{y})^{2}$ . Bond order $= \frac{1}{2} (8 - 4) = 2$ . Diamagnetic. Interesting note: $C_{2}$ 's double bond is made of two $π$ -bonds (no $σ$ from $2 p_{z}$ ), unlike most double bonds which have one $σ$ and one $π$ .

Nitrogen ( $N_{2}$ ) - 14 electrons

Configuration: $K K (σ 2 s)^{2} (σ^{*} 2 s)^{2} (π 2 p_{x})^{2} (π 2 p_{y})^{2} (σ 2 p_{z})^{2}$ . Bond order $= \frac{1}{2} (10 - 4) = 3$ . Diamagnetic. Very stable - the triple bond is the highest bond order among neutral diatomics and gives $N_{2}$ its exceptional bond dissociation energy of $\sim 946 kJ mo l^{- 1}$ .

Oxygen ( $O_{2}$ ) - 16 electrons

Uses the standard order. Configuration: $K K (σ 2 s)^{2} (σ^{*} 2 s)^{2} (σ 2 p_{z})^{2} (π 2 p_{x})^{2} (π 2 p_{y})^{2} (π^{*} 2 p_{x})^{1} (π^{*} 2 p_{y})^{1}$ . Bond order $= \frac{1}{2} (10 - 6) = 2$ . Two unpaired electrons $\Rightarrow$ paramagnetic. This paramagnetism of $O_{2}$ is MOT's most celebrated success - it explains why liquid oxygen sticks to a magnet, a fact VBT cannot easily account for.

Fluorine ( $F_{2}$ ) - 18 electrons

Configuration: $K K (σ 2 s)^{2} (σ^{*} 2 s)^{2} (σ 2 p_{z})^{2} (π 2 p_{x})^{2} (π 2 p_{y})^{2} (π^{*} 2 p_{x})^{2} (π^{*} 2 p_{y})^{2}$ . Bond order $= \frac{1}{2} (10 - 8) = 1$ . Diamagnetic. Weak single bond.

Neon ( $N e_{2}$ ) - 20 electrons

All bonding and antibonding MOs up to $σ^{*} 2 p_{z}$ are filled. Bond order $= \frac{1}{2} (10 - 10) = 0$ . Molecule does not exist.

Species	Total electrons	Bond order	Magnetic
$H_{2}^{+}$	1	0.5	Paramagnetic
$H_{2}$	2	1	Diamagnetic
$H e_{2}^{+}$	3	0.5	Paramagnetic
$H e_{2}$	4	0	Does not exist
$L i_{2}$	6	1	Diamagnetic
$B e_{2}$	8	0	Does not exist
$B_{2}$	10	1	Paramagnetic
$C_{2}$	12	2	Diamagnetic
$N_{2}$	14	3	Diamagnetic
$O_{2}$	16	2	Paramagnetic
$F_{2}$	18	1	Diamagnetic
$N e_{2}$	20	0	Does not exist

7. Molecular Ions of Oxygen and Nitrogen

Oxygen family: $O_{2}$ , $O_{2}^{+}$ , $O_{2}^{-}$ , $O_{2}^{2 -}$

Starting from $O_{2}$ configuration and adding/removing electrons from the highest-occupied ( $π^{*} 2 p$ ) MOs:

Species	Electrons	$π^{*} 2 p$ occupancy	Bond order	Bond length	Magnetic
$O_{2}^{+}$ (dioxygenyl)	15	$(π^{*})^{1}$	2.5	112 pm (shortest)	Paramagnetic
$O_{2}$	16	$(π^{*})^{2}$	2.0	121 pm	Paramagnetic
$O_{2}^{-}$ (superoxide)	17	$(π^{*})^{3}$	1.5	128 pm	Paramagnetic
$O_{2}^{2 -}$ (peroxide)	18	$(π^{*})^{4}$	1.0	149 pm (longest)	Diamagnetic

Bond energy order: $O_{2}^{+} > O_{2} > O_{2}^{-} > O_{2}^{2 -}$ . As electrons are added to the antibonding $π^{*}$ MO, the bond becomes weaker and longer.

Solved Example 1

Compare the bond energies of

O_{2}

O_{2}^{+}

, and

O_{2}^{-}

Solution:

Bond order determines bond strength. $O_{2}^{+}$ has bond order 2.5 (removing one electron from the antibonding $π^{*}$ MO strengthens the bond), $O_{2}$ has 2.0, and $O_{2}^{-}$ has 1.5 (adding one electron to antibonding weakens it). So bond energy: $O_{2}^{+} > O_{2} > O_{2}^{-}$ .

Nitrogen family: $N_{2}$ , $N_{2}^{+}$ , $N_{2}^{-}$

Species	Electrons	Bond order	Magnetic
$N_{2}^{+}$	13	2.5	Paramagnetic
$N_{2}$	14	3.0	Diamagnetic
$N_{2}^{-}$	15	2.5	Paramagnetic

Solved Example 2

Compare bond energies of

N_{2}

N_{2}^{+}

, and

N_{2}^{-}

Solution:

$N_{2}$ has bond order 3 (both $N_{2}^{+}$ and $N_{2}^{-}$ have 2.5). So $N_{2}$ has the largest bond energy. Between the two ions, $N_{2}^{+}$ (electron removed from bonding $σ 2 p_{z}$ ) and $N_{2}^{-}$ (electron added to antibonding $π^{*} 2 p$ ) both give the same bond order. However, $N_{2}^{+}$ has one fewer electron overall, reducing inter-electronic repulsion and giving a slightly stronger bond than $N_{2}^{-}$ . So bond energy: $N_{2} > N_{2}^{+} > N_{2}^{-}$ .

8. Heteronuclear Diatomic Molecules

For heteronuclear diatomics (CO, NO, HF, etc.), the atomic orbitals of the two atoms differ in energy. The MO diagram becomes skewed - bonding MOs have more contribution from the more electronegative atom, and antibonding MOs from the less electronegative atom. Detailed MO diagrams for CO and NO are complex due to hybridisation of oxygen and carbon orbitals, but bond order and magnetic behaviour can still be worked out by counting total electrons and treating the sequence similar to isoelectronic $N_{2}$ (for CO and $C N^{-}$ ) or $O_{2}$ -related (for NO).

Species	Total electrons	Isoelectronic with	Bond order	Magnetic
$CO$	14	$N_{2}$	3	Diamagnetic
$C N^{-}$	14	$N_{2}$	3	Diamagnetic
$N O^{+}$	14	$N_{2}$	3	Diamagnetic
$NO$	15	-	2.5	Paramagnetic
$N O^{-}$	16	$O_{2}$	2	Paramagnetic (2 unpaired in $π^{*}$ )
$CN$	13	-	2.5	Paramagnetic

Solved Example 3

Which of

C N^{-}

O H^{-}

N O^{-}

C O^{+}

has the same bond order as

N_{2}

Solution:

$N_{2}$ has 14 electrons, 10 bonding, 4 antibonding, so bond order 3. $C N^{-}$ has 14 electrons (isoelectronic with $N_{2}$ ) and identical bond order 3. The others have different electron counts and different bond orders.

Solved Example 4

Explain why

N O^{+}

is more stable than NO but

C O^{+}

is less stable than CO.

Solution:

In $NO$ (15 electrons), the highest-occupied MO is antibonding ( $π^{*} 2 p$ ). Removing this electron to give $N O^{+}$ (14 electrons) reduces antibonding population and increases bond order from 2.5 to 3. Hence $N O^{+}$ is more stable.

In $CO$ (14 electrons), all lower MOs are filled and the highest-occupied MO is bonding ( $σ$ ). Removing an electron to give $C O^{+}$ (13 electrons) empties a bonding MO, decreasing bond order from 3 to 2.5. Hence $C O^{+}$ is less stable than CO.

9. Why MOT is Powerful

MOT succeeds where VBT struggles:

Paramagnetism of $O_{2}$ : MOT correctly predicts two unpaired electrons in $π^{*}$ MOs. VBT predicts a diamagnetic double bond.
Existence of odd-electron molecules: $H_{2}^{+}$ (bond order 0.5), $H e_{2}^{+}$ (0.5), NO (2.5) - all explained by MO occupancy, unproblematic for MOT.
Non-existence of certain molecules: $H e_{2}$ , $B e_{2}$ , $N e_{2}$ all have zero bond order, so cannot form. VBT gives no clear reason.
Fractional bond orders: $O_{2}^{-}$ (1.5), $O_{2}^{+}$ (2.5), superoxide, peroxide - all naturally described.
Bond length and bond energy trends: Directly follow from bond order via MO occupancy.

Common Mistakes to Avoid

Watch out

Using the wrong MO order. For $B_{2}$ , $C_{2}$ , $N_{2}$ , use the modified order (with $π 2 p$ below $σ 2 p_{z}$ ). For $O_{2}$ , $F_{2}$ , $N e_{2}$ , use the standard order. This is the biggest MOT mistake.
Forgetting inner shell KK. Always account for the two filled inner-shell MOs $(σ 1 s)^{2} (σ^{*} 1 s)^{2}$ when writing configurations for period-2 molecules. They contribute equally to bonding and antibonding and don't affect bond order but should still be shown.
Confusing sigma with pi occupation. $σ 2 p_{z}$ holds 2 electrons max; the two degenerate $π 2 p$ MOs together hold 4 max (with Hund's rule filling one each before pairing).
Assuming higher bond order always means paramagnetic or diamagnetic. Magnetic behaviour depends only on unpaired electrons, not on bond order. $N_{2}$ (BO 3) is diamagnetic but $B_{2}$ (BO 1) is paramagnetic.
Confusing MOT with VBT. In VBT electrons stay localised in atomic orbitals of two bonded atoms. In MOT they belong to the whole molecule. Both can coexist in different contexts.
Miscounting electrons in ions. Add electrons for negative charge, subtract for positive. $O_{2}^{+}$ has 15, $O_{2}^{-}$ has 17, $O_{2}^{2 -}$ has 18.

Frequently Asked Questions

Q1. What is a molecular orbital and how does it differ from an atomic orbital?

A molecular orbital is a region of space around two or more nuclei where an electron is likely to be found. It is polycentric, meaning it extends over multiple atoms. An atomic orbital is centred on just one nucleus (monocentric). Molecular orbitals are formed by combining atomic orbitals via LCAO and are filled following the same rules (Aufbau, Pauli, Hund) as atomic orbitals.

Q2. What is the difference between a bonding and an antibonding molecular orbital?

A bonding MO forms by constructive interference of the combining atomic orbital wave functions ( $ψ_{A} + ψ_{B}$ ). It has lower energy than the parent atomic orbitals and greater electron density between the nuclei, stabilising the molecule. An antibonding MO forms by destructive interference ( $ψ_{A} - ψ_{B}$ ) and has higher energy, with a node between the nuclei where electron density is zero.

Q3. Why does $H e_{2}$ not exist?

$H e_{2}$ would have 4 electrons: configuration $(σ 1 s)^{2} (σ^{*} 1 s)^{2}$ . Bond order $= (2 - 2) /2 = 0$ , meaning no net bond holds the two atoms together. However, $H e_{2}^{+}$ (3 electrons) has bond order 0.5 and does exist transiently in discharge tubes.

Q4. Why is $O_{2}$ paramagnetic?

$O_{2}$ has 16 electrons. The two highest-occupied MOs are the degenerate $π^{*} 2 p_{x}$ and $π^{*} 2 p_{y}$ with one electron each, unpaired, following Hund's rule. Two unpaired electrons make $O_{2}$ paramagnetic. This is one of MOT's most famous predictions; VBT could not easily explain it.

Q5. What is the bond order in superoxide $O_{2}^{-}$ ?

$O_{2}^{-}$ has 17 electrons, one more than $O_{2}$ . This extra electron goes into a $π^{*} 2 p$ MO, giving occupancy $(π^{*})^{3}$ . Bond order $= (10 - 7) /2 = 1.5$ . Superoxide is paramagnetic (one unpaired electron in $π^{*}$ ) with a longer, weaker O-O bond than $O_{2}$ .

Q6. Why do $B_{2}$ , $C_{2}$ , and $N_{2}$ use a different MO order than $O_{2}$ , $F_{2}$ , and $N e_{2}$ ?

In lighter diatomics, the energy gap between the $2 s$ and $2 p$ atomic orbitals is small enough that $s$ - $p$ mixing occurs. This mixing raises the energy of $σ 2 p_{z}$ above the $π 2 p$ MOs. In heavier diatomics ( $O_{2}$ onwards), the $2 s$ - $2 p$ gap is large and mixing is negligible, so $σ 2 p_{z}$ stays below $π 2 p$ .

Q7. Compare stability and bond order of $N_{2}$ , $N_{2}^{+}$ , and $N_{2}^{-}$ .

$N_{2}$ : 14 electrons, bond order 3, diamagnetic - most stable. $N_{2}^{+}$ : 13 electrons (removed from bonding $σ 2 p_{z}$ ), bond order 2.5, paramagnetic. $N_{2}^{-}$ : 15 electrons (added to antibonding $π^{*} 2 p$ ), bond order 2.5, paramagnetic. Between the ions, $N_{2}^{+}$ is slightly more stable than $N_{2}^{-}$ due to lower inter-electron repulsion.

Q8. Which species is isoelectronic with $N_{2}$ and has the same bond order?

Any species with 14 electrons and comparable orbital arrangement: $CO$ , $C N^{-}$ , and $N O^{+}$ are all isoelectronic with $N_{2}$ and have bond order 3.

Q9. Why is $N O^{+}$ more stable than NO?

NO has 15 electrons, with the last one in an antibonding $π^{*} 2 p$ MO. Removing that electron to form $N O^{+}$ (14 electrons) empties the antibonding MO and increases bond order from 2.5 to 3, giving a stronger, shorter bond. So $N O^{+}$ is more stable than NO.

Q10. How does MOT explain fractional bond orders?

Fractional bond orders arise naturally when an odd number of electrons is distributed between bonding and antibonding MOs. For example, in $H_{2}^{+}$ one electron sits in $σ 1 s$ giving bond order 0.5. In $O_{2}^{-}$ the antibonding $π^{*}$ MOs hold three electrons, giving bond order 1.5. Classical Lewis theory cannot describe these, but MOT handles them straightforwardly.

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1. The MOT Approach

Salient features of MOT

2. Formation of Molecular Orbitals - LCAO Method

Conditions for combining atomic orbitals

3. Types of Molecular Orbitals

Sigma (σ) MOs

Pi (π) MOs

4. Energy Level Diagrams

Standard order (for O2​, F2​, Ne2​)

Modified order (for B2​, C2​, N2​)

5. Bond Order, Bond Length, Bond Energy, and Magnetic Behaviour

6. Application to Homonuclear Diatomic Molecules

Hydrogen (H2​) - 2 electrons

Hydrogen molecule ion (H2+​) - 1 electron

Helium molecule (He2​) - 4 electrons

Helium molecule ion (He2+​) - 3 electrons

Lithium (Li2​) - 6 electrons

Beryllium (Be2​) - 8 electrons

Boron (B2​) - 10 electrons

Carbon (C2​) - 12 electrons

Nitrogen (N2​) - 14 electrons

Oxygen (O2​) - 16 electrons

Fluorine (F2​) - 18 electrons

Neon (Ne2​) - 20 electrons

7. Molecular Ions of Oxygen and Nitrogen

Oxygen family: O2​, O2+​, O2−​, O22−​

Nitrogen family: N2​, N2+​, N2−​

8. Heteronuclear Diatomic Molecules

9. Why MOT is Powerful

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is a molecular orbital and how does it differ from an atomic orbital?

Q2. What is the difference between a bonding and an antibonding molecular orbital?

Q3. Why does He2​ not exist?

Q4. Why is O2​ paramagnetic?

Q5. What is the bond order in superoxide O2−​?

Q6. Why do B2​, C2​, and N2​ use a different MO order than O2​, F2​, and Ne2​?

Q7. Compare stability and bond order of N2​, N2+​, and N2−​.

Q8. Which species is isoelectronic with N2​ and has the same bond order?

Q9. Why is NO+ more stable than NO?

Q10. How does MOT explain fractional bond orders?

Previous year questions on Molecular Orbital Theory

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Sigma ( $σ$ ) MOs

Pi ( $π$ ) MOs

Standard order (for $O_{2}$ , $F_{2}$ , $N e_{2}$ )

Modified order (for $B_{2}$ , $C_{2}$ , $N_{2}$ )

Hydrogen ( $H_{2}$ ) - 2 electrons

Hydrogen molecule ion ( $H_{2}^{+}$ ) - 1 electron

Helium molecule ( $H e_{2}$ ) - 4 electrons

Helium molecule ion ( $H e_{2}^{+}$ ) - 3 electrons

Lithium ( $L i_{2}$ ) - 6 electrons

Beryllium ( $B e_{2}$ ) - 8 electrons

Boron ( $B_{2}$ ) - 10 electrons

Carbon ( $C_{2}$ ) - 12 electrons

Nitrogen ( $N_{2}$ ) - 14 electrons

Oxygen ( $O_{2}$ ) - 16 electrons

Fluorine ( $F_{2}$ ) - 18 electrons

Neon ( $N e_{2}$ ) - 20 electrons

Oxygen family: $O_{2}$ , $O_{2}^{+}$ , $O_{2}^{-}$ , $O_{2}^{2 -}$

Nitrogen family: $N_{2}$ , $N_{2}^{+}$ , $N_{2}^{-}$

Q3. Why does $H e_{2}$ not exist?

Q4. Why is $O_{2}$ paramagnetic?

Q5. What is the bond order in superoxide $O_{2}^{-}$ ?

Q6. Why do $B_{2}$ , $C_{2}$ , and $N_{2}$ use a different MO order than $O_{2}$ , $F_{2}$ , and $N e_{2}$ ?

Q7. Compare stability and bond order of $N_{2}$ , $N_{2}^{+}$ , and $N_{2}^{-}$ .

Q8. Which species is isoelectronic with $N_{2}$ and has the same bond order?

Q9. Why is $N O^{+}$ more stable than NO?