The MO Toolkit for Diatomics
In molecular orbital theory, atomic orbitals on two combining atoms merge by linear combination to form an equal number of molecular orbitals. Each pair of combining atomic orbitals yields one bonding molecular orbital of lower energy and one antibonding molecular orbital of higher energy. The bonding orbital concentrates electron density between the nuclei; the antibonding orbital carries a node there.
For a homonuclear diatomic, the two atoms are identical, so their orbitals contribute equally. The $1s$ orbitals give $\ce{\sigma}1s$ and $\ce{\sigma}^*1s$; the $2s$ orbitals give $\ce{\sigma}2s$ and $\ce{\sigma}^*2s$; and the three $2p$ orbitals give one sigma pair ($\ce{\sigma}2p_z$, $\ce{\sigma}^*2p_z$) along the bond axis plus two degenerate pi pairs ($\pi 2p_x = \pi 2p_y$ and $\pi^*2p_x = \pi^*2p_y$) perpendicular to it.
Three quantities are extracted from the filled configuration. Stability is judged by whether bonding electrons ($N_b$) outnumber antibonding electrons ($N_a$). The strength of the bond is captured by the bond order, and the presence of any unpaired electron decides the magnetic character.
| Quantity | Rule (NCERT 4.7.5) | Consequence |
|---|---|---|
| Stability | $N_b > N_a$ → stable; $N_b < N_a$ → unstable | Net bonding influence must dominate |
| Bond order | $\text{b.o.} = \tfrac{1}{2}(N_b - N_a)$ | Positive value → molecule exists |
| Bond nature | b.o. = 1, 2, 3 → single, double, triple | Matches classical bond count |
| Bond length | Increases as bond order decreases | Higher b.o. → shorter, stronger bond |
| Magnetism | All MOs doubly filled → diamagnetic; any singly filled → paramagnetic | $\ce{O2}$ is the classic paramagnetic case |
Two Energy Orderings: Before and After N2
A subtlety that NEET examiners exploit repeatedly is that the order of molecular orbital energies is not the same across the entire second period. The crossover occurs at oxygen. For the lighter molecules — $\ce{B2}$, $\ce{C2}$, $\ce{N2}$ and their neighbours — extensive mixing between the $2s$ and $2p_z$ orbitals pushes the $\ce{\sigma}2p_z$ orbital above the two $\pi 2p$ orbitals.
For $\ce{B2}$, $\ce{C2}$, $\ce{N2}$ (and lighter):
$$\ce{\sigma}1s < \ce{\sigma}^*1s < \ce{\sigma}2s < \ce{\sigma}^*2s < (\pi 2p_x = \pi 2p_y) < \ce{\sigma}2p_z < (\pi^*2p_x = \pi^*2p_y) < \ce{\sigma}^*2p_z$$
For $\ce{O2}$, $\ce{F2}$ (and heavier):
$$\ce{\sigma}1s < \ce{\sigma}^*1s < \ce{\sigma}2s < \ce{\sigma}^*2s < \ce{\sigma}2p_z < (\pi 2p_x = \pi 2p_y) < (\pi^*2p_x = \pi^*2p_y) < \ce{\sigma}^*2p_z$$
The single difference is the position of $\ce{\sigma}2p_z$ relative to the $\pi 2p$ pair. For $\ce{O2}$ and $\ce{F2}$ the $\ce{\sigma}2p_z$ drops below the pi orbitals. Because $\ce{O2}$ and $\ce{F2}$ each have enough electrons to fill these orbitals fully or nearly so, the two orderings happen to give identical configurations for those two molecules — but for $\ce{B2}$, $\ce{C2}$ and $\ce{N2}$ the lighter-molecule ordering is essential to predicting the right number of unpaired electrons.
Period-1 Molecules: H2 and He2
The first period offers the cleanest illustration of how bond order decides existence. Each hydrogen atom brings one $1s$ electron, so $\ce{H2}$ has two electrons, both in the bonding $\ce{\sigma}1s$ orbital.
$\ce{H2}$ : $(\ce{\sigma}1s)^2$
$$\text{b.o.} = \tfrac{1}{2}(N_b - N_a) = \tfrac{1}{2}(2 - 0) = 1$$
A single covalent bond, bond length 74 pm, bond dissociation energy 438 kJ mol⁻¹. No unpaired electrons, so $\ce{H2}$ is diamagnetic.
Helium tells the opposite story. Each helium atom has the configuration $1s^2$, so $\ce{He2}$ would have four electrons, filling both $\ce{\sigma}1s$ and $\ce{\sigma}^*1s$. The antibonding pair exactly cancels the bonding pair, and the bond order falls to zero.
$\ce{He2}$ : $(\ce{\sigma}1s)^2 (\ce{\sigma}^*1s)^2$
$$\text{b.o.} = \tfrac{1}{2}(2 - 2) = 0$$
Zero bond order means no net bond; $\ce{He2}$ is unstable and does not exist. By the same argument $\ce{Be2}$, with $(\ce{\sigma}1s)^2 (\ce{\sigma}^*1s)^2 (\ce{\sigma}2s)^2 (\ce{\sigma}^*2s)^2$, also has bond order zero and does not exist.
Zero bond order means non-existence, not weak existence
NEET 2025 paired the false claim "a hypothetical molecule with bond order zero is quite stable" with the false claim "bond length increases as bond order increases." Both are wrong. A zero or negative bond order means the molecule is unstable and does not form, and bond length decreases as bond order rises.
Bond order $= 0$ → molecule does not exist (e.g. $\ce{He2}$, $\ce{Be2}$). Higher bond order → shorter bond, higher bond enthalpy.
Light Period-2 Molecules: Li2, B2, C2, N2
Beyond beryllium, the $2p$ orbitals begin to fill and the molecules become stable. Lithium contributes a single $2s$ electron each, giving $\ce{Li2}$ six electrons. Its closed K-shell $(\ce{\sigma}1s)^2 (\ce{\sigma}^*1s)^2$ is abbreviated KK, leaving the two $\ce{\sigma}2s$ electrons to provide the bond.
Carbon, with $1s^2 2s^2 2p^2$, gives $\ce{C2}$ twelve electrons. Under the lighter-molecule ordering, the four valence $2p$ electrons fill the degenerate $\pi 2p_x$ and $\pi 2p_y$ orbitals completely before the $\ce{\sigma}2p_z$ orbital is touched. This produces an unusual double bond made entirely of two pi bonds.
$\ce{Li2}$ : $\text{KK}\,(\ce{\sigma}2s)^2 \Rightarrow \text{b.o.} = \tfrac{1}{2}(4-2) = 1$, diamagnetic. Known to exist in the vapour phase.
$\ce{C2}$ : $\text{KK}\,(\ce{\sigma}2s)^2 (\ce{\sigma}^*2s)^2 (\pi 2p_x^2 = \pi 2p_y^2) \Rightarrow \text{b.o.} = \tfrac{1}{2}(8-4) = 2$, diamagnetic. The double bond is two pi bonds.
$\ce{N2}$ : $\text{KK}\,(\ce{\sigma}2s)^2 (\ce{\sigma}^*2s)^2 (\pi 2p_x^2 = \pi 2p_y^2)(\ce{\sigma}2p_z)^2 \Rightarrow \text{b.o.} = \tfrac{1}{2}(10-4) = 3$, diamagnetic.
The nitrogen result is the headline case. With ten bonding and four antibonding electrons, $\ce{N2}$ achieves a bond order of three — a triple bond. NCERT records its bond enthalpy as 946 kJ mol⁻¹, one of the highest known for any diatomic, which is why molecular nitrogen is so chemically inert. All its molecular orbitals are doubly occupied, so $\ce{N2}$ is diamagnetic.
New to bonding and antibonding orbitals? Start with Molecular Orbital Theory before working through these configurations.
Oxygen and Fluorine: O2 and F2
Oxygen is where molecular orbital theory earns its reputation. Each oxygen atom is $1s^2 2s^2 2p^4$, so $\ce{O2}$ has sixteen electrons. After all bonding levels and the antibonding $\ce{\sigma}^*2s$ are filled, the last two electrons must occupy the two degenerate antibonding pi orbitals. By Hund's rule they enter $\pi^*2p_x$ and $\pi^*2p_y$ singly, with parallel spins.
$\ce{O2}$ : $\text{KK}\,(\ce{\sigma}2s)^2 (\ce{\sigma}^*2s)^2 (\ce{\sigma}2p_z)^2 (\pi 2p_x^2 = \pi 2p_y^2)(\pi^*2p_x^1 = \pi^*2p_y^1)$
$$\text{b.o.} = \tfrac{1}{2}(10 - 6) = 2$$
A double bond — but with two singly occupied $\pi^*$ orbitals, $\ce{O2}$ has two unpaired electrons and is paramagnetic, exactly as experiment shows.
The diagram below tracks the sixteen electrons of $\ce{O2}$ up the energy ladder. Two electrons sit alone in the antibonding pi level — the structural reason oxygen is drawn to a magnet, a fact no simple Lewis structure can explain.
Fluorine continues the trend by adding two more electrons, which pair up the $\pi^*$ orbitals completely. With every molecular orbital now doubly filled, $\ce{F2}$ is diamagnetic with bond order one — a single bond, isoelectronic in bond order with the peroxide ion.
$\ce{F2}$ : $\text{KK}\,(\ce{\sigma}2s)^2 (\ce{\sigma}^*2s)^2 (\ce{\sigma}2p_z)^2 (\pi 2p_x^2 = \pi 2p_y^2)(\pi^*2p_x^2 = \pi^*2p_y^2)$
$$\text{b.o.} = \tfrac{1}{2}(10 - 8) = 1$$
All MOs doubly occupied → diamagnetic. NCERT notes $\ce{F2}$ and $\ce{O2^{2-}}$ share bond order 1.
Diatomic Ions and Bond-Order Trends
Ions are handled by adding or removing electrons from the parent neutral configuration. The decisive point is which orbital changes occupancy. For the oxygen family, the frontier orbitals are the antibonding $\pi^*2p$ pair, so every electron added there weakens the bond and every electron removed from there strengthens it.
| Species | Total e⁻ | π* occupancy | Bond order | Magnetic |
|---|---|---|---|---|
| $\ce{O2^+}$ | 15 | $\pi^*2p^1$ | $\tfrac{1}{2}(10-5)=2.5$ | Paramagnetic (1 unpaired) |
| $\ce{O2}$ | 16 | $\pi^*2p^2$ (unpaired) | $\tfrac{1}{2}(10-6)=2.0$ | Paramagnetic (2 unpaired) |
| $\ce{O2^-}$ (superoxide) | 17 | $\pi^*2p^3$ | $\tfrac{1}{2}(10-7)=1.5$ | Paramagnetic (1 unpaired) |
| $\ce{O2^{2-}}$ (peroxide) | 18 | $\pi^*2p^4$ | $\tfrac{1}{2}(10-8)=1.0$ | Diamagnetic |
| $\ce{N2^+}$ | 13 | — | $\tfrac{1}{2}(9-4)=2.5$ | Paramagnetic (1 unpaired) |
For $\ce{N2^+}$, the electron is removed from the highest filled bonding orbital ($\ce{\sigma}2p_z$ in the light-molecule ordering), dropping the bond order from 3 to 2.5 and creating one unpaired electron. This is why ionising the famously inert $\ce{N2}$ weakens its bond.
The O2 / O2⁺ / O2⁻ / O2²⁻ bond-order ladder
Examiners love asking you to rank these four. Because each added electron occupies an antibonding $\pi^*$ orbital, more electrons mean lower bond order:
Bond order: $\ce{O2^+}\,(2.5) > \ce{O2}\,(2.0) > \ce{O2^-}\,(1.5) > \ce{O2^{2-}}\,(1.0)$. Bond length runs in the exact reverse: $\ce{O2^{2-}} > \ce{O2^-} > \ce{O2} > \ce{O2^+}$.
Master Table of Homonuclear Diatomics
The complete picture for period-1 and period-2 homonuclear diatomics is collected below. Read it as a single sweep: bond order rises to a maximum at $\ce{N2}$, then falls toward $\ce{F2}$ as antibonding levels fill, while bond length moves inversely.
| Molecule | e⁻ | Valence MO configuration | Bond order | Magnetic nature | Bond length trend |
|---|---|---|---|---|---|
| $\ce{H2}$ | 2 | $(\ce{\sigma}1s)^2$ | 1 | Diamagnetic | Short (74 pm) |
| $\ce{He2}$ | 4 | $(\ce{\sigma}1s)^2(\ce{\sigma}^*1s)^2$ | 0 | — | Does not exist |
| $\ce{Li2}$ | 6 | $\text{KK}(\ce{\sigma}2s)^2$ | 1 | Diamagnetic | Long (single bond) |
| $\ce{Be2}$ | 8 | $\text{KK}(\ce{\sigma}2s)^2(\ce{\sigma}^*2s)^2$ | 0 | — | Does not exist |
| $\ce{B2}$ | 10 | $\text{KK}(\ce{\sigma}2s)^2(\ce{\sigma}^*2s)^2(\pi 2p_x^1=\pi 2p_y^1)$ | 1 | Paramagnetic (2 unpaired) | Decreasing → |
| $\ce{C2}$ | 12 | $\text{KK}(\ce{\sigma}2s)^2(\ce{\sigma}^*2s)^2(\pi 2p_x^2=\pi 2p_y^2)$ | 2 | Diamagnetic | Decreasing → |
| $\ce{N2}$ | 14 | $\text{KK}(\ce{\sigma}2s)^2(\ce{\sigma}^*2s)^2(\pi 2p_x^2=\pi 2p_y^2)(\ce{\sigma}2p_z)^2$ | 3 | Diamagnetic | Shortest (triple) |
| $\ce{O2}$ | 16 | $\text{KK}(\ce{\sigma}2s)^2(\ce{\sigma}^*2s)^2(\ce{\sigma}2p_z)^2(\pi 2p^4)(\pi^*2p_x^1=\pi^*2p_y^1)$ | 2 | Paramagnetic (2 unpaired) | Increasing → |
| $\ce{F2}$ | 18 | $\text{KK}(\ce{\sigma}2s)^2(\ce{\sigma}^*2s)^2(\ce{\sigma}2p_z)^2(\pi 2p^4)(\pi^*2p^4)$ | 1 | Diamagnetic | Long (single bond) |
Boron deserves a footnote: under the light-molecule ordering, its two valence $2p$ electrons enter the degenerate $\pi 2p$ orbitals singly, making $\ce{B2}$ paramagnetic with bond order 1 — a result that would be predicted incorrectly if the $\ce{\sigma}2p_z$ orbital were placed below the pi orbitals. This is the practical payoff of knowing both energy orderings.
Across the second period the pattern is symmetric about $\ce{N2}$. Bond order climbs $\ce{Li2}(1) \to \ce{B2}(1) \to \ce{C2}(2) \to \ce{N2}(3)$ as bonding orbitals fill, then descends $\ce{N2}(3) \to \ce{O2}(2) \to \ce{F2}(1)$ as the antibonding $\pi^*$ orbitals take electrons. Bond enthalpy tracks bond order, while bond length runs opposite.
Homonuclear diatomic bonding in one screen
- Bond order $= \tfrac{1}{2}(N_b - N_a)$; positive → molecule exists, zero/negative → it does not.
- $\ce{He2}$ and $\ce{Be2}$ have bond order 0 and do not exist; $\ce{H2}$, $\ce{Li2}$, $\ce{F2}$ have bond order 1.
- $\ce{N2}$ has bond order 3 (triple bond, 946 kJ mol⁻¹) and is diamagnetic; $\ce{C2}$ has bond order 2 from two pi bonds.
- $\ce{O2}$ is paramagnetic with two unpaired $\pi^*$ electrons — MO theory's signature success.
- Bond order ladder: $\ce{O2^+}(2.5) > \ce{O2}(2) > \ce{O2^-}(1.5) > \ce{O2^{2-}}(1)$; bond length is the reverse.
- For $\ce{B2}$, $\ce{C2}$, $\ce{N2}$ the $\ce{\sigma}2p_z$ lies above $\pi 2p$; for $\ce{O2}$, $\ce{F2}$ it lies below.