Chemistry · Chemical Bonding · NEET
Two things separate them: energy and the node. The bonding MO has energy LOWER than the two original atomic orbitals, so it is more stable. The antibonding MO (written with a star, like sigma* or pi*) has energy HIGHER than the atomic orbitals, so it is less stable. Also, the bonding MO has NO node between the two nuclei (electron density is high in the middle, gluing the atoms), while the antibonding MO HAS a node between the nuclei (almost no electron density there, so atoms are pushed apart). For NEET, remember: bonding pulls atoms together, antibonding pushes them apart.
Electron waves can add or subtract. In the bonding MO the two waves reinforce each other (constructive interference), so electron density builds up between the nuclei. This negative charge sits between the two positive nuclei and holds them together, which lowers the energy. In the antibonding MO the waves cancel (destructive interference), leaving a node (zero density) between the nuclei. Now the nuclei repel each other more, so the energy is higher. NCERT states this directly: the bonding MO has lower energy and greater stability than the antibonding MO.
The bonding MO has NO node between the two nuclei — electron density is largest right in the middle. The antibonding MO HAS one node exactly between the two nuclei, a flat plane where the chance of finding the electron is zero. This is why a NEET 2023 question said 'the pi* antibonding molecular orbital has a node between the nuclei' — that statement is correct. Tip: node between nuclei = antibonding; no node between nuclei = bonding.
The star (*) simply marks an ANTIBONDING orbital. So sigma is a bonding sigma orbital, and sigma* (say 'sigma star') is the antibonding one. Same for pi and pi*. When you write a molecular orbital configuration, every starred orbital is antibonding and every unstarred one is bonding. This matters for bond order: Bond order = (bonding electrons − antibonding electrons) / 2. If you miss a star, you get the bond order wrong.
They are NOT always empty. Antibonding orbitals are real orbitals; electrons fill them once the lower bonding orbitals are full, following the same rules (lowest energy first, Pauli, Hund). For example, O2 has electrons in its pi* antibonding orbitals, and that is exactly why O2 is paramagnetic. In Ne2, the highest occupied MO (HOMO) is the antibonding sigma*2p (NEET 2026). Electrons in antibonding orbitals cancel the effect of bonding electrons, which lowers the bond order.
Count electrons and compute bond order = (Nb − Na)/2, where Nb = bonding electrons and Na = antibonding electrons. If bonding electrons win, bond order is positive and the molecule can exist (like H2, bond order 1). If antibonding electrons fully cancel the bonding ones, bond order is 0 and the molecule does NOT form. That is why He2 and Be2 do not exist — their bonding and antibonding electrons exactly cancel.
Which one of the following statements is incorrect related to Molecular Orbital Theory?
The highest occupied molecular orbital (HOMO) for Ne2 is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Lower. The bonding molecular orbital is lower in energy than the atomic orbitals that made it, so it is more stable. The antibonding orbital is higher in energy and less stable.
One node between the two nuclei — a plane where the electron probability is zero. The bonding sigma orbital has no such node between the nuclei.
They decrease it. Bond order = (bonding electrons − antibonding electrons)/2, so each antibonding electron cancels a bonding electron and lowers the bond order.
Constructive interference makes the bonding orbital (waves add, density builds between nuclei). Destructive interference makes the antibonding orbital (waves cancel, a node forms).
MO theory is a repeat NEET topic. Knowing which orbital is bonding vs antibonding lets you write MO configurations, find bond order, and predict if a molecule is paramagnetic or even exists — questions asked almost every year.
