--- title: "Bonding orbitals and instability of H3" subject: "Electrical Engineering Material" chapter: 1 key: "semester_3/electrical_engineering_material:h3-molecule-not-stable" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/h3-molecule-not-stable site: BCT Engineering --- # Bonding orbitals and instability of H3 Electrical Engineering Material · Chapter 1: Theory of Metals - Asked in: 2080 Bhadra, 2+6 marks; 2072 Chaitra, 6 marks (asked 2 times) - Video: [Bonding and Antibonding Molecular Orbitals](https://www.youtube.com/watch?v=B2bYPD-1yGE) — Andrey K - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/h3-molecule-not-stable - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1 ## Question What is bonding and anti-bonding molecular orbitals? The formation of H₃ molecule is not stable, justify on the basis of molecular orbital bonding theory. ## Short answer A bonding orbital (σ) concentrates charge between nuclei and lowers energy; an antibonding orbital (σ*) has a node and raises energy. In H₃ the two lowest-energy electrons fill σ; the third must enter σ*, raising the total energy and canceling the bond gain, so H₃ is unstable compared to H₂ + H. ## Answer #### Bonding and antibonding molecular orbitals When two atomic orbitals $\psi_A$ and $\psi_B$ overlap, they combine (LCAO) into two molecular orbitals: | Point | Bonding orbital ($\sigma$) | Antibonding orbital ($\sigma^*$) | |---|---|---| | Combination | $\psi_A + \psi_B$ (in phase) | $\psi_A - \psi_B$ (out of phase) | | Charge between nuclei | Large (waves add) | Zero at mid-point (node) | | Energy | Lower than atomic level | Higher than atomic level | | Effect | Holds atoms together | Pushes atoms apart | Each molecular orbital can hold at most two electrons, with opposite spins (Pauli exclusion principle). #### Why H₃ is not stable **H₂:** Each H atom has one 1s electron. The two 1s orbitals form $\sigma$ and $\sigma^*$. Both electrons go into the bonding orbital $\sigma$ with opposite spins, and $\sigma^*$ stays empty. The energy falls by about 4.5 eV, so H₂ is stable. **Adding a third H atom:** The third electron cannot enter $\sigma$, because $\sigma$ is already full with two electrons of opposite spin (Pauli principle). It must go into the higher-energy **antibonding** orbital $\sigma^*$. ```text Energy | sigma* _^_ <- 3rd electron forced here | / \ | 1s __ --- --- __ 1s | \ / | sigma _^v_ <- full (2 electrons) | | H2 + H: bonding gain is cancelled by the | energy cost of the sigma* electron ``` - The electron in $\sigma^*$ puts its charge **outside** the region between the nuclei, so it causes repulsion and raises the energy. - The energy gained from the bonding pair is largely cancelled by the energy cost of the antibonding electron, so the total energy of H₃ is **not lower** than that of H₂ + H. - A system always goes to the state of lowest energy, so H₃ breaks up into a stable H₂ molecule and a free H atom. (In a full three-orbital treatment of linear H₃, the third electron goes into a non-bonding/antibonding level above the bonding one, and the result is the same: H₃ has no energy minimum and is not a stable molecule.) #### Same idea for He₂ He has two 1s electrons, so He₂ would have 2 electrons in $\sigma$ and 2 in $\sigma^*$. Bond order $= \frac{1}{2}(2 - 2) = 0$, so He₂ does not form. In general, a molecule is stable only if bonding electrons outnumber antibonding electrons enough to lower the total energy. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which describes the antibonding orbital? A. In-phase combination with a high charge between the nuclei B. Combination psi_A - psi_B with a node at the midpoint and higher energy C. Combination with lower energy than the atomic level D. An orbital that holds 4 electrons Answer: B. The antibonding orbital is out of phase, has zero density at the midpoint and pushes atoms apart. (Key point: Bonding vs antibonding) 2. Where must the third electron of H3 go? A. The sigma orbital, as a third electron B. The sigma orbital with spin up C. The antibonding sigma* orbital, because sigma is full (Pauli principle) D. It escapes completely Answer: C. sigma already holds two electrons of opposite spin, so the third goes to higher-energy sigma*. (Key point: Third electron) 3. Why is H3 not stable? A. The antibonding electron cancels the bonding gain, so energy is not lowered below H2 + H B. H3 has bond order 3 C. Three protons attract too strongly D. Its energy is the lowest possible Answer: A. The energy cost of the sigma* electron largely cancels the bonding gain, so H3 breaks up into H2 and H. (Key point: Energy outcome) 4. Why does He2 not form? A. Bond order is 2 B. Bond order = (2 - 2)/2 = 0 as it has 2 bonding and 2 antibonding electrons C. He has no 1s orbital D. Helium has only one electron Answer: B. Equal bonding and antibonding electrons give zero net bonding. (Key point: He2 bond order) ## Related in Electrical Engineering Material - [H2 molecule by molecular orbital theory](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Ah2-molecule-mo-theory) — Electrical Engineering Material, chapter 1, asked 3× - [Two helium atoms: bonding and antibonding orbitals](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Ahelium-atoms-bonding-orbitals) — Electrical Engineering Material, chapter 1 - [LCAO and stability of H2 molecule](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Alcao-h2-molecule-stable) — Electrical Engineering Material, chapter 1