--- title: "H2 molecule by molecular orbital theory" subject: "Electrical Engineering Material" chapter: 1 key: "semester_3/electrical_engineering_material:h2-molecule-mo-theory" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/h2-molecule-mo-theory site: BCT Engineering --- # H2 molecule by molecular orbital theory Electrical Engineering Material · Chapter 1: Theory of Metals - Asked in: 2082 Kartik (new course), 6 marks; 2081 Bhadra, 6 marks; 2069 Chaitra, 6 marks (asked 3 times) - Video: [Molecular Orbital Theory Diagram | Energy Level Diagram](https://www.youtube.com/watch?v=CAqvjOG3pkM) — Najam Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/h2-molecule-mo-theory - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1 ## Question Explain the formation of H₂ molecule using molecular orbital bonding theory and draw the necessary diagrams. ## Short answer Two 1s atomic orbitals combine into a bonding (σ) and an antibonding (σ*) molecular orbital. Both electrons fill the low-energy bonding orbital with opposite spins, lowering the energy by about 4.5 eV at a separation of 0.074 nm, giving a stable H₂ molecule. ## Answer According to **molecular orbital (MO) theory**, when two atoms come close, their atomic orbitals combine (Linear Combination of Atomic Orbitals, LCAO) to form molecular orbitals that belong to the whole molecule. The H₂ molecule forms because its two electrons can occupy a bonding orbital of lower energy than the separate 1s orbitals. #### Molecular orbitals of H₂ Let $\psi_{1s}(r_A)$ and $\psi_{1s}(r_B)$ be the 1s orbitals of atoms A and B. They combine in two ways: $$ \psi_\sigma = \psi_{1s}(r_A) + \psi_{1s}(r_B) \qquad \text{(bonding)} $$ $$ \psi_{\sigma^*} = \psi_{1s}(r_A) - \psi_{1s}(r_B) \qquad \text{(antibonding)} $$ ```text bonding psi_s antibonding psi_s* .--. .--. .--. / \_/ \ / \ / ^ \ ------/------\------- 0 A (overlap) B A \ /B '--' |psi_s|^2 high between A,B |psi_s*|^2 = 0 at mid-point ``` - In $\psi_\sigma$ the waves add between the nuclei, so electron density builds up there. This negative charge attracts both protons and holds them together. Energy is **lowered**. - In $\psi_{\sigma^*}$ the waves cancel at the mid-point (a node), there is little charge between the nuclei, and the protons repel. Energy is **raised**. #### Energy level diagram ```text Energy | sigma* (empty) | / ____ \ | 1s __ --- --- __ 1s | (H_A) ^ \ / ^ (H_B) | \ _^v_ / | sigma (2 electrons, | opposite spins) ``` The two electrons (one from each H atom) go into $\sigma$ with opposite spins (Pauli principle). The antibonding orbital stays empty. #### Energy versus separation ```text E ^ \ | \ sigma* (repulsive, no minimum) | \___ | ''----.....______________ 0+-------------------------------------> R | \ ____...---- | \ _---' sigma (bonding) | \ _-' | '--' <- minimum: R0 = 0.074 nm | binding energy = 4.5 eV ``` The bonding curve has a minimum at $R_0 = 0.074$ nm. At this separation the H₂ molecule is about **4.5 eV** lower in energy than two separate H atoms, so H₂ is stable. Bond order $= \frac{1}{2}(2 - 0) = 1$, a single covalent bond. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are the bonding and antibonding orbitals of H2 formed from 1s orbitals? A. Bonding: psi_A - psi_B; antibonding: psi_A + psi_B B. Both are psi_A + psi_B C. Bonding: psi_A + psi_B; antibonding: psi_A - psi_B D. Both are psi_A - psi_B Answer: C. In-phase addition gives bonding; out-of-phase subtraction gives antibonding. (Key point: LCAO) 2. Why does the bonding orbital lower the energy? A. Electron density builds up between the nuclei, attracting both protons B. There is a node at the midpoint C. The electrons move away from the nuclei D. The protons are removed Answer: A. Waves add between the nuclei, and this negative charge attracts both protons together. (Key point: Why bonding binds) 3. How are H2's two electrons arranged? A. Both in sigma* B. One in sigma, one in sigma* C. Both in sigma with parallel spin D. Both in sigma with opposite spins; sigma* empty Answer: D. By the Pauli principle two electrons of opposite spin occupy the lower sigma orbital. (Key point: Electron filling) 4. What is the bond order and why is H2 stable? A. Bond order 1; molecule is about 4.5 eV lower in energy than two H atoms at R0 = 0.074 nm B. Bond order 0; energy is the same as separate atoms C. Bond order 2; energy is higher D. Bond order 1; the bonding curve has no minimum Answer: A. Bond order = (2 - 0)/2 = 1, and the bonding curve has a minimum at 0.074 nm with 4.5 eV binding energy. 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