Molecular Orbital Theory: PECTAA Class 11 Chemistry, Punjab Board

Draw the Lewis structure of O2 and every electron is paired. Yet liquid oxygen sticks between the poles of a magnet, which needs unpaired electrons. The older picture cannot explain this. Molecular orbital theory can, and it also tells you why He2 never forms.

Chapter 3 of the PECTAA Chemistry 11 book is Chemical Bonding. Section 3.9 covers molecular orbital theory (MOT) and section 3.9.1 gives the molecular orbital diagrams of some diatomic molecules. This note is for FSc Part 1 (1st year, Class 11) students in Lahore, Faisalabad, Sargodha, Rawalpindi, Multan, Gujranwala and the other Punjab boards. Gujrat district comes under the Gujranwala board. PECTAA writes the books. Your own board sets your paper.

What to know

The postulates of MOT, in short.

1. When a molecule forms, atomic orbitals of the joining atoms overlap to make new molecular orbitals, which belong to the whole molecule.
2. Orbitals overlap through lobes of suitable sign of the wave function. One lobe of a p orbital is marked + and the other −.
3. Two atomic orbitals always give two molecular orbitals:
   • same-sign overlap gives a bonding orbital (σ or π), lower in energy than the parent orbitals;
   • opposite-sign overlap gives an antibonding orbital (σ* or π*), higher in energy.
4. Bond order = (bonding electrons − antibonding electrons) / 2. It tells how many bonds join the two atoms.

s–s overlap (H2). Two 1s orbitals give σ1s (bonding) and σ*1s (antibonding). The bonding orbital is symmetrical about the line joining the nuclei. The antibonding orbital has its electron density away from the region between the nuclei. H2 puts both electrons in σ1s, so its bond order is (2 − 0)/2 = 1.

p orbital overlap.

• Head-on overlap along the bond axis (say x) gives σ2pₓ and σ*2pₓ.
• Sideways overlap of parallel pᵧ or p_z orbitals gives π orbitals. Bonding π orbitals have zero electron density on the nuclear axis (a nodal plane). Antibonding π* orbitals have the least density between the nuclei.

Three molecules from the book.

• He2. Each He is 1s². Four electrons fill σ1s (2) and σ*1s (2). Bond order = (2 − 2)/2 = 0, so He2 does not form.
• N2. Each N is 1s² 2s² 2pₓ¹ 2pᵧ¹ 2p_z¹. The valence order of energy in the book is σ2s < σ2s < π2pᵧ = π2p_z < σ2pₓ, then the antibonding π and σ* orbitals. The ten valence electrons fill up to σ2pₓ. Bonding valence electrons = 8, antibonding = 2, so bond order = (8 − 2)/2 = 3: one σ bond and two π bonds.
• O2. For oxygen the book places σ2pₓ below the two π2p orbitals. Twelve valence electrons give 8 bonding and 4 antibonding, so bond order = (8 − 4)/2 = 2: one σ and one π bond, a double bond O=O. The last two electrons go singly into π2pᵧ and π2p_z, so O2 has two unpaired electrons.

Paramagnetism. A paramagnetic substance has one or more unpaired electrons and is attracted by a magnetic field. More unpaired electrons give stronger paramagnetism. The book shows liquid oxygen held between the poles of a strong magnet. MOT explains this through the two unpaired π* electrons.

Key points

• Two in, two out: two atomic orbitals always make two molecular orbitals.
• Bond order 0 means no molecule. He2 is the book's example.
• Unpaired electrons → paramagnetic. O2 is the classic case.

Worked examples

Example 1. Bond order of H2. Two electrons in σ1s and none in σ*1s: (2 − 0)/2 = 1, a single bond.

Example 2. Why N2 has a triple bond. Valence electrons: 8 bonding (σ2s², π2pᵧ², π2p_z², σ2pₓ²) and 2 antibonding (σ*2s²). Bond order = (8 − 2)/2 = 3.

Example 3. Is O2 diamagnetic or paramagnetic? Paramagnetic. Its two highest electrons sit singly in π2pᵧ and π2p_z, so two electrons are unpaired.

Example 4. Bond order of He2⁺ (one electron removed). Electrons: σ1s² σ*1s¹. Bond order = (2 − 1)/2 = ½. A bond order above zero means the ion can exist.

Common mistakes in the exam

1. Saying antibonding orbitals are lower in energy. They are higher than the parent atomic orbitals.
2. Forgetting to halve. Bond order is the difference divided by 2.
3. Pairing the π electrons in O2.* They go singly into the two π* orbitals, which is why O2 is paramagnetic.
4. Mixing the σ2pₓ position. For N2 the book puts σ2pₓ above the π2p pair; for O2 below it.
5. Calling π orbitals symmetrical about the axis. That describes σ orbitals. Bonding π orbitals have a nodal plane on the axis.

Exam technique

• Draw the MO diagram with atomic orbitals on both sides and molecular orbitals in the middle, as in the book's figures.
• Fill electrons from the lowest level, one per orbital before pairing in equal-energy orbitals.
• Write the bond order formula, then the numbers, then the meaning (single, double, triple bond).
• Check your board's notice for the topics in your paper.

Practice questions

1. How many molecular orbitals form from two atomic orbitals? Answer: Two: one bonding and one antibonding.
2. Find the bond order of N2. Answer: (8 − 2)/2 = 3.
3. Which type of overlap gives π orbitals? Answer: Sideways overlap of parallel p orbitals.
4. What is the bond order of He2, and what does it mean? Answer: 0. The molecule does not form.
5. How many unpaired electrons does O2 have? Answer: Two, one in each π* orbital.

Ready to practise this? Continue in 1st Year Chemistry.