Valence Bond Theory and Orbital Overlap
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Sigma and pi bonds from valence bond theory for JEE: how a covalent bond forms when half-filled orbitals overlap, the energy of two H atoms (lowest at 74 pm, 435 kJ/mol released), positive, negative and zero overlap, head-on and sideways overlap, why sigma is stronger than pi, why a double bond cannot rotate, and counting sigma and pi bonds in ethene, ethyne and acrylonitrile.
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Transcript
0:00Introduction
Hello everyone! Okay, one small puzzle to start. Take two carbon atoms joined by a single bond. One end can turn freely about that bond, like a wheel on its axle. Now join the same two carbons with a double bond. Suddenly, the turning stops. The rotation is locked! Just one extra bond, and no more turning? Strange, right? Today we see how overlapping orbitals make bonds, and this puzzle will solve itself. Counting sigma and pi bonds comes up often in JEE, so let's begin.
0:37How a covalent bond forms
So, here is valence bond theory, in one line. Each atom brings an orbital that is half-filled, that means it holds just one electron. When the two atoms come close, these two orbitals overlap; they share the same region of space. Okay? Now, the two electrons, one from each atom, pair up in that shared region. What must be true of their spins? Opposite spins, yes, exactly! Only electrons of opposite spin can pair up like this. Now let's watch a bond form in the simplest case. Two hydrogen atoms, far apart, start coming closer. What happens to the energy? It falls, correct! As the hydrogen atoms approach, the energy keeps going down, until it is lowest at a distance of seventy four picometres. On screen, H is hydrogen, and p m means picometres. Clear? At that point, four hundred and thirty five kilojoules per mole of energy is released. k J means kilojoules, and mol to the power minus one means per mole. And if you push them any closer, the energy rises steeply. So the two atoms settle at seventy four picometres. With me?
1:52More overlap, stronger bond
Now, the most important rule of this whole topic. The greater the overlap, the stronger the bond. The double arrow on screen means implies, or simply, so. We will use this rule again and again. Okay? See, each lobe of an orbital carries a sign, plus or minus, called its phase. When two lobes of the same phase overlap, the overlap is positive, and that is bonding. Right? When the lobes have opposite phases, the overlap is negative. And when there is no net overlap at all, for example, two orbitals at right angles to each other, the overlap is zero. Only positive overlap is bonding. Got it? Quick question. Say the line joining the two nuclei is the z axis. One atom brings a p x orbital, the other a p y orbital. The small letter below the p tells you its direction, okay? What kind of overlap do they give? Zero, exactly! p x points along x, and p y points along y, so the two are at right angles. Zero overlap, so no bond. Clear?
2:59Head-on and sideways overlap
Now, look at the figure. The dashed line is the internuclear axis, the line joining the two nuclei. When orbitals meet head-on, right along this axis, we get a sigma bond. Sigma is the Greek letter on screen, like a small o with a tail. Okay? Here are the orbitals that meet head-on. s with s, the first drawing. s with p, the second. And two p orbitals pointing at each other along the axis, end-on, the third drawing; we call that axial p p overlap. The little dash between the letters just means, with. Hybrid orbitals also make sigma bonds; we meet them later in this chapter. Right? Now look at the last drawing. Two p orbitals, standing parallel, side by side. They cannot meet head-on. So, can they still overlap? Yes, they can, sideways! Sideways overlap of parallel p orbitals gives a pi bond. Pi is the Greek letter you see, like a little table with two legs. Got it? And notice where this overlap sits. Not on the axis, but above and below it, as the figure shows. This picture explains what comes next. With me?
4:10Sigma versus pi
Alright, now use our rule: more overlap, stronger bond. Which one is stronger, sigma or pi? Sigma, yes, correct! Head-on overlap along the axis is much bigger than the sideways overlap of parallel p orbitals. More overlap, stronger bond. So a sigma bond is stronger than a pi bond. Okay? Now, a double bond is two bonds between the same two atoms. So it must be two sigma bonds, right? No, no, careful! Look: a single bond is one sigma. Then a double bond is one sigma plus one pi. And a triple bond is one sigma plus two pi. The sigma forms first, head-on along the axis; any extra bond overlaps sideways, as pi. So, only one sigma between any two atoms. Clear? Now, our puzzle from the start! Rotation is free about a sigma bond, but not about a pi bond. The reason: turning one carbon would twist the parallel p orbitals apart, and break the pi overlap. That is why the double bond locks the rotation. Mystery solved! And here is a topper's trick for counting. Every link between two atoms gives exactly one sigma. And every extra line in that link is one pi. So count the links for sigma, and the extra lines for pi. That's it!
5:35Example 1: sigma and pi bonds in ethene and ethyne
Example one. Count the sigma and pi bonds in ethene and in ethyne. Ethene is C H two, double bond, C H two. The small two means two hydrogens, and the two lines mean a double bond. Ethyne is H C, triple bond, C H: three lines, a triple bond. Pause the video, and try it first. Done? Okay, let's check. Ethene first. Each carbon holds two hydrogens, so four C H links; the dash between C and H means one bond. Plus one carbon carbon link. Every link gives one sigma: four plus one, five sigma bonds. Right? Now the carbon carbon double bond. Its sigma is already counted, and its extra line is one pi bond. So ethene has one pi bond. Okay? Now ethyne. Each carbon holds just one hydrogen, so two C H links, plus one carbon carbon link. That makes three sigma bonds. Clear? And the triple bond: one sigma, already counted, plus two extra lines, so two pi bonds. So, ethene has five sigma and one pi, and ethyne has three sigma and two pi. Simple, right?
6:48Example 2: sigma and pi bonds in acrylonitrile
Now a harder one, example two. Count the sigma and pi bonds in acrylonitrile. Its formula: C H two, double bond, C H, single bond, C, triple bond, N, where N stands for nitrogen. Pause the video, and try it. Alright, a quick count first. Three C H bonds are hidden in the formula. Then two in the double bond, one single, and three in the triple bond. Nine bonds in all. So, nine sigma bonds, right? Oh no, careful! This is the trap. Only one sigma per link. So list the links: three C H, then the carbon carbon double, the carbon carbon single, and the carbon nitrogen triple. Okay? Three plus one plus one plus one: six sigma bonds. With me? Now the extra lines. One pi in the carbon carbon double bond, and two pi in the carbon nitrogen triple bond. That is three pi bonds. Got it? So the answer: six sigma and three pi. And look, six plus three is nine, exactly the total number of bonds we counted. Nice check, isn't it?
8:02Quick revision: overlap, sigma vs pi, counting bonds
So, quick revision. First, a covalent bond forms when half-filled orbitals overlap, and the more the overlap, the stronger the bond. Sigma is head-on overlap along the axis; pi is sideways overlap of parallel p orbitals. Each link has one sigma, the extra lines are pi, and pi bonds lock the rotation. Now go and try the practice problems on this page. You will find them easy. All the best!
More videos in Unit 3 · Chemical Bonding & Molecular Structure
- The Kossel–Lewis Approach and the Octet Rule
- Formation of Ionic Bonds
- Lattice Enthalpy and the Born–Haber Cycle
- Covalent Character in Ionic Bonds: Fajans' Rules
- Lewis Structures and Formal Charge
- Coordinate Bonds
- Resonance
- Bond Parameters
- Electronegativity, Polarity and Dipole Moment
- The VSEPR Model
- Shapes with Bond Pairs Only
- Shapes with Lone Pairs
- Comparing Bond Angles
- Counting Lone Pairs; Isostructural and Isoelectronic Species
- Hybridisation with s and p Orbitals
- Hybridisation Involving d Orbitals
- pπ–pπ and pπ–dπ Bonding; Changes of Hybridisation
- Formation of Molecular Orbitals
- Energy Order and Electronic Configurations
- Bond Order, Stability and Magnetism
- Bond Length, Bond Energy and Heteronuclear Species
- Hydrogen Bonding
- Van der Waals Forces
- Metallic Bonding
- Overview
- Set 1 (Ionic Bonding, Lewis Structures and Polarity)
- Set 2 (Shapes, Lone Pairs and Hybridisation)
- Set 3 (Molecular Orbitals and Intermolecular Forces)
- Overview
- Part A: JEE Main Level
- Part B: Harder Problems
- Part C: Concept Checks