# Electron Structure and Chemical Bonding

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## Contents

### Bond Enthalpy

Amount of energy required to break the bonds of the reactant molecules - Amount of energy released when the bonds of the products form

If we know how much energy is stored in the chemical bonds, we can calculate the energy change of a reaction. It is important to realize that bonds are broken on the reactant side of the equation, and bonds are formed on the product side. This will give us an equation to use that is very similar to our Hess's Law formula, but with one important different - when using Hess's Law, we subtract the reactants from the products. But when using bond enthalpies, we will subtract the products from the reactants. Be careful to remember these differences.

To solve bond enthalpy questions, you'll need to be able to draw the structural formulas of models, something you likely learned in Chemistry 20. You'll need to know which atoms are bonded together, and if single, double, or triple bonds are involved. We'll keep our molecules pretty simple here.

If the four hydrogen atoms in a methane molecule (CH4) were bound to the three 2p orbitals and the 2s orbital of the carbon atom, the H-C-H bond angles would be 90° for 3 of the hydrogen atoms and the 4th hydrogen atom would be at 135° from the others. Experimental evidence has shown that the bond angles in methane are not arranged that way but are 109.5° giving the overall shape of a tetrahedron. The tetrahedral structure makes much more sense in that hydrogen atoms would naturally repel each other due to their negative electron clouds and form this shape. If you think electron-electron repulsion isn't significant, try walking through a wall! There is plenty of space for your nuclei to pass through the nuclei of the wall material but ouch, it just doesn't work that way.

Example: Using bond enthalpies, provided in the table below, calculate the heat of reaction, ΔH, for: ½ H2(g) + ½ Cl2(g) → HCl(g)

Given the following bond enthalpies:H — H -> 436 17:02, 17 August 2009 (UTC)~Cl — Cl -> 243 kJ 17:02, 17 August 2009 (UTC)~H — Cl -> 433 kJ

(you can find a more complete list of bond enthalpies in the bond enthalpy table)

Solution:

Draw structural formulas for all molecules. Pretty simple for these molecules: Chemical formula Structural Formula ~~H2: H — H ~~Cl2: Cl — Cl ~~HCl: H — Cl

Using the balancing coefficients in the balanced equation and the structural formulas, determine how much energy is required to break all of the bonds of the reactants, and how much energy is released when all of the product bonds form:

H - H 1× 436 = 436

Cl - Cl 1 × 243 = 243

H-Cl 1 × 433 = 433

We still need to use the balancing coefficients from the balanced equation: For example, 1 mole of H - H bonds requires 436 kJ but in our balanced equation only ½ mole of H - H bonds is involved, which will require only 218 kJ (436 × 0.5):

ΔH = Σ (bonds broken) – Σ (bonds formed) ~~ ΔH = [½(436) + ½(243)] – [(433)] ~~ ΔH = 339.5 – 433 = __-93.5 kJ __

### Hybridization of Atomic Orbitals and the Shape of Molecules

Experimental evidence has also shown that the H-N-H bond angles in ammonia (NH3) are 107° and the H-O-H bond angles in water are 105°. It is clear from these bond angles that the non-bonding pairs of electrons occupy a reasonable amount of space and are pushing the hydrogen atoms closer together compared to the angles found in methane. Lone pair-lone pair repulsion is greater than bond pair-bond pair repulsion.

The valence shell electron-pair repulsion model (VESPR) was devised to account for these molecular shapes. In this model, atoms and pairs of electrons will be arranged to minimize the repulsion of these atoms and pairs of electrons. Since the non-bonded electron pairs are held somewhat closer to the nucleus than the attached hydrogen atoms, they tend to crowd the hydrogen atoms. Thus ammonia exists as a distorted tetrahedron (trigonal pyramidal) rather than a trigonal plane and water also exists as a distorted tetrahedron (bent) rather than a linear molecule with the hydrogen atoms at a 180° bond angle.

This concept proposes that since the attached groups are not at the angles of the p orbitals and their atomic orbitals would not have maximum overlap (to form strong bonds) the s and p orbitals will be hybridized to match the bond angles of the attached groups.

The number of these new hybrid orbitals must be equal to the numbers of atoms and non-bonded electron pairs surrounding the central atom!

Hybridization Involving d-Orbitals

As we discussed earlier, some 3rd row and larger elements can accommodate more than eight electrons around the central atom. These atoms will also be hybridized and have very specific arrangements of the attached groups in space. The two types of hybridization involved with d orbitals are sp3d and sp3d2.

For a more complete approach of hybridization (with schemes and drawings), please consult the following website