Lesson 8.5 Bond Length, Order, and Energy

Lesson 8.5 Bond Length, Order, and Energy

Lesson 8.5 Bond Length, Order, and Energy

Suggested Reading
  • Zumdahl Chapter 8 Section 8.1, 8.2, 8.8
Essential Question
  • What are the properties of chemical bonds?

Learning Objective
  • State and explain the relationship between the number of bonds, bond length and bond strength.
  • Define the term average bond enthalpy.
  • Explain the concept of bond energy.
  • Explain in terms of average bond enthalpies, why some reaction are exothermic and others are endothermic.
  • Calculate the heats of reaction for bond energies.
Bond Length and Bond Order
Bond order and bond length indicate the type andstrength ofcovalent bondsbetween atoms. Bond order and length are inversely proportional to each other. When bond order is increased, bond length is decreased.
Bond Order
Bond order is the number of bonding pairs of electrons between two atoms.In acovalent compound, asingle bond has a bond order of one, a double bond has a bond order of two, a triple bond has a bond order of three, and so on.
To determine the bond order between two covalently bonded atoms, follow these steps:
  1. Draw theLewis structure.
  2. Figure out the type of bond between the two atoms.
Example:Determine the bond order for cyanide: CN-.
Solution:
1) Draw the Lewis Structure.

2) Figure out the type of bond between the two atoms.
Since thereare 3dashes, that means that it is atriple bond. Atriple bond means that there is a bond order of 3.
When thereare more than two atoms inthe molecule, follow these steps to determine the bond order:
  1. Draw the Lewis Structure.
  2. Count the total number of bonds.
  3. Count the number of pairs of bonded atoms.
  4. Divide the number of bonds by the total number of pairs of bonded atoms.
Example: Determine the bond order for nitrate: NO3.
1) Draw the Lewis Structure.

2) Count the total number of bonds.
The total number of bonds is 4.
3) Count the number of pairs of bonded atoms.
The number of pairs of bonded atoms is 3.
4) Divide the number of bonds by the number of pairs of bonded atoms.
4/3= 1.33 The bond order is 1.33
Ahigh bond order causes more attraction between electrons. A higher bond order also means that the atomsare held together tighter. This correlation is the same for a low bond order. With a lower bond order, there is less attraction between electrons and this causes the atoms to be held together looser. Bond order alsoindicatesthe stability of the bond. The higher the bond order, the more electrons holding the atoms together, and therefore the greater the stability.
Trends in the Periodic Table
Bond order increases across aperiod and decreases down a group. Thus, it follows the same general trend as effective nuclear charge. Hmm....
Bond Length

Bond length is the distance between the nuclei in a covalent bond. Bond lengths are determined experimentally using x-ray diffraction or the analysis of molecular spectra. Recall that Watson and Crick based their model of DNA on the x-ray diffraction studies carried out by Rosalind Franklin. Knowing the bond length in a particular molecule can sometimes provide a clue about the type of bonding present.
The bond length varies for each type of bond and a comparison of bond lengths is shown below in Table 2.
Table 2 - Bond Length Notation
08 T05
Therefore, the bond length of a triple bond < double bond < single bond.
The length of the bond can be inferred from the bond order. The higher the bond order, thestronger the pull between the two atoms and the shorter the bond length. Generally, the length of the bond between two atoms is approximately the sum of the covalent radii of the two atoms, X + Y. Bond length is given in picometers or nanometers.
The bond energy (bond dissociation energy) is a measure of the amount of energy needed to break apart one mole of covalently bonded gaseous atoms. The SI units used to describe bond energy are kilojoules per mole of bonds (kJ/mol). The bond dissociation energy is a measure of the strength of a particular bond and is essentially the enthalpy change for a gas-phase reaction in which bonds break. Recall that the enthalpy change, ∆H, is the heat absorbed or released during a chemical reaction.
When a chemical reaction occurs, molecular bonds are broken and other bonds are formed to make different molecules. For example, the bonds of two water molecules are broken to form hydrogen and oxygen.
2H2O → 2H2+ O2.
Bonds do not break and form spontaneously-an energy change is required. The energy input required to break a bond is known as bond energy. Bond energy serves a very important purpose in describing the structure and characteristics of a molecule. Conversely, energy is released when bonds are formed.
When a bond is strong, there is a higher bond energy because it takes more energy to break a strong bond. This correlates with bond order and bond length. When the bond order is higher, bond length is shorter, and the shorter the bond length the greater the bond energy. Think about it this way: it is easy to snap a pencil, but if you keep snapping the pencil it gets harder each time since the length of the pencil decreases. A higher bond energy (or a higher bond order or shorter bond length) means that a bond is less likely to break apart. In other words, it is more stable than a molecule with a lower bond energy. With Lewis Structures then, the structure with the higher bond energy is more likely to occur.
Enthalpy & Bond Energy
Enthalpy is related to bond energy because an energy change is required to break bonds. More specifically, bond energy measures the energy that is added to the system to break bonds. We can use bond energy to determine if a reaction is endothermic or exothermic.
  • If the reactants have weak bonds, while the products have strong bonds then the reaction is exothermic (enthalpy change < 0). There is a small amount of energy needed to break the bond (smaller bond energy) and a bigger energy released when strong bonds form. A negative enthalpy change means that the system released energy.
  • If the reactants have strong bonds, but the products have weak bonds its an endothermic reaction. (enthalpy change > 0). The energy required to break the reactant bonds is greater than the energy released when the product bonds form.
Average Bond Energy
The same bond can appear in different molecules, but it will have a different bond energy in each molecule because the other bonds in the molecule will affect the bond energy of the specific bond. So the bond energy of C-H in methane is slightly different than the bond energy of C-H in ethane. We can calculate a more general bond energy by finding the average of the bond energies of a specific bond in different molecules to get the average bond energy. Thus, when solving problems using bond energies, we use the average bond energies.
Average Bond Energies (kj/mol):
Bond Energy (kj/mol) / Bond / Bond Energy
(kj/mol) / Bond / Bond Energy (kj/mol) / Bond
297 / H-I / 347 / C-C / 163 / N-N
364 / H-Br / 611 / C=C / 418 / N=N
368 / H-S / 837 / C:::C / 946 / N:::N
389 / H-N / 305 / C-N / 222 / N-O
414 / H-C / 615 / C=N / 590 / N=O
431 / H-Cl / 891 / C:::N
436 / H-H / 360 / C-O
464 / H-O / 736 / C=O
565 / H-F / 339 / C-Cl
151 / I-I / 142 / O-O
159 / F-F / 498 / O=O
193 / Br-Br
243 / Cl-Cl
Example: Using Bond Energies to Calculate the Enthalpy Change for a Reaction
Calculate the enthalpy change for the following reaction:H2(g)+I2(g)→2HI(g)
First look at the equation and determine what bonds exist.
There's an H-H bond, I-I bond, and 2 H-I bonds.
Then examine the bond breakagewhich is located in the reactant side and find the average bond energy in the table above:
1 mol H-H bonds→1 mol x 436 kj/mol = 436 kJ
1 mol I-I bonds → 1 mol x 151 kj/mol = 151 kJ
The sum of the bonds broken is 587 kJ.
Then we look at the bond formation which is on the product side :
2 mol H-I bond → 2 mol x 297 kj/mol = 594 kJ
The sum of the bonds formed is 594 kJ.
The net change of the reaction is therefore 587-594= -7 kJ. Since it's a negative number you know that the reaction is exothermic.
In general, the enthalpy of reaction is approximately equal to the sum of the bond energies for bonds broken minus the sum of the bond energies for bonds formed.
∆H = ∑B.E.bonds broken - ∑B.E.bonds formed

HOMEWORK: Practice exercise 8.12, book questions pg. 385 questions 78,79,80