Lesson 8.5 Bond Length, Order, and Energy
Suggested Reading
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Zumdahl Chapter 8 Section 8.1, 8.2, 8.8
Essential Question

What are the properties of chemical bonds?
Learning Objective
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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 and strength of covalent
bonds between 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 a covalent compound, a single 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.
2.
Draw the Lewis structure.
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 there are 3 dashes, that means that it is a triple bond. A triple bond
means that there is a bond order of 3.
When there are more than two atoms in the molecule, follow these
steps to determine the bond order:
1.
2.
3.
4.
Draw the Lewis Structure.
Count the total number of bonds.
Count the number of pairs of bonded atoms.
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
A high bond order causes more attraction between electrons. A higher
bond order also means that the atoms are 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 also indicates the 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 a period 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
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, the stronger 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.
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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
Bond
Bond
Energy
(kj/mol)
Bond
H-I
(kj/mol)
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
431
436
H-C
H-Cl
H-H
615
891
360
C=N
C:::N
C-O
590
N=O
464
565
151
H-O
H-F
I-I
736
339
142
C=O
C-Cl
O-O
297
159
F-F
193
Br-Br
243
Cl-Cl
498
O=O
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 breakage which 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