Bond Strength – Bond Length

Relationship between bond order, strength and length

We have learned that covalent bonds can come as either single, double or triple bonds. We also saw in the last section that we identify these different types of bonds by their bond order.

In what way does a bond change as the bond order increases? It turns out that both the length and strength of a bond change with bond order.

Bond

# of electrons

Bond Order

Bond Strength

Bond Length

Single

2

1

Weakest

Longest

Double

4

2

   
Triple

6

3

Strongest

Shortest

One analogy you can use is to think about atoms as nerf balls and bonds as rubber bands. The rubber bands act as the force which holds the balls together, as we increase the number of rubber bands the balls are squished closer together and it takes more force to pull them apart.

In a molecule as you increase the number of electrons shared between two atoms, you increase the bond order, increase the strength of the bond, and decrease the distance between nuclei.

We experimentally measure bond lengths using x-ray and neutron diffraction, which was mentioned briefly in chapter 6.

We experimentally measure bond strength by measureing the amount of energy needed to break a bond:

Cl2(g) 2Cl(g) DH = 242 kJ

So we say that the enthalpy (strength) of a Cl-Cl single bond is 242 kJ/mol or

(242 103 J/mol)/(6.02 1023 molecules/mol) = 4.02 10-19 J/bond

Important points to remember about bond enthalpies:

Lets look at the strengths and lengths of various bonds.

Bond

Bond Order

Bond Enthalpy

Bond Length

C-C

1

348 kJ/mol

1.54

C=C

2

614 kJ/mol

1.34

CC

3

839 kJ/mol

1.20

N-N

1

163 kJ/mol

1.47

N=N

2

418 kJ/mol

1.24

NN

3

941 kJ/mol

1.10

We can use the idea of bond enthalpy and Hess’ Law to estimate enthalpies:

DHrxn = S (enthalpy of bonds broken) – S (enthalpy of bonds formed)

Example

Estimate the enthalpy for the formation of water using bond enthalpies.

First I write the balanced chemical equation and determine the number and type of bonds on each side of the arrow.

H2(g) + O2(g) H2O(g)

First we need to draw the Lewis dot structures and determine which bonds are present in each molecule. Then remember to multiply the number of bonds in each molecule by the coefficient in front of the molecule.

I look up the appropriate bond enthalpies in table 8.3.

DHrxn = 436 kJ/mol + (495 kJ/mol) – 2(463 kJ/mol) = -243 kJ/mol

If we measure the enthalpy of the formation of water in a calorimeter we obtain –242 kJ/mol, so we see that we can get a pretty good estimate of the enthalpy of a reaction by using the bond enthalpy approach. More importantly we see the fundamental basis for reaction enthalpy comes from the energy associated with forming and breaking bonds.

Note Bond enthalpies are averages not exact values. Furthermore they are derived only for gaseous molecules, not liquids, solids or aqueous compounds.