Average Bond Order - Online Tutor, Practice Problems & Exam Prep

Understanding the relationship between bond order, bond strength, and bond length is crucial in chemistry. Bond order indicates the number of chemical bonds between a pair of atoms, with single, double, and triple bonds having bond orders of one, two, and three, respectively. The bond strength is directly proportional to the bond order—higher bond orders result in stronger bonds. Conversely, bond length is inversely proportional to bond order; as the bond order increases, the bond becomes shorter. Recognizing these relationships helps predict the physical properties of molecules and their reactivity.

Average Bond Order

Here we're going to say that the average bond order is just the average number of chemical bonds between a pair of elements. Now a single bond has a bond order of 1, a double bond has a bond order of 2. So what triple bond would have a bond order of 3?

Now remember, as the average bond order increases, that means the strength of the bond will increase and the length of the bond will decrease. So we'd say average bond order and bond strength are directly proportional. Both increase together and they are inversely proportional to bond length, which decreases as they increase.

Average Bond Order Example 1

Here it says what is the average bond order of the sulfur oxygen bonds within the sulfide ion. So within the sulfide ion we have one sulfur that is basically double bonded to one oxygen and single bonded to the other two. The steps we take here is given only the molecular formula. Then draw one of the resonance structures. If multiple resonance structures are given, choose only one.

So let's just choose this first one. We're going to count the total number of bonds between the surrounding elements involved, and then we're going to divide the number of bonds by the total number of those surrounding elements. So how many total bonds do we have between sulfur and oxygen? 1234. So we have 4 total bonds and we're going to divide it by the total number of surrounding elements. Who are the surrounding elements? Oxygen. So divided by three surrounding elements, so that's going to be 43 or 1.33.

So the average bond order of a sulfur to oxygen bond within the sulfide ion is 1.33. What does that mean? Well, each oxygen for sure is single bonded to the sulfur. But that pesky π bond, because of resonance, it doesn't fully exist with this oxygen or this oxygen or this oxygen. The possibility is it could be with either of them, so they're going to share that one bond. So that one bond is cut into three. So each one has a third possession of that π bond. So that's why it comes out to 1.33.

But just remember, just count the total number of bonds and divide it by the number of surrounding elements involved, and you'll always be able to calculate your average bond order.

What is the bond order of the phosphate–oxygen bonds within the phosphate ion, PO₄^3–?

• A. 1.25
• B. 2
• C. 2.5
• D. 1

Here’s what students ask on this topic:

How can I find the average bond order?

To find the average bond order, you need to consider the number of bonding and antibonding electrons within a molecule's molecular orbitals. Bond order gives you an indication of the stability and strength of a bond between atoms. Here's a simple step-by-step process to calculate it:

1. Determine the Bonding Electrons: Count the total number of electrons in the bonding molecular orbitals. These are the orbitals that stabilize the molecule when they are filled with electrons.
2. Determine the Antibonding Electrons: Count the total number of electrons in the antibonding molecular orbitals. Antibonding orbitals, typically denoted with an asterisk (*), destabilize the molecule when they are filled with electrons.
3. Calculate the Bond Order: Subtract the number of antibonding electrons from the number of bonding electrons, and then divide the result by two. The formula looks like this:

$\mathrm{Bond}\mathrm{Order}=\frac{(\mathrm{Number}\mathrm{of}\mathrm{Bonding}\mathrm{Electrons}-\mathrm{Number}\mathrm{of}\mathrm{Antibonding}\mathrm{Electrons})}{2}$

A higher bond order typically indicates a stronger bond. For example, a bond order of 1 represents a single bond, 2 represents a double bond, and 3 represents a triple bond. If you're dealing with resonance structures, the bond order will be an average of these structures, reflecting the delocalization of electrons.

How do you calculate average bond order?

To calculate the average bond order in a molecule, you first need to understand the concept of bond order itself. Bond order represents the number of chemical bonds between a pair of atoms. For example, a single bond has a bond order of 1, a double bond has a bond order of 2, and a triple bond has a bond order of 3.

For molecules that have resonance structures (different ways of drawing the molecule that show different bond arrangements), the average bond order can be calculated by taking the sum of the bond orders for a given bond in all resonance structures and dividing by the number of resonance structures.

Here's a step-by-step process:

1. Draw all possible resonance structures for the molecule.
2. Count the total number of bonds of a particular type between two atoms across all the resonance structures.
3. Divide this total by the number of resonance structures to get the average bond order.

For example, in the carbonate ion (CO₃^2-), there are three resonance structures, each with one double bond and two single bonds between the carbon and oxygen atoms. The average bond order for the C-O bonds would be calculated as follows:

$(2+1+1)/3=\frac{4}{3},\text{or approximately}1.33$

What is the average bond order?

The average bond order refers to the number of chemical bonds between a pair of atoms, averaged over similar types of bonds or across a molecule. It's a way to describe the strength and stability of a bond in molecular orbital theory. To calculate the average bond order, you divide the total number of bonding electrons by the total number of molecular orbitals involved in bonding.

For example, in the molecular ion O₂⁺, there are a total of 15 electrons. In the molecular orbital diagram, 8 electrons occupy bonding orbitals (2 in σ_2s, 2 in σ*_2s, 2 in σ_2p, and 2 in Π_2p), and 7 electrons occupy antibonding orbitals (3 in Π*_2p and 4 in σ*_2p). The bond order is calculated as (number of bonding electrons - number of antibonding electrons) / 2. So for O₂⁺, it would be (8 - 7) / 2 = 0.5. This indicates a weaker bond compared to O₂, which has a bond order of 2.