# Simplest Method to find Bond Order

Bond Order questions are very common in CSIR NET and other entrance exams. In this post, we will discuss about this topic and at the end I will share a very useful trick that helps to solve questions based on bond order very quickly.

## Bond Order

Bond order is a number that tells us how strong the bond between two atoms forming a molecule is. If the bond order is high, the bond is strong and so the molecule is stable. If the bond order is low, the molecule is unstable.

## How do we calculate Bond Order? (Usual Method)

There are many methods to finding bond order. One of the most famous and most used formulas states:

where, Nb = No. of electrons in Bonding M.O.
and Na = No. of electrons in Anti-Bonding M.O.

This is USUAL method for finding bond order, but this method is NOT good for solving problems. Why?

1. You need to properly find the number of electrons in bonding molecular orbitals and antibonding molecular orbitals.
2. It takes a lot of time.

## Simple Method of Calculating Bond Order:

With a little memorization and practice, you can find out the bond order of a species just by looking at it. The method that we will be using IS NOT applicable everywhere, but it works in 90% of  CSIR NET, IIT JAM, and Other Entrance Examination's problems. So, you see the worth in learning it.

The Basic Principle Behind this Trick is: N2 has 14 electrons and its bond order is 3. Every electron added or subtracted to 14, reduces the bond order by 0.5.

For example, if the number of electrons is 15, then the bond order is 3 – 0.5 = 2.5. Done!

Here's a Table which shows this relation easily:

1. Just count the total number of electrons.
2. Add number of Negative Charges and Subtract number of Positive Charges on molecule.
3. Find the difference of total number of electrons from 14. Call it ‘n’.
4. Bond order = 3  – 0.5n

## Few Examples:

Q.1. Find the Bond Order of CO molecule?

Total number of electrons = 6 + 8 = 14.
B.O. = 3.
Done! simple.

Q2. Find the B.O. of O22-.
Total number of electrons = 8 x 2 + 2(for the – charge) = 18.
B.O. = 3 – 0.5 x 4 = 1

Q3. Which of the following has highest bond order?
a) O22+
b) O22-
c) F22-
d) O2+

Just count the total number of electrons for each.
O22+ has 14 electrons. B.O = 3
O22- has 18 electrons. B.O = 1.
You can find out the B.O. for the other two, and conclude that the answer is (a).

Q4. Which of the following orders regarding the bond order is correct?
1. O2 > O2 > O2+
2. O2 < O2 < O2+
3. O2 > O2 < O2+
4. O2 < O2 >O2+
Once again, we count the number of electrons for each of these species.
For O2 no. of electrons = 17. B.O. = 1.5
For O2, no. of electrons = 16. B.O. = 2
For O2+, no. of electrons = 15. B.O. = 2.5

NOTE: This method will work for any species that have between 8 and 20 electrons. Most problems asked in Entrance Exams have electrons between 8 and 20, therefore it is very handy.

## Application Of Bond Order:

With Calculation of Bond Order you can also predict Bond Length, Bond Strength, Stability of Molecule etc.
As:
1. Bond Order is directly proportional to Bond Strength, means a molecule having larger value of Bond Order will be more strongly held and will require more energy to break it's bonds.
2. Bond Order is Inversely proportional to Bond Length, means a molecule having large value of Bond Order will have smalller bond length than a molecule having less value of Bond Order.
3. Molecules Having Bond Order value of ZERO have no existance, e.g. He2 molecule have no significance becoz its bond order is zero. 