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.
Let us start with some introduction.

## 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, N

_{b}= No. of electrons in Bonding M.O.

and N

_{a}= 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?

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

**Also See: Hybridisation In Five Simple Steps**

## 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:

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

**N**_{2}**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:**__You can follow these steps:__

- Just count the total number of electrons.
- Add number of Negative Charges and Subtract number of Positive Charges on molecule.
- Find the difference of total number of electrons from 14. Call it ‘n’.
**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 O**

_{2}^{2-}.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) O**

_{2}^{2+}**b) O**

_{2}^{2-}**c) F**

_{2}^{2-}**d) O**

_{2}^{+}Just count the total number of electrons for each.

O

_{2}

^{2+}has 14 electrons. B.O = 3

O

_{2}

^{2-}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?**

**O**_{2}^{–}> O_{2}> O_{2}^{+}**O**_{2}^{–}< O_{2}< O_{2}^{+}**O**_{2}^{–}> O_{2}< O_{2}^{+}**O**_{2}^{–}< O_{2}>O_{2}^{+}

For O

_{2}

^{–}no. of electrons = 17. B.O. = 1.5

For O

_{2}, no. of electrons = 16. B.O. = 2

For O

_{2}

^{+}, no. of electrons = 15. B.O. = 2.5

So, the answer is

**(b).**

**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:

**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.**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.**Molecules Having Bond Order value of ZERO have no existance**, e.g. He_{2}molecule have no significance becoz its bond order is zero.

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