Bond order from molecular orbital theory: Class 11 practice
Bond order = ½(bonding electrons − antibonding electrons). Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Electronic configuration
The idea behind bond order from molecular orbital theory
Bond order = ½(bonding electrons − antibonding electrons). Higher bond order means a shorter, stronger bond; bond order 0 means the molecule doesn't exist. Unpaired electrons make a species paramagnetic.
Step-by-step method
- Count total valence-shell electrons (include charge).
- Fill MOs in order.
- Bond order = ½(Nb − Na); check unpaired electrons.
Worked example 1 (Basic)
Find the bond order of O₂⁺ and say whether it is paramagnetic.
- Bonding 10, antibonding 5 (counting from 1s/2s as applicable)
- Bond order = ½(10 − 5) = 2.5
- Has unpaired electrons → paramagnetic
- Answer: 2.5, paramagnetic
Worked example 2 (Standard)
Find the bond order of B₂ and say whether it is paramagnetic.
- Bonding 6, antibonding 4 (counting from 1s/2s as applicable)
- Bond order = ½(6 − 4) = 1
- Has unpaired electrons → paramagnetic
- Answer: 1, paramagnetic
Worked example 3 (Challenge)
Find the bond order of N₂ and say whether it is paramagnetic.
- Bonding 10, antibonding 4 (counting from 1s/2s as applicable)
- Bond order = ½(10 − 4) = 3
- All paired → diamagnetic
- Answer: 3, diamagnetic
Common mistakes and misconceptions
- Forgetting to add/remove electrons for ions
- Calling O₂ diamagnetic
Practice questions
- (Basic) Find the bond order of C₂ and say whether it is paramagnetic.
- (Basic) Find the bond order of H₂ and say whether it is paramagnetic.
- (Standard) Find the bond order of He₂ and say whether it is paramagnetic.
- (Standard) Find the bond order of N₂⁺ and say whether it is paramagnetic.
- (Challenge) Find the bond order of O₂⁻ and say whether it is paramagnetic.
- (Challenge) Find the bond order of F₂ and say whether it is paramagnetic.
Answer key with full working
- Q1: Bonding 8, antibonding 4 (counting from 1s/2s as applicable) → Bond order = ½(8 − 4) = 2 → All paired → diamagnetic → Answer: 2, diamagnetic
- Q2: Bonding 2, antibonding 0 (counting from 1s/2s as applicable) → Bond order = ½(2 − 0) = 1 → All paired → diamagnetic → Answer: 1, diamagnetic
- Q3: Bonding 2, antibonding 2 (counting from 1s/2s as applicable) → Bond order = ½(2 − 2) = 0 → All paired → diamagnetic → Answer: 0, diamagnetic
- Q4: Bonding 9, antibonding 4 (counting from 1s/2s as applicable) → Bond order = ½(9 − 4) = 2.5 → Has unpaired electrons → paramagnetic → Answer: 2.5, paramagnetic
- Q5: Bonding 10, antibonding 7 (counting from 1s/2s as applicable) → Bond order = ½(10 − 7) = 1.5 → Has unpaired electrons → paramagnetic → Answer: 1.5, paramagnetic
- Q6: Bonding 10, antibonding 8 (counting from 1s/2s as applicable) → Bond order = ½(10 − 8) = 1 → All paired → diamagnetic → Answer: 1, diamagnetic
Syllabus note
This skill is listed under "Chemical Bonding" for Class 11 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
Frequently asked questions
Which chapter is bond order from molecular orbital theory in?
"Chemical Bonding", Class 11 Chemistry (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.