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Bond Order - UNSOLVED PRACTICE SET

Class 11

Chapter: Chemical Bonding and Molecular Structure | Topic: Bond Order

Study Material.
Class 11

BOND ORDER - UNSOLVED PRACTICE SET

Topic: Bond Order

Time: 40 mins | Marks: 30 | Difficulty: Medium

Multiple Choice Questions

Q1. Bond order is defined as:

  1. The total number of bonds between two atoms
  2. Half the difference between the number of bonding and antibonding electrons
  3. The energy required to break a bond
  4. The length of a bond

Q2. A bond order of zero means:

  1. A very strong bond
  2. A very weak bond
  3. No bond exists between the atoms
  4. A coordinate bond

Q3. The bond order of Nβ‚‚ is:

  1. 1
  2. 2
  3. 3
  4. 4

Q4. If a species has a bond order of 1.5, it indicates:

  1. A single bond
  2. A double bond
  3. A bond intermediate between single and double
  4. No bond

Q5. The bond order of O₂⁺ compared to Oβ‚‚ is:

  1. Lower
  2. Higher
  3. The same
  4. Zero

Q6. Which of the following has the highest bond order?

  1. Oβ‚‚
  2. O₂⁺
  3. O₂⁻
  4. O₂²⁻

Short Answer Questions

Q7. Define bond order. How is it related to bond strength and bond length?

Q8. Calculate the bond order of Oβ‚‚ using molecular orbital theory. Is Oβ‚‚ paramagnetic or diamagnetic? 

Q9. Why does Nβ‚‚ have a higher bond order than Oβ‚‚? What does this imply about their relative stability?

Q10. The bond order of NO is 2.5. Explain what a fractional bond order means in terms of molecular stability.

Q11. Your friend tells you that her relationship with her study partner is "2.5 out of 3" β€” stronger than a casual friendship (1) but not as committed as best friends (3). How is this exactly analogous to the bond order in NO, where the fractional value indicates a bond strength between a double bond and a triple bond?

Q12. Arrange the following in order of increasing bond order: Oβ‚‚, O₂⁺, O₂⁻, O₂²⁻. Explain your reasoning. 

Long Answer Questions

Q13. Explain the concept of bond order. How is it calculated using Molecular Orbital Theory? Discuss the relationship between bond order and:

(a) Bond length

(b) Bond energy

(c) Magnetic properties

Illustrate with the examples of Nβ‚‚, Oβ‚‚, and Fβ‚‚.

Q14. Describe how bond order helps predict the stability and properties of molecular species. For the following species, calculate bond order and predict stability:

(a) Hβ‚‚

(b) Heβ‚‚

(c) Liβ‚‚

(d) Beβ‚‚

Why does Heβ‚‚ not exist as a stable molecule while He₂⁺ does?

Q15. During a chemistry olympiad, you encounter the following problem:

(a) The bond order of CO is 3, the same as Nβ‚‚. Yet CO is polar while Nβ‚‚ is non-polar. Explain why bond order alone cannot predict molecular polarity.

(b) The superoxide ion O₂⁻ has a bond order of 1.5. Draw its MO configuration and show the calculation.

(c) The peroxide ion O₂²⁻ has a bond order of 1.0. Compare its bond length with Oβ‚‚ (bond order 2) and explain.

(d) A mystery diatomic species has a bond order of 2.5 and is paramagnetic. Identify the species and write its MO configuration.

Numerical / Application-Based Problems

Q16. Calculate the bond order for the following species using molecular orbital theory. Show your calculations and predict magnetic behaviour:

(a) Nβ‚‚

(b) N₂⁺

(c) N₂⁻

(d) O₂²⁻

Q17. The following table gives information about diatomic species of the second period:

SpeciesTotal ElectronsBond OrderBond Length (pm)Magnetic Nature
Liβ‚‚6?267?
Bβ‚‚10?159?
Cβ‚‚12?124?
Nβ‚‚14?110?
Oβ‚‚16?121?
Fβ‚‚18?142?

(a) Complete the table by calculating bond order and predicting magnetic nature.

(b) Plot a graph of bond order vs. bond length for these species. What trend do you observe?

(c) Nβ‚‚ has the shortest bond length despite Oβ‚‚ having a lower bond order than expected. Explain this anomaly.

(d) Predict the bond order and bond length of Neβ‚‚. Does it exist?

Q18. Consider the following diatomic species and their ions:

(a) For the species CN, CO, and NO⁺ (all 14-electron species):

(i) Write the MO configuration for each.

(ii) Calculate the bond order.

(iii) Predict which has the highest bond dissociation energy.

(b) The species NO has 15 electrons. Compare its bond order with NO⁺ and NO⁻.

(c) Explain why isoelectronic species (same number of electrons) tend to have similar bond orders but may differ in other properties.


Total: 30 Marks | Time: 40 mins

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