Anand Classes explains the Bond Order and Electronic Configurations of Heteronuclear Diatomic Molecules such as NO, NO⁺, CN, CN⁻, and CO, based on the Molecular Orbital Theory (MOT). Students will learn how molecular orbitals are formed from atomic orbitals of different atoms and how this affects bond order, bond length, stability, and magnetic character. The post includes step-by-step electronic configurations, bond order calculations, and energy-level diagrams for each molecule. This concept is essential for mastering Class 11, Class 12, JEE, and NEET Chemistry. It also features MCQs, Q&A, Assertion-Reason, and Case Study questions for concept clarity and practice. Click the print button to download study material and notes.
Molecular Orbital Electronic Configurations of Some Common Heteronuclear Molecules
The configurations of heteronuclear molecules (containing different atoms) can be written in a similar manner as in the case of homonuclear molecules.
What is Electronic Configuration and Bond Order of Nitric Oxide (NO) ?
Total number of electrons = $7 + 8 = 15$
Configuration:
$$NO : KK \, (\sigma 2s)^2 \, (\sigma^* 2s)^2 \, (\pi 2p_x)^2 \, (\pi 2p_y)^2 \, (\sigma 2p_z)^2 \, (\pi^* 2p_x)^1$$
$$\text{Bond order} = \frac{N_b – N_a}{2} = \frac{8 – 3}{2} = 2.5$$
What is Electronic Configuration and Bond Order of Nitric Oxide Cation (NO+) ?
Configuration:
$$NO^+ : KK \, (\sigma 2s)^2 \, (\sigma^* 2s)^2 \, (\pi 2p_x)^2 \, (\pi 2p_y)^2 \, (\sigma 2p_z)^2$$
$$\text{Bond order} = \frac{8 – 2}{2} = 3$$
What is Electronic Configuration and Bond Order of Cyanide Radical (CN) ?
Total number of electrons = $6 + 7 = 13$
Configuration:
$$CN : KK \, (\sigma 2s)^2 \, (\sigma^* 2s)^2 \, (\pi 2p_x)^2 \, (\pi 2p_y)^2 \, (\sigma 2p_z)^1$$
Bond order:
$$\text{Bond order} = \frac{7 – 2}{2} = 2.5$$
What is Electronic Configuration and Bond Order of Cyanide Anion (CN–) ?
Configuration:
$$CN^- : KK \, (\sigma 2s)^2 \, (\sigma^* 2s)^2 \, (\pi 2p_x)^2 \, (\pi 2p_y)^2 \, (\sigma 2p_z)^2$$
Bond order:
$$\text{Bond order} = \frac{8 – 2}{2} = 3$$
What is Electronic Configuration and Bond Order of Carbon Monoxide (CO) ?
Total number of electrons = $6 + 8 = 14$
Configuration:
$$CO : KK \, (\sigma 2s)^2 \, (\sigma^* 2s)^2 \, (\pi 2p_x)^2 \, (\pi 2p_y)^2 \, (\sigma 2p_z)^2$$
Bond order:
$$\text{Bond order} = \frac{8 – 2}{2} = 3$$
Key Takeaways
- NO → bond order = 2.5 (paramagnetic, due to one unpaired electron)
- NO+ → bond order = 3 (more stable than NO)
- CN → bond order = 2.5 (radical species)
- CN– → bond order = 3 (very stable ion)
- CO → bond order = 3 (strong triple bond, similar to N2)
Short Answer Type (SAT) Questions
Q1. Why is NO paramagnetic while NO+ is diamagnetic?
Answer:
- NO has 15 electrons → 1 unpaired electron in $\pi^*2p$ orbital → paramagnetic.
- NO+ has 14 electrons → all orbitals paired → diamagnetic.
Q2. Compare the bond orders of CN and CN–. Which is more stable?
Answer:
- Bond order of CN = $\dfrac{7-2}{2} = 2.5$
- Bond order of CN⁻ = $\dfrac{8-2}{2} = 3$
Thus, CN⁻ has higher bond order and is more stable.
Q3. Why is CO similar to N2 in bonding?
Answer:
Both CO and N₂ have 14 total electrons and similar MO configurations.
Bond order of each = 3 → strong triple bond and short bond length.
Multiple Choice Questions (MCQs)
Q1. The bond order of NO molecule is:
(a) 2
(b) 2.5
(c) 3
(d) 1.5
Answer: Correct Option (b)
Explanation:
Bond order = $(8-3)/2 = 2.5$
Q2. Which of the following species is paramagnetic?
(a) NO+
(b) CO
(c) CN–
(d) NO
Answer: Correct Option (d)
Explanation:
NO has 1 unpaired electron in $\pi^*$ orbital → paramagnetic.
Q3. Which heteronuclear molecule has the highest bond order?
(a) NO
(b) NO+
(c) CN
(d) CO
Answer: Correct Option (b) and (d)
Explanation:
Bond orders: NO = 2.5, NO+ = 3, CN = 2.5, CO = 3.
Highest = NO+ and CO (bond order = 3).
Assertion-Reason Type Questions
Q1.
Assertion (A): CN– is more stable than CN.
Reason (R): CN– has a higher bond order than CN.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, R is false.
(d) A is false, R is true.
Answer: Correct Option (a)
Explanation:
Bond order CN = 2.5, CN– = 3 → higher stability of CN–.
Q2.
Assertion (A): CO molecule is diamagnetic.
Reason (R): All electrons in CO are paired in its MO configuration.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, R is false.
(d) A is false, R is true.
Answer: Correct Option (a)
Explanation:
CO has 14 electrons, all paired in MOs → diamagnetic.
Case Study Question
Case Study:
Molecular orbital theory explains the bonding in heteronuclear diatomic molecules like NO, NO+, CN, CN–, and CO.
These species have similar MO filling but differ in total electron count, which affects their stability, magnetic properties, and bond order.
Questions:
- Calculate the bond order of NO, NO+, CN, CN–, and CO.
- Which of these species are paramagnetic?
- Arrange the species in increasing order of bond strength.
- Why is CO more stable than NO?
Answers:
- Bond orders:
- NO = 2.5
- NO+ = 3
- CN = 2.5
- CN– = 3
- CO = 3
- Paramagnetic species → NO, CN (due to unpaired electron).
- Increasing bond strength:
$$NO = CN < NO^+ = CN^- = CO$$ - CO is more stable than NO because it has a full pairing of electrons (diamagnetic) and a bond order of 3.
