MO diagram of NO: why NO+ has a shorter bond and a higher stretching frequency
By Vimal Swarnkar (VKC) · UPSC Chemistry · Civil Services, Combined Geo-Scientist and IFoS
Take one electron away from a molecule and its bond gets stronger. It sounds wrong until you draw the MO diagram and see where that electron was sitting. In this post I will show you how often UPSC asks the NO family, which book to use, how I plan the answer, and how the answer should look in your booklet.
Draw the molecular orbital (MO) diagram of NO molecule. The experimental bond dissociation energy of NO is 626 kJ mol−1 while that of NO+ is 1047 kJ mol−1: rationalize it. NO can also act as a reactive radical: how?
NO has 11 valence electrons: (σ2s)2(σ*2s)2(σ2p)2(π2p)4(π*2p)1. Bond order = ½(8 − 3) = 2.5, with one unpaired electron, so NO is paramagnetic and a radical. Ionisation removes the antibonding π* electron, so NO+ has bond order 3, is diamagnetic and isoelectronic with N2 and CO. A higher bond order means a stronger, shorter, stiffer bond: D = 626 → 1047 kJ mol−1, r = 115 → 106 pm, ν̃ = 1876 → about 2344 cm−1. Adding an electron instead gives NO− (bond order 2, about 126 pm).
Watch the video
Day 05 of my Daily Answer Writing series on YouTube: I draw the MO diagram and write this 20-mark answer the way it should go into your booklet.
A word from me
Students usually remember the bond order of NO, 2.5, but lose marks on the diagram itself. They draw N and O at the same height, as if NO were N2, or they forget to say which orbital the electron leaves from. The examiner reads the diagram first. If the oxygen orbitals sit lower, the levels are labelled and the single π* electron is drawn clearly, half the marks are already in the bag before you write a line.
So practise the diagram, not the formula. Draw the N2/CO-type diagram once from Book 1 below, then redo it for NO, NO+, NO−, O2+, O2, O2− and O22−, just moving electrons in and out of π*. Seven species, one picture. If something does not add up, write to me on Telegram or at [email protected].
100 marks in five years
From 2021 to 2025, MO theory of diatomic molecules came 9 times, for 100 marks, in Civil Services Paper I (2022, 2024 twice and 2025) and IFoS Paper I (2022, 2023 twice and 2025 twice). NO and NO+ were asked 2 times directly: Civil Services 2025 Q4(a) for 20 marks and IFoS 2023 Q2(b) for 15 marks. The other 7 times it came as the same method on a different molecule: CO (Civil Services 2024 Q2(b)(ii)), O2 and its ions (Civil Services 2024 Q2(b)(i), IFoS 2025 Q2(b)(ii)), O2 and CO together (IFoS 2023 Q2(c)), B2 (IFoS 2025 Q2(c)(ii)), orbital symmetry and overlap (Civil Services 2022 Q1(b)) and BeH2 (IFoS 2022 Q3(a)). The Geo-Scientist papers did not ask it in these years, but they touch NO in spectroscopy (Paper II 2023 Q8, rotational lines of NO), so the diagram is worth knowing there too.
Syllabus and books
In Civil Services it is Paper I, Unit 2, Chemical bonding. IFoS asks it in Paper I. The Civil Services syllabus reads:
“… molecular orbital theory (LCAO method); bonding in H2+, H2, He2+ to Ne2, NO, CO, HF and CN−; comparison of valence bond and molecular orbital theories, bond order, bond strength and bond length.”
The syllabus names NO itself, and the last phrase, bond order, bond strength and bond length, is exactly what the 626 against 1047 kJ mol−1 part tests.
You do not need all of these books. Any one of them is enough for this question.
- Book 1: Shriver and Atkins’ Inorganic Chemistry (5th edition), Chapter 2, Section 2.9 “Heteronuclear diatomic molecules” and Section 2.10 “Bond properties”, with Example 2.6 (bond orders of O2, O2−, O22−), Example 2.7 (bond order against bond length) and the graph of bond enthalpy against bond order. If you have this book, this is the one I would use.
- Book 2: Miessler, Fischer and Tarr, Inorganic Chemistry (5th edition), Section 5.3 “Heteronuclear Diatomic Molecules” for the CO-type diagram, and Table 8.10, which gives N–O = 115 pm in NO (“bond order approximately 2.5; paramagnetic”) and 106 pm in NO+ (“isoelectronic with CO”).
- Book 3: J. D. Lee, Concise Inorganic Chemistry, the chapter on the covalent bond, MO treatment of NO. Good if you want the simpler drawing.
How to build the answer
Read the tree from left to right. Every leaf is a point to write, in that order. The highlighted words are the ones the examiner looks for.
Given information
- N 2s22p3 + O 2s22p4 = 11 valence e⁻
Assumptions
- LCAO: only valence AOs combine
- O orbitals lower (more electronegative)
- Same symmetry, similar energy
- Aufbau, Pauli, Hund
Key law
- B.O. = ½(Nb − Na)
Diagram
- labelled MO diagram, 1 e⁻ in π*
Calculation
- NO: ½(8 − 3) = 2.5
- NO+: ½(8 − 2) = 3
- NO−: ½(8 − 4) = 2
Values
- r 115 → 106 pm; ν̃ 1876 → ~2344 cm−1
- cycle: 626 + 1314 − 894 = 1046
Check
- NO+ isoelectronic with N2, CO
Reading the result
- electron lost from an antibonding MO
- odd e⁻: paramagnetic, radical
Principle
- higher B.O. → shorter, stronger, stiffer bond
Application
- NO as signalling molecule; NO+ salts; linear vs bent nitrosyls
Final answer
- B.O. 2.5 vs 3, double underlined
Two pictures that earn marks
What gains marks
- The electron count in one line: 5 + 6 = 11 valence electrons.
- A labelled diagram with the O orbitals lower than the N orbitals, and the single electron drawn in π*.
- The configuration and B.O. = ½(Nb − Na) worked for NO (2.5) and NO+ (3).
- The words antibonding electron removed; NO+ isoelectronic with N2 and CO.
- The energy check: D(NO+) = D(NO) + IE(O) − IE(NO) = 626 + 1314 − 894 = 1046 kJ mol−1.
- Bond length and IR: 115 → 106 pm, ν̃ 1876 → about 2344 cm−1, with k ∝ ν̃2.
- Why NO is a radical, and one application: signalling molecule, or linear versus bent nitrosyl complexes.
Match the length to the marks
I always tell my students to let the marks decide the length. For 5 marks, I would draw the diagram and write the two bond orders and one line on paramagnetism, about half a page. For 10 marks, as in IFoS 2023 Q2(c) on O2 and CO, add the numbered assumptions, the configurations and a small table, the two pages below. For 15 to 20 marks, as in IFoS 2023 Q2(b) and Civil Services 2025 Q4(a), add the energy cycle, the bond lengths and IR data, and the radical chemistry of NO.
How it looks in the booklet
This is the full answer in the order of the tree above. Page 1 builds the diagram and the bond orders, and page 2 explains the energies, the radical and the conclusion. For the full 20 marks of Civil Services 2025, I would add one more half page on the radical reactions.
Q.)Draw the MO diagram of NO. D(NO) = 626 kJ mol−1 while D(NO+) = 1047 kJ mol−1: rationalize. How can NO act as a reactive radical? [CSE 2025], 20M
Ans.Given: N (Z = 7): 2s22p3; O (Z = 8): 2s22p4. So NO has 5 + 6 = 11 valence electrons.
For a heteronuclear diatomic molecule, assume that
- i) MOs form by LCAO of the valence 2s and 2p orbitals; the 1s cores do not take part.
- ii) O is more electronegative, so its AOs lie lower; bonding MOs have more O character, antibonding MOs more N character.
- iii) only AOs of the same symmetry about the bond axis and similar energy combine.
- iv) electrons fill by aufbau, Pauli and Hund’s rule.
So, we get the configuration
So, putting the counts of (i) in (ii), we get
| Species | Valence e⁻ | π* e⁻ | B.O. | Unpaired | Nature |
|---|---|---|---|---|---|
| NO+ | 10 | 0 | ½(8 − 2) = 3 | 0 | dia |
| NO | 11 | 1 | ½(8 − 3) = 2.5 | 1 | para |
| NO− | 12 | 2 | ½(8 − 4) = 2 | 2 | para |
All other symbols have their usual meaning.
Why D(NO+) > D(NO): ionisation (IE = 894 kJ mol−1, 9.26 eV) removes the electron from π*2p, an antibonding MO. Losing an antibonding electron raises the B.O. from 2.5 to 3, so the bond becomes stronger.
Check by an energy cycle: D(NO+) = D(NO) + IE(O) − IE(NO)
which matches the experimental 1047 ✓.
Bond length and IR: r(N–O) falls from 115 pm (NO) to 106 pm (NO+); ν̃(N–O) rises from 1876 to about 2344 cm−1. Since k ∝ ν̃2, the force constant rises about 1.56 times (≈ 1550 → 2420 N m−1).
NO as a radical: the single π* electron makes NO paramagnetic and reactive: 2NO + O2 → 2NO2; 2NO + Cl2 → 2NOCl; NO loses this electron easily to give NO+ (in NO+BF4−, NO+HSO4−).
Reading the result:
- i) NO+ has 10 valence electrons, isoelectronic with N2 and CO: a triple bond, diamagnetic.
- ii) NO has one electron more, in an antibonding MO, so its bond is weaker and longer.
- iii) NO− adds a second π* electron: B.O. 2, bond longer still (about 126 pm).
Principle: bond order is the deciding number. A higher B.O. gives a shorter, stronger and stiffer bond; electrons removed from antibonding MOs strengthen the bond.
Application: NO is a biological signalling molecule (vasodilation); as a ligand, linear M–NO behaves like NO+ and bent M–NO like NO−.
So, NO: B.O. 2.5, one unpaired e⁻ (radical); NO+: B.O. 3, so D 1047 > 626 kJ mol−1 and r 106 < 115 pm
The same question in other exams
IFoS 2023, Paper I, Q2(b), 15 marks: MO diagram of NO; magnetic behaviour and bond order; stability of NO+. The same diagram: B.O. 2.5, paramagnetic; NO+ B.O. 3, diamagnetic and more stable.
Civil Services 2024, Paper I, Q2(b)(ii), 10 marks: MO diagram of CO. Ten valence electrons, B.O. = ½(8 − 2) = 3, diamagnetic. The HOMO is a σ lone pair on carbon, which is why CO binds metals through C.
Civil Services 2024, Paper I, Q2(b)(i), 10 marks: bond orders of O2, O2−, O22−, O2+: 2, 1.5, 1, 2.5; most stable O2+. Same rule: each π* electron changes the B.O. by ½.
IFoS 2023, Paper I, Q2(c), 10 marks: bond orders of O2 and CO: 2 (paramagnetic) and 3 (diamagnetic).
IFoS 2025, Paper I, Q2(c)(ii), 5 marks: B2 has B.O. 1 and is paramagnetic: s–p mixing puts π2p below σ2p, so the last two electrons go singly into π2p.
Every version is the same three steps: count the valence electrons, fill the right MO diagram (s–p mixing for B2, C2, N2, CO, NO+), then use B.O. = ½(Nb − Na) and count the unpaired electrons.
Questions students ask
Why is the bond in NO+ stronger than in NO?
NO has 11 valence electrons, and the last one sits in an antibonding π*2p orbital, so its bond order is ½(8 − 3) = 2.5. Ionisation removes exactly this antibonding electron, so NO+ has bond order ½(8 − 2) = 3. The bond dissociation energy rises from 626 to 1047 kJ/mol and the bond shortens from 115 to 106 pm. This was asked in UPSC Civil Services 2025, Chemistry Paper I, Q4(a), for 20 marks.
What is the bond order of NO, NO+ and NO−?
NO+ has 10 valence electrons and bond order 3; NO has 11 and bond order 2.5; NO− has 12 and bond order 2. Each electron added to the π* level lowers the bond order by one half.
Why is NO paramagnetic and a free radical?
With 11 valence electrons, NO has one unpaired electron in the π*2p orbital. An unpaired electron makes a molecule paramagnetic and lets it react like a radical: NO combines rapidly with O2 to give NO2, and it gives up the electron easily to form the nitrosonium ion NO+.
Why does NO+ have a higher IR stretching frequency than NO?
A higher bond order means a stiffer bond, with a larger force constant k. Since the wavenumber is proportional to the square root of k divided by the reduced mass, and the reduced mass is the same, ν(N–O) rises from 1876 cm−1 in NO to about 2344 cm−1 in NO+.
Which species is isoelectronic with NO+?
N2 and CO. All three have 10 valence electrons, bond order 3 and no unpaired electron. NO+ also behaves like CO as a ligand: linear metal nitrosyls are counted as NO+.
How often does UPSC ask MO diagrams and bond order?
From 2021 to 2025 MO theory of diatomics came 9 times, for 100 marks, in Civil Services Paper I and IFoS Paper I. NO and NO+ were asked directly in Civil Services 2025 (20 marks) and IFoS 2023 (15 marks); CO and O2 species were asked in Civil Services 2024 and IFoS 2023 and 2025.
Read the complete answers
By VKC, your mentor. Integrated M.Sc. Chemistry, IIT Roorkee. CSIR NET Chemical Sciences, All India Rank 4. New posts are announced first on Telegram.