Microwave spectrum of CO: bond length from the line spacing, and what 13C does to it
By Vimal Swarnkar (VKC) · UPSC Chemistry · Civil Services, Combined Geo-Scientist and IFoS
One number, the gap between two lines, gives you the length of a chemical bond to a tenth of a picometre. It is the most reliable 10-mark numerical in UPSC spectroscopy, and the examiner keeps coming back to the same molecule, CO. In this post I will show you how often UPSC asks it, which book to use, how I plan the answer, and how it should look in your booklet.
The pure rotational spectrum of 12C16O shows lines spaced 3.8424 cm−1 apart.
(a) Calculate the rotational constant B, the moment of inertia and the C–O bond length.
(b) Predict the wavenumber of the J = 3 ← 2 line, the line spacing for 13C16O (same bond length), and the J value of the most populated rotational level at 300 K.
(c) Why do H2 and CO2 show no pure rotational spectrum?
(12C = 12.0000, 13C = 13.0034, 16O = 15.9949 u; 1 u = 1.6605 × 10−27 kg; h = 6.626 × 10−34 J s; c = 2.998 × 1010 cm s−1; k = 1.381 × 10−23 J K−1)
For a rigid rotor εJ = BJ(J + 1) and absorption lines fall at 2B(J + 1), so the spacing is 2B. (a) B = 3.8424/2 = 1.9212 cm−1; μ = 6.8562 u = 1.1385 × 10−26 kg; I = h/8π²cB = 1.457 × 10−46 kg m²; r = √(I/μ) = 113.1 pm. (b) J = 3 ← 2 at 6B = 11.53 cm−1. 13C16O has the same r but μ′ = 7.1724 u, so B′ = 1.8365 cm−1 and the spacing falls to 3.673 cm−1. Jmax = √(kT/2hcB) − ½ ≈ 7 at 300 K. (c) A pure rotational spectrum needs a permanent dipole moment; H2 and CO2 have none.
Watch the video
Day 10 of my Daily Answer Writing series on YouTube: I draw the level ladder and solve all three parts the way they should go into your booklet.
Video coming soon. It goes up on the VKC YouTube channel and is announced first on Telegram.
A word from me
This is the numerical I most want my students to get full marks on, because nothing in it is hard. The marks are lost on units. Students take c as 3 × 108 while B is in cm−1, or they put μ in atomic mass units into I = μr2, and a correct method ends at r = 3.6 pm or 3.6 m. The examiner sees the wrong answer first.
So write one line before you calculate: B in cm−1, c in cm s−1, μ in kg, I in kg m2. Then check your answer against what you know: a bond between two second-row atoms is 100 to 150 pm. And remember the one-line reason for every isotope question: the bond length does not change, only the mass. If something does not add up, write to me on Telegram or at [email protected].
105 marks in five years
From 2021 to 2025, rotational (microwave) spectroscopy came 15 times, for 105 marks: 3 times in Civil Services Paper II (25 marks), 4 times in IFoS Paper II (25 marks) and 8 times in Geo-Scientist Mains Paper II (55 marks). CO itself was asked 5 times: Civil Services 2023 Q8(b)(i) (rotational constants of 12C16O and 13C16O, 10 marks), Civil Services 2025 Q5(d)(ii) (why 13C narrows the spacing, 5 marks), IFoS 2025 Q8(a)(i) (the same 3.8424 cm−1, B and bond length, 10 marks), IFoS 2022 Q5(d) (bond length from 3.84 cm−1, 5 marks) and Geo-Scientist 2025 Q1(n) (CO or CO2, which is active, 5 marks).
The other 10 times it was the same method on another molecule: HCl, NO, H2, a general rigid rotor, or “which of these molecules is microwave active”. It came again in 2026: Civil Services Paper II Q8(a)(i), and Geo-Scientist Paper II Q9 and Q15(a).
Syllabus and books
In Civil Services it is Paper II, Spectroscopy, and IFoS has the same line in Paper II. The Civil Services syllabus reads:
“Principle and applications in structure elucidation: (i) Rotational: diatomic molecules; isotopic substitution and rotational constants.”
The Geo-Scientist Mains Paper II syllabus is even more direct: “rigid rotor model, selection rules, spectrum, characteristic features of spectral lines. Determination of bond length, effect of isotopic substitution.” That is this question, part by part.
You do not need all of these books. Any one of them is enough for this question.
- Book 1: C. N. Banwell and E. M. McCash, Fundamentals of Molecular Spectroscopy (4th edition), Chapter 2 “Microwave spectroscopy”: the rigid diatomic molecule, intensities of spectral lines (with Jmax) and the effect of isotopic substitution, all worked on CO. If you have this book, this is the one I would use.
- Book 2: Atkins' Physical Chemistry, the chapter on rotational spectroscopy: rotational energy levels of linear rotors, selection rules and the appearance of the spectrum.
- Book 3: G. Aruldhas, Molecular Structure and Spectroscopy, the chapter on microwave spectroscopy. Many solved numericals of the UPSC type.
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
- spacing 3.8424 cm−1
- masses, h, c, k
Assumptions
- rigid rotor: r fixed
- isotopes: same r
- ΔJ = ±1, dipole needed
Key law
- εJ = BJ(J + 1)
- B = h/8π²cI; spacing 2B
Diagram
- level ladder + stick spectrum
(a)
- B = 1.9212 cm−1
- μ = 6.8562 u; I = 1.457 × 10−46 kg m²
- r = √(I/μ) = 113.1 pm
(b)
- 6B = 11.53 cm−1
- B ∝ 1/μ: 2B′ = 3.673 cm−1
- Jmax = √(kT/2hcB) − ½ ≈ 7
(c)
- permanent dipole needed
Principle
- spacing → B → I → r
- heavier isotope: lines move down
Application
- radio astronomy: CO maps gas clouds
- 13C/12C ratio
Final answer
- r = 113.1 pm; 3.673 cm−1
Pictures that earn marks
What gains marks
- Spacing = 2B, so B = 1.9212 cm−1; reduced mass converted to kg (2 marks).
- I = 1.457 × 10−46 kg m2 and r = 113.1 pm, with units throughout (3 marks).
- J = 3 ← 2 at 6B = 11.53 cm−1; 13CO spacing 3.673 cm−1, with the words same bond length (2 marks).
- Jmax ≈ 7 from NJ ∝ (2J + 1)e−hcBJ(J+1)/kT (1 mark).
- Gross selection rule: permanent dipole moment; H2 and CO2 inactive, studied by rotational Raman (2 marks).
Match the length to the marks
For 5 marks, as in IFoS 2022 Q5(d) or Civil Services 2025 Q5(d)(ii), write B and r, or the isotope reason, in about half a page. For 10 marks, as in Civil Services 2023 Q8(b)(i) and this question, add the level ladder, the isotope shift and the selection rule: the two pages below. For 15 marks, add the derivation of Jmax and a line on centrifugal distortion (εJ = BJ(J + 1) − DJ2(J + 1)2).
How it looks in the booklet
This is the full answer in the order of the tree above. Page 1 sets up the rigid rotor and finds B, I and r; page 2 does the isotope shift, the population maximum and the selection rule, and closes with the principle and one application.
Q.)12C16O lines spaced 3.8424 cm−1: (a) B, I, r; (b) J = 3 ← 2 line, spacing for 13C16O, Jmax at 300 K; (c) why no spectrum for H2, CO2? [Day 10], 10M
Ans.Treat CO as a rigid rotor (r fixed); selection rules: ΔJ = ±1 and a permanent dipole.
(a) B = 3.8424/2 = 1.9212 cm−1
μ = (12.0000 × 15.9949)/27.9949 = 6.8562 u = 1.1385 × 10−26 kg
I = h/(8π²cB) = 6.626 × 10−34/(8π² × 2.998 × 1010 × 1.9212) = 1.457 × 10−46 kg m²
(b) J = 3 ← 2 (J = 2 in (ii)): ν̃ = 2B × 3 = 6B = 11.53 cm−1
13C16O: μ′ = (13.0034 × 15.9949)/28.9983 = 7.1724 u. Isotopes change mass, not electrons, so r is the same and B ∝ 1/μ:
B′ = 1.9212 × 6.8562/7.1724 = 1.8365 cm−1
Population NJ ∝ (2J + 1)e−hcBJ(J+1)/kT; d/dJ = 0 gives Jmax = √(kT/2hcB) − ½ = 7.37 − 0.5 = 6.87 ≈ 7
(c) Gross selection rule: a permanent dipole moment. H2 and linear O=C=O have μ = 0, so they are microwave inactive (studied by rotational Raman).
Conclusion: line spacing gives B, then I and r; a heavier isotope raises μ, lowers B and moves every line down. Application: CO lines map gas clouds in radio astronomy; 13CO lines give the 13C/12C ratio.
So, B = 1.9212 cm−1, r(C–O) = 113.1 pm; 6B = 11.53 cm−1; 13CO spacing 3.673 cm−1; Jmax ≈ 7
The same question in other exams
Civil Services 2023, Paper II, Q8(b)(i), 10 marks: first lines of 12C16O and 13C16O at 3.84235 and 3.67337 cm−1. The first line (J = 1 ← 0) is at 2B, so B = 1.92118 and 1.83669 cm−1. Their ratio, 1.0460, equals μ′/μ = 1.0461: proof that the bond length is unchanged.
Civil Services 2025, Paper II, Q5(d), 10 marks: which of CH4, H2O, NH3, BCl3, XeF4 are rotationally active (only H2O and NH3, which have a dipole), and why 13C narrows the CO spacing (μ rises, B falls).
IFoS 2025, Paper II, Q8(a)(i), 10 marks: the first rotational line of CO is 3.8424 cm−1; B = 1.9212 cm−1, r ≈ 1.131 Å. Exactly part (a), in CGS units (h in erg s), so B comes out the same.
Geo-Scientist 2021, Paper II, Q15(a), 10 marks: the J = 3 → 4 line is at 0.50 cm−1, so 8B = 0.50 and the J = 6 → 7 line is at 14B = 0.875 cm−1. Part (b) of this question in reverse.
Geo-Scientist 2023, Paper II, Q8, 10 marks: from the rotational lines of 14N16O and 15N16O, the ratio of frequencies equals μ15/μ14, which gives the atomic mass of 15N ≈ 15.0 u. The isotope rule used backwards.
Every version is the same three lines: spacing (or first line) = 2B; B = h/8π²cI; I = μr2. For an isotope, keep r fixed and scale B by 1/μ. For “is it active?”, ask only whether the molecule has a permanent dipole.
Questions students ask
How do you find the bond length of CO from its microwave spectrum?
Lines of a rigid diatomic rotor are spaced 2B apart. For 12C16O the spacing is 3.8424 cm−1, so B = 1.9212 cm−1. Then I = h/(8π²cB) = 1.457 × 10−46 kg m², the reduced mass is 6.8562 u = 1.1385 × 10−26 kg, and r = √(I/μ) = 1.131 × 10−10 m, that is 113.1 pm.
Why does replacing 12C by 13C decrease the line spacing of CO?
Isotopic substitution changes the masses but not the electrons, so the bond length stays the same. The reduced mass rises from 6.8562 to 7.1724 u, the moment of inertia rises, and B = h/(8π²cI) falls in the same ratio: 1.9212 to 1.8365 cm−1, so the spacing falls from 3.842 to 3.673 cm−1. This was asked in UPSC Civil Services 2023 and 2025, Chemistry Paper II.
What is the most populated rotational level of CO at 300 K?
The population of level J is proportional to (2J + 1) exp(−hcBJ(J + 1)/kT). Setting its derivative to zero gives J(max) = √(kT/2hcB) − 1/2 = 7.37 − 0.5 ≈ 7 for CO at 300 K.
Why do H2 and CO2 not show a pure rotational spectrum?
The gross selection rule for a microwave (pure rotational) spectrum is that the molecule must have a permanent electric dipole moment. H2 is homonuclear and CO2 is linear and symmetric, so both have zero dipole moment and are microwave inactive. Their rotational constants are measured by rotational Raman spectroscopy instead.
What is the wavenumber of the J = 3 ← 2 line of CO?
Absorption lines fall at 2B(J + 1), where J is the lower level. For J = 3 ← 2, the line is at 2B × 3 = 6B = 6 × 1.9212 = 11.53 cm−1.
How often does UPSC ask rotational spectroscopy?
From 2021 to 2025, rotational (microwave) spectroscopy came 15 times for 105 marks: 3 times in Civil Services Paper II, 4 times in IFoS Paper II and 8 times in Geo-Scientist Mains Paper II. CO itself was asked 5 times, including Civil Services 2023 Q8(b)(i) and IFoS 2025 Q8(a)(i).
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.