Spectroscopy MCQs – 40 Practice Questions with Answers and Explanations

Master the analytical techniques that appear in every modern chemistry exam: chromatography, mass spectrometry, infrared spectroscopy and NMR. These 40 original practice questions are free to use, and every answer comes with a clear explanation, including worked Rf and relative atomic mass calculations. Ideal for IGCSE students meeting chromatography for the first time and for A-Level, IB and AP students tackling full spectroscopic analysis.

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Section 1: Separation and Chromatography

Q1. In a paper chromatography experiment, a spot of dye moves 4.2 cm from the start line while the solvent front moves 7.0 cm. What is the Rf value of the dye?

  • (A) 0.42
  • (B) 0.60
  • (C) 1.67
  • (D) 2.80
Show Answer

Answer: B – Rf = distance moved by spot / distance moved by solvent front = 4.2 / 7.0 = 0.60. Rf is always less than 1.

Q2. A student spots a mixture onto chromatography paper so that the spots sit below the level of the solvent in the tank. Why will the experiment fail?

  • (A) The mixture dissolves directly into the solvent instead of travelling up the paper
  • (B) The solvent front moves too quickly for separation to occur
  • (C) The paper absorbs the solvent and stops it rising
  • (D) The spots become too concentrated to see
Show Answer

Answer: A – If the spots are below the solvent level, the substances dissolve straight into the solvent in the tank and are washed off the paper, so no separation happens. Spots must start above the solvent level.

Q3. Amino acids are colourless. Which locating agent is sprayed onto a chromatogram to make amino acid spots visible as purple marks?

  • (A) Iodine vapour
  • (B) Universal indicator
  • (C) Ninhydrin
  • (D) Bromine water
Show Answer

Answer: C – Ninhydrin reacts with amino acids on gentle heating to give purple spots. Locating agents are needed whenever the substances being separated are colourless.

Q4. A food colouring is tested by paper chromatography in two different solvents and gives just one spot in each. What does this suggest?

  • (A) The colouring is a mixture of two dyes
  • (B) The colouring contains a locating agent
  • (C) The colouring is insoluble in both solvents
  • (D) The colouring is a pure substance
Show Answer

Answer: D – A pure substance produces a single spot. Checking in a second solvent strengthens the conclusion, because two different substances might share the same Rf in one solvent by coincidence.

Q5. Which statement about Rf values is correct?

  • (A) A given compound has the same Rf value in the same solvent and stationary phase under the same conditions
  • (B) Rf values are always greater than 1
  • (C) Rf depends on how much of the substance is spotted onto the paper
  • (D) Every compound has a unique Rf value in every solvent
Show Answer

Answer: A – Rf is a ratio of distances, so it does not depend on the amount spotted. It is only constant when the solvent, stationary phase and temperature are kept the same.

Q6. In thin-layer chromatography (TLC), what is the stationary phase?

  • (A) The solvent moving up the plate
  • (B) The glass or plastic backing sheet
  • (C) A thin layer of silica gel or alumina coated on the plate
  • (D) The locating agent sprayed on afterwards
Show Answer

Answer: C – In TLC the stationary phase is a thin layer of an adsorbent such as silica gel on a plate, and the solvent (mobile phase) moves over it. TLC is usually faster than paper chromatography and needs only tiny samples.

Q7. What is the purpose of two-way paper chromatography?

  • (A) To separate spots that overlap in the first solvent by turning the paper 90 degrees and running it in a second solvent
  • (B) To dry the paper more quickly between runs
  • (C) To double the Rf value of every spot
  • (D) To test the same mixture at two different temperatures
Show Answer

Answer: A – Some mixtures, such as the amino acids from a protein hydrolysate, cannot be fully separated in one solvent. Running the chromatogram again at right angles in a different solvent spreads the spots out in two dimensions.

Q8. A chromatogram contains colourless spots that fluoresce. Which method would reveal them without spraying any chemical?

  • (A) Heating the paper strongly
  • (B) Viewing the plate under ultraviolet (UV) light
  • (C) Adding more solvent to the tank
  • (D) Leaving the paper in sunlight for a day
Show Answer

Answer: B – Many colourless compounds absorb UV light or fluoresce under it, so the spots show up as dark or glowing patches on a UV-treated plate. This is a common non-destructive locating method.

Q9. A pigment moves 3.6 cm up a TLC plate while the solvent front reaches 9.0 cm. What is its Rf value?

  • (A) 0.25
  • (B) 2.50
  • (C) 0.40
  • (D) 5.40
Show Answer

Answer: C – Rf = 3.6 / 9.0 = 0.40. Remember to divide the spot distance by the solvent-front distance, not the other way round.

Q10. Why is the start line on chromatography paper drawn in pencil rather than pen?

  • (A) Pencil lines are easier to erase afterwards
  • (B) Pen ink would dissolve in the solvent and move up the paper, contaminating the chromatogram
  • (C) Pencil graphite catalyses the separation
  • (D) Pen ink reacts with the locating agent
Show Answer

Answer: B – Pen ink is itself a mixture of dyes that would dissolve and travel with the solvent, giving extra spots. Graphite pencil does not dissolve, so the line stays in place.

Section 2: Mass Spectrometry

Q11. What is the correct order of the stages inside a mass spectrometer?

  • (A) Vaporisation, acceleration, ionisation, detection, deflection
  • (B) Ionisation, vaporisation, detection, acceleration, deflection
  • (C) Acceleration, ionisation, vaporisation, deflection, detection
  • (D) Vaporisation, ionisation, acceleration, deflection, detection
Show Answer

Answer: D – The sample is first vaporised, then ionised (usually by electron impact), then the ions are accelerated, deflected by a magnetic field according to their m/z value, and finally detected.

Q12. In the mass spectrum of an organic compound, the molecular ion peak (M+) is useful because it gives the:

  • (A) Number of carbon atoms in the molecule
  • (B) Boiling point of the compound
  • (C) Relative molecular mass (Mr) of the compound
  • (D) Number of hydrogen atoms in the molecule
Show Answer

Answer: C – The molecular ion is the whole molecule with one electron removed, so its m/z value equals the Mr of the compound. It is usually the peak with the largest m/z value.

Q13. What is the base peak in a mass spectrum?

  • (A) The peak with the largest m/z value
  • (B) The smallest peak that can be detected
  • (C) The peak produced by the molecular ion
  • (D) The tallest peak, assigned a relative abundance of 100
Show Answer

Answer: D – The base peak is the most abundant ion (often a stable fragment). All other peak heights are measured relative to it as percentages.

Q14. A compound containing a CH3CO group shows a strong fragment peak at m/z = 43. Which ion is most likely responsible?

  • (A) CH3+
  • (B) C2H5+
  • (C) CO2+
  • (D) CH3CO+
Show Answer

Answer: D – CH3CO+ has mass 12 + 3 + 12 + 16 = 43. (The C3H7+ fragment also has m/z 43, which is why fragment peaks must be interpreted alongside other evidence.)

Q15. A compound contains one chlorine atom. What pattern of peaks is expected for its M and M+2 peaks, given that 35Cl and 37Cl occur in a ratio of about 3:1?

  • (A) M : M+2 in the ratio 3 : 1
  • (B) M : M+2 in the ratio 1 : 1
  • (C) M : M+2 in the ratio 1 : 3
  • (D) Only one peak, because chlorine has one isotope
Show Answer

Answer: A – Molecules containing 35Cl form the M peak and those containing 37Cl form the M+2 peak, two mass units higher. Their heights follow the 3:1 isotope ratio.

Q16. A compound contains one bromine atom. Bromine has two isotopes, 79Br and 81Br, in almost equal abundance. What M : M+2 pattern is expected?

  • (A) 3 : 1
  • (B) 1 : 1
  • (C) 9 : 6 : 1
  • (D) 2 : 1
Show Answer

Answer: B – Because the two bromine isotopes are roughly 50:50, the M and M+2 peaks are of almost equal height. This 1:1 pair is a classic fingerprint for one bromine atom.

Q17. An element X has two isotopes: 35X with 75% abundance and 37X with 25% abundance. What is the relative atomic mass (Ar) of X?

  • (A) 36.0
  • (B) 35.0
  • (C) 35.5
  • (D) 36.5
Show Answer

Answer: C – Ar = (35 × 75 + 37 × 25) / 100 = (2625 + 925) / 100 = 3550 / 100 = 35.5. This is exactly how the Ar of chlorine arises.

Q18. Chlorine gas, Cl2, is analysed in a mass spectrometer. What is the expected ratio of the peaks at M (70), M+2 (72) and M+4 (74)?

  • (A) 3 : 1 : 0
  • (B) 1 : 1 : 1
  • (C) 6 : 3 : 1
  • (D) 9 : 6 : 1
Show Answer

Answer: D – With two chlorine atoms, the combinations are 35Cl35Cl (9 parts), 35Cl37Cl (6 parts) and 37Cl37Cl (1 part), giving a 9:6:1 pattern.

Q19. In the deflection stage of a mass spectrometer, which ions are deflected the most by the magnetic field?

  • (A) Ions with the largest m/z value
  • (B) Ions with the smallest m/z value
  • (C) Ions with the greatest abundance
  • (D) All ions are deflected equally
Show Answer

Answer: B – Deflection depends on the mass-to-charge ratio. Light ions (small m/z) are deflected most, which is how the instrument separates ions of different masses.

Q20. High-resolution mass spectrometry can distinguish between C2H4, N2 and CO, even though all three have a nominal mass of 28, because:

  • (A) Their exact masses differ very slightly, and the instrument measures mass to several decimal places
  • (B) They produce ions with different charges
  • (C) They have different boiling points
  • (D) They absorb infrared radiation differently
Show Answer

Answer: A – Exact masses are not whole numbers (for example CO is 27.9949 and N2 is 28.0061). A high-resolution instrument can separate these tiny differences and identify the molecular formula.


Section 3: Infrared Spectroscopy

Q21. A carbonyl group, C=O, in aldehydes, ketones and carboxylic acids absorbs infrared radiation in which wavenumber range?

  • (A) 3200-3600 cm-1
  • (B) 1000-1200 cm-1
  • (C) 1680-1750 cm-1
  • (D) 2500-3000 cm-1
Show Answer

Answer: C – The C=O stretch gives a strong, sharp absorption at about 1680-1750 cm-1. Its presence is one of the most useful clues in an IR spectrum.

Q22. Which absorption is characteristic of the O-H bond in an alcohol?

  • (A) A broad absorption at 3200-3600 cm-1
  • (B) A sharp absorption at 1720 cm-1
  • (C) A medium absorption at 1640 cm-1
  • (D) No absorption, because O-H does not absorb infrared
Show Answer

Answer: A – Hydrogen bonding makes the alcohol O-H absorption broad, typically between 3200 and 3600 cm-1. A sharp peak near 1720 cm-1 would indicate C=O instead.

Q23. An unknown compound shows a very broad absorption from 2500 to 3000 cm-1 and a strong peak at 1715 cm-1. Which functional group is present?

  • (A) Alcohol only
  • (B) Alkene only
  • (C) Amine
  • (D) Carboxylic acid
Show Answer

Answer: D – The very broad 2500-3000 cm-1 absorption is the O-H of a carboxylic acid, and the 1715 cm-1 peak is its C=O. The combination of both peaks is diagnostic of a -COOH group.

Q24. What is the main use of the fingerprint region of an IR spectrum (below about 1500 cm-1)?

  • (A) It shows the exact mass of the molecule
  • (B) It gives a pattern unique to one compound, which can be matched against a database to identify it
  • (C) It counts the number of hydrogen atoms present
  • (D) It measures the concentration of the sample
Show Answer

Answer: B – The fingerprint region contains many overlapping absorptions from complex bending vibrations. The overall pattern is unique to a particular compound, like a fingerprint.

Q25. Which absorption would confirm that a hydrocarbon contains a C=C double bond?

  • (A) A strong peak at 1720 cm-1
  • (B) A medium absorption at about 1620-1680 cm-1
  • (C) A broad peak at 3300 cm-1
  • (D) A peak at 1050 cm-1
Show Answer

Answer: B – The C=C stretch absorbs at roughly 1620-1680 cm-1. A peak at 1720 cm-1 would suggest a carbonyl group instead.

Q26. A compound shows a broad absorption at 3300 cm-1 but no absorption between 1680 and 1750 cm-1. The compound is most likely:

  • (A) An alcohol
  • (B) A carboxylic acid
  • (C) A ketone
  • (D) An ester
Show Answer

Answer: A – The broad 3300 cm-1 peak shows an O-H group, but the absence of any C=O peak rules out acids, ketones and esters, leaving an alcohol.

Q27. The N-H bond in amines and amides absorbs in which region?

  • (A) 1680-1750 cm-1
  • (B) 2500-3000 cm-1
  • (C) 3300-3500 cm-1
  • (D) 900-1100 cm-1
Show Answer

Answer: C – N-H stretching absorptions appear around 3300-3500 cm-1, in a similar region to O-H but usually sharper and less broad.

Q28. Carbon dioxide is a greenhouse gas partly because its molecules:

  • (A) Absorb infrared radiation, causing their bonds to vibrate more strongly
  • (B) Reflect ultraviolet radiation back into space
  • (C) React with water vapour in the atmosphere
  • (D) Absorb visible light and appear coloured
Show Answer

Answer: A – Greenhouse gases such as CO2, methane and water vapour absorb infrared radiation radiated from the Earth, increasing molecular vibration and warming the atmosphere.

Q29. Some modern breathalysers use infrared spectroscopy to test drivers. They work by measuring:

  • (A) The colour change of acidified dichromate
  • (B) The mass spectrum of the breath sample
  • (C) The electrical conductivity of the breath
  • (D) How much infrared radiation is absorbed by the ethanol in the breath
Show Answer

Answer: D – The instrument measures absorption at wavenumbers characteristic of ethanol (for example its C-H and O-H bonds). The more ethanol present, the more IR radiation is absorbed.

Q30. A liquid hydrocarbon shows absorptions at 2850-2960 cm-1 (C-H) but no peaks for C=C, C=O or O-H. The compound is most likely:

  • (A) An alkene
  • (B) An alcohol
  • (C) An alkane
  • (D) A carboxylic acid
Show Answer

Answer: C – Only C-H absorptions are present, with no double bond, carbonyl or hydroxyl peaks. The substance must be a saturated alkane.

Section 4: NMR Spectroscopy

Q31. Tetramethylsilane (TMS) is used as the reference standard in NMR spectroscopy. Its signal is assigned a chemical shift, δ, of:

  • (A) 0 ppm
  • (B) 1 ppm
  • (C) 7 ppm
  • (D) 100 ppm
Show Answer

Answer: A – TMS gives a single sharp peak (all 12 hydrogens are equivalent) and is assigned δ = 0. Chemical shifts of all other signals are measured relative to it.

Q32. How many different hydrogen environments are there in ethanol, CH3CH2OH?

  • (A) 1
  • (B) 3
  • (C) 2
  • (D) 6
Show Answer

Answer: B – The CH3 hydrogens, the CH2 hydrogens and the O-H hydrogen are in three different environments, so ethanol shows three 1H NMR signals.

Q33. How many signals appear in the 1H NMR spectrum of propanone, CH3COCH3?

  • (A) 3
  • (B) 2
  • (C) 1
  • (D) 6
Show Answer

Answer: C – The two CH3 groups are in identical environments, so all six hydrogens give just one signal.

Q34. How many different hydrogen environments are there in ethanoic acid, CH3COOH?

  • (A) 1
  • (B) 3
  • (C) 4
  • (D) 2
Show Answer

Answer: D – The CH3 hydrogens form one environment and the -COOH hydrogen is in a second, very different environment (far downfield, near δ 11).

Q35. In the 1H NMR spectrum of ethanol, the CH3 signal is split into a triplet. Using the n+1 rule, this is because the CH3 group is next to:

  • (A) A CH2 group containing 2 hydrogens on the adjacent carbon
  • (B) The O-H hydrogen
  • (C) Three equivalent hydrogens on the same carbon
  • (D) No hydrogens at all
Show Answer

Answer: A – Splitting follows the n+1 rule, where n is the number of hydrogens on the neighbouring carbon. With 2 neighbouring hydrogens, the signal splits into 2+1 = 3 peaks, a triplet.

Q36. An ethyl group, -CH2CH3, in a molecule typically gives which splitting pattern?

  • (A) A singlet and a doublet
  • (B) A quartet (CH2) and a triplet (CH3) with an integration ratio of 2:3
  • (C) Two singlets of equal height
  • (D) A doublet and a doublet of equal size
Show Answer

Answer: B – The CH2 sees 3 neighbouring H (quartet) and the CH3 sees 2 neighbouring H (triplet). This quartet-triplet combination is the classic signature of an ethyl group.

Q37. A 1H NMR spectrum has two signals with an integration ratio of 3:2. What does the integration ratio tell you?

  • (A) The ratio of chemical shifts of the two signals
  • (B) The ratio of the masses of the two fragments
  • (C) The relative numbers of hydrogens in each environment, here 3 hydrogens compared with 2
  • (D) The number of neighbouring hydrogens causing splitting
Show Answer

Answer: C – The area under each signal (its integration) is proportional to the number of hydrogens in that environment. A 3:2 ratio often means a CH3 group and a CH2 group.

Q38. The hydrogen of an aldehyde group, -CHO, produces a 1H NMR signal at approximately:

  • (A) δ 1 ppm
  • (B) δ 3.5 ppm
  • (C) δ 5 ppm
  • (D) δ 9-10 ppm
Show Answer

Answer: D – Aldehyde protons are strongly deshielded by the carbonyl group and appear far downfield at about δ 9-10, a very distinctive position. Carboxylic acid protons appear even further, near δ 11.

Q39. An unknown compound shows a 1H NMR peak that disappears when the sample is shaken with D2O. The peak was due to:

  • (A) An O-H (or N-H) hydrogen, which exchanges with deuterium
  • (B) A CH3 group
  • (C) The TMS standard dissolving
  • (D) A carbon-carbon double bond
Show Answer

Answer: A – Hydrogens attached to oxygen or nitrogen exchange rapidly with D2O and are replaced by deuterium, which does not show in the 1H spectrum. This D2O shake test identifies -OH and -NH peaks.

Q40. Propan-1-ol and propan-2-ol are isomers. How can they be distinguished using 13C NMR?

  • (A) They cannot be distinguished by 13C NMR
  • (B) Propan-2-ol shows 2 peaks (its two CH3 carbons are equivalent) while propan-1-ol shows 3 peaks
  • (C) Propan-1-ol shows 1 peak and propan-2-ol shows 3 peaks
  • (D) Both show exactly 2 peaks
Show Answer

Answer: B – In propan-2-ol the two CH3 groups are in identical carbon environments, giving 2 13C signals. In propan-1-ol all three carbons are different, giving 3 signals.

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