These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. All compounds labelled with a letter are fictional, and their data are invented for practice. Approximate region boundaries for every question: microwave 1 mm to 1 m; infrared 700 nm to 1 mm; visible 400 nm to 700 nm; ultraviolet 10 nm to 400 nm. 1 mm = 10⁻³ m, 1 μm = 10⁻⁶ m, 1 nm = 10⁻⁹ m.
Question 1 (multiple choice · foundation)
Which list puts the regions in order of increasing energy per photon?
- (A) ultraviolet, visible, infrared, microwave
- (B) microwave, infrared, visible, ultraviolet
- (C) infrared, microwave, visible, ultraviolet
- (D) microwave, visible, infrared, ultraviolet
Answer and explanation
Answer: (B). Energy per photon rises as wavelength falls. Microwaves have the longest wavelength of the four and ultraviolet the shortest.
- (A) is the right sequence but in order of decreasing energy (increasing wavelength).
- (C) swaps microwave and infrared. Infrared photons carry more energy than microwave photons.
- (D) swaps infrared and visible. Infrared lies just beyond the red, low-energy end of the visible region.
Question 2 (multiple choice · foundation)
A gaseous molecule absorbs a photon of infrared radiation. Which statement best describes the change in the molecule?
- (A) An electron moves to a higher electronic energy level.
- (B) The molecule moves to a higher vibrational energy level.
- (C) A covalent bond breaks and the molecule splits into two fragments.
- (D) An electron is removed, leaving a positive ion.
Answer and explanation
Answer: (B). Infrared photons match the gaps between vibrational levels, so the bonds stretch and bend with more energy.
- (A) describes an electronic transition, which needs visible or ultraviolet photons.
- (C) needs photons carrying at least the bond energy (a few hundred kJ mol⁻¹ per mole of photons); infrared photons carry far less.
- (D) is ionization. It needs even more energy, such as the ultraviolet or X-ray photons used in photoelectron spectroscopy.
Question 3 (multiple choice · core)
A fictional compound absorbs radiation at the four wavelengths below. Which absorption corresponds to the largest energy gap in the molecule?
- (A) 3.0 mm
- (B) 9.5 μm
- (C) 420 nm
- (D) 2.6 μm
Answer and explanation
Answer: (C). In metres the wavelengths are 3.0 × 10⁻³, 9.5 × 10⁻⁶, 4.2 × 10⁻⁷ and 2.6 × 10⁻⁶ m. The shortest wavelength, 420 nm, has the most energetic photon, so it matches the largest gap. It is in the visible region, so it is an electronic transition.
- (A) has the smallest number but the longest wavelength: it is a microwave, rotational transition with the smallest gap. Choosing it means the units were not converted.
- (B) is infrared (vibrational), and it has a longer wavelength than (D), so it is the smaller of the two vibrational gaps.
- (D) is also infrared (vibrational). Its gap is larger than (B)’s but much smaller than an electronic gap.
Question 4 (data analysis · core)
A fictional gaseous compound, K, is made of polar molecules. Its absorption measurements are:
| Absorption | Wavelength |
|---|---|
| 1 | 1.6 mm |
| 2 | 3.4 μm |
| 3 | 9.1 μm |
| 4 | 210 nm |
(a) For each absorption, name the region of the spectrum and the type of transition. (b) Two absorptions involve the same type of transition. Identify them and suggest why the molecule shows two. (c) List the four absorptions in order of increasing energy gap.
Worked solution and suggested Marlbridge rubric
(a)
| Absorption | Wavelength in m | Region | Transition |
|---|---|---|---|
| 1 | 1.6 × 10⁻³ | microwave | rotational |
| 2 | 3.4 × 10⁻⁶ | infrared | vibrational |
| 3 | 9.1 × 10⁻⁶ | infrared | vibrational |
| 4 | 2.1 × 10⁻⁷ | ultraviolet | electronic |
(b) Absorptions 2 and 3 are both vibrational. A molecule with several bonds can vibrate in more than one way (different bonds stretching, or bond angles bending), and each kind of vibration has its own ladder of levels with its own gap.
(c) Longest wavelength = smallest gap: 1 (1.6 mm) < 3 (9.1 μm) < 2 (3.4 μm) < 4 (210 nm).
| Point | What earns it |
|---|---|
| 1 | Absorption 1 microwave/rotational and absorption 4 ultraviolet/electronic |
| 1 | Absorptions 2 and 3 both infrared/vibrational |
| 1 | A reason for two vibrational absorptions: different bonds or different kinds of vibration (stretch and bend) |
| 1 | Correct order 1 < 3 < 2 < 4 |
Accept “visible/UV” for absorption 4 only if “electronic” is stated. Do not award (c) if 2 and 3 are in the wrong order: 3.4 μm is the shorter wavelength, so it is the larger gap.
Question 5 (constructed response · core)
A solution of a fictional metal complex, M, looks blue in white light. Its visible spectrum has one strong absorption, centred at 610 nm (orange light). A solution of a fictional organic compound, P, is colourless, but its spectrum shows a strong absorption at 260 nm.
(a) Identify the type of transition responsible for the absorption by M, and explain why the solution looks blue. (b) Compound P has an electronic transition but no colour. Explain why. (c) Which compound has the larger gap between the electronic levels involved? Justify your answer.
Worked solution and suggested Marlbridge rubric
(a) 610 nm is in the visible region, so the absorption is an electronic transition: an electron in M moves to a higher energy level. White light contains all visible wavelengths. M removes mainly the orange light, and the light that reaches your eye is the remainder, which looks blue (roughly the complement of orange).
(b) 260 nm is in the ultraviolet, outside the visible range. P absorbs no significant visible light, so all visible wavelengths pass through and the solution looks colourless. Its electronic gap is simply too large for visible photons.
(c) P. A shorter wavelength means a more energetic photon, and the absorbed photon’s energy equals the gap. 260 nm is shorter than 610 nm, so P’s gap is larger.
| Point | What earns it |
|---|---|
| 1 | M’s absorption is electronic (visible region) |
| 1 | Blue colour explained as the light not absorbed (complement of orange) |
| 1 | P absorbs only in the UV, so no visible light is removed |
| 1 | P has the larger gap, linked to shorter wavelength meaning higher photon energy |
Question 6 (argumentation · core)
A student writes: “Infrared radiation has a longer wavelength than ultraviolet radiation, so each infrared photon carries more energy. That means infrared light is better at breaking C–H bonds than ultraviolet light.”
Data: a mole of 4.00 μm infrared photons carries 29.9 kJ; a mole of 220 nm ultraviolet photons carries 544 kJ. The average C–H bond energy is about 413 kJ mol⁻¹.
(a) Identify the error in the first sentence. (b) Use the data to evaluate the student’s conclusion. (c) State what a 4.00 μm photon does do when a molecule containing C–H bonds absorbs it.
Worked solution and suggested Marlbridge rubric
(a) The relationship is reversed. Longer wavelength means lower frequency and less energy per photon, so each infrared photon carries less energy than each ultraviolet photon.
(b) The conclusion is wrong. A mole of the infrared photons carries only 29.9 kJ, about 7% of the 413 kJ needed to break a mole of C–H bonds. A mole of the ultraviolet photons carries 544 kJ, more than the bond energy. So a single UV photon can supply enough energy to break a C–H bond, but a single IR photon cannot.
(c) It raises the molecule to a higher vibrational level: a C–H bond stretches or bends with more energy. (4.00 μm is in the infrared.)
| Point | What earns it |
|---|---|
| 1 | Longer wavelength → lower photon energy (reverses the student’s claim) |
| 1 | Compares 29.9 kJ mol⁻¹ with 413 kJ mol⁻¹: IR photon energy is far too small |
| 1 | Compares 544 kJ mol⁻¹ with 413 kJ mol⁻¹: UV photon energy is enough |
| 1 | IR absorption causes a vibrational transition |
Question 7 (explanation · stretch)
The ground emits most of its energy as infrared radiation. Carbon dioxide in the air absorbs some of this infrared, yet CO₂ is a colourless gas that lets most visible sunlight pass straight through it.
Explain both observations in terms of energy levels and transitions in CO₂ molecules.
Model answer and suggested Marlbridge rubric
Model answer. CO₂ molecules have quantized vibrational levels. The gaps between some of these levels match the energies of infrared photons, so CO₂ absorbs those photons and its bonds vibrate with more energy. Visible photons carry more energy than infrared photons, but they are absorbed only if they match a gap. Their energies are far larger than CO₂’s vibrational gaps and smaller than its electronic gaps, which need ultraviolet photons. With no matching gap, visible light passes through, so CO₂ is colourless.
| Point | What earns it |
|---|---|
| 1 | Infrared absorption by CO₂ is a vibrational transition, because IR photon energies match vibrational gaps |
| 1 | A photon is absorbed only if its energy matches a gap (quantized levels) |
| 1 | CO₂’s electronic gaps are larger than visible photon energies (they need UV), so visible light is not absorbed and the gas is colourless |
Do not award the third point for “visible light has too little energy” alone. Visible photons carry more energy than the infrared photons CO₂ does absorb, so the answer must say they match no gap: too large for the vibrational gaps and too small for the electronic gaps.
How did you do?
- Q1 or Q3 wrong: redo the ranking rule in Figure 1 of the study guide, and always convert units first.
- Q2 or Q4 wrong: re-learn “microwave → rotation, infrared → vibration, UV/visible → electronic” with Figure 2.
- Q5, Q6 or Q7 incomplete: your answer needs the matching idea: a photon is absorbed only if its energy equals a gap.
Then tick off the topic checklist.