Spectrophotometry Fundamentals Questions and Answers

Answer: C

Ultraviolet and visible photons carry enough energy to promote electrons between molecular orbitals, which is why coloured compounds and conjugated systems absorb there. Vibrational transitions need infrared photons, rotational ones microwave, and nuclear spin transitions radio frequencies in a magnetic field. That each spectral region probes a different kind of motion, in order of the energy each requires, organises the whole of spectroscopy into one picture. Working one numerical case through both cases side by side fixes the distinction better than any rule of thumb.

Answer: D

Delocalising electrons over a longer system lowers the energy needed to promote one, and a smaller gap means a longer wavelength. The absorptivity does rise as well, which increases the intensity rather than shifting the band. Scattering and bond counting are not the mechanism. That colour in organic compounds tracks the extent of conjugation follows directly, and it is why dyes are built around long conjugated chains. Most errors here come from carrying a result across from a neighbouring case where it does not hold.

Answer: A

A chromophore is the structural feature - a carbonyl, a nitro group, an aromatic ring - that gives rise to the absorption band. Solvent, detector and wavelength are separate things. Because the same chromophore absorbs at a broadly similar wavelength in different molecules, its presence can be inferred from a spectrum, which is what makes ultraviolet absorption useful qualitatively as well as quantitatively. Naming the cost of a choice alongside its benefit is what turns a fact into a design decision.

Answer: B

An auxochrome such as a hydroxyl or amino group has no absorption band of its own in the usual range but shifts and intensifies the chromophore's when attached to it. It does not absorb alone, quench fluorescence or act primarily on solubility, though it may affect the latter. A group that modifies a property without possessing it is worth a name of its own, and the pair of terms together describes most of what determines an organic compound's colour. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Answer: C

A bathochromic or red shift moves the band to longer wavelength and hence lower energy, while hypsochromic is the shift the other way. Changes in intensity are hyperchromic and hypochromic. Four terms covering two axes of change, and since a solvent change or a structural modification typically moves a band on both axes at once, all four are needed to describe what happened.

Answer: D

A polar solvent stabilises whichever state has the larger dipole moment more, so the gap between them changes and the band moves. Path length is set by the cell, refractive index affects reflection rather than the transition energy, and chemical reaction is a separate problem. Because the direction of the shift reveals which state is the more polar, solvent studies are used to identify the kind of transition involved. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: A

A solvent absorbing strongly at the wavelength of interest leaves too little light for a reliable measurement, and although a blank corrects the mean it cannot restore the lost signal-to-noise. Reaction, absorptivity changes and scattering are separate concerns. Each solvent has a cut-off wavelength below which it is unusable, and choosing a solvent means checking that cut-off against the band to be measured. A limit that survives making every external component ideal is an intrinsic one, and worth knowing for that reason.

Answer: B

Ordinary glass absorbs below roughly 340 nanometres, so quartz cells are needed for ultraviolet work. Scattering, fluorescence and refractive index changes are minor by comparison. That the cell itself becomes the limiting component in one part of the range, while being irrelevant in another, is a good illustration of why an instrument's usable range is decided by whichever component runs out first.

Answer: C

Taking the ratio of the uncertainty to the reading:

\[ \frac{0.001}{0.4} = 0.0025 \]

giving about 0.25 per cent. At a reading of 0.04 the same noise would be 2.5 per cent, which is why very dilute samples are measured in longer cells rather than simply read as they are. Expressing an absolute uncertainty as a fraction of the reading is what shows where the useful measuring range actually lies.

Answer: D

A lamp's output and a detector's response both drift for some minutes after switch-on, and on a single-beam instrument that drift appears directly in every reading. Grating expansion is negligible, the compartment is already at room temperature, and the electronics settle in seconds. Warm-up time is quoted in a manual precisely because the drift is predictable and finite, and ignoring it is one of the commonest causes of results that will not repeat from one session to the next.