UV Visible Spectroscopy Questions and Answers

Answer: C

Holmium oxide gives a series of narrow, well-characterised bands whose positions are certified, so the instrument's scale can be checked at several points at once. Uniform absorption would check the photometric scale rather than the wavelength one, and neither transmittance nor refractive index is the point. Checking a scale at several points rather than one reveals whether the error is an offset or a stretch, which decides how it must be corrected. Naming the cost of a choice alongside its benefit is what turns a fact into a design decision.

Answer: D

With identical contents in both paths the trace should be flat at zero, so any structure in it is instrumental - lamp changeover steps, grating artefacts, cell mismatch or noise. The cut-off shows in a blank against air, wavelength accuracy needs a standard, and the sample is not involved. Measuring a difference that ought to be zero is the most direct way to see what an instrument contributes on its own, and it should precede any difficult measurement. The shared properties explain why these belong to one family, and the differences explain why more than one exists.

Answer: A

Electronic bands in solution are tens of nanometres wide and a molecule shows only one or two, so the spectrum carries far less structural information than an infrared or a nuclear magnetic resonance spectrum. It is non-destructive, many organic compounds absorb well, and resolution is adequate for what structure exists. What the technique does reveal - the presence and extent of conjugation - it reveals quickly and at low concentration, which is why it survives as a complement rather than a rival.

Answer: B

With absorptivities in the tens of thousands, micromolar and even nanomolar concentrations are measurable quickly and cheaply, which is why the technique underpins so much routine assay and detection work. Identification, molecular weight and stereochemistry are the business of infrared, mass spectrometry and other methods. Knowing what a technique is good at, rather than what it can be made to do, is what decides where it belongs in an analytical scheme. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: C

A great many pharmaceutical and biological analytes contain conjugated systems, and a fixed-wavelength absorbance detector is cheap, robust and sensitive enough. It certainly does not respond equally to everything, which is a real limitation, it measures no masses, and the mobile phase must be transparent at the working wavelength. A detector that suits most of the analytes most of the time, at low cost, is often preferred to a universal one that costs ten times as much.

Answer: D

A mobile phase absorbing at the working wavelength leaves too little light for the small peaks that matter, and gradient elution would produce a drifting baseline as its composition changes. Reactivity, retention and cell damage are separate concerns. That the choice of solvent is constrained by the detector as well as by the separation is a coupling between two parts of a method that catches people out when a wavelength is changed without revisiting the eluent. Working one numerical case through both cases side by side fixes the distinction better than any rule of thumb.

Answer: A

The first two are correct and the third is the reverse: pi to pi star transitions in conjugated systems are strongly allowed, with absorptivities in the thousands or tens of thousands. Their strength is exactly why the technique is so sensitive for such compounds. A statement that inverts the intensity of the one transition that matters most would make the whole method look useless, which is a clue that it is wrong. Framing the question as which quantity the change acts upon makes the answer follow without calculation. Most errors here come from carrying a result across from a neighbouring case where it does not hold.

Answer: B

The first two are correct and describe the opposite shifts that distinguish the two transitions; the third overstates, since solvent interaction affects intensity as well as position. That the two transitions move in opposite directions with polarity is the most reliable diagnostic available for telling them apart, and it costs only a second spectrum in a different solvent.

Answer: C

Both are true and the second is the mechanism: spreading electrons over more atoms raises the highest occupied orbital and lowers the lowest unoccupied one, so less energy suffices and the wavelength lengthens. The intensity rises too, which is a separate consequence of the same delocalisation. One structural change moving a band on both axes is the general pattern for conjugated systems.

Answer: D

The reason is true and is precisely why the assertion is false: an auxochrome is defined as a group with no accessible band of its own that nevertheless modifies a chromophore's. A group that did absorb strongly alone would be a chromophore. The two terms divide the substituents into those that produce absorption and those that only alter it, and that division is the whole point of having two words.