UV Visible Spectroscopy Questions and Answers

Answer: C

Hyperchromic means the band grows more intense without necessarily moving, and hypochromic means it weakens. Wavelength shifts are bathochromic and hypsochromic. Narrowing and new bands have no such names. Keeping the two axes separate matters because the mechanisms differ: intensity reflects transition probability while position reflects the energy gap, and a substituent can change either independently.

Answer: D

Delocalising electrons over more atoms raises the highest occupied orbital and lowers the lowest unoccupied one, so the gap shrinks and a longer wavelength suffices. The absorptivity does rise as well, which is a separate hyperchromic effect. Electron count and scattering are not the mechanism. That colour in organic compounds tracks conjugation length follows directly, and it is why dyes are built around long conjugated chains. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Answer: A

Enough conjugation lowers the transition energy until the band falls between 400 and 700 nanometres, and the compound then appears the complement of what it absorbs. Solubility, fastness and affinity are separately engineered with other substituents. That a structural feature chosen for one property brings the others along as problems to be solved is the ordinary situation in molecular design.

Answer: B

A pi to pi star excited state is more polar than the ground state, so a polar solvent lowers its energy further and the gap narrows. Refractive index affects reflection rather than transition energy, and the solvent's own absorption is a separate limit. Because the direction of the shift reveals which state is more polar, solvent studies are used to identify what kind of transition a band represents.

Answer: C

A polar or hydrogen-bonding solvent interacts strongly with the lone pair in the ground state, lowering its energy and widening the gap - the opposite of what happens to a pi to pi star band. The excited state here is less polar rather than more. That the two transitions move in opposite directions with solvent polarity is the most reliable way to tell them apart, and it costs only a second spectrum in a different solvent. Naming the cost of a choice alongside its benefit is what turns a fact into a design decision.

Answer: D

Every solvent absorbs strongly below some wavelength, and beneath that cut-off too little light reaches the detector for a useful measurement whatever the blank correction. Solubility, refractive index and fluorescence are separate considerations. Water and acetonitrile reach below 200 nanometres while acetone is useless below about 330, so the solvent has to be chosen against the band to be measured rather than for convenience alone. Most errors here come from carrying a result across from a neighbouring case where it does not hold.

Answer: A

Water is usable to about 190 nanometres, since it has no accessible electronic transitions in the range. Acetone's carbonyl absorbs strongly to about 330, benzene's ring to about 280, and chloroform to about 245. A solvent's cut-off follows directly from the chromophores it contains, so a molecule with no double bonds and no lone pairs in conjugation is transparent furthest - which is why saturated hydrocarbons and water dominate ultraviolet work. A limit that survives making every external component ideal is an intrinsic one, and worth knowing for that reason.

Answer: B

Ordinary glass contains species that absorb below roughly 340 nanometres, so the cell itself would swallow the light before the sample saw it. Scattering and fluorescence are minor, and path length is set by the cell's dimensions rather than its material. That the cell becomes the limiting component in one part of the range while being irrelevant in another illustrates how an instrument's usable range is set by whichever part runs out first. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Answer: C

Multiplying the three terms of the Beer-Lambert law:

\[ A = 15000 \times 1 \times 2\times10^{-5} = 0.30 \]

giving 0.30, comfortably inside the range where photometric error is least. A compound with an absorptivity in the tens of thousands can therefore be determined at micromolar concentrations, which is what makes the technique so sensitive for conjugated systems.

Answer: D

Rearranging the law for concentration:

\[ \begin{aligned} c &= \frac{A}{\varepsilon b} \\ &= \frac{0.60}{12000} \\ &= 5\times10^{-5} \ \mathrm{M} \end{aligned} \]

which is 50 micromolar. Converting molar to micromolar means multiplying by a million, and slipping a factor of ten there is the commonest error in this calculation - worth checking by asking whether the answer is a plausible working concentration.