Spectroscopic Techniques Questions and Answers

Answer: C

Correct answer: 1-Q, 2-P, 3-R

The three techniques probe three different kinds of transition, in descending order of photon energy. Ultraviolet radiation promotes electrons between molecular orbitals. Infrared radiation, less energetic, excites vibrations. Radiofrequency radiation, less energetic still, flips nuclear spins in a magnetic field. The ordering follows directly from \( E = h\nu \) and the spacing of the corresponding energy levels.

Answer: D

Correct answer: 1-Q, 2-P, 3-R

Three of the most useful diagnostic bands. A broad absorption near 3400 is the hydrogen-bonded O-H stretch, broadened by the range of bonding environments. A strong sharp band near 1700 is the carbonyl, intense because the bond is highly polar. A weak band near 2200 is a carbon-carbon triple bond, weak because a nearly symmetrical alkyne undergoes little dipole change.

Answer: A

Correct answer: 2, 3, 4, 1

Photon energy rises with frequency, so the order follows the spectrum: radiofrequency is the least energetic, then infrared, then visible, then ultraviolet. The same order gives what each region can do - flip a nuclear spin, excite a vibration, and promote an electron, the last requiring the most energy of the four.

Answer: B

Correct answer: 1 and 2 only are correct

Statements 1 and 2 are correct: absorbance is linear in concentration through Beer’s law, and a vibration absorbs in the infrared only if it changes the dipole moment. Statement 3 is precisely backwards - the chemical shift is expressed in parts per million so that it does not depend on the instrument’s field, which is what makes shift values transferable between spectrometers.

Answer: C

Correct answer: using absorbance rather than transmittance as the measured quantity

Using absorbance is not a deviation but the correct practice, since it is absorbance that is linear in concentration. The other three are genuine causes of departure from Beer’s law: stray light sets a ceiling on the apparent absorbance, a polychromatic beam averages over absorptivities that differ across the band, and association or dissociation at high concentration changes the absorbing species itself.

Answer: D

Correct answer: a Raman active vibration

Raman activity requires a change in polarisability, not in dipole moment - that is the whole distinction between the two techniques. Any infrared active vibration, whether a stretch or a bend, requires a dipole change by definition, and the asymmetric stretch of carbon dioxide is a particular example of one. In a centrosymmetric molecule the two requirements are mutually exclusive, so a mode active in one is inactive in the other.

Answer: A

Correct answer: carboxylic acid

The two bands together settle it. A broad absorption at 3400 is a hydrogen-bonded O-H, and a strong band at 1710 is a carbonyl; a compound containing both is a carboxylic acid. A simple ketone would show the carbonyl without the hydroxyl, a primary amine would show N-H bands near 3300 without a carbonyl, and a terminal alkyne would show a sharp C-H near 3300 with a weak band near 2100 and no carbonyl at all.

Answer: B

Three protons split by two neighbours give the triplet and two protons split by three give the quartet, which is the signature of a chain ending in a methyl next to a methylene. An isopropyl group would give a doublet and a septet instead.

Answer: C

Correct answer: both the assertion and the reason are true and the reason explains the assertion

Both are true and the reason is the mechanism. Infrared activity requires the vibration to change the dipole moment. In the symmetric stretch both oxygen atoms move outward together, so the two C-O bond dipoles remain equal and opposite at every instant and the resultant stays zero. With no oscillating dipole there is nothing for the radiation to couple to, and the mode is silent in the infrared - though it appears clearly in the Raman spectrum.

Answer: D

Correct answer: both the assertion and the reason are true and the reason explains the assertion

Both are true and the reason accounts for the assertion. The molecular ion is formed by removing a single electron, whose mass is negligible, so the ion retains the whole molecule and its mass-to-charge ratio gives the molecular mass. The caution implicit here is that the peak must be identified as the molecular ion: in many compounds it fragments so readily that it is weak or missing, and a fragment mistaken for it would give the wrong answer.