Molecular Spectroscopy Questions and Answers

Answer: A

The molecule that does not absorb microwave radiation is CO\(_{2}\).

A pure rotational (microwave) spectrum needs a permanent electric dipole moment, so that the rotating molecule presents an oscillating dipole to the radiation. CO₂ is linear and symmetric, O=C=O, and its two bond dipoles cancel, so it has no permanent dipole and shows no microwave spectrum.

H₂O is bent and strongly polar, and the heteronuclear diatomics CO and NO each have a permanent dipole, so all three absorb in the microwave region.

Answer: B

The spectra show a singlet in (ii) and doublets in (i) and (iii).

A Mossbauer doublet comes from quadrupole splitting, which needs an electric field gradient at the iron nucleus. In K₄[Fe(CN)₆] the iron is low-spin Fe(II), t_2g⁶, a spherically symmetric configuration in a regular octahedron of identical ligands, so there is no field gradient and a single line appears.

FeSO₄·7H₂O contains high-spin Fe(II), t_2g⁴e_g², whose unevenly filled t_2g set produces a large field gradient and a wide doublet. Sodium nitroprusside is low-spin but its ligand set (five CN^- and one NO⁺) is not cubic, so it too shows a doublet. That rules out every option giving (i) or (iii) as a singlet, or (ii) as a doublet.

Answer: A

The incorrect statement is (A): the two compounds do not show equal quadrupole splitting.

Quadrupole splitting measures the electric field gradient at the iron nucleus, and in both compounds the I = 3/2 excited state splits into two levels, so each gives a doublet. FeCl₃.6H₂O contains high-spin Fe(III), d⁵, whose half-filled t_2g³e_g² shell is spherical, so the valence electrons add no field gradient and only the lattice (ligand) contribution acts: the splitting is small. FeCl₂.4H₂O contains high-spin Fe(II), d⁶, whose extra electron sits in a non-spherical d orbital and adds a large valence contribution, so its splitting is much larger. The splittings are therefore unequal, which makes (B) true, and (D) states exactly this valence versus lattice difference.

Statement (C) is true in substance: both salts are high-spin, and the pair is one Fe(III) and one Fe(II) complex. As printed, 'respectively' pairs Fe(II) with the first formula, FeCl₃.6H₂O, which is the wrong way round; the question evidently means only the pair of oxidation states, so (A), false on any reading, is the answer. The splitting values themselves are not quoted here; the argument rests on the d⁵ versus d⁶ electronic structure.

Answer: C

Both molecules give pure rotational spectra in the pair CH₂Cl₂ and CHCl₃.

A molecule shows a pure rotational (microwave) spectrum only if it has a permanent dipole moment, because the rotating dipole is what interacts with the radiation. CH₂Cl₂ (C_2v) and CHCl₃ (C_3v) both have permanent dipole moments.

CH₄ and CCl₄ are regular tetrahedra whose bond dipoles cancel, so they are microwave inactive. Every other pair contains at least one of these two non-polar molecules.

Answer: C

The correct statement is K. E of photoelectrons decrease when the oxidation state of the surface is increased.

In XPS the kinetic energy of an ejected electron is KE = hν − BE − φ. A higher oxidation state leaves the atom more positive, holding its core electrons more tightly; the binding energy rises (a chemical shift of a few eV), so at fixed photon energy the kinetic energy falls.

XPS ejects mainly core electrons, not valence electrons (A). Intensity changes the number of electrons, not their kinetic energy (B). And with a monochromatic source, the number of photoelectrons depends on intensity, not on frequency (D).

Answer: A

Nitrobenzene has 36 normal modes.

A non-linear molecule of N atoms has 3N − 6 vibrational modes. Nitrobenzene, C₆H₅NO₂, has 6 + 5 + 1 + 2 = 14 atoms:

\[ 3(14) - 6 = 42 - 6 = 36 \]

14 is just the number of atoms, and 31 and 9 correspond to no standard count for a 14-atom non-linear molecule.

Answer: B

Focus on the 2000 - 1800 cm⁻¹ region, where the carbonyl stretches lie.

Terminal CO ligands on a low-valent metal such as Mo(0) absorb between about 2050 and 1850 cm⁻¹. The cis and trans isomers give different patterns there: cis-[Mo(CO)₄(PEt₃)₂] (C_2v) has four IR-active CO stretches, while the trans isomer (D_4h) has essentially one strong band. Watching the cis bands fade and the single trans band grow follows the isomerization directly.

3500 - 3000 cm⁻¹ holds O-H and N-H stretches, absent here. 1600 - 1400 cm⁻¹ holds C=C and C-H bending bands, and 1200 - 800 cm⁻¹ is the fingerprint region of the ethyl groups; neither changes clearly between the isomers.

Answer: B

It is called Rayleigh scattering.

When light passes through a sample, most of the scattered light comes out at exactly the incident wavelength, because the photons exchange no energy with the molecules; this elastic scattering is Rayleigh scattering. Its strength varies as 1/λ⁴, which is why the sky is blue.

A very small fraction, roughly one photon in a million, is scattered inelastically, gaining or losing a vibrational or rotational quantum. That is Raman scattering. Its lines at lower frequency (the molecule gains energy) are the Stokes lines, and those at higher frequency (the molecule gives up energy) are the anti-Stokes lines. None of these is at the incident wavelength.

Answer: A

The pattern corresponds to (A), upper and lower potential curves with the same equilibrium distance.

By the Franck-Condon principle, an electronic transition is vertical: the nuclei do not move during it, so the most intense vibrational band is the one whose upper-state level lies directly above the ground-state equilibrium geometry.

In (A) the excited-state minimum sits directly above the ground-state minimum, so the vertical transition lands on v′ = 0. The 0-0 band is the strongest and the higher bands (0-1, 0-2 and so on) get steadily weaker, just as the figure shows: intensity highest at the first line and falling away.

When the upper curve is displaced to a longer or shorter distance, as in (B), (C) and (D), the vertical transition reaches a higher v′, so the intensity first rises to a maximum in the middle of the progression; in (D) it reaches above the dissociation limit and gives a continuum.

Answer: B

The keyed selection rule is Δv = 0, ±1, ±2, ...

For a harmonic oscillator the vibrational Raman selection rule is Δv = ±1, and in a real, anharmonic molecule weak overtones with Δv = ±2, ±3 also appear. Textbooks such as Banwell write the rule for vibrational Raman spectra as Δv = 0, ±1, ±2, ..., where Δv = 0 is the Rayleigh (and pure rotational) scattering at the exciting line, and the key follows that form.

Options (A) and (C) are printed identically and omit Δv = 0. Δv = ±2 alone is the rotational Raman rule (ΔJ = ±2) misapplied to vibration.