Electronic Spectra of Molecules Questions and Answers

Answer: A

The best match is (i) - T; (ii) - P,S,T; (iii) - Q,R; (iv) - S.

• Visible light ejects electrons from metal surfaces with low work functions: the photoelectric effect (T).
• X-rays come from transitions between inner (core) electron levels (P), have wavelengths comparable to atomic spacings and so reveal crystal structure (S), and also cause photoemission (T).
• Gamma rays come from transitions between nuclear energy levels (Q), and above 1.022 MeV they can create electron-positron pairs (R).
• Thermal neutrons have de Broglie wavelengths of about an angstrom and are used for crystal (and magnetic) structure determination (S).

The other options credit visible light with core transitions or crystal structure determination, or X-rays with pair production, or thermal neutrons with the photoelectric effect, none of which fits.

Answer: C

The region is the Infrared.

Infrared sources, such as a heated Globar or Nernst glower, are weak, and infrared detectors are relatively insensitive and noisy. Fourier transform spectroscopy overcomes this: an interferometer records all wavelengths at once (the multiplex or Fellgett advantage) and passes far more light than a slit (the throughput or Jacquinot advantage). That is why FTIR is the standard infrared method.

In the ultraviolet and visible, sources are bright and photomultiplier detectors very sensitive, so dispersive instruments suffice; X-ray spectroscopy uses crystal diffraction.

Answer: C

The identity that is not correct is sp\(^{3}\)-square planar - PtCl\(_{4}^{2-}\).

Square planar complexes such as \(PtCl_{4}^{2-}\) use \(dsp^{2}\) hybrid orbitals: one d, one s and two p orbitals pointing to the corners of a square. \(sp^{3}\) hybridisation gives a tetrahedral arrangement, not a square planar one.

The others are correct: \(BCl_{3}\) is trigonal planar with \(sp^{2}\) boron, \(CO_{2}\) is linear with sp carbon, and \(CH_{4}\) is tetrahedral with \(sp^{3}\) carbon.

Answer: C

The intense lines are for transitions between the vibrational levels of two electronic states at a constant internuclear separation.

The Franck-Condon principle (spelt 'Frank-Condon' in the paper) says that an electronic transition is so fast compared with nuclear motion that the nuclei do not move during it. Transitions are therefore vertical on a potential energy diagram, at fixed internuclear distance, and the strongest lines join vibrational levels of the two electronic states whose wave functions overlap most at that distance.

Monoatomic gases have no vibrational structure. Transitions within one electronic state are vibrational, not electronic, transitions. The v = 0 to v = 0 line is strongest only when the two potential curves have nearly the same equilibrium separation, so it is not the general rule.

Answer: A

The included terms are the kinetic energy of the electrons and the electron-nucleus, electron-electron and nucleus-nucleus Coulomb energies.

In the Born-Oppenheimer approximation the nuclei are so much heavier than the electrons that they are treated as fixed while the electronic problem is solved. The nuclear kinetic energy is therefore dropped from the electronic Hamiltonian, which keeps:

• the kinetic energy of the electrons;
• the electron-nucleus attraction;
• the electron-electron repulsion;
• the nucleus-nucleus repulsion, which is a constant for a given nuclear arrangement.

The options with the kinetic energy of the nuclei instead of the electrons have it the wrong way round, and the option with no kinetic energy at all leaves the electrons without the term that gives them their quantum energy levels.