Infrared Spectroscopy Questions and Answers

Answer: A

Infrared photons carry tenths of an electron volt, enough to raise a vibrational level but not to move an electron. Electronic transitions need ultraviolet or visible photons, pure rotation far less energy, and nuclear spin transitions a millionth as much. That the energy required orders the spectroscopies from X-ray down to radio is the idea that places each technique in the sequence. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Answer: B

Infrared spectra are plotted in wavenumbers, reciprocal centimetres, which are directly proportional to photon energy. Nanometres serve the ultraviolet and visible, megahertz the magnetic resonance experiment, and electron volts are used in physics rather than in routine spectroscopy. A unit chosen so that a scale becomes linear in energy makes vibrational transitions evenly spaced, which is why the convention differs. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: C

The mid infrared runs from 4000 down to about 400 reciprocal centimetres, corresponding to 2.5 to 25 micrometres. The ultraviolet and visible range is quoted in nanometres and covers quite different transitions. That the axis conventionally runs from high wavenumber on the left to low on the right, the reverse of a wavelength plot, is a convention worth noting before two spectra are compared. The shared properties explain why these belong to one family, and the differences explain why more than one exists.

Answer: D

A vibration absorbs infrared radiation only if it changes the molecule's dipole moment, so that the oscillating field has something to couple to. Polarisability change is the Raman selection rule, which is why the two techniques are complementary. Double bonds and stretches are not required. That a symmetric molecule can vibrate vigorously and absorb nothing follows directly, and it is why nitrogen and oxygen are transparent in the infrared. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: A

Both oxygens move outward together, so the two bond dipoles cancel throughout the motion and the net dipole stays zero. The asymmetric stretch, where one bond lengthens as the other shortens, does change the dipole and absorbs strongly. Linearity and equal force constants are what make the cancellation exact rather than being the rule itself. A mode that is silent in the infrared is generally active in the Raman, which is why the two techniques are run together.

Answer: B

A homonuclear diatomic has no dipole moment and stretching it changes nothing, so the single vibration is infrared inactive. The bond certainly vibrates, the mass is not the issue, and the frequency does fall within range. That the two principal atmospheric gases are both invisible in the infrared is what allows the technique to be used in open air at all, and it is also why carbon dioxide and water vapour dominate the atmospheric background. Most errors here come from carrying a result across from a neighbouring case where it does not hold.

Answer: C

Of the three degrees of freedom per atom, three describe translation and three rotation, leaving 3N minus 6 vibrational modes. A linear molecule has only two rotational degrees of freedom and so has 3N minus 5. Counting the modes tells you how many bands to expect at most, though degeneracy, inactivity and overlap always reduce the number actually seen.

Answer: D

Applying the non-linear count:

\[ 3N - 6 = 9 - 6 = 3 \]

giving three - a symmetric stretch, an asymmetric stretch and a bend, all three of which are infrared active because water is bent and polar. Carbon dioxide, with the same number of atoms but linear, has four instead, of which only two are active. Geometry decides both the count and how many of them can be seen.

Answer: A

Stretching a bond opposes the full bond strength and so needs more energy than bending it does, putting stretches at higher wavenumber. That is why the region above 1500 reciprocal centimetres is dominated by stretches and the fingerprint region below it by bends and skeletal modes. Knowing which kind of motion a region contains halves the work of reading an unfamiliar spectrum. A limit that survives making every external component ideal is an intrinsic one, and worth knowing for that reason.

Answer: B

The harmonic oscillator gives the frequency as the square root of the force constant divided by the reduced mass, so a stiff bond between light atoms vibrates fastest. Bond length, dipole moment and electronegativity influence the force constant indirectly rather than entering the expression. Two quantities, one measuring stiffness and one measuring inertia, account for the position of every band in the group frequency region. Most errors here come from carrying a result across from a neighbouring case where it does not hold.