Magnetic Resonance Spectroscopy Questions and Answers

Answer: A

The characteristic feature is \(g_{\parallel} \gt g_{\perp}\).

In frozen solution, Cu(II) (d⁹) is surrounded by water molecules in a tetragonally elongated octahedron, the Jahn-Teller distortion. The single hole lies in the \(d_{x^{2}-y^{2}}\) orbital, and for that ground state spin-orbit coupling gives an axial spectrum with \(g_{\parallel} \gt g_{\perp} \gt 2.0023\) (typically about 2.4 and 2.1).

\(g_{\parallel} \lt g_{\perp}\) is the signature of a \(d_{z^{2}}\) ground state, as in a tetragonally compressed geometry. Equal g values would mean an isotropic, undistorted environment, which the Jahn-Teller effect rules out for d⁹. Three different g values describe a rhombic site, not the characteristic axial spectrum of hydrated copper(II).

Answer: B

The septet has intensities 1 : 3 : 6 : 7 : 6 : 3 : 1.

The CD₃ carbon couples to three deuterium nuclei, each with spin I = 1. The number of lines is \(2nI + 1 = 2 \times 3 \times 1 + 1 = 7\). Each deuterium has three equally likely spin states (-1, 0, +1), so the intensities are the coefficients of \((1 + x + x^{2})^{3}\):

\[ (1 + x + x^{2})^{3} \to 1, 3, 6, 7, 6, 3, 1 \]

This is the familiar septet of the solvent peak of acetone-d₆. 1 : 6 : 15 : 20 : 15 : 6 : 1 is the binomial pattern for six spin-1/2 nuclei, as in the CH of an isopropyl group in ¹H NMR, not three deuteriums. The other two sequences do not come from any set of equivalent nuclei.

Answer: A

The multiplicities are I- pentet, II- quartet, III- doublet and IV- singlet.

Apply the n + 1 rule to the nuclei with spin 1/2 that couple to each hydride:

• (I) The Ni-H couples to four equivalent ³¹P nuclei (I = 1/2, 100% abundant): 4 + 1 = 5, a pentet.
• (II) The Si-H couples to the three protons of the Si-CH₃ group (three bonds apart): a quartet.
• (III) Each H of PH₃ couples to the single ³¹P: a doublet; the three H are equivalent and do not split each other.
• (IV) The two Zr-H hydrides are equivalent, and Zr has no abundant spin-1/2 isotope (and the Cp* methyl protons are too remote): a singlet.

Each of the other options mis-counts at least one of these neighbours, for example treating PH₃ as a singlet by ignoring phosphorus or giving the zirconium hydride a coupling partner it does not have.

Answer: D

The EPR spectrum shows 8 hyperfine lines.

VO²⁺ contains V(IV), d¹, with one unpaired electron. Vanadium-51, nearly 100 percent abundant, has nuclear spin I = 7/2, and the electron couples to it to give 2I + 1 lines:

\[ 2\left(\tfrac{7}{2}\right) + 1 = 8 \]

In solution at room temperature the anisotropy averages out and the eight-line pattern of the vanadyl ion is seen. 3 and 5 would correspond to nuclei with I = 1 or 2, and 7 to I = 3.

Answer: C

The EPR spectrum of the phenyl radical shows 18 lines.

The unpaired electron is in an sp² σ orbital at the carbon that lost the hydrogen, and it couples to three sets of equivalent protons: the two ortho, the two meta and the one para hydrogen. Each set of n equivalent protons splits every line into n + 1:

\[ \begin{aligned} N &= (2+1)(2+1)(1+1) \\ &= 3 \times 3 \times 2 = 18 \end{aligned} \]

8 would be the count for a radical with three inequivalent protons, and 28 or 3 do not correspond to the proton sets present. The ortho, meta and para coupling constants are all different, so none of the lines coincide.

Answer: B

Each set consists of 6 hyperfine lines.

High-spin Mn²⁺ is d⁵ with S = 5/2. Zero-field splitting separates its six M_S levels into Kramers doublets, and the allowed ΔM_S = 1 transitions give five fine-structure lines. Each line is split by coupling to the ⁵⁵Mn nucleus, I = 5/2, into

\[ 2I + 1 = 2 \times \frac{5}{2} + 1 = 6 \]

lines, giving the familiar six-line Mn²⁺ pattern. Thirty is the total number of lines (5 × 6), not the number in each set, and 5 is the number of fine-structure sets.

Answer: A

The spectra are a doublet and a triplet.

With the O-H protons decoupled, only the protons bonded directly to phosphorus split the ³¹P signal, through the large one-bond P-H coupling. Each signal shows n + 1 lines for n equivalent protons of spin 1/2.

• Phosphorous acid, HPO(OH)₂, has one P-H bond: n = 1 gives a doublet.
• Hypophosphorous acid, H₂PO(OH), has two P-H bonds: n = 2 gives a triplet.

Option C reverses the two, and options B and D treat the acids as having the same number of P-H protons.

Answer: C

The spectrum shows a doublet of doublet and doublet of triplet.

Wilkinson's catalyst, RhCl(PPh₃)₃, is square planar with two mutually trans phosphines and one phosphine trans to chloride. The two trans phosphines are equivalent: each signal is split by ¹⁰³Rh (I = 1/2) into a doublet and by the single unique phosphorus into a further doublet, giving a doublet of doublets of intensity 2.

The unique phosphine is split by rhodium into a doublet and by the two equivalent cis phosphines into a triplet, giving a doublet of triplets of intensity 1. Options with two simple doublets or triplets ignore either the Rh-P or the P-P coupling.

Answer: D

The molecule without a temperature-dependent ¹⁹F spectrum is SF₆.

In octahedral SF₆ all six fluorines are equivalent and the molecule is rigid, so its ¹⁹F spectrum is a single line at every temperature.

The other three are fluxional. In trigonal bipyramidal PF₅ and PCl₂F₃, axial and equatorial fluorines differ, but Berry pseudorotation exchanges them; at room temperature the signals average, and only on cooling can separate environments appear. T-shaped ClF₃ has two kinds of fluorine whose signals are resolved at low temperature and coalesce on warming as the fluorines exchange.

Answer: C

The spectrum shows 24 lines.

The unpaired electron couples with the ²⁷Al nucleus (I = 5/2) and with three equivalent protons (I = 1/2). The number of lines from n equivalent nuclei of spin I is 2nI + 1.

\[ \begin{aligned} \text{Al}: &\ 2(1)(5/2) + 1 = 6 \\ \text{3 H}: &\ 2(3)(1/2) + 1 = 4 \\ \text{total} &= 6 \times 4 = 24 \end{aligned} \]

Six is the aluminium sextet alone, four the proton quartet alone, and 18 would come from treating the three protons as six-fold rather than as a quartet.