NMR and ESR Spectroscopy Questions and Answers

Answer: D

The material that shows no ESR spectrum is Zn.

Electron spin resonance needs unpaired electron spins. Zinc (Z = 30) has the configuration [Ar]3d¹⁰4s²: completely filled shells, no unpaired electrons, and it is diamagnetic, so it gives no ESR signal.

V ([Ar]3d³4s²), Cr ([Ar]3d⁵4s¹) and Fe ([Ar]3d⁶4s²) all have partly filled 3d shells and unpaired electrons, the condition for a resonance absorption between spin states in a magnetic field.

Answer: A

An NMR spectrometer operates in the Radio frequency region.

Nuclear spin energy levels split in a magnetic field by very small amounts. For protons in fields of a few tesla, the resonance frequency is tens to hundreds of megahertz, for example 400 MHz at 9.4 T, which lies in the radio frequency range.

Microwaves are used for electron spin resonance (ESR) and for rotational spectra, ultraviolet radiation excites electronic transitions, and audio frequencies are far too low for nuclear resonance in practical fields.

Answer: D

The usual reference is tetramethylsilane (TMS), printed 'Tetramethylsilame'.

TMS, Si(CH\(_{3}\))\(_{4}\), has twelve equivalent protons that give one sharp, strong signal. Silicon is less electronegative than carbon, so these protons are highly shielded and their signal lies upfield of almost all organic protons; it is assigned \(\delta = 0\) ppm. TMS is also chemically inert, soluble in most organic solvents and volatile, so it is easily removed.

Ethyl alcohol and methyl alcohol give several signals, including an OH peak that shifts with conditions, so they are unsuitable references. Carbon tetrachloride has no protons at all; it is used as a solvent, not a reference.

Answer: D

The resonant frequency is about 100 MHz.

NMR resonance occurs when \(h\nu = g_{N}\beta_{N}B\):

\[ \begin{aligned} \nu &= \frac{g_{N}\beta_{N}B}{h} \\ &= \frac{5.58 \times 5.05 \times 10^{-27} \times 2.3}{6.6 \times 10^{-34}} \\ &= 9.8 \times 10^{7}\ \text{Hz} \approx 100\ \text{MHz} \end{aligned} \]

That matches the rule that protons resonate at about 42.6 MHz per tesla, so a 2.35 T magnet makes a '100 MHz' spectrometer. 1000 MHz would need about 23 T, and the kHz and 1 MHz options are far too low.

Answer: A

The relative population is essentially 1.0.

The energy gap between the two nuclear spin states and the thermal energy are

\[ \begin{aligned} \Delta E &= g_{N}\mu_{N}B = 5.256 \times 5.05 \times 10^{-27} \\ &= 2.65 \times 10^{-26}\ \text{J} \\ k_{B}T &= 1.38 \times 10^{-23} \times 300 = 4.14 \times 10^{-21}\ \text{J} \end{aligned} \]

\[ \begin{aligned} \frac{N_{upper}}{N_{lower}} &= e^{-\Delta E/k_{B}T} \\ &= e^{-6.4 \times 10^{-6}} \approx 0.999994 \end{aligned} \]

The two populations differ by only a few parts per million, which is why NMR signals are weak. A ratio of 2.0, 2.5 or 0.5 would need a gap comparable to kT, about a million times larger than the nuclear Zeeman splitting at 1 T.

Answer: A

The resonance frequency is 40 MHz.

The NMR condition is \(h\nu = g_{N}\mu_{N}B\):

\[ \begin{aligned} \nu &= \frac{5.256 \times 5.0504 \times 10^{-27} \times 1.0}{6.626 \times 10^{-34}} \\ &= \frac{2.655 \times 10^{-26}}{6.626 \times 10^{-34}} \approx 4.0 \times 10^{7}\ \text{Hz} \end{aligned} \]

That is 40 MHz. 4.0 MHz and 0.60 MHz are out by powers of ten, and 60 MHz is closer to the proton's 42.6 MHz per tesla but does not follow from these data.

Answer: B

The field is 321.5 mT.

The ESR condition is \(h\nu = g\mu_{B}B\):

\[ \begin{aligned} B &= \frac{h\nu}{g\mu_{B}} \\ &= \frac{6.626 \times 10^{-34} \times 9 \times 10^{9}}{2 \times 9.2732 \times 10^{-24}} \\ &= \frac{5.963 \times 10^{-24}}{1.855 \times 10^{-23}} \approx 0.3215\ \text{T} \end{aligned} \]

That is 321.5 mT, the familiar X-band value near 0.32 T. 600.5 mT and 60.50 mT do not follow, and 400.0 mT would need a frequency of about 11.2 GHz.

Answer: C

The g value is 4.13.

Rearrange the resonance condition \(h\nu = g\mu_{B}B\):

\[ \begin{aligned} g &= \frac{h\nu}{\mu_{B}B} \\ &= \frac{6.626 \times 10^{-34} \times 9.4 \times 10^{9}}{9.274 \times 10^{-24} \times 0.163} \\ &= \frac{6.228 \times 10^{-24}}{1.512 \times 10^{-24}} \approx 4.12 \end{aligned} \]

which matches 4.13 within rounding. A free-electron line (g about 2.0) at 9.4 GHz would sit near 0.336 T; the line here occurs at about half that field, so g is about twice as large.

Answer: D

The field is 0.36 Tesla.

Resonance occurs when the photon energy equals the gap between the \(m_{J} = +1/2\) and \(-1/2\) levels, \(\Delta E = g_{J}\mu_{B}B\,\Delta m_{J}\) with \(\Delta m_{J} = 1\):

\[ \begin{aligned} h\nu &= g_{J}\mu_{B}B \\ B &= \frac{6.626 \times 10^{-34} \times 10^{10}}{2 \times 9.274 \times 10^{-24}} \\ &= \frac{6.626 \times 10^{-24}}{1.855 \times 10^{-23}} \approx 0.36\ \text{T} \end{aligned} \]

0.71 T is what one gets by leaving out \(g_{J} = 2\), and 0.18 T by doubling it again; 1.2 T has no basis in the data.