In the 1H NMR spectrum, multiplicity of the signal (bold and underlined H atom) in the following species is(I) [HNi(OPEt3)4]+(II) Ph2Si(Me)H(III) PH3(IV) (Cp*)2ZrH2 (Cp* = pentamethylcyclopentadienyl)

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.

The 31P{OH} NMR spectra of HPO(OH)2 and H2PO(OH) (I = 3p, 1H = 1/2) shows respectively :

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.

Magnetic Resonance Spectroscopy Questions and Answers

Q: The characteristic feature of an electron spin resonance (ESR) spectrum of frozen aqueous solution of CuSO4·5H2O at 77 K is

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).

Q: The 13C NMR spectrum of acetone-d6 has a signal at 30 ppm as a septet in the intensity ratio

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.

Q: In the EPR spectrum of an aqueous solution of VOSO4 at room temperature, the total number of hyperfine splitting signals is

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.

Q: The X-band EPR spectra of phenyl radical will exhibit.................lines.

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.

Q: The EPR spectra of [Mn(H2O)6]2+ (55Mn=I=5/2) with zero field splitting show five set of lines. Each set further consist of ........... ...hyperfine lines (At. No. Mn = 25)

Each set consists of 6 hyperfine lines.High-spin Mn2+ is d5 with S = 5/2. Zero-field splitting separates its six MS levels into Kramers doublets, and the allowed ΔMS = 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.

Q: The 31P {H} NMR spectrum of complex (Rh(PPh3)3Cl] will exhibit :

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.

Q: Which of the following molecules will NOT exhibit a temperature dependent 19F NMR spectrum ?

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.

Q: The number of EPR lines observed in AlH3 radical (27Al, nuclear spin = 5/2) will be :

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.

The characteristic feature of an electron spin resonance (ESR) spectrum of frozen aqueous solution of CuSO₄·5H₂O at 77 K is

A.\(g_{\parallel} \gt g_{\perp}\)
B.\(g_{\parallel} \lt g_{\perp}\)
C.\(g_{\parallel} = g_{\perp}\)
D.\(g_{x} \ne g_{y} \ne g_{z}\)

Exam Year(s)

GATE Chemistry - 2017,

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).

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The ¹³C NMR spectrum of acetone-d₆ has a signal at 30 ppm as a septet in the intensity ratio

A.1 : 6 : 15 : 20 : 15 : 6 : 1
B.1 : 3 : 6 : 7 : 6 : 3 : 1
C.1 : 2 : 3 : 5 : 3 : 2 : 1
D.1 : 3 : 7 : 10 : 7 : 3 : 1

Exam Year(s)

GATE Chemistry - 2017,

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.

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In the ¹H NMR spectrum, multiplicity of the signal (bold and underlined H atom) in the following species is

(I) [HNi(OPEt₃)₄]⁺
(II) Ph₂Si(Me)H
(III) PH₃
(IV) (Cp*)₂ZrH₂ (Cp* = pentamethylcyclopentadienyl)

A.I- pentet, II- quartet, III- doublet and IV- singlet
B.I- pentet, II- singlet, III- singlet and IV- doublet
C.I- triplet, II- triplet, III- doublet and IV- doublet
D.I- singlet, II- quartet, III- singlet and IV- singlet

Exam Year(s)

GATE Chemistry - 2023,

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.

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In the EPR spectrum of an aqueous solution of VOSO₄ at room temperature, the total number of hyperfine splitting signals is

Exam Year(s)

GATE Chemistry - 2024,

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.

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The X-band EPR spectra of phenyl radical will exhibit.................lines.

A.3
B.8
C.18
D.28

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

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.

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The EPR spectra of [Mn(H₂O)₆]²⁺ (⁵⁵Mn=I=5/2) with zero field splitting show five set of lines. Each set further consist of ........... ...hyperfine lines (At. No. Mn = 25)

A.5
B.6
C.30
D.1

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

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.

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The ³¹P{OH} NMR spectra of HPO(OH)₂ and H₂PO(OH) (I = ³p, ¹H = 1/2) shows respectively :

A.doublet and triplet
B.triplet and triplet
C.triplet and doublet
D.doublet and doublet

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

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.

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The ³¹P {H} NMR spectrum of complex (Rh(PPh₃)₃Cl] will exhibit :

A.Two triplets
B.Two doublets
C.Doublet of doublet and doublet of triplet
D.Triplet of doublet and triplet of triplet

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

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.

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Which of the following molecules will NOT exhibit a temperature dependent ¹⁹F NMR spectrum ?

A.\(\mathrm{PF_{5}}\)
B.\(\mathrm{ClF_{3}}\)
C.\(\mathrm{PCl_{2}F_{3}}\)
D.\(\mathrm{SF_{6}}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

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.

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10.

The number of EPR lines observed in AlH₃ radical (²⁷Al, nuclear spin = 5/2) will be :

A.6
B.4
C.24
D.18

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

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.

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