A solution and solid state study of niobium complexes University of ...
A solution and solid state study of niobium complexes University of ... A solution and solid state study of niobium complexes University of ...
Chapter 3 Electrons, neutrons and protons are considered to posses spinning properties and when these spins are paired against each other, the overall spin of the atom can be determined. The rules for determining the overall spin (I) of a nucleus are as follows: i. If the number of protons and neutrons are both even, the nucleus has no spin. ii. If the number of protons plus the number of neutrons is odd, the nucleus has a half-integer spin (1/2, 3/2, 5/2). iii. If the number of protons and the number of neutrons are both odd, the nucleus has an integer spin (1, 2, 3). According to quantum mechanics, a nucleus of spin I will have 2I + 1 possible orientations. 2 Hence, a nucleus ( 1 H) with spin 1/2 will have 2 possible orientations. When no external magnetic field is applied, the spin orientations are of equal energy. In the presence of a magnetic field, the energy levels are split. Each level has a magnetic quantum number (m) as indicated in Figure 3.1. Energy Figure 3.1: Energy levels for a nucleus with spin quantum number 1/2. The spin state +1/2 has the lowest energy as it is aligned with the applied magnetic field, whereas the –1/2 spin state has higher energy because it is opposed to the applied field. The two spin states are separated by an energy difference, ∆E, which is dependent on the strength of the magnetic field as well as the size of the nuclear magnetic moment. The difference in energy between the levels can be determined from: ∆E = 2 P.J. Hore, Nuclear Magnetic Resonance, Oxford University Press, New York, 52, 1995. (3.1) 43
Chapter 3 with γ = magnetogyric ratio, h = Planck’s constant and B = strength of the magnetic field at the nucleus. The spinning nucleus generates a small magnetic field which causes it to possess a magnetic moment (µ): µ = (3.2) The magnetic moment has both magnitude and direction as defined by its axis of spin. In the presence of a magnetic field, the nucleus’s axis of spin will precess around the magnetic field. The frequency of this precession is called the Larmor frequency and it is identical to the transition frequency: E = -µB cosθ (3.3) where θ is the angle between the direction of the applied field and the axis of nuclear rotation. The angle will change if the nucleus absorbs energy. In a nucleus with a spin 1/2, the absorption of radiation will “flip” the magnetic moment so that it will oppose the applied magnetic field. Figure 3.2: Axis of rotation precessing around the magnetic field. The magnetic field experienced at the nucleus is not equal to the applied magnetic field due to shielding from the electrons around the nucleus. If the electron density around the nucleus is high, the induced field experienced by the nucleus is stronger due to the electrons. The reverse is true if the electron density decreases. Deshielding occurs due to electron withdrawing groups and shielding results from electron donating groups. 44
- Page 3 and 4: Table of contents Abbreviations and
- Page 5 and 6: 3.4 Infrared Spectroscopy (IR) ....
- Page 7 and 8: Abbreviations and Symbols Abbreviat
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- Page 11 and 12: Opsomming 93 Nb KMR is met sukses g
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- Page 15 and 16: Synopsis... 2. Literature Review of
- Page 17 and 18: Chapter 2 Niobium resembles tantalu
- Page 19 and 20: 2.1.2 Uses Chapter 2 Niobium has a
- Page 21 and 22: 2.2 Separation of Nb and Ta 2.2.1 M
- Page 23 and 24: Chapter 2 Buachuang et al. 16 repor
- Page 25 and 26: Chapter 2 Niobium oxide surfaces ex
- Page 27 and 28: 2.4.6 Water absorption Chapter 2 Th
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- Page 31 and 32: Chapter 2 containing niobium as the
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- Page 35 and 36: Chapter 2 Figure 2.5: Structure of
- Page 37 and 38: 2.6.3.3 [NbCl3O(ttbd) - ] Chapter 2
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- Page 43 and 44: Chapter 2 Reactions of dialkylamid
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- Page 49 and 50: EtO EtO EtO EtO Cl Nb Cl Cl Nb Cl R
- Page 51 and 52: Chapter 2 The hemicarbonate formed
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- Page 57 and 58: Chapter 3 electromagnetic spectrum
- Page 59 and 60: 3.5.1 Bragg’s law Chapter 3 Bragg
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- Page 75 and 76: Chapter 4 longer bonds (C1-C2, C2-C
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Chapter 3<br />
with γ = magnetogyric ratio, h = Planck’s constant <strong>and</strong> B = strength <strong>of</strong> the magnetic<br />
field at the nucleus. The spinning nucleus generates a small magnetic field which<br />
causes it to possess a magnetic moment (µ):<br />
<br />
µ =<br />
(3.2)<br />
The magnetic moment has both magnitude <strong>and</strong> direction as defined by its axis <strong>of</strong><br />
spin. In the presence <strong>of</strong> a magnetic field, the nucleus’s axis <strong>of</strong> spin will precess<br />
around the magnetic field. The frequency <strong>of</strong> this precession is called the Larmor<br />
frequency <strong>and</strong> it is identical to the transition frequency:<br />
E = -µB cosθ (3.3)<br />
where θ is the angle between the direction <strong>of</strong> the applied field <strong>and</strong> the axis <strong>of</strong> nuclear<br />
rotation. The angle will change if the nucleus absorbs energy. In a nucleus with a<br />
spin 1/2, the absorption <strong>of</strong> radiation will “flip” the magnetic moment so that it will<br />
oppose the applied magnetic field.<br />
Figure 3.2: Axis <strong>of</strong> rotation precessing around the magnetic field.<br />
The magnetic field experienced at the nucleus is not equal to the applied magnetic<br />
field due to shielding from the electrons around the nucleus. If the electron density<br />
around the nucleus is high, the induced field experienced by the nucleus is stronger<br />
due to the electrons. The reverse is true if the electron density decreases.<br />
Deshielding occurs due to electron withdrawing groups <strong>and</strong> shielding results from<br />
electron donating groups.<br />
44