TEMA 4: Electronic Spectroscopy

1. 1Why do we use electronic spectroscopy?

   1. AWe want to get information about the rotational and vibrational states.
   2. BWe want to get information about the rotational and electronic states.
   3. CWe want to get information about the electronic and vibrational states.
   4. DWe want to get information about the electronic structure and molecular information in some cases.
2. 2Electronic Spectroscopy shows up at the ultraviolet and visible regions of the electromagnetic spectrum (10.000-100.000cm-1).

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3. 3The bands registered correspond to transitions between ro-vibrational energy levels.

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4. 4Electronic Spectroscopy allows us to get information from atoms that are neither active in RAMAN nor IR.

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5. 5When a transition takes place, there are noticeable changes in both vibrational and rotational states.

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6. 6Regarding the electronic states in polielectronic atoms, we have to assume that...

   1. AElectrons keep occupying individual spin-orbitals, with effective nuclear charges (due to shielding )
   2. BElectrons keep occupying individual spin-orbitals, without effective nuclear charges (due to shielding )
7. 7Electronic states are obtained directly from...

   1. Athe electronic Schrödinger equation.
   2. Bfrom the eigen values of the Shrödinger equation derivative with respect the time.
8. 8The electronic structure is expressed through...

   1. Adiagrams of electronic states.
   2. Ba list of unoccupied orbitals.
   3. Cits electronic configuration.
9. 92s and 2p are both examples of subshells, while 1s is a shell.

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10. 10Z*(2s)=Z*(2p)

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11. 11For electronic configurations with a full subshell there are plenty of ways to place the electrons.

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12. 12In the case of C, there is more than one way to place the electrons, because the subshell is not full.

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13. 13If we pack 2 electrons in the same orbital, there'll be less repulsion rather than packing them in separated orbitals.

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14. 14If we didn't take into consideration the repulsion between electrons, states would have the same energy.

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15. 15States or microestates are defined by its Ms, ML and Mn value.

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16. 16Which are the values of n,l and ml of the following state?
    Take into consideration that we are studying the ground state configuration of C.

    

    1. An=2, l=1, ml= -1,0,1
       n=2, l=0, ml= 0
       n=2, l=0, ml= -1,0,1
    2. Bn=2, l=1, ml= -2,0,1
       n=1, l=0, ml= 0
       n=2, l=1, ml= -1,0,1
    3. Cn=2, l=1, ml= -1,0,1
       n=2, l=0, ml= 0
       n=1, l=0, ml= 0
    4. Dn=2, l=1, ml= -1,0,1
       n=1, l=0, ml= 0
       n=2, l=1, ml= -1,0,1
17. 17We use this formula in order to calculate...

    

    1. Athe number of occupied orbitals.
    2. Bthe number of states associated to a particular electronic configuration.
    3. CThe number of states associated to all possible electronic configurations.
18. 18ML and MS are the result of the summatory of every ms and ml.

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19. 19The configuration: (2s)1 (2p)1 has 14 states.

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20. 20In atomic spectroscopy, in order to know the relative order of the energy of the different states, we should determine the S atomic term.

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21. 21The states that belong to different spectral terms have different energy since...

    1. Athe magnitude of the intraelectronic repulsion is equal for each term.
    2. Bthe magnitude of the dipole moment is different for each term.
    3. Cthe magnitude of the interelectronic repulsion is different for each term.
22. 22An atomic term contains (2S+1)(2L+2) microstates.

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23. 23Configurations with closed subshells contains spin-paired electrons, which means that there is a unique microstate.

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24. 24Configurations with open subshells, where electrons are not equivalents, the exclusion principle does apply.

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25. 25To determine atomic termis in configurations with open subshells, values of L and S goes from:
    L=[l1+l2] to [l1-l2]
    S=[s1+s2] to [s1-s2]

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26. 26In open subshells, equivalent electrons (same subshell) has a unique microstate.

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27. 27Considering the configuration: 2p1 3p1, we could...

    1. Ahave a spectral term with L=2.
    2. Bnot have a spectral term with L=2
28. 283D spectral term has...

    1. A10 states.
    2. B5 states.
    3. C9 states.
    4. D15 states.
29. 29If we sum all the states that com from the atomic terms of a certain configuration, we will obtain the total number of states of that electronic configuration.

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30. 30We have more microstates in non-equivalent electrons open shells rather than in equivalent open shells.

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31. 31When we build tables of Ms and Ml, what we are trying to do is find all the possible states of the configuration, and then gather them in the different atomic terms.

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32. 32All the states inside an atomic termal are degenerated in regards the energy.

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33. 33From these atomic terms, the most stable on is...
    1S, 3P, 1D

    1. A1S
    2. B3P
    3. C1D
34. 34Regarding the selection rules in order to have allowed transitions AS=0 and AL=0,1.

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35. 35Regarding Hund's Empirical Rule, if there is more than one term with the largest S, then the term with the largest S and the lowest L is more stable.

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36. 36(GREEN SLIDES) Spin-orbit Interaction arises because of the dynamic moment caused by the orbital motion around the nucleus, and the dynamic moment caused by the spin angular moment.

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37. 37(GREEN SLIDES) The spin-orbit interaction denpds on the total angular momentum of the electron J .

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38. 38(GREEN SLIDES) It is acceptable to compute the total angular momentum (J) by summing the total electronic angular momentum (L) and the total spin angular momentum (S) when the Spin-Orbit interaction is small.

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39. 39(GREEN SLIDES) Values of J go from the sum to the [rest] (els corxets indica valor absolut) of L and S values. 2J+1 are the states in a certain level, corresponding to the possible values of Mj.

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40. 40(GREEN SLIDES)
    3P has 3 spin-orbit interaction energy levels, which are 3P2 (5 states), 3P1(3 states) and 3P0(1 state. )

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41. 41(GREEN SLIDES) The State is a sum of H^º + H^repulsion + H^spin-orbit + H^B.

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42. 42The selection rules for the allowed transitions have to accomplish: AS=0, AL=-1,0,1 ,AJ=-1,0,1 ,AMj=-1,0,1. There is 2 exceptions: Transition J= 0 --> 0 / Transition Mj= 0-->0 if AJ=0
    

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43. 43For diatomic molecules with little Z, it is not necessary to apply the Born-Oppenheimer approximation.

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44. 44The Slater Determinant is ...

    1. Aan antisimetrized product of molecular orbitals.
    2. Ba simetrized product of molecular orbitals.
    3. Can antisimetrized sum of molecular orbitals.
    4. Da simetrized sum of molecular orbitals.
45. 45We use the Slater Determinant in order to accomplish the antisymmetry principle.

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46. 46We could say that a HOMO-LUMO transition is an electronic transition because one electron is shifted from one occupied to a virtual one.

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47. 47(GREEN SLIDES) Atomic structures build upon the hydrogenoid atom (energies and orbitals) while diatomics rely on the only solvable molecule within Born-Oppenheimer (H2).

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48. 48(GREEN SLIDES) In monoatomic molecules, the orbitals are named with greek letter (sigma, pi, delta), while in diatomic the orbitals are named with s,p,d,f.

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49. 49(GREEN SLIDES) In diatomic molecules, states are eigenfunctions of the Hamiltonian and Lz, not L^2.

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50. 50(GREEN SLIDES)
    -m: quantizes the angular momentum component on the internuclear axis .
    -landa: equals the number of nodal planes that contain the internuclear axis.

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51. 51An homonuclear molecule would have a "D infinit h" symmetry group, and it has an inversion center, so there are gerade and ungerade orbitals.

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52. 52Heteronuclear diatomic molecules would have a symmetry group of "C infinit H".

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53. 53(GREEN SLIDES) Spectral terms are actually the same as atomic terms but L=^ and these are named with capital greek letters.

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54. 54(GREEN SLIDES) As well as in monoatomic molecules, in diatomic molecules, the closed shells have only one possible microstate (1SIGMAg), while in open shells you can do the table and get the spectral terms that contain different microstates.

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55. 55(GREEN SLIDES) Transitions allowed selection rules are the same for both di and monoatomic molecules, in exception that in diatomic we have as well the transition g-->u.

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56. 56All transitions between vibrational states are allowed, there's no selection rule governing the vibrational and rotational part of the transition.

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57. 570<--0 and 1<--1 would be examples of progressions .
    0<--0 and 1<--0 would be examples of sequences.

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58. 58The intensity depends on two factors: the transition dipole moment and the population of vibrational levels.

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59. 59The Dipole moment has both nuclear and electronic contribution.

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60. 60Molecular wavefunction has one component: the electronic one.

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Beschreibung

Quiz am TEMA 4: Electronic Spectroscopy, erstellt von Montse Montané Garcia am 17/06/2018.

Autor

Montse Montané Garcia

Sprache

English

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