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Chemistry 2 > Elementary quantum mechanics & bonding > Flashcards

Elementary quantum mechanics & bonding Flashcards

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1
Q

Planck constant (h)

A

h = 6.626 x 10^(-34) Js

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2
Q

Photoelectric effect of Red light: Are e- ejected, why?

A

No, e- are not ejected no matter how bright, because there is not enough energy to exceed the threshold to eject e-

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3
Q

Photoelectric effect of Green light: Are e- ejected, why?

A

Yes, e- ejected no matter how dim, always JUST enough energy to exceed the threshold to eject e-

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4
Q

Photoelectric effect of Blue light: Are e- ejected, why?

A

Yes, e- ejected no matter how dim, always MORE than enough energy to exceed the threshold to eject e-

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5
Q

Rydberg equation

A

1/λ = R(H) x [1/(n(1))^2 − 1/(n(2))^2]

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6
Q

Wavelenght, wavenumber and frequency

A

Wavelength = λ
Wave number = 1/λ
Frequency: v = c/λ

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7
Q

What is wave-particle duality

A

Light is both wave-like and particle-like

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8
Q

Experiment that shows wave-like properties

A

The double-slit experiment where diffraction and interference occur

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9
Q

Experiment that shows particle-like properties

A

Photoelectric effect of Compton scattering on X-rays due to e-

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10
Q

Photon momentum (p)

A

p = E/c = hv/c = h/λ

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11
Q

Heisenberg uncertainty principle description

A

A quantum object cannot simultaneously have an exact position and exact momentum with arbitrary precision

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12
Q

Heisenberg uncertainty principle, relations

A

[(uncertainty in linear momentum parallel to axis q)(uncertainty in position along q axis)] is greater than or equal to [0.5(h/2π)]

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13
Q

2 classic types of waves

A
  1. Travelling (flowy)

2. Stationary (switchy)

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14
Q

Stationary wave description

A
  1. Oscillates in time, but remains @ stationary point
  2. Waves reflected at boundary interfere such that only stationary waves remain
  3. Characterised by no. of nodes
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15
Q

What is a node (on a stationary wave)?

A

Fixed positions where u(x,t)=0 at all times

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16
Q

How are quantum objects characterized?

A

By a wavefunction

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17
Q

Born interpretation of a wavefunction

A

Square of a wavefunction is proportional is proportional to the probability density of finding the particle at that point in space

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18
Q

How to obtain the probability of finding a particle @ x

A

Multiplying the probability density by a volume element

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19
Q

What are the conditions for an acceptable wavefunction?

A
  1. Continuous
  2. Continuous 1st derivative
  3. Single valued
  4. Finite
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20
Q

Why does the quantum system of a particle in a 1D box have no potential energy?

A

Because only kinetic energy is possible inside the box, as the particle does not exist outside the boxes boundaries

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21
Q

Wave equation for motion (1D box) - What values must the wavelength at 0 and L be

A

Zero, as the particle does not exist outside the box

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22
Q

Wave equation for motion (1D box) - New wavefunction superimposing both opposing wavefunctions caused my interference

A 
Ψ(x) = 2A cos(kx) OR
Ψ(x) = 2iA sin(kx)
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23
Q

Wave equation for motion (1D box) - Ψ(x) = 2Ai sin(kx) and possible values of n

A 
Ψ(x) = 2Ai sin(kx)
Sin(nπ) = 0, Ψ(0) = 0, Ψ(L) = 0

Ψ(L) = 2iA sin(kL)
2iA sin(kL) = 0
Sin(kL) = 0
kL = nπ, where n = 1, 2, 3,…

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24
Q

Particle in a 1D box - Energy quantization and what energies cannot occur

A
  1. Energy of particle can only take certain discrete values
  2. Energy is quantized with quantum no. n
  3. E=0 (n=0) cannot occur as the particle always has some energy, zero-point energy even at T=0K.
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25
Q

What is zero-point energy and why do we have it?

A

Particle always has some energy, so zero-point energy is the lowest reachable energy level.

This is because E=0 would mean no motion and therefore complete localization which is impossible for quanta.

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26
Q

Quantum system of Hydrogen atom’s interactions - Key points

A
  1. Potential energy < 0 because it is an attractive interaction
  2. Interaction has spherical symmetry
  3. Uses 3D Schrodinger equation
  4. Uses Spherical polar coordinates
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27
Q

Spherical polar coordinates - Key points

A

(r, θ, ϕ)

Can be split into a radial term (r) and an angular term (θ, ϕ):
Ψ(r, θ, ϕ) = R(r)*Y(θ, ϕ)

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28
Q

Spherical polar coordinates - r

A

r = Distance from centre, 0 ≤ r ≤ ∞

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29
Q

Spherical polar coordinates - θ

A

θ = Co-latitude, 0 ≤ θ ≤ π

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30
Q

Spherical polar coordinates - ϕ

A

ϕ = Azimuth, 0 ≤ θ ≤ 2π

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31
Q

Quantum system of Hydrogen atom - Solutions of Schrodinger equation KEY POINTS

A 
n = principle quantum no., = 1, 2, 3,...
l = Orbital angular momentum quantum no., = 0,..., n-1
ml = Magnetic orbital quantum no. = -l,..., l
32
Q

Quantum systems - What is an orbital

A

Wavefunction of an electron

33
Q

Degenerate orbitals

A

Several orbitals of the same energy

Same n but have different l and ml

34
Q

Radial part of orbital general form

A

Radial part = Rn,l(r)

Rn,l(r) = normalization factorrpolynomial in r*decaying exponential in r

35
Q

Radial part of orbital - Key points

A
  1. Radial part determines the spatioal extent of the wavefunction, essentially whether or not the e- spends more/less of its time close/far from the nucleus
  2. For larger r, radial part is directly proportional to exponential decay
  3. Only s-orbitals have a finite value (start above 0) at the nucleus
36
Q

Radial nodes

A

Orbital has (n-l-1) nodes, so the 1st time a particular l occurs (e.g. 1s, 2p, 3d) there are no nodes

Following on from the 1st occurrences of l, each higher l has 1 more node

Location of node found by setting bridging term to 0 and rearranging for r

37
Q

Radial distribution function (rdf)

A

Probability of finding the e- in spherical shell of thickness dr at distance r

38
Q

Radial distribution function - Probability density

A

(wavefunction)^2

39
Q

Radial distribution function - Probability

A

Probability = Probability density*Volume

P(r) = R(r)^2 * dV
= R(r)^2 * 4π(r)^2 dr
= 4π * (r)^2 R(r) dr

40
Q

Radial distribution function - Properties

A
  1. Contains factor (r)^2 so the rdf is 0 at the nucleus
  2. R(r)^2→0 exponentially for r→∞, as does the rdf
  3. (r)^2 increases with r, while R(r)^2 decays with r, hence the rdf goes through 1+ maximums
  4. Where R(r), and also R(r)^2, has a node, the rdf = 0
41
Q

Spherical harmonics - Properties

A
  1. Function of angular part of orbital
  2. Determines the shape of the orbital
  3. Real-valued functions can be defined by cartesian coordinates (x, y, z)
  4. There are (2l + 1) angular functions for each value of l
  5. s-functions have no angular dependence, they are spherically symmetrical
42
Q

Angular nodes

A

An orbital has l angular nodes where Y(θ, ϕ) = 0

43
Q

Kinetic energy of e- in H atomic orbitals

A
  1. Radial part is more highly curved, w/ 1 radial node, than the node of 2p, 0 nodes, therefore 2s has more RADIAL KINETIC ENERGY
  2. Angular part of 2s less curved than that of 2p, therefore, 2p has more ANGULAR KINETIC ENERGY
44
Q

Effect of nuclear charge (Z) on wavefunction for H-like ions

A

Higher Z leads to:

  1. Contraction of orbitals
  2. Lower energies for the e-
45
Q

Effective nuclear charge, Zeff

A

Any 1 electron does not ‘feel’ the full nuclear charge (Z), but an effective Zeff. Net attraction reduced by the repulsion of the other electrons.

Zeff < Z

46
Q

Stern-Gerlach experiment

A

Beam of Ag atoms, w/ 1 unpaired e-, in inhomogeneous magnetic field
Observation: Atoms are deflected exactly 2 ways
Interpretation: The atom’s magnetic moment can be orientated in 2 ways

47
Q

Spin quantum no. (s)

A

Quantisation of spin angular momentum

48
Q

Magnetic spin quantum no. (ms)

A

Quantisation of the projection onto the z-axis.

ms is restricted to (2s + 1) values

49
Q

Quantisation of e- spin

A

The electron has spin 1/2
2 allowed orientations w/ respect to Z: ms = +1/2 or -1/2
Called spin up (alpha) or spin down (beta)

50
Q

Shielding

A

Filled inner orbitals shield the +ve charge of the nucleus that an outer electron experiences

51
Q

Penetration

A

An e- in an outer orbital that has a significant probability of being found inside an inner orbital

52
Q

Why does 2s fill before 2p?

A
  • 2s is less shielded than 2p because it penetrates (1s)2 shell, so it therefore experiences HIGHER Zeff than 2p.
  • 2s is therefore lower in energy, due to greater Zeff attraction
  • so, 2s fills before 2p
53
Q

Summary of shielding AND penetration effects

A
  1. Inner e- partially shield outer e- form full Z
  2. If a valence orbital penetrates the core, the shielding is weaker and attraction is stronger
  3. Weaker shielding = Higher Zeff = Lower energy = Shell fills first
54
Q

Hund’s rule

A

e- occupying degenerate orbitals prefer to have the same value of ms

Configuration with the highest no. of unpaired spins is preferable

55
Q

Atomic radii trends: Increasing down group

A

Orbitals w/ successively higher principle quantum no. n are occupied

w/ each new period, the no. of fully filled shells increases, complete shells shield the best

56
Q

Atomic radii trends: Decreasing across period

A

W/ increasing nuclear charge e- are attracted more strongly to nucleus, making atoms more compact

57
Q

Ionisation energy trends: Decreasing down group

A

Shielding becomes more effective

Outermost e- is therefore bound more weakly

58
Q

Ionisation energy trends: Increasing across period

A

Nuclear charge increases, so the e- are bonded more strongly

59
Q

Ionisation energy anomalies: Li

A

Lower than the avg. trend

First time 2s is occupied
Well shielded by complete (1s)2 shell, making it easier to remove

60
Q

Ionisation energy anomalies: B

A

First time 2p is occupied

p electrons are less strongly bound than s electrons.
As s is lower in energy than p, making the outermost p electron easier to remove

61
Q

Ionisation energy anomalies: O

A

First time any of the 2p orbitals are doubly occupied

Electron-electron repulsion is higher for an e- in a doubly occupied orbital.
This lowers the binding energy, making the e- in the double occupied orbital easier to remove

62
Q

Treating a polyene as a 1D box: Conj. Pi system structure

A

6 pz orbitals in C atoms overlap, forming a molecular orbital that extends across the length of molecule

Occupied by 6e- which can move over entire length of molecule

63
Q

Treating a polyene as a 1D box: Orbital assumptions

A

Interaction between e- does not change the solutions

Simply use orbitals for 1e- in a box, and occupy 6e-

64
Q

Treating a polyene as a 1D box: UV-Vis absorption

A

Total En = E4 - E3 = 0.79 * (10)^-18 J
Wavelength = hc/ Total En = 2.5 * (10)^-7 m = 250nm

In experiment:
Lowest UV absorption of 1.3.5-Hexatriene is at wavelength = 258nm

65
Q

Combining separate atoms in a molecule:

Electrons in the bonding MO

A
  • Higher probability of being close to both nuclei & building up a density between nuclei
  • Internuclear density & nuclei attract each other, STABILISING
  • Density pulls nuclei together, BONDING
  • More stable than separate atoms
66
Q

Combining separate atoms in a molecule:

Electrons in the antibonding MO

A
  • 0 probability of being close to both nuclei & deplete e- density
  • Nuclei pulled apart by density outside internuclear region
  • Node means the e- has high KE, which is STABILISING
  • Less stable than separate atoms
67
Q

Calculating bond order

A

BO = [(no. of e- in bonding MO)-(no. of e- in antibonding MO)]/2

68
Q

Sigma-bonding MOs

A
  • Rotationally symmetric w/ respect to rotation about bond axis
  • Formed by overlapping s-orbital or from p-orbitals orientated along the bond axis
69
Q

Pi-bonding MOs

A
  • Antisymmetric w/ respect to nodal plane containing bond axis
  • Formed from overlapping p-orbitals orientated perpendicular to bond axis
70
Q

Pi-antibonding MOs

A
  • Antisymmetric w/ respect to nodal plane containing bond axis
  • Additional nodal plane perpendicular to bond axis
71
Q

Paramagnetism and example

A
  • O2 is paramagnetic
  • Paramagnetic = material that becomes magnetised by an external magnetic field and is weakly attracted to external field
72
Q

Diamagnetism and example

A
  • N2 is paramagnetic

- Diamagnetic = Material is weakly repelled by an external magnetic field

73
Q

Para/diamagnetism experiment

A
  • Pour O2 and N2 between the poles of a magnet

- O2 holds between the 2 poles

74
Q

Para/diamagnetism experiment: why does O2 hold

A
  • O2 holds between the 2 poles because:
  • 2 parallel-spin unpaired e- in pi-antibonding orbitals
  • parallel spins lead to magnetic dipole moment
  • this allows O2 molecules to align and be attracted by external field
75
Q

Producing an MO diagram

A
  1. Are the possible overlaps possible? matching energy, phase and symmetry?
  2. What will the shape of the diagram be? Overlap of orbitals w/ same/different energies?
  3. Include bonding, antibonding and non-bonding orbitals
76
Q

Producing an MO diagram: Shape of overlapping orbitals of SAME energies

A

Equal contribution to MO from both AO, same weight and same coefficient

77
Q

Producing an MO diagram: Shape of overlapping orbitals of DIFFERENT energies

A
  • Dominant contribution to MO from energetically closer AO
  • Bonding MO resembles the lower-lying AO
  • Antibonding MO resembles the higher-lying AO

Chemistry 2 (3 decks)

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