Elementary quantum mechanics & bonding Flashcards

Question

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

Answer 1

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.

Answer 2

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

Answer 3

(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)

Answer 4

r = Distance from centre, 0 ≤ r ≤ ∞

Answer 5

θ = Co-latitude, 0 ≤ θ ≤ π

Answer 6

ϕ = Azimuth, 0 ≤ θ ≤ 2π

Answer 7

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

Answer 8

Wavefunction of an electron

Answer 9

Several orbitals of the same energy | Same n but have different l and ml

Answer 10

Radial part = Rn,l(r) Rn,l(r) = normalization factor*r*polynomial in r*decaying exponential in r

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

(wavefunction)^2

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

Higher Z leads to: 1. Contraction of orbitals 2. Lower energies for the e-

Answer 21

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

Answer 22

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

Answer 23

Quantisation of spin angular momentum

Answer 24

Quantisation of the projection onto the z-axis. | ms is restricted to (2s + 1) values

Answer 25

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)

Answer 26

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

Answer 27

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

Answer 28

- 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

Answer 29

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

Answer 30

e- occupying degenerate orbitals prefer to have the same value of ms Configuration with the highest no. of unpaired spins is preferable

Answer 31

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

Answer 32

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

Answer 33

Shielding becomes more effective Outermost e- is therefore bound more weakly

Answer 34

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

Answer 35

Lower than the avg. trend First time 2s is occupied Well shielded by complete (1s)2 shell, making it easier to remove

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

Interaction between e- does not change the solutions Simply use orbitals for 1e- in a box, and occupy 6e-

Answer 40

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

Answer 41

- 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

Answer 42

- 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

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

- Antisymmetric w/ respect to nodal plane containing bond axis - Additional nodal plane perpendicular to bond axis

Answer 47

- O2 is paramagnetic - Paramagnetic = material that becomes magnetised by an external magnetic field and is weakly attracted to external field

Answer 48

- N2 is paramagnetic | - Diamagnetic = Material is weakly repelled by an external magnetic field

Answer 49

- Pour O2 and N2 between the poles of a magnet | - O2 holds between the 2 poles

Answer 50

- 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

Answer 51

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

Answer 52

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

Answer 53

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

Elementary quantum mechanics & bonding Flashcards

(77 cards)