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NST Chemistry > Shapes and structures II > Flashcards

Shapes and structures II Flashcards

1
Q

photoelectron spectroscopy is used to determine…

A

how much energy is needed to remove a single e- from an atom
by bombarding sample w X-rays of photons of known energy

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2
Q

define degenerate (orbitals)

A

orbitals that have the same energy
eg. there are 5 degenerate 3d orbitals

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3
Q

four quantum numbers

A

principle, n (shell)
angular, l = 0 to n-1 (s, p, d - orbital)
magnetic, m = +l to -l, integers (orientation)
also m(s), spin

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4
Q

for a wave function ψ, what is given by ψ^2?

A

the probability density - a measure of the probability of finding an e- in a given position

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5
Q

simple form of Schrodinger’s eqn

A

En = -Rh (z^2/n^2)

En = energy of wave function
Rh = Rhyberg constant (it can be written out in terms of a bunch of other things)
z = nuclear charge of 1e- system
n = principle quantum number

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6
Q

the energy of an orbital of a 1e- system depends only on __

A

n, the principle quantum number
as shown in schrodinger’s

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7
Q

how can a wave function be expressed in terms of polar coordinates?

A

ψ can be in terms of cartesian coordinates OR polar (r,θ, ø)
RADIAL defined by n, l, function of r
ANGULAR defined by l, m, function of θ, ø

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8
Q

what does the RDF show?

A

radial distribution function:
the probability of finding an e- at a given radius, summed over all angles
(sum of e-dens for a thin shell)

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9
Q

RDF eqn

A

RDF = (R(r))^2 x 4πr^2
wave function x area of spherical surface

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10
Q

difference between ψ^2 and RDF

A

ψ^2 shows probability of an e- being in a certain volume element at a set of coordinates (position)
RDF shows probability of an e- being at a certain RADIUS

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11
Q

all s orbitals…

A

have spherical symmetry - the value of the wave function depends only on r, not on θ, ø

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12
Q

define node

A

where wave function is 0

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13
Q

if orbitals have the same n and l, then…

A

they have the same radial parts (same nodes etc)
R depends only on n, l

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14
Q

how do radial and angular nodes appear in density plots?

A

radial = circles
angular = planes

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15
Q

total number of nodes in a H orbitals rule

A

n-1 total orbitals
depends only on principal quantum number

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16
Q

number of angular nodes rule

A

= angular number, l
eg. s: l=0, 0 angular nodes
p: l=1, 1 angular node
d: l=2

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17
Q

rule for number of radial nodes

A

total - angular
n-1-l

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18
Q

name and describe the five 3d orbitals

A

3d(xy), 3d(xz), 3d(yz), 3d(x2-y2), 3d(z2)
all four lobes except 3d(z^2), which has two nodal cones + donut shape

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19
Q

describe orbital approximation

A

in multi-e- systems, the effect of other electrons on one particular electron can be modified into Zeff nuclear charge

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20
Q

in multi-e- systems, explain why orbitals with the same principal quantum number but different angular are no longer degenerate.

A

due to penetration of the orbitals, so that they experience more of the nuclear charge
eg. 2s penetrates more than 2p
the orbitals are also more contracted overall due to greater nuclear charge
(think of this on the RDF vs r graph)

so ordering energy levels becomes complicated

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21
Q

trend in Zeff going across the table

A

Zeff increases, as e- in the same shell do not shield each other very well
so + nuc charge > +e-

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22
Q

trend in orbital energy going across the table

A

decreases as Zeff increases

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23
Q

LCAO

A

linear combination of atomic orbitals, either in or out of phase
to form molecular orbitals by assuming nuclei are at a fixed distance apart and AO interact

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24
Q

bonding orbital arises from ___ combination
anti-bonding arises from __

A

Bonding from in-phase combination, lower energy
anti-bonding from out-of-phase combination

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25
Q

why does in-phase combination result in a lower energy MO than out of phase?

A

1MO > 2AOs, more space = more uncertainty in position = less uncertainty in momentum = lower Ek of e- in MO.
also, anti bonding node results in greater repulsion of nuclei

26
Q

the more e- in bonding MOs, …

A

the stronger the bond

27
Q

how does separation of nuclei affect the energies of the MOs?

A

more internuclear separation –> less interaction between AOs –> smaller difference in energy between bonding and anti bonding MOs

28
Q

what happens when same #e- in each of the bonding and ab MOs?

A

total energy is always positive
ab MO raised more than bonding MO lowered
hence unstable, molecule falls apart to minimise energy

29
Q

calculation of bond order

A

1/2 (#e- in bonding MOs - #e- in ab MOs)

30
Q

identify sigma and pi MOs

A

sigma: axis of rotation about internuclear axis, sign does not change on path perpendicular to axis

31
Q

how to identify u and g MO labels

A

ONLY if molecule has a centre of inversion!!!!!
g = even = no change of sign passing through centre of inversion
u = uneven = changes sign

32
Q

compare end-on (sigma) vs side on (pi) overlap for 2p orbitals

A

at typical bond length separation, end-on overlap is better, so sigma orbitals are raised / lowered in energy more than pi

33
Q

paramagnetic molecule

A

has unpaired electrons

34
Q

THREE factors determining how well AOs overlap

A
  1. suitable symmetry (orientation of orbitals)
  2. close in energy
  3. size of orbitals - two small orbitals interact more (greater raise/lower in energy)
35
Q

define covalent bond in terms of AOs and MOs

A

when two atoms contribute equally to form a MO (ie. same/v similar energy), a covalent bond is formed

36
Q

define ionic bond in terms of AOs / MOs

A

two atoms involved have different energy and unequal contributions to the MO (e- not shared equally), the bond is polarised and has more ionic character

37
Q

explain s-p mixing

A

two MOs close in energy can further interact
same symmetry: eg. 2sσg and 2pσg, and 2sσu and 2pσu

38
Q

overall effect of s-p mixing

A

for all first row homonuclear diatomic up to and including N, the bonding σ orbital is HIGHER in energy than the bonding pi orbitals

39
Q

describe LCAO for three atoms

A

3 AOs –> 3 MOs
bonding, non-bonding and anti bonding
each MO can still hold 2e-
2e- in bonding MO –> 3-centre-2-electron bond

40
Q

how are sp3 HAOs created

A

combining 3x p orbitals and 1x s orbital

41
Q

sp2 and sp hybridisation

A

sp2: 1x s orbital, 2x p orbitals –> 3 sp2 HAOs + unchanged p orbital

sp: 1x s, 1x p –> 2sp, 2 unchanged p

42
Q

what does it mean if an orbital has more/less s or p character?

A

p-orbitals point at 90˚ to each other, so reducing bond angle = more p-character, remaining sp3 HAOs have more s character

43
Q

hybridisation states for square planar shape

A

sp2d or p2d2
four ligands needs four AOs to make four HAOs
three p orbitals, the d(z2) and d(x2-y2) orbitals point along exes toward ligands, hence suitable to interact

44
Q

hybridisation states for trigonal bipyramidial

A

sp3d or spd3

45
Q

hybridisation for octahedral

A

sp3d2

46
Q

describe the pi system in a benzene ring

A

six 2p orbitals (leftover, pointing out of the plane, as Cs are sp2 hybridised) combine to give six MOs
each MO holds 2e- so the three lowest energy MOs are occupied

all conjugated pi systems can be considered this way

47
Q

how to predict the MOs formed by combining out of plane p orbitals

A

using the sine wave rule

because higher energy MOs have more nodes

48
Q

describe the sine wave rule

A

start and end one bond length outside end, start with n-1 nodes for n C’s, reduce to 0
height at each C = coefficient for contribution from each p orbital

49
Q

how to determine number of electrons in the pi system

A

COUNT.
total available e- minus e- already in sigma bonds and lone pairs

50
Q

where in the MO does the nucleophile attack?

A

attacks into the LUMO
at the position where there is the greatest e- density as predicted by the sine wave rule

51
Q

how can the shapes of molecules be predicted from MOs?

A

basically trial and error for each of the MOs
eg. assume linear, measure energy, alter, measure again. lowest energy configuration most favourable

52
Q

a reaction between a HOMO of one reactant and the LUMO of another results in

A

a net lowering of energy of the most energetic electrons

53
Q

the best HOMO-LUMO interaction is…

A

the one where orbitals are closest in energy

54
Q

ordering of orbitals by energy

A

highest to lowest energy:
sigma * anti bonding
pi * anti bonding
non bonding
pi bonding
sigma bonding

55
Q

nucleophilic addition to carbonyl general mechanism

A

Nu- attacks carbonyl (LUMO) C, forming new bond and simultaneously breaking C-O pi bond (O now has -ve charge, where the e- density has shifted)

56
Q

how is the angle of Nu- attack rationalised by MOs

A

compare net constructive/destructive interaction between the HOMO and LUMO - most constructive = best angle

57
Q

why do nucleophiles not readily attack the C=C pi* orbital?

A

C=C has no dipole moment, hence no initial electrostatic attraction bringing species together.
C-C pi* higher energy than C-O pi*, hence worse match with the HOMO of the Nu-
C is also much worse at holding -ve charge than the O

58
Q

Nucleophilic substitution in terms of MOs

A

Nu- attacks into a sigma* MO, breaking a sigma bond

59
Q

how to predict relative energies of C-X vs C-Y MOs

A

compare the energy match and size match between the two AOs
greater difference in energy / size = poorer interaction, MOs likely lower in energy

60
Q

how do MOs explain inversion during a nucleophilic substitution reaction?

A

MOs show that the best orientation of approach of the Nu- HOMO to the LUMO to maximise constructive orbital interaction is from the SIDE

61
Q
A

NST Chemistry (16 decks)

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