Inorganic and Materials Chem 2 Flashcards

1
Q

what is the basic idea behind metallic bonding, how can we get an estimate of the strength of metallic bonding

A
  • metallic bonding is what is seen in metals
  • we can obtain an estimate of the strength by looking at the energy required to vaporise the metal
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2
Q

what are the three main metallic structures, what do we mean when we speak about bonds in metals/ why is this slightly problematic

A
  • fcc, bcc, hcp
  • when we use the word ‘bonds’ for metals, we don’t necessarily mean the standard 2-centre 2-electron bond that we tend to deal with
  • it usually just means the number of nearest neighbours
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3
Q

how can we think of metallic bonding, use lithium as an example to show the difference between metallic and simple covalent bonding

A
  • we can consider it in the same way as normal covalent bonding
  • normally lithium is a metal but under certain conditions it can be a diatomic
  • Li2 has a bond strength of 238kJ/mol, i.e. formation of 1 mole is 119kJ/mol
  • for metallic lithium, enthalpy of atomisation is about 159kJ/mol
  • this shows for some elements it is more energetically favourable to have 8 slightly weaker bonds than one stronger bond
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4
Q

using orbital sizes explain why the left side of the periodic table and the bottom of the p-block are metallic

A
  • from left to right, Zeff increases and orbital size decreases, smaller orbitals have more effective overlap, eventually it is efficient enough for localised bonding
  • on the right and at the bottom, orbital sizes are greater, this means that bonding is weaker so it is better to have less-directional, delocalised bonding - metals
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5
Q

what happens to the VDW bonding radius and metallic bonding radius and their ratio moving down a group (particularly halogens)

A
  • they both increase but the VDW radius increases slower so their ratio gets closer to 1
  • in a true metal, the ratio is 1
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6
Q

what happens to the separation of MO energies when moving from Li2 to Li3 Li4 etc. up to infinity, how does this contribute to a ‘band’ theory

A
  • the energy difference between the most B MO and most AB MO increases as the number of atoms in the chain increases
  • at infinity the energy difference is twice that of in Li2
  • this is because in Li2 all atoms only form one B or AB interaction
  • whereas in at an infinite chain length all atoms form 2 B or AB interactions
  • the separation of the orbital energies also decreases so as you tend to infinity, we end up with a virtually continuous range of MO energies - bands
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7
Q

what do we mean by orbital bands, what are its limitations

A
  • when we have a huge metallic lattice e.g. 10^23 atoms, the spacing between individual COs (crystal orbitals) is so small that there is effectively a continuous range of energies from the lowest bonding CO to the highest antibonding CO
  • these continuous sections are called bands

NOTE: there is not really a continuous range of energy levels within a band but it is close

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8
Q

what does a density of states graph show

A
  • you can calculate the number of energy levels within a given energy range
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9
Q

define band width

what does it depend on

A

band width is the difference in energy between the lowest bonding orbital and highest antibonding orbital

i.e. the energy range of the band

  • it depends on the degree of overlap possible between the AOs, (greater overlap for smaller orbitals etc.)
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10
Q

when considering energy and orbital bands rather than discrete orbitals, where are the net bonding and net antibonding regions

A
  • the net bonding region is the bottom half of the band
  • the net antibonding region is the top half of the band
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11
Q

what can we say about crystal orbitals and the movement of electrons, what condition does this put on conductivity

A
  • COs come in degenerate pairs
  • one corresponds to electrons moving one way and the other corresponds to electrons moving the other way
  • this means for a completely filled band, there is no net movement of electrons so no conductivity
  • if a band is partially empty, when an electric field is applied, some electrons near the top of the filled part of the band can move into the vacant higher energy orbitals and there is a net movement of electrons
  • hence, a metal conducts if it has a partially filled band
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12
Q

when are there deviations from our predicted filled bands = monoatomic gas rule

A
  • consider Be, you expect all of the 2s orbital band to be filled leading to net AB and hence monoatomic gas
  • This does not occur because at the equilibrium bond length there is an overlap of the 2s and 2p bands
  • this means the highest energy 2s band electrons move to the lowest energy 2p band places
  • this leads to a net lowering of energy
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13
Q

at what point does an element with a filled valence shell become conductive

A
  • at the point where the ‘HOMO’ band e.g. 2s for Be just starts to overlap with the ‘LUMO’ band e.g. 2p for Be
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14
Q

define/classify metals

A
  • have delocalised bonding with each atom typically having eight to twelve nearest neighbours. The presence of partially filled bands or overlapping bands allow conductivity
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15
Q

define/classify semimetals

A
  • typically have structures with fewer nearest neighbours than metals and are usually poorer conductors of electricity than true metals.
  • either due to presence of V small band gap or bands which overlap to a very small degree
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16
Q

what occurs to the electrical conductivity of Metals, Semimetals and Semiconductors with a change in temperature

A
  • metals and semimetals have lower conductivity at higher temps
  • semiconductors have higher conductivity at higher temps
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17
Q

define/classify semiconductors

A
  • conduct electricity poorly but the conductivity increases with temperature and when irradiated with light of a suitable wavelength
  • band gap of 0.1-4eV
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18
Q

define/classify non-metals

A
  • electrical insulators
  • filled conduction band has large energy gap to vacant conduction band
  • energy gap usually > 4eV
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19
Q

what does the trend of metals to non-metals follow and why

A
  • the trend follows the trend in the sizes of valence orbitals
  • on crossing a period, Zeff increases, valence orbital size decreases, better overlap
  • this gives more localised bonding, fewer nearest neighbours
  • on descending a group, n increases so valence orbital gets bigger so tends towards metals
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20
Q

consider H2 molecules, explain when it has simple B and AB MOs and when we can consider bands, is solid H2 conductive?

A
  • H2 molecules have B and AB MOs, when gaseous, the atoms are very far apart and do not easily interact
  • when the pressure is increased and they are forced together, the MOs from different molecules start to interact to form new orbitals at different energies
  • eventually we obtain bands
  • the lower energy sigma band and higher energy sigma* band are totally separated with a large energy gap so it is a conductor (unless under extreme pressure)
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21
Q

explain what the form of the MOs at the top and bottom of the sigma and sigma* band in solid hydrogen

A
  • our solid structure still contains molecules but all aligned regularly like in a solid
  • bottom of sigma band = many H2 molecules each with bonding MOs all aligned such that there are no nodes between molecules
  • top of sigma band = many H2 molecules with bonding MOs but with the maximum number of nodes between molecules
  • bottom of sigma* band = many H2 molecules with antibonding MOs but with no nodes between molecules
  • top of sigma* band = many H2 molecules with antibonding Mos and the max number of nodes between molecules
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22
Q

how can we consider more complicated structures diamond using a band theory like for solid hydrogen

A
  • in reality it all gets v complicated, v quickly for 3D structures
  • but we can take a simplified view and do the same thing
  • i.e. with all sigma in phase at bottom of sigma band etc.
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23
Q

explain the structure of graphite

A
  • in graphite there are hexagonal layers of sp2 carbons, ABAB structure
  • each have one p-orbital contributing to a pi network over the whole molecule
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24
Q

what orbitals does each atom have in graphite, which bands does this form

A
  • each atom has
    three sigma B AOs
    one pi B MO
    one pi* AB MO
    three sigma* AB MOs

this forms bands of
filled sigma band
filled pi band
empty pi* band
empty sigma* band

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25
Q

considering the band system of graphite (sketch out) explain why graphite conducts

A
  • the delocalised pi system itself doesn’t make graphite conductive
  • But when sketching out the bands we can see that the filled pi band and the empty pi* band just touch
  • this means electrons can easily be promoted to higher energy levels so graphite will conduct in the pi plane
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26
Q

what type of material is graphite

A
  • a semimetal
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27
Q

what is the purpose of doping

A
  • doping is where materials can have improved qualities by treating them with other materials, dopants
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28
Q

explain how/why graphite is doped and what qualities this improves

A
  • graphite is usually doped to improve it’s electrical conductivity
  • this is done by either doping with atoms which will donate electrons to give a partially filled pi* band e.g. potassium
  • or with atoms which will accept electrons to give a partially filled pi band, e.g. bromine
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29
Q

what other effects does doping graphite cause

A
  • the added atoms are usually between layers increasing interlayer spacing
  • it often changes the structure to an AAAA structure
  • it increases both conductivity in the plane and perpendicular to the plane
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30
Q

what are the 4 main ways that we can describe bonding

A

1) LCAO, linear combination of atomic orbitals

2) HAOs

3) VB - Valence Bond model, keep track of all valence electrons using Lewis Structures

4) VSEPR - Valence shell electron pair repulsion - a kind of extension to HAOs but without much focus on known orbitals

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31
Q

Briefly explain the basis of VB

A
  • VB (valence bond) model is a simple way to keep track of valence electrons
  • always considers them as localised pairs
  • either bonding pairs or non-bonding (lone) pairs
  • what we use in curly arrow mechanisms
  • resonance structures sometimes required
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32
Q

state the basic idea behind VSEPR

A
  • the basic idea is valence shell electron pair repulsion
  • it does not consider valence shells/orbitals as MO theory would, it is the idea that all valence shell electron pairs repel each other
33
Q

give rule 1 of VSEPR

A
  • the most favourable arrangement of a given number of electron pairs in the valence shell of an atom is that which maximises the distance between them
34
Q

give rule 2 of VSEPR

A
  • a non-bonding pair of electrons occupies more space than a bonding pair
35
Q

give rule 3 of VSEPR

A
  • the ‘size’ of a bonding pair between the central atom and a ligand decreases with the increasing electronegativity of the ligand
36
Q

give rule 4 of VSEPR

A
  • the two or three electron pairs of a double or triple bond occupy more space than the one electron pair of a single bond
37
Q

considering bonding pairs only i.e. no lone pairs, give the shapes for
Bonding pairs:
2,3,4,5,6,7,8

A

2 = linear
3 = trigonal planar
4 = tetrahedral
5 = trigonal bipyramidal
6 = octahedral
7 = mono-capped octahedron
8 = square antiprism

38
Q

what are some problems that can occur with rule 1

A
  • often for a given number of ligands, several different arrangements are possible
  • or there are some molecules where VSEPR doesn’t help at all
39
Q

what is the physical justification of rule 2 of VSEPR

A
  • when an electron pair is shared between two atoms, it is more restrained or localised than when it is only associated with one atom
40
Q

what are some problems with rule 2 of VSEPR

A
  • sometimes the lone pair repulsion doesn’t seem to count at all
  • these are exceptions though
41
Q

what is the physical justification of rule 3 of VSEPR

A
  • in a bond to a very electronegative ligand such as fluorine, the electron pair will be pulled towards the ligand so will occupy less space
42
Q

what are some problems with rule 3 of VSEPR

A
  • sometimes it just doesn’t work
  • especially when comparing fluorine with hydrogen as hydrogen is small anyway
43
Q

what is fluxionality

A
  • bond vibrations and distortions can lead to interconversion between different geometries
  • this can be rapid if the energy difference is sufficiently small and the activation energy is available
44
Q

give an example of a molecule of when fluxionality is very prominent, explain how we know using NMR

A

PF5 is a good example

  • it should be trigonal bipyramidal
  • this would give 2 19F NMR peaks and a triplet of quartets on the 31P NMR spectrum
  • at higher temps (or even RTP) PF5 interconverts rapidly between trigonal bipyramidal and square based pyramidal
  • this gives a single doublet in the 19F NMR and a sextet in the 31P NMR
  • this is because we see the ‘average’ of the two
45
Q

what is the name of the mechanism by which fluxionality occurs

A

Berry pseudorotation

46
Q

what are the limitations of VSEPR theory, when do we need to use MO instead

A
  • VSEPR is ok but it’s just a a model with a lack of theoretical foundation
  • in particular it doesn’t work that well for transition metals
  • when we find that it doesn’t work, we should use MO theory instead
47
Q

what is the main premise of VSEPR, give an example where it doesn’t work and what should we do instead

A
  • the main premise of VSEPR is that each bond is a 2e2c shared pair
  • in some molecules this is not possible as there are simply not enough valence electrons e.g. B2H6
  • in this case, we must use MO theory
48
Q

in B2H6, explain how VSEPR doesn’t work, but MO theory explains it instead

A
  • using VSEPR, we would have 16 valence electrons
  • this doesn’t work
  • what we find is using MO theory, each terminal B-H bond is a 2e2c bond as expected
  • but the central B-H bonds are 2e3c bonds
  • the form of these bonds can be predicted using the sine rule for delocalised systems
49
Q

what is a hypervalent compound

A
  • it is a compound where if using the 2c2e VSEPR bond system, you would expect it to have too many valence electrons
50
Q

explain the hypervalent bonding in PF5

A
  • we think of the phosphorous as sp2 hybridized, each sp2 HAO overlaps with F to make standard sigma bonds on the equatorially
  • axially the bonding is that the remaining p orbital on the phosphorus overlaps, with the two remaining fluorines to form 3 MOs: bonding, non-bonding, antibonding
  • there are a total of 4 electrons in this system, filling the bonding and non-bonding orbitals
  • we only think about the 2 bonding electrons to give a 3c2e system axially
51
Q

what can we say about the positions of lone pairs in hypervalent compounds

A
  • they will always be equatorially
  • this is because if they were axially, there would be a p-orbital containing 4 electrons, this cannot happen
52
Q

what can we say about the lengths of 2c2e bonds compared to bonds in 3c2e systems

A
  • 3c2e bonds are longer
53
Q

why do period 2 p-block elements (first row p-block) not form hypervalent compounds

A
  • they are simply too small to fit more than four other atoms around them
54
Q

why do we need to be careful when considering bond strengths

A
  • bond strengths can change a lot depending on if they’re an average or for a particular molecule
  • they are always positive because bonds require energy to be broken
55
Q

how can we calculate bond strengths using enthalpies of reaction, formation etc.

A
  • set up a Hess cycle
  • deduce the enthalpy of atomisation for a molecule
  • assuming the molecule only contains one type of bond, you can just divide the atomisation enthalpy by the number of bonds
56
Q

what are the three main factors on bond strength

A

1) size of orbitals (smaller orbitals = better overlap)

2) degree of ionic character (more ionic character = stronger bond)

3) repulsion around the central atom - the bond strength generally decreases as the number of atoms around the central one increases

57
Q

state the main 3 trends in homonuclear bonds of periods 2,3,4

A

1) strongest bonds are for group 14 (C, Si…) NOT G15 solids (N2) as G14 have no antibonding electrons

2) atomisation energies: P2>P3>P4 (some exceptions)

3) atomisation energies either side of carbon are fairly equal as you move out (B = N, Be = O etc.)

58
Q

explain why N triple bond N is not the strongest bond strength in the first p row (period 2)

A
  • if you draw out the MO energy diagram for N2 you find that N2 does have an occupied antibonding orbital (sigma u*)
  • this is not the case for carbon
59
Q

explain the general bonding of the group 14 elements

What allotropes are formed

What hybridisation do they form

Which MOs are filled

A
  • all solids at RTP, have same structure as diamond (different bond lengths)
  • consider each atom to be sp3
  • sp3 HAOs from each atom overlap form bonding sigma and antibonding sigma* MOs
  • for n carbon atoms
    n(4 x sp3) HAOs —> 4n MOs, 2n bonding, 2n antibonding

as we only have 4n electrons they completely fill the bonding MOs only

60
Q

explain a good way to think about bonding/non-bonding MOs when considering the number of bonds/lone pairs formed for each group

A
  • we can often think of it as the antibonding MO electrons ‘cancelling out’ the bonding MO electrons to form a non-bonding pair
61
Q

explain the bonding of group 15 elements

rationalise the number of bonding/ lone pairs

A
  • each element has 5 valence electrons
  • the 5th goes into an antibonding orbital
  • thus we expect fewer bonds than G14

overall we have 1e- in antibonding, 4e- in bonding

we use the cancelling logic to give

3e- in bonding, 2e- in non-bonding

thus each G15 element forms 3 bonds, 1 lone pair

62
Q

what is the standard allotrope of most group 15 elements

A
  • each (apart from nitrogen) has a rhombohedral allotrope made from hexagonal layers
63
Q

explain the bonding/rationalise number of bonding/lone pairs in group 16 elements

A
  • each element has 6 valence electrons
  • two more than G14

4 e- in bonding
2 e- in antibonding

use the cancellation to give

2e- in bonding
4e- as lone pairs

so 1 bond, 2 lone pairs

64
Q

what is the most stable form of sulphur and a possible form of lots of other G16 elements

what other forms is common (most dense form)

A
  • the orthorhombic alpha form
  • consists of eight membered ring
  • the most dense form consists of 6 membered rings (similar to diamond structure)
65
Q

what form does tellurium exist in (what other elements can exist in this form)

A
  • it exists in a single crystalline form consisting of a network of spiral chains
  • sulphur and selenium
66
Q

explain the bonding/rationalise number of bonding/lone pairs in group 17 elements

A
  • 7 valence electrons
  • 4 e- in bonding
  • 3e- in antibonding

so effectively 6 e- in non-bonding
1e- in bonding

  • thus, it forms 1 bond and has 3 lone pairs
67
Q

what is the most common form of G17 elements

A
  • individual X2 units arranged in layers
68
Q

what can we say about the bonding and structure in isoelectronic materials

A
  • we can use the same arguments as before
  • for example, KSi forms Si- and K+ ions
  • the Si- ions are isoelectronic to the G15 elements so we see similar bonding
  • it forms anionic sheets of covalently bonded Si atoms with layers of K+ ions in between
69
Q

explain why O2 and N2 are anomalous in their groups (existing as diatomic gases)
explain in terms of bond enthalpies

A
  • the N-N, O-O and F-F sigma bonds are anomalously weak
  • this means whereas its more energetically favourable for elements like P to form extensive solid lattices with each atom forming 3 bonds, 1 lp, this is not the case for nitrogen
  • it is more energetically favourable for N and O to form diatomics due to the fact that their pi bonds are much stronger than their sigma bonds (unusual)
70
Q

explain in terms of lone pair repulsion and bonding/antibonding orbitals why the bond enthalpies for N-N, O-O and F-F are anomalously weak

A
  • common answer = lone pair-lone pair repulsion weakens these bonds far more than for others due to short bond length

OR can be explained with MOs:
- for each of the molecules, HOMO = antibonding pi*
- these are raised in energy more than the pi MOs are lowered
- so when they’re filled, they cancel out the pi completely and some of the sigma bonding, weakening the bond

  • this effect becomes less as you move down the group
71
Q

how can we calculate the change in enthalpy for an ionic bond at an infinite separation

A

ionisation energy for cation + electron attachment enthalpy of anion

72
Q

what is wrong with our model for ionic bond enthalpy at infinite separation, what attractional energy term can we include

A
  • our model never predicts an exothermic process
  • we know it doesn’t always occur at infinite separation
  • we include an electric attraction energy term
    Eattraction = Qq/4piEo r
  • this term helps but suggests the lowest energy point is at no separation
73
Q

what is the problem with our model for ionic bonding including an attractional energy term, how do we further fix this

A
  • it suggests the lowest energy point is at no separation
  • we haven’t accounted for nuclear repulsion, Erepulsive
  • Erepulsive is directly proportional to 1/r^n
  • n is the Born exponent NOT the principal quantum number
74
Q

what do we get if we combine Eattractive, Erepulsive

Eionpair

A
  • we end up with a graph that does have an equilibrium bond separation
  • adding together the terms gives

Eionpair = (-NAQq / 4pieo*ro) (1-1/n)

75
Q

How can we derive Elattice from Eionpair, what is the constant that arises

A
  • we must consider each of the atoms with which it interacts, e.g. in NaCl it has 6 nearest neighbours which it is attracted to
  • 12 neighbours that are repulsive at sqrt(2) away etc.

this gives

Elattice = Eionpair (6-12/sqrt(2) + 8/sqrt(3)….)

the bit in the brackets converges and is the Madelung constant

76
Q

how can the madelung constant be incorporated into the expression for Eionpair to give the Born-lande expression for Elattice

A

multiply through by it to give

Elattive = (-MNA / 4piE0) (Z+Z- / r0) ( 1-1/n)

77
Q

how can experimental values for lattice enthalpy be found

A
  • using a Born-Haber cycle
78
Q

do the experimental and calculated values for lattice enthalpy agree?

when they don’t, what is the general reason

A
  • generally the values agree well
  • where they don’t, it is usually because the covalent character hasn’t been accounted for
  • e.g. especially in compounds with Tl+ or Ag+
79
Q

explain the Kapustinskii equation / the simplifications that it does for the Born-Lande eq.

A

Madelung constant / n(ions) is approx constant at 0.87

so NAM / 4pi*Eo can be replaced in the Born-Lande eq. by -1.075

  • NOTE: radius in pm, ro = r- + r+, energy in KJ/mol

n(ions) = number of ions in one molecule e.g. NaCl = 2
MgCl2 = 3