inorganic bonding - molecular orbital theory

Question 1

what is molecular orbital theory?

Answer 1

the idea that during bonding, electrons are placed into multicentre molecular orbitals, and are delocalised over the whole molecule

Question 2

give 2 approximations that MO theory requires

Answer 2

1- orbital approximation - that the overall wavefunction of a molecule with N electrons can be written as the product of N single-electron wavefunctions
2- the linear combination of atomic orbitals - as atomic orbitals are single electron wavefunctions that can interact with each other, MOs are formed from weighted sums of atomic orbitals (in-phase by addition, out-of-phase by subtraction)

Question 3

molecular orbital definition

Answer 3

the single electron wave functions in molecules

Question 4

what is the purpose of a normalisation constant, N?

Answer 4

for the linear combination of atomic orbitals, a normalisation constant must be used to ensure the wave function remains a single electron wavefunction

Question 5

bonding orbital definition

Answer 5

the molecular orbital created from the in-phase combination of similar orbitals

Question 6

what does it mean if electron density is enhanced between the nuclei?

Answer 6

bond formation

Question 7

what does it mean if electron density is reduced between nuclei?

Answer 7

bond is broken/no bond is formed

Question 8

antibonding orbital definition

Answer 8

molecular orbital created from out-of-phase combination of orbitals

Question 9

what orbitals are given from the in-phase and out-of-phase combinations of 2 1s orbitals?

Answer 9

1 orbitals are formed:
in-phase gives σg bonding MO
out-of-phase gives σu* antibonding MO

Question 10

what does σ mean in terms of orbitals?

Answer 10

refers to single/standard bonding/antibonding MOs - a σ orbital has cylindrical symmetry, it can be rotated about the internuclear axis with no change in appearance

Question 11

what do g and u mean in terms of orbitals?

Answer 11

this refers to symmetry about an inversion centre, they are parity labels
g = gerade, meaning the orbital stays the same on inversion (symmetrical)
u = ungerade, meaning the orbital changes on inversion (not symmetrical)

Question 12

what does * mean in terms of orbitals?

Answer 12

this indicated the presence of a nodal plane between the nuclei, this means it is an antibonding orbital

Question 13

give the notation for single bonding and antibonding orbtials

Answer 13

bonding - σg
antibonding - σu*

Question 14

describe the energetic differences between bonding and antibonding MOs

Answer 14

bonding MOs are always at a lower energy than the initial atomic orbitals, indicating stability
antibonding MOs are always at a higher energy to the initial atomic orbitals

Question 15

what does π mean in terms of orbitals?

Answer 15

π refers to directional interactions of p-orbitals, such as sideways overlap - indicated double/triple bonds

Question 16

how do π orbitals differ to σ orbitals?

Answer 16

π bonding orbitals are asymmetric when inverted
π antibonding orbitals are symmetric when inverted
- this is opposite to σ orbitals
therefore bonding = πu and antibonding = πg*

Question 17

how do p-orbitals interact to form MOs?

Answer 17

p-orbitals are directional so they don’t all interact the same way:
pz orbitals interact in-phase to form σg bonding orbitals, and out-of-phase to form σu* antibonding orbitals
py and px orbitals can interact in-phase to form πu bonding orbitals, and out-of-phase to form πg* antibonding orbitals

Question 18

what is the difference between π orbitals created from sideways overlap of px orbitals and py orbitals?

Answer 18

there is no difference, other than direction
both sets of πu/πg* MOs are degenerate

Question 19

describe the energetic differences between π and σ bonding and antibonding MOs

Answer 19

π interactions are less efficient than σ interactions, so π MOs are stabilised/destabilised to a lesser degree than σ MOs

Question 20

can 2 different types of p-orbitals interact to form π bonds?

Answer 20

no, must be px+px/py+py, cannot be px+py/pz
this is because p-orbitals are directional, so there is no net interaction between 2 different types of p-orbital as they cannot overlap without the interaction cancelling out (essentially, fully in-phase or out-of-phase interaction is not possible)

Question 21

π-bonds are most common for period 2 elements, why?

Answer 21

π interactions are more sensitive to distance than σ interactions, and π-bonds/double bonds are much shorter, so best suit period 2 elements which have small atomic radii

Question 22

how is bond order calculated?

Answer 22

no. filled bonding orbitals - no. filled antibonding orbitals

Question 23

how are MOs filled?

Answer 23

electrons are places in MOs using aufbau + pauli principle + hunds first rule

Question 24

why can UV radiation break bonds, in terms of molecular orbitals?

Answer 24

UV radiation is of high enough energy that it is able to promote an electron from the bonding to antibonding orbitals, thus the excited species is left with a bond order of 1 less than before, indicating that a bond was broken
e.g. bond order 1 -> bond order 0

Question 25

how can bond order be used to determine if molecules can exist?

Answer 25

a molecule can only exist if bond order between atoms is greater than 0 - if bond order = 0 or less, there is no stabilisation upon formation of the molecule, so it will not happen - does not have to be a whole number! just has to be >0

Question 26

the bond dissociation enthalpy for Li2 is less than for H2, why?

Answer 26

Li2 involved interaction between 2s orbitals, whereas H2 uses 1s orbital interactions, 2s orbitals are more diffuse than 1s so their interaction in Li2 is less than the interaction in H2, so less energy is needed to break the bond

Question 27

how can H2 become an electrical conductor?

Answer 27

electrons can hop in between H2+ ions when is it in the form of a metallic fluid, free flowing electrons means it can carry a charge - this metallic fluid is formed from H2 at very high pressures

Question 28

HOMO definition

Answer 28

highest occupied molecular orbital

Question 29

LUMO definition

Answer 29

lowest occupied molecular orbital

Question 30

paramagnetism definition

Answer 30

an atoms weak attraction to a magnet

Question 31

why does paramagnetism occur?

Answer 31

occurs when parallel electron spins are able to align with an applied magnetic field, so they behave as tiny magnets

Question 32

give one use of paramagnetism

Answer 32

paramagnetism of O2 can be used to measure O2 content in air by measuring magnetism of the gas - this is used in premature baby incubators

Question 33

diamagnetism definition

Answer 33

an atoms weak repulsion from a magnet - this is a much weaker effect than paramagnetism

Question 34

when does diamagnetism occur?

Answer 34

occurs in atoms/molecules with fully paired electrons

Question 35

why do peroxide O2^2- ions and superoxide ions O2^- have different BDEs?

Answer 35

they have different bond orders: peroxide bond order = 1 superoxide bond order = 1.5 so superoxide bond is stronger, so must have a greater BDE

Question 36

outline the relationship between bond order and bond strength

Answer 36

the greater the bond order, the stronger the bond

Question 37

isoelectronic definition

Answer 37

refers to 2 atoms/molecules which the same electronic structure + no. of valence electrons

Question 38

s-p mixing definition

Answer 38

the interaction between σ orbitals derived from s and p atomic orbitals

Question 39

how does s-p mixing affect the energies of MOs?

Answer 39

stabilises s based σu and establishes p based σ g therefore pushing σg orbital above πu orbitals π orbitals are unaffected by s-p mixing as they are of the wrong symmetry to interact with the σ orbitals

Question 40

name 1 compound in which s-p mixing takes place

Answer 40

N2 - this is why σ and π orbitals are out of typical order also occurs in O2 and F2 but it is much weaker, and so isn't enough to move orbitals out of order

Question 41

how does s-p mixing change across a period (L ->R) ?

Answer 41

across the period the s-p gap increases so s-p mixing becomes less significant - the s-p gap increases because increasing nuclear charge pulls electrons towards nucleus, Zeff increases across a period, which stabilises s and p orbitals, but s orbitals are stabilised to a greater extent than p as s is more penetrating

Question 42

how are energies for individual MOs calculated?

Answer 42

ionisation energy is measured using photoelectron spectroscopy - the sample is bombarded with photons and kinetic energy of expelled electrons is measured photon energy = IE + electron kinetic energy

Question 43

give the 3 possible interactions of d-orbitals

Answer 43

end on interaction gives σ orbitals edge on interaction gives π orbitals face on interaction gives δ orbitals

Question 44

how can stability of atomic orbitals be determined for heteronuclear diatomics?

Answer 44

the more electronegative atom has its valence orbitals at lower energy

Question 45

give the wavefunction for a heteronuclear diatomic molecule

Answer 45

bonding orbital: Ψ = N[Cxϕx + Cyϕy] for molecule X-Y, where Cx and Cy are unequal constants antibonding : Ψ* = N*[C'xϕx - C'yϕy] where C'x and C'y are unequal constants

Question 46

how are Cx and Cy affected by electronegativity?

Answer 46

if Y is more electronegative than X, Cy > Cx, meaning the more electronegative atom contributes more to the bonding orbital, distribution of electrons is towards Y

Question 47

how are C'x and C'y affected by electronegativity?

Answer 47

if Y is more electronegative than X, C'x > C'y, meaning the more electropositive atom contributes more to the antibonding orbital

Question 48

how are g/u MO labels changed for heteronuclear diatomics?

Answer 48

they are not used because inversion changes the positions of the nuclei (no symmetry)

Question 49

how are atoms in molecules different to free atoms?

Answer 49

atoms in molecules can change the shapes/energies of their atomic orbitals to provide the best overlap + maximise bonding

Question 50

how is energy conserved in MO theory?

Answer 50

total energy of new molecular orbital set must = total energy of original atomic orbitals

Question 51

why is it difficult to extend MO theory from diatomic to polyatomics?

Answer 51

this is because it is difficult to identify which atomic orbitals have the correct symmetry to interact together, this is explained using group theory - exception is linear molecules

Question 52

describe the bonding in a linear molecule AX2, where A is a group 2 element, using a valence bond theory approach

Answer 52

electronic configuration ends in s2, since molecule structure is linear, we know sp hybridisation will occur as 2 sp hybrid orbitals are 180 apart so s + pz -> 2 sp hybrids, + px + py new directional sp hybrid orbitals engage with X atoms forming 2 σ bonds and remaining p orbitals stay unoccupied

Question 53

describe the bonding in a linear molecule AX2, where A is a group 2 element, using an MO theory approach

Answer 53

combinations of X orbitals can interact with A orbitals of the same symmetry, in-phase combinations have the correct symmetry to interact with A orbitals to give 2 sets of bonding/antibonding orbitals, px and py cannot interact with X orbitals so form non bonding orbitals MO diagram shows 2 lowest bonding orbitals are both occupied, one for each A-X bond, although these orbitals are different, the 2 A-X bonds are identical as electron density from both bonding orbitals contributes to each bond

Question 54

what is the purpose of partial molecular orbital schemes?

Answer 54

can provide good understanding of bonding in a molecule, especially useful for explaining bonding in electron poor/rich compounds that do not obey the octet rule

Question 55

hypervalent compounds definition

Answer 55

compounds in which the central atom breaks the octet rule - elements in period 3+ are capable of forming these

Question 56

how are hypervalent compounds accounted for by valence bond theory?

Answer 56

d-orbitals are included in the hybridisation, allowing for geometries in which the octet has been expanded

Question 57

what is the hybridisation in square planar complexes?

Answer 57

dsp2, involving dx2y2 / px / py / s

Question 58

what is the hybridisation in trigonal bipyramidal complexes?

Answer 58

dsp3, involving dz2 / all p and s

Question 59

what is the hybridisation in square pyramidal complexes?

Answer 59

dsp3, involving dx2y2 / all p and s

Question 60

what is the hybridisation in octahedral complexes?

Answer 60

d2sp3, involving dx2y2 / dz2 / all p and s

Question 61

why is MO theory preferred for modelling the bonding in hypervalent compounds?

Answer 61

valence bond theory approach can result in strange non whole number bond orders

Question 62

how are hypervalent compounds accounted for using MO theory?

Answer 62

e.g. PF5 = trigonal bipyramidal molecule with 10 ve- equatorial bonds are based on sp2 (px+py) hybridisation of central P atom interacting with F p-orbitals, average bond order = 1 axial bonded F atoms interact with P both via pz orbitals (left over from hybridisation)

Question 63

explain bridging of group 13 atoms like B using valence bond theory

Answer 63

B has 3 valence electrons each B = sp3 hybridised as tetrahedral 2 sp3 hybrid orbitals interact with with H 1s orbital forming B-H σ bonds - uses up 2 electrons from B remaining electron from B in 1 remaining sp3 orbital, the other is empty - both hybrid orbitals are involved with bridging H atoms, each bridge containing 1 electron (electrons are delocalised around the bridging)

Question 64

explain bridging of group 13 atoms like B using MO theory

Answer 64

bonding orbital formed by in-phase combination of B and H orbitals antibonding orbital formed by out-of-phase combination of B and H orbitals out of phase combinations of B orbitals do not have the correct symmetry to interact with H 1s orbital therefore 2 leftover B sp3 orbitals from both B atoms form the bonding, antibonding and non bonding orbitals for bridge bonds, bond order = 1 total = 0.5 per bond