Orbitals

Question 1

promotion energy

Answer 1

energy required to promote e- from ground state to valence state (= one type of excited state configuration used for bonding)

Question 2

localised

Answer 2

e- are confined to a particular bond or atom

Question 3

Lewis approach to bonding

Answer 3

pairs of e- are localised in bonds or as non-bonding “lone pairs” on atoms

each bond is formed by a pair of e- shared by 2 atoms

Question 4

octet rule

Answer 4

most main group atoms will tend to end up with ns2 np6 electronic configuration

Question 5

importance of Lewis model

Answer 5

1st structure to describe bonding in structures of diatomics/polyatomics

introduced ideal of e- being shared

introduced bond order

Question 6

problems with Lewis model

Answer 6

can’t easily explain radicals

difficult to explain expansion of octet

can’t explain why O2 = paramagnetic

Question 7

paramagnetism

Answer 7

contains unpaired e-

weakly attracted by externally applied magnetic field

Question 8

valence bond theory (VBT)

Answer 8

localised quantum mechanical approach to describe bonding in molecules

provides mathematical justification for Lewis model of e- pairs making bonds between atoms

thinks of bonds formed between 2 e- waves (bond will form if there is sufficient overlap of appropriate orbitals on 2 atoms + these orbitals are populated by a max. of 2 e-)

Question 9

sigma bond

Answer 9

symmetrical about internuclear axis

Question 10

pi bond

Answer 10

have node on internuclear axis + sign changes across axis

Question 11

double bond

Answer 11

σ bond and π bond

Question 12

triple bond

Answer 12

σ bond and 2 π bonds

Question 13

diamagnetism

Answer 13

all e- paired

repelled by magnetic field

Question 14

formation of π bond

Answer 14

2px orbitals interact to form a π-bonding orbital (lies above and below plane)

2py orbitals interact in a similar way, giving a π-bonding orbital that lies at 90° to px

Question 15

bond length and bond order

Answer 15

higher bond order = shorter bond length = higher bond energy/stronger bond

Question 16

sp3 hybridisation

Answer 16

occurs in molecules with tetrahedral shape

exhibits sp3 hybridisation on central atom

Question 17

sp2 hybridisation

Answer 17

occurs in molecules with trigonal planar configuration

hybridised orbitals = σ bonds

unhybridised orbitals = π bonds

Question 18

sp hybridisation

Answer 18

linear geometry

[examples]
BeH2
CO2 - sp on C and sp2 on O

Question 19

PF5 hybridisation

Answer 19

exceeds octet

nS and nP only provide 4 - need nd orbitals (ndz2) = 5 sp3dz2

2 sets - x2 axial (180 degrees from each other)
+ x3 equatorial (120 from each other and 90 from axial bonds)

*not equivalent

Question 20

octahedral geometry

Answer 20

d2sp3 hybridisation - 6 equivalent bonds 90 degrees from each other

Question 21

MOs

Answer 21

orbitals that encompass entire molecule (constructed from AOs)

Question 22

basic rules for molecular orbital theory

Answer 22

nAOs = nMOs

to combine, AOs must have appropriate symmetry and energy

each MO must be normal and orthogonal (90) to every other MO

MO formed from weighted sums of AOs

in-phase = adding atoms
out-of-phase = subtracting atoms

Question 23

energy of bonding and anti-bonding MO

Answer 23

bonding MO = lower in energy = more stable (increased e- density between overlap due to overlap reduces repulsion)

anti-bonding MO = higher in energy = less stable (decreased e- density between overlap due to overlap increases repulsion)

Question 24

forces in anti-bonding orbital

Answer 24

attractive forces outweighed by repulsive forces

net of raising energy (reduced stability) relative to 2 atoms being apart

Question 25

forces in chemical bond

Answer 25

e- held between 2 nuclei

attractive forces outweigh repulsive forces

net reduced energy (higher stability) relative to 2 atoms being apart

Question 26

bond dissociation energy

Answer 26

net of lowering energy

Question 27

bond order = 0

Answer 27

molecule doesn’t exist

Question 28

which p orbitals can’t interact?

Answer 28

px and py/pz - don’t have correct symmetry

Question 29

which has less overlap - π or σ bonds?

Answer 29

π (as they’re 90 from each other)

Question 30

relative energies for bonding

Answer 30

σ2p < π2p < π2p < σ2p

π = more sensitive to distance

Question 31

isoelectronic

Answer 31

species with same no. of e- and same bond order

Question 32

what happens to bond order as bond length increases?

Answer 32

decreases

Question 33

sp mixing

Answer 33

energy difference between s and p = so small
that they are close enough in energy to mix

occurs in boron, carbon and nitrogen

Question 34

evidence for sp mixing

Answer 34

normal MO diagram for Boron would indicate that it’s diamagnetic = wrong

Question 35

sp mixing - degree of interaction

Answer 35

similar energy = large interaction

large energy difference = small interaction

separation between nS and nP orbitals increases as atomic number increases

Question 36

reason for stabilisation of s and p orbitals across period (left to right)

Answer 36

1 element to right = addition of proton in nucleus and additional e-

increases nuclear charge pulls orbitals closer to nucleus but increased electron repulsion pushes them further out

increased attraction = more important since e- with same quantum number don’t shield each other from nuclear charge

s orbital = stabilised to greater extent than p with increased z as it’s more penetrating

Zeff increases across row

Question 37

penetration

Answer 37

2p e- = more effectively shielded than 2s

2s has radial node (2p doesn’t) - presence of node means there’s an area of e- density relatively close to nucleus for 2s e-

2s penetrates 1s e- ∴ 2s has higher Zeff than 2p

2s = more stabilised

*e- which experience greater penetration experience less shielding ∴ increased Zeff but shield other e- more effecitvely

Question 38

consequences of sp mixing

Answer 38

π = same
σ = moves higher in energy (less bonding)

Question 39

why 2s orbitals is at lower energy than 2p for lithium whereas 2s + 2p orbitals are at the same energy for Li2+ ?

Answer 39

[Li]
1s orbital = occupied (shields 2s/2s from Z)
because of radial node, 2s penetrates 1s e- density ∴ lower in energy

[Li2+]
only contains a single e-
no shielding

Question 40

evidence for MO diagrams

Answer 40

[photonelectron spectroscopy]

to construct accurate MO diagrams, we need to determine energies for individual MOs

IE measured using Photoelectron Spectroscopy (PSE)

sample is bombarded with photons and Eke of expelled e- is measured

Question 41

MO diagram for heteronuclear diatomic molecules (XY)

Answer 41

bonding MO - more electronegative atom contributes more to bonding orbital

antibonding MO - more electropositive atom contributes more to anti-bonding orbital

*more electronegative atom sits lower in energy

Question 42

when do you get sp mixing for hetereonuclear diatomic molecules?

Answer 42

2s AOs will only interact if atoms are next to each other in the periodic table

Question 43

which is more stable - cyclic or linear arrangement of H3+?

Answer 43

although cyclic has a b.o. of 1/3 and linear = 1/2, cyclic is more stable due to lower energy in bonding MO

Question 44

what are Walsh diagrams used for?

Answer 44

to estimate electronic configuration that will provide lowest overall energy + be more stable by changing shape

Question 45

why are d orbitals not always used in hybridisation

Answer 45

v. high in energy compared in relation to s/p orbitals