Orbitals Flashcards
promotion energy
energy required to promote e- from ground state to valence state (= one type of excited state configuration used for bonding)
localised
e- are confined to a particular bond or atom
Lewis approach to bonding
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
octet rule
most main group atoms will tend to end up with ns2 np6 electronic configuration
importance of Lewis model
1st structure to describe bonding in structures of diatomics/polyatomics
introduced ideal of e- being shared
introduced bond order
problems with Lewis model
can’t easily explain radicals
difficult to explain expansion of octet
can’t explain why O2 = paramagnetic
paramagnetism
contains unpaired e-
weakly attracted by externally applied magnetic field
valence bond theory (VBT)
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-)
sigma bond
symmetrical about internuclear axis
pi bond
have node on internuclear axis + sign changes across axis
double bond
σ bond and π bond
triple bond
σ bond and 2 π bonds
diamagnetism
all e- paired
repelled by magnetic field
formation of π bond
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
bond length and bond order
higher bond order = shorter bond length = higher bond energy/stronger bond
sp3 hybridisation
occurs in molecules with tetrahedral shape
exhibits sp3 hybridisation on central atom
sp2 hybridisation
occurs in molecules with trigonal planar configuration
hybridised orbitals = σ bonds
unhybridised orbitals = π bonds
sp hybridisation
linear geometry
[examples]
BeH2
CO2 - sp on C and sp2 on O
PF5 hybridisation
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
octahedral geometry
d2sp3 hybridisation - 6 equivalent bonds 90 degrees from each other
MOs
orbitals that encompass entire molecule (constructed from AOs)
basic rules for molecular orbital theory
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
energy of bonding and anti-bonding MO
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)
forces in anti-bonding orbital
attractive forces outweighed by repulsive forces
net of raising energy (reduced stability) relative to 2 atoms being apart
forces in chemical bond
e- held between 2 nuclei
attractive forces outweigh repulsive forces
net reduced energy (higher stability) relative to 2 atoms being apart
bond dissociation energy
net of lowering energy
bond order = 0
molecule doesn’t exist
which p orbitals can’t interact?
px and py/pz - don’t have correct symmetry
which has less overlap - π or σ bonds?
π (as they’re 90 from each other)
relative energies for bonding
σ2p < π2p < π2p < σ2p
π = more sensitive to distance
isoelectronic
species with same no. of e- and same bond order
what happens to bond order as bond length increases?
decreases
sp mixing
energy difference between s and p = so small
that they are close enough in energy to mix
occurs in boron, carbon and nitrogen
evidence for sp mixing
normal MO diagram for Boron would indicate that it’s diamagnetic = wrong
sp mixing - degree of interaction
similar energy = large interaction
large energy difference = small interaction
separation between nS and nP orbitals increases as atomic number increases
reason for stabilisation of s and p orbitals across period (left to right)
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
penetration
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
consequences of sp mixing
π = same
σ = moves higher in energy (less bonding)
why 2s orbitals is at lower energy than 2p for lithium whereas 2s + 2p orbitals are at the same energy for Li2+ ?
[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
evidence for MO diagrams
[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
MO diagram for heteronuclear diatomic molecules (XY)
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
when do you get sp mixing for hetereonuclear diatomic molecules?
2s AOs will only interact if atoms are next to each other in the periodic table
which is more stable - cyclic or linear arrangement of H3+?
although cyclic has a b.o. of 1/3 and linear = 1/2, cyclic is more stable due to lower energy in bonding MO
what are Walsh diagrams used for?
to estimate electronic configuration that will provide lowest overall energy + be more stable by changing shape
why are d orbitals not always used in hybridisation
v. high in energy compared in relation to s/p orbitals
