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PT IA Chemistry > The Shapes And Structures Of Molecules Pt. 2 > Flashcards

The Shapes And Structures Of Molecules Pt. 2 Flashcards

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1
Q

Do electrons in atoms with more than two electrons have the same energy, how can we record how much energy is needed to remove an electron

A
  • no they have different energies
  • we can use photoelectron spectroscopy
  • this bombarded a sample with x-ray photons and measuring the KE of the removed electron to tell the difference
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2
Q

How many electrons can an orbital hold

A

2 of opposite spins

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3
Q

What are orbitals of the same energy described as

A

Degenerate

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4
Q

What 3 things do we need to distinguish between atomic orbitals and what do we use to express these

A
  • which ‘shell’ it is in
  • whether it’s an s,p,d,f orbital
  • which of the orbitals in each subshell it’s in
  • each of these is expressed by a quantum number
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5
Q

What is the principal quantum number, what form does it take and what can it determine

A

The principal quantum number, n, specifies which shell is being referred to

n takes an integer value 1,2,3…

For a one electron system, n alone determines the energy of the electron

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6
Q

What is the angular momentum number, what form does it take

A

The angular momentum number, l, specifies which type of subshell orbital we are referring to (s,p,d,f…)

  • l takes integer values from 0 to (n-1)

Each value of l has a different letter associated with it

Value of l: 0,1,2,3,4,5
Letter used: s,p,d,f,g,h

E.g. where n = 3, l = 0,1,2, these are s,p,d

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7
Q

What can the value of the angular momentum number l, take and what is the equation

A

The value of l determines the orbital angular momentum of the electron

Angular momentum = (h/2(pi)) sqrt(l(l+1))

Orbitals with different l numbers have different spatial arrangements

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8
Q

What is the magnetic quantum number

A

The magnetic quantum number, ml, is the number that tells us which orbital in the subshell it is

ml takes integer steps from +l to -l

E.g. for a P orbital, l = 1
So ml = -1, 0, 1
Hence there are 3 p orbitals

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9
Q

What is the intrinsic angular momentum, s, of all electrons

A

s = 1/2

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10
Q

what additional information/number do we need to be able to identify an individual electron in an orbital rather than simply the orbital alone

A
  • the orientation of the electron’s intrinsic angular momentum, ms

ms = +1/2 or -1/2
for spin up (up arrow) or spin down (down arrow) respectively

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11
Q

what is a useful analogy for thinking about the orbital angular momentum of an electron and the intrinsic angular momentum of an electron

A
  • we can think of the orbital angular momentum as the angular momentum the electron has due to its movement within an orbital
  • we can think of the intrinsic angular momentum as the momentum an electron has due to its spin on its own axis
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12
Q

summarise the purpose of each of the letters when specifying the state of an electron

A
  • n specifies the electron energy
  • l specifies the magnitude of the orbital angular momentum (s,p,d etc.)
  • ml specifies the orientation of the orbital angular momentum (which s,p,d orbital)
  • ms specifies the orientation of the spin angular momentum (spin up or spin down)
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13
Q

what are two things that MUST happen when describing electrons

A
  • in an orbital if there are two electrons they MUST have opposite spin
  • any electron in an atom MUST have a unique set of 4 quantum numbers
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14
Q

what can the true properties of an electron be described using

A

a wavefunction represented by psi

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15
Q

what is ψ

A

a wavefunction, its a function of coordinates that describes the probability of finding the electron e.g. (xyz)

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16
Q

how do we write our wavefunction and what do the different bits represent

A

ψ[n,l,ml] (x,y,z)

ψ represents the wavefunction
[n,l,ml] shows that the specific function changes depending on the orbital, where our orbital is given by n,l,ml

(x,y,z) are examples of coordinates where our wavefunction is a function of coordinates

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17
Q

give a brief overview of the born interpretation of the wavefunction

A
  • the wavefunction, ψ, gives numbers associated with different coordinates
  • ψ ^2 or (ψ)(ψ*) gives the probability density
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18
Q

how can we calculate wavefunctions and energies associated with them

A

using the Schrodinger equation

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19
Q

when can we find exact solutions to the Schrodinger equation and what do these solutions represent

A
  • we can solve Schrodinger’s equation exactly for any, one-electron system
  • the solutions give the atomic orbitals
  • there are a variety of solutions depending on the quantum numbers of the orbital
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20
Q

What is the equation for the energy, En of a wavefunction (in a one electron system)

A

En = (- (mee^4)/(8Eo*h^2)) (z^2/n^2)

where
me = mass of electron
e = charge on an electron
h = Planck’s constant
Eo = permittivity of free space
z = nuclear charge of atom
n = principal quantum number

this can be simplified to

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

where Rh = a constant

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21
Q

what are the key features to note about the equation for the energy of a wavefuntion

A
  • the energy of an orbital depends on n only
  • this means that 2s and 2p have the same energy, and 3s, 3p and 3d all have the same energy unless in a multi-electron system
  • the predicted energies are also negative because zero energy represents an electron totally free from the nucleus
  • this makes sense because as n –> infinity
    En –> 0
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22
Q

what is a physical interpretation of the constant Rh

A

Rh = ionisation energy for 1H atom from 1s shell

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23
Q

what coordinates are usually used in the Schrodinger equation and why

A
  • spherical polar coordinates
    (r,θ,φ)
    r = radius
    θ = polar angle
    φ = azimuthal angle
  • this simplifies our equation from

psi [n,l,ml] (r,θ,φ) = R [n,l] (r) x Y [l,ml] (θ, φ)

where R [n,l] (r) is the radial part of the wavefunction, dependent on n and l and defined in terms of r

and Y is the angular part of the wavefunction dependent on l and ml and defined in terms of θ and φ

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24
Q

what is the only factor that a 1s orbital depends on

A

radius, note: this is only the distance from the nucleus, it does not take into account displacement

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25
Q

what does the wavefunction against radius graph look like for a 1s orbital (hydrogen atom)

A
  • a decreasing exponential with radius
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26
Q

what do density plots and contour plots show

A
  • density plots show regions where the wavefunction has higher values
  • the higher the value of the wavefunction, the darker the shading on the density plot
  • contour plots show a ‘slice’ through the cube-space and have lines joining certain points with the same wavefunction value
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27
Q

what does an isosurface plot show

A
  • if you join up all the positions within the cube plot with the same particular value of the wavefunction then you get an isosurface plot
  • it is in 3D
  • exactly what it looks like depends on the value of the wavefunction chosen and the particular orbital
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28
Q

what is an RDF, (radial distribution function)

A
  • shows how the total electron density varies at a distance r from the nucleus, summed over all angles of φ and θ
  • although it can be thought of as the electron density in a thin shell, it depends on r^2 not r^3 because we work off surface area not volume
  • it is the product of φ^2 and r^2

RDF = [R(r)]^2 x (4pi*r^2)

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29
Q

what is the difference between an RDF and phi^2

A
  • phi^2 tells us the probability of finding an electron at a small volume delta(v) at a set of coordinates
  • RDF gives probability of finding an electron at a given radius r from the nucleus
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30
Q

what do the density plots and contour plots look like for the 1s orbital

A
  • a gradually less shaded circle
  • the contour plot is circles at different radii, where the value of the wavefunction increases at radii closer to the nucleus
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31
Q

what does the isosurface plot look like for the 1s orbital (hydrogen atom)

A
  • spheres of differing radii depending on the value of the wavefunction chosen
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32
Q

what does the RDF look like for the 1s orbital (hydrogen atom)

A
  • a line which rises from the origin to a maximum at 1 Bohr radius
  • it then drops and becomes asymptotic to 0 as the e^-r term becomes dominant
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33
Q

what are two features of all s orbitals

A
  • although they may be different shapes, all s orbitals have spherical symmetry
  • all s orbitals have no φ or θ terms so only depend on r
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34
Q

what is the Bohr radius and what is its symbol

A
  • a(subscript)o is the symbol
  • it represents the radius at which an electron is most likely to be found for a hydrogen atom in its ground state
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35
Q

what does the wavefunction against radius plot look like for a 2s orbital (hydrogen atom)

A
  • a decreasing curve which drops below the x axis then rises to become asymptotic to the x axis but still negative
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36
Q

what does the RDF look like for a 2s obital (hydrogen atom)

A
  • starts at 0, rises to a small peak, drops back to 0 at bohr radius 2
  • rises up to a large peak at approx 4.5 bohr radii, then drops back down to become asymptotic to 0
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37
Q

what is the significance of the point where the wavefunction crosses the x axis/ goes +ve to -ve

A
  • its the point where there’s no chance of finding an electron, i.e. a radial node
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38
Q

what does the density plot of a 2s orbital look like (in a hydrogen atom)

A
  • a dark sphere immediately around the nucleus, then a light patch in a ring around that at the radial node (at 2 bohr radii)
  • then it becomes daker again in a ring before fading back to light
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39
Q

what do the 2p orbitals (and every orbital apart from s) depend on

A
  • the radial parts of the orbitals (the parts as a function of r) don’t depend on ml, so they’re the same for px, py and pz etc.
  • the angular parts of the orbitals (the parts as a function of θ and φ) do depend on ml so change for px, py and pz etc.
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40
Q

what is a feature that all p orbitals have

A
  • they all have 1 angular node
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41
Q

what does the (radial) wavefunction-radius and RDF plot for a 2p orbital look like
(in a hydrogen atom)

A
  • the wavefunction-radius plot rises slightly from 0 to a peak then slowly drops and is asymptotic to 0 again
  • the RDF rises from 0 concavely then hits a large peak at about 4.2 bohr radii and falls pack down to be asymptotic to 0
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42
Q

what do the 2px, 2py and 2pz isosurface plots look like, give their angular nodes in each case (in a hydrogen atom)

A
  • dumbells
  • 2px is along the x-axis
  • 2py is along the y-axis
  • 2pz is along the z-axis

angular nodes are at
2px, φ = 90
2py, φ = 0,180
2pz, θ = 90

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43
Q

what does the density plot for 2p orbitals look like (in a hydrogen atom)

A
  • two dark oval shapes either side of the centre
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44
Q

what does the wavefunction-radius plot and RDF look like for 3s (in a hydrogen atom) and what are the key features

A
  • wavefunction-radius plot starts very high, drops down to pass through the x-axis at 2 bohr radii, it then slowly rises back up passing back through the x-axis at approx. 7 bohr radii and becomes asymptotic to the x-axis
  • the RDF starts at 0 has a small peak at 0<r<2 (bohr radii) then hits 0 at r=2, theres another peak at 2<r<7, then another root at r = 7 then a large peak and it becomes asymptotic to the x-axis
  • the points where the wavefunction plot has a root is where the RDF has drops to 0, these are the radial nodes
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45
Q

what do the (radial) wavefunction - radius and RDF plots look like for a 3p orbital (in a hydrogen atom)

A
  • the R(r) function rises from 0, then hits a peak and slowly decreases, passing through a root(r = 6), it has a small negative peak then remains negative but becomes asymptotic to the x-axis
  • the RDF starts at 0, there’s a peak, it drops back to 0 at r=6, it then has a large peak and drops back to become asymptotic to the x-axis
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46
Q

what is an additional feature of a 3p orbital compared to a 2p orbital (in a hydrogen atom), link this to the RDF plot

A
  • a 3p orbital contains a radial node as well as an angular node, whereas the 2p orbital only contains an angular node
  • this makes sense because the RDF plot of the 2p orbital never reaches 0 (apart from at origin) but the RDF plot of 3p does, (at r=6)
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47
Q

what do the isosurface and density plots of a 3p orbital look like (in a hydrogen atom)

A
  • a smaller 3D oval shape, followed by a radial node, then a larger 3D oval shape on top, this occurs in both directions along the x,y, and z directions
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48
Q

what are the similarities and differences between the different types of 3d orbitals (in a hydrogen atom)

A
  • the radial parts of all 5 3d orbitals are the same but the angular part depends on ml so changes
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49
Q

what does the radial wavefunction-radius plot and RDF look like for the 3d orbitals (in a hydrogen atom)

A
  • the wavefunction-radius plot is a small peak that peaks around 6 bohr radii then drops down to become asymptotic to the x-axis
  • the RDF is similar but peaks later and has a larger peak
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50
Q

what is the group of three 3d orbitals that have similar properties (in a hydrogen atom), and what are their features

A
  • 3dxy, 3dxz, and 3dyz
  • each have 2 angular nodes and no radial nodes
  • each look like 4, 3D oval shapes positioned between the axes in the plane given by the letters in the orbital name
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51
Q

what does the isosurface plot for 3d(x^2 -y^2) look like (in a hydrogen atom)

A
  • 4, 3D oval shapes positioned on the x and y axes, pointing along the axes
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52
Q

what does the isosurface plot for 3d(z^2) look like (in a hydrogen atom), what is the form of the angular node for this orbital

A

a ring passing around the z-axis in the x-y plane at z = 0, two 3D oval shapes above and below the the origin on the z-axis

the angular node is in the form of two cones, theta = 54.7, 125.3

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53
Q

what is a feature of d orbitals (in a hydrogen atom) and what are the forms of the angular nodes for each orbital

A
  • each d orbital has two angular nodes
  • the angular nodes define planes for all but the 3d(z^2) orbital which is cones
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54
Q

what does the total number of nodes in hydrogen orbitals depend on,
give the relationships between n,l and numbers of nodes

A

only depends on the principal quantum number, n

total nodes = n-1
radial nodes = (n-1) - l
angular nodes = l

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55
Q

Why can’t we solve the Schrodinger for multi-electron systems and what approximations can we make and how

A
  • there’s attraction between the electrons and the nucleus, there’s also repulsion between nearby electrons, this makes it too complicated
  • good approximations can be made using the orbital approximation
  • this is done by observing the electron system from the ‘point of view’ of one electron
  • the effect of all the other electrons can be averaged out to give a modified potential/charge centered on the nucleus
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56
Q

what is an assumption we make when doing orbital approximations, what does this lead to

A
  • we can make a modified charge from the effects of all the other electrons and that this is spherically symmetric and centered on the nucleus
  • this means that the wavefunctions for each electron have the same form as those in hydrogen
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57
Q

what are the different screening effects of electrons in different subshells

A
  • the 1s electrons screen the 2s electrons very well from the nuclear charge, approx. 0.85 of a proton charge each
  • the 2s electrons hardly screen the nuclear charge
  • the 1s electrons partially screen each other, approx 0.3 of a proton charge each
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58
Q

what is the difference between subshell energies in hydrogen vs in multi-electron systems

A
  • in hydrogen the 2s, 2p etc. electrons have the same energy/ are degenerate
  • in multi-electron systems, the 2s, 2p and 3s, 3p, 3d electrons are NOT degenerate
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59
Q

what is a way to observe how the degeneracy of 2s, 2p etc. electrons is lost

A
  • observe a radial distribution function for an atom e.g. lithium
  • we can observe that the 2s orbital ‘penetrates’ the 1s orbital more than the 2p orbital does, this can be worked out by viewing the amount of overlap of the radial functions
  • this higher penetration means the 2s orbital experiences a greater nuclear charge than the 2p orbital so the electron in the 2s orbital has a more negative (lower) energy
  • this can also be derived from En = -Rh Z^2/n^2 where z is the nuclear charge
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60
Q

how easy is it to predict the ordering of energy levels in multi-electron systems

A
  • the effects of orbital penetration can become so pronounced that it becomes very difficult to predict the energy order in larger atoms without the aid of a computer
  • even for isoelectronic atoms the ordering can be very different
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61
Q

why is it meaningless to say ‘the energy of a 4s orbital is lower than the energy of a 3d orbital’

A
  • the energy of an atom with a particular electron configuration depends on the energies of all the electrons it contains
  • removing or adding an electron can change the energy of all the electrons present
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62
Q

why does effective nuclear charge increase across a period

A
  • electrons in the same shell do not shield each other very well (only about 30-35%)
  • so when both a proton and an electron are added to an atom, the effect of adding a proton has the greater effect so Zeff increases
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63
Q

what are the key points to remember about the orbital energy vs atomic number graph

A
  • orbital energy decreases along a period due to increasing Zeff and different degrees of penetration and shielding
  • core electrons have very low energy and take little part in reactions
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64
Q

what does it actually mean when a bond is formed in a homonuclear diatomic

A
  • the diatomic molecule is lower in energy than the two atoms separately
  • this energy changes with distance between the atoms
  • the point at which the energy is at a minimum is the bond length
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65
Q

how can the form of molecular orbitals be visualised, what is the equation for this

A

the molecular orbitals (MO’s) can be visualised by combining the atomic orbitals (AO’s) of the constituent atoms

MO = c1 AO1 + c2 AO2

where c1 and c2 are coefficients which define the ‘proportions’ of the two atomic orbitals in the molecular orbital

66
Q

what are the two ways that our wavefunctions can be combined

A
  • constructively or destructively
  • this acts very similarly to normal constructive/destructive interference in waves
67
Q

how do we combine our wavefunctions in homonuclear diatomics

A
  • we can take the value of our two wavefunctions at each point and add them together
  • doing this over all space forms the molecular orbital
68
Q

how can we represent the combining of wavefunctions constructively and destructively

A
  • draw out the wavefunction-radius plots and add the values of the two functions at each point
  • draw out the isoelectric plots with different shading for the positive/negative values of the wavefunction and combine in a similar way
  • simplify the isoelectric plot to 2D ones and combine similarly
69
Q

What are the energies of the resultant molecular orbitals when combining constructively/destructively compared to each other and the original AOs

A
  • the in-phase combination (MO1) is lower in energy than the isolated AO
  • the out-of-phase combination (MO2) is higher in energy than the isolated AO
70
Q

what are the two main reasons that the in-phase MO is lower in energy than the original AO

what does this lead to

A
  • when the orbital is occupied there’s a greater electron density between the nuclei so there’s an attraction to both nuclei, it also shields the nuclei from each other
  • an electron in the in-phase MO is more delocalised than in the AO, this means it has less kinetic energy
  • this means that occupancy of this orbital gives rise to bonding, it is the ‘bonding molecular orbital’
71
Q

what are the three main reasons that the out-of-phase MO is higher in energy than the original AO

what does this lead to

A
  • when this orbital is occupied there is more electron density outside the inter-nuclear region than in it
  • these electrons exert an outwards force on the nuclei which moves them apart
  • eventually the atoms would totally separate
  • equally the MO contains a node which restricts the electron and increases its KE
  • this leads to the species ‘falling apart’ as greater displacements are lower energy so its called the ‘anti-bonding orbital’
72
Q

what is an energy level diagram and what are the significant parts to it

A
  • increasing energy on Y
  • one AO on the left
  • one AO on the right
  • two lines, one higher, one lower representing the molecular orbitals
  • the higher line is the antibonding orbital
  • the lower line is the bonding orbital
  • the higher line is more displaced than the lower line
73
Q

what are the two important things to note about forming molecular orbitals

A
  • combining two atomic orbitals gives two molecular orbitals, one bonding, one antibonding/ generally combining n AOs gives n MOs
  • one molecular orbital (bonding) is lower in energy than the atomic orbitals, one molecular orbital (antibonding) is higher in energy than the atomic orbitals
74
Q

what are the similarities between atomic orbitals in atoms and molecular orbitals in molecules

A
  • just as in atoms, electrons are ‘fed’ into MOs, two spin paired electrons in each
  • they add from the lowest energy up
75
Q

how can we analyse the energies of the bonding and antibonding orbitals at different bond lengths and what are the main points to note

A
  • it’s possible to calculate the energies of the bonding and antibonding orbitals at any distance by combining the two 1s orbitals in and out of phase at those distances
  • at larger internuclear separations the difference in energy between the bonding and antibonding MOs decreases, this is because the further apart the nuclei are, the less the orbitals interact
76
Q

what is the explanation using the energy-bond length curve as for why there’s an equilibrium distance for the bonding MO but not for the antibonding

A
  • there’s an energy minimum at a finite bond length for the bonding MO
  • the lowest energy for the antibonding MO occurs as the bond length tends to infinity
  • this means if there’s only an electron in the antibonding MO (or more electrons in it) then the molecule will fall apart
77
Q

what can we say about the total energy of a molecule where there’s the same number of electrons in the bonding and antibonding orbitals

A
  • the energy will be positive at any separation
  • this means at any separation energy can be lowered by increasing bond length so the molecule falls apart
78
Q

explain why He2 does not exist but He2(+) does using bonding and antibonding MOs and stability with respect to dissociation

A
  • He2 contains 4 electrons
  • 2 will be in the bonding MO and 2 will be in the antibonding MO
  • this means overall it will have a positive energy (at any separation) so will break down into two helium atoms as this will be lower energy
  • it is said to be ‘unstable with respect to helium atoms’
  • He2(+) only contains 3 electrons
  • 2 electrons in the bonding MO and 1 in the antibonding MO
  • so overall there is an equilibrium separation where the energy is negative so it is ‘stable with respect to dissociation’
79
Q

What is bond order and how can we calculate it

A

Bond order tells us the difference between the number of electrons in the bonding and antibonding orbitals

bond order = 1/2 ((number of electrons in bonding MOs) - (number of electrons in antibonding MOs))

80
Q

how do 2s AOs combine to form MOs

A
  • they interact and combine in exactly the same way that 1s orbitals do
  • we can ignore the radial nodes because these are too close to the nuclei to be relevent
81
Q

how do two p orbitals combine and what is the extra thing we need to consider

A
  • p orbitals combine in broadly the same way but we now need to consider which direction they point in and how they overlap, we can use the same methods of 2D isosurface plots with different shadings for +ve or -ve values of the wavefunction
82
Q

How do two (identical) pz orbitals combine, how do they appear in-phase or out-of-phase

A
  • pz orbitals will combine in the same way as s orbitals, end-on
  • this is because they lie along the internuclear axis, because by convention the internuclear axis is the z-axis, it has the most symmetry
  • when in-phase they appear as a large broad section between the nuclei of one shading and smaller than normal teardrop sections outside of the nuclei of the opposite shading
  • when out-of-phase they appear as two small teardrop shapes of opposite shadings between the nuclei and two large teardrop shapes of opposite shading outside the nuclei
83
Q

how do two (identical) px or py orbitals combine, how do they appear in-phase or out-of-phase

A
  • py and px orbitals will combine side on (they combine in the same way just rotated by 90 degrees)
  • when in phase it appears like a standard pi bond, two broad oval sections above and below (or in front and behind) the axis of opposite shadings
  • when out of phase it appears as small teardrop shapes going outwards from the nuclei at angles outward facing (i.e. up left, down left, up right, down right), or the same but facing in/out of page
84
Q

when is an MO labelled pi or sigma

A
  • depends on the symmetry around the internuclear axis, if it can be traversed in a plane perpendicular to the internuclear axis and:
85
Q

when is an MO labelled pi or sigma

A

depends on the symmetry around the internuclear axis, if it can be traversed in a plane perpendicular to the internuclear axis and:

  • phase doesn’t change then its sigma, i.e. the internuclear axis does not contain a nodal plane
  • phase does change then its pi, i.e. the internuclear axis does contain a nodal plane
86
Q

what sort of MOs are formed when 2s,2px,2py or 2pz orbitals are combined

A
  • 2s = sigma
  • 2pz = sigma
  • 2px, 2py = pi
87
Q

when is an MO labelled with u or g

A
  • When a molecule contains a centre of inversion, ONLY IN HOMONUCLEAR DIATOMICS (or symmetrical molecules)
  • g if the sign of the wavefunction is the same after moving through the centre of inversion
  • u if the sign of the wavefunction changes after moving through the centre of inversion
88
Q

when is an MO labelled with a star (*)

A
  • if it is an antibonding orbital (formed from out-of-phase interactions)
89
Q

what is a method we can use to assess the strength of the interaction between two orbitals and how does it work

A
  • we can use the overlap integral
  • this is found by multiplying together the values of two AOs at a particular point then taking the integral of this product over all space
  • a graph of the ‘overlap’ at different internuclear separations can be plotted
90
Q

what can we derive about the strengths of different p MOs from their overlap graphs

A
  • the overlap between end on p orbitals (pz) is greater at typical bond lengths than the overlap between side-on p orbitals (px and py)
  • this means at typical bond lengths the energy of a bonding MO for end-on overlaps of p-orbitals is lower than a bonding MO for side-on overlaps of p-orbitals
  • equally, at typical bond lengths, the energy of an antibonding MO for end-on overlaps of p-orbitals is higher than an antibonding MO for side-on overlaps of p-orbitals
91
Q

what are some things that we should consider when forming an energy level MO diagram for larger homonuclear diatomics e.g. O2

A
  • the 1s orbitals will not interact as they are too contracted from greater Zeff
  • the valence electron shells are the main shells which will interact
  • the pz orbitals will interact to form sigma MOs and the other p orbitals will interact to form pi MOs, the sigma MOs are further displaced vertically up/down than the pi MOs
92
Q

why does O2 have a bond order of 2 but F2 have a bond order of 1

A
  • O2 has 16 electrons (we can generally ignore 1s MOs to make it 12), 10 are in bonding orbitals 1s, 2s, 2p(sigma), 2p(pi), 6 are in antibonding MOs 1s* , 2s* , 2p(pi)* , 2p(sigma)*

bond order = 1/2 (10-6) = 2

  • F2 has two more electrons which fills the 2p(pi)* antibonding orbital giving

bond order = 1/2 (10-8) = 1

93
Q

what are the three main factors which determine how well two AOs combine

A

(1) the constituent orbitals must have suitable symmetry to interact

(2) the constituent AOs must be close in energy for significant bonding or antibonding interactions to occur

(3) even if the orbitals are close in energy, the degree of overlap will change depending on the sizes of the orbitals

94
Q

give an example/ further explain the first factor that affects how well AOs combine, the suitable symmetry factor

A
  • two orbitals with similar symmetry will interact better e.g. pz and 2s will interact better than px/py and 2s as the overlap possibilities are better
95
Q

give an example/ further explain the second factor that affects how well AOs combine, the similar energy factor

A
  • the constituent AOs must be close in energy for significant bonding or antibonding interactions to occur
  • hence for a diatomic such as O2 there is no significant interaction between the 1s AO of one and the 2s AO of another
96
Q

give an example/ further explain the third factor that affects how well AOs combine, the size of orbital factor

A
  • even if the orbitals are close in energy, the degree of overlap will vary depending on the sizes of the orbitals
  • the interaction between two large orbitals is less than the interaction between two small orbitals

e.g. two 1s orbitals in H2 have a strong interaction whereas two 2s orbitals in Li2 have a weaker interaction as the 2s orbitals are bigger than the 1s orbitals

97
Q

what are two general features of an MO that are different when two AOs of different energies combine (compared to = energy)

A
  • the resultant bonding MO is not lowered in energy, nor the resultant antibonding MO raised in energy as much as when the two combining AOs have the same energy
  • the two AOs no longer contribute equally to the MOs
98
Q

how would an AO/MO diagram look different for two AOs of the same energy interacting vs two AOs of (slightly) different energies

A
  • for two AOs of the same energy, the antibonding MO is much higher, and the bonding MO much lower, in energy than the AO
  • the MOs are also symmetrical in shape
  • for two AOs of different energy, the antibonding MO is not much higher than the higher energy AO and the bonding MO is not much lower than the lowest energy AO
  • the MOs are also not symmetrical in shape, the antibonding MO will have a greater ‘cloud’ around the atom with the higher energy AO, and the bonding MO will have a greater ‘cloud’ around the atom with the lower energy
99
Q

what makes an atom have lower energy AOs

A
  • generally strongly linked to electronegativity
  • effectively how close the AOs are on average to the nucleus
100
Q

what might an AO/MO diagram look like for two AOs of VERY different energy interacting

A
  • the MOs are so shifted they effectively are the same as the initial AO positions
  • what happens in this situation is the electrons are often in different MOs in the product than the energy of the AOs in the initial atoms (even though the AOs and MOs are essentially the same energy)
  • all of the electrons still fill as normal but not traditionally in bonding/antibonding orbitals
  • this means one atom effectively gains an electron whilst the other loses one
  • leading to an ionic bond
101
Q

what types of bond/things can we tell about the bonding between two atoms from the difference in energy of their AOs

A
  • where the energies of the AOs are VERY different e.g. LiF then the electrons become unevenly shared and we gain ions held together with a strong electrostatic attraction - AN IONIC BOND
  • when two AOs contribute equally to form an MO the electrons are shared between the atoms in what is termed A COVALENT BOND
  • when the electrons are shared unequally (giving polarised bond) because of slightly different energy AOs we say the bond has IONIC CHARACTER
102
Q

what is something to note about bonds formed when MOs are formed from different energy AOs

A
  • they are not necessarily weaker than equally shared AO bonds
103
Q

what things should we consider when forming the MOs for heteronuclear diatomics, why do some AOs not interact with each other

A
  • we should consider the energy levels of the AOs in different atoms
  • e.g. in HF the 1s of the fluorine is so low in energy it will never interact
  • we can then look at only the valence electrons and observe which AOs are able to interact
104
Q

what is something that we must remember when forming MOs in heteronuclear diatomics between an s and a p orbital, what is formed

A
  • if an s orbital and a p orbital from different molecules are similar enough in energy to interact then they will, this pretty much only happens when there is a large difference in electronegativity
  • however only the pz orbital is able to because the other orbitals do not have the appropriate symmetry
  • this means 1 bonding MO, 1 antibonding MO are formed and 2 non-bonding MOs are at the same energy as the initial AOs and are where lone pairs come from
105
Q

what is sp mixing (in homonuclear diatomics) and what method might we take to consider it

A
  • there is no reason to assume that in homonuclear diatomics only orbitals of the same type can interact
  • s orbitals can mix with p orbitals etc.
  • we could generate more accurate MOs by considering how all the suitable AOs contribute to a given MO
  • or we can form initial MOs then consider which MOs might interact to form and improved set of MOs
106
Q

what rules can we use when considering which MOs in sp mixing (in homonuclear diatomics) can combine

A
  • the same rules as for combining AOs
  • we use our symmetry labels to determine which MOs have suitable symmetry to interact

e.g. 2s(sigma)(g) and 2p(sigma)(g) can interact, as can 2s(sigma)(u) and 2p(sigma)(u)

i.e. they must have the same type, letter and * (or lack of)

107
Q

how can we work out what the result of the sp mixing (from the MOs that can further interact) will be (in homonuclear diatomics)

A
  • we know any 2p orbitals will be higher in energy that the 2s orbitals
  • therefore we know that the bonding MO will be mostly due to the 2s and partly due to the 2p
  • and the antibonding orbital will be mostly due to the 2p and partly due to the 2s

NOTE: these apply to both bonding and antibonding initial MOs

(this can all be done pictorially as before)
- small sub-AO/MO diagrams can be made

  • this can be recombined with the initial AO/MO diagram to form a refined model
108
Q

do the p(sigma)(g) and s(sigma)(g) or the p(sigma)(u* ) and s(sigma)(u* ) have closer energies and what does this lead to

A
  • the p(sigma)(g) and s(sigma)(g) are closer in energy
  • this means they have a stronger interaction
109
Q

if sp mixing occurs, how can we now label our MO energy diagram and what do the numbers mean

A
  • we can use the same sigma(g), sigma(u), pi(u) and pi(g) can be used but given they are now formed from a and p orbitals, we can no longer include the s and p labels
  • their label is determined by their symmetry only
  • so they are just labelled 1sigma(g), 1sigma(u*) etc.

NOTE: these numbers are not the principal quantum numbers, just the numbers counting up from the bottom

110
Q

after sp mixing, some of the MOs might only be very weakly bonding or antibonding, e.g. in N2, in this case what can we consider them as

A
  • non-bonding orbitals or lone pairs
111
Q

what is the main factor on the amount of sp mixing that occurs, what occurs to the sp mixing as you move across a period

A
  • how close in energy the s and p orbitals are
  • moving across a period, the nuclear charge increases, both 2s and 2p orbitals decrease in energy but 2s decreases more
  • this means the difference in energy is greater
  • so sp mixing has a less dominant effect
112
Q

in which homonuclear diatomics is sp mixing significant and in which is it not

A
  • significant in: Li2, Be2, B2, C2, N2
  • not significant in: O2, F2
113
Q

what does the MO formation look like for larger molecules such as H3+

A
  • (assuming linear geometry) we can combine 3 1s orbitals
  • as usual this will lead to 3 molecular orbitals
  • our bonding orbital is one large region over all 3 hydrogens
  • we have a non bonding orbital which is effectively 2 separate 1s orbitals on the terminal hydrogens and the middle without anything (wrong symmetry)
  • we also have an anti-bonding orbital where the two outer orbitals are the same phase but the middle orbital is a different phase
  • as always these MOs can only hold two electrons, H3+ only has 2 electrons hence it has 2 electrons in the bonding MO over all 3 hydrogens
  • NOTE: actual geometry is different
114
Q

why don’t we typically calculate all the MOs for large molecules in the same ways as we do for small molecules

A
  • it rapidly becomes too complicated to do this in large molecules e.g. butane has 26 AOs so 26 different MOs, each spread over all atoms and with different energies
  • it also doesn’t give us a very useful picture of the molecule
115
Q

what is a solution when doing MOs for molecules so that you don’t have to calculate all the MOs for the molecule as a whole, what is an issue we encounter when doing this and what method do we use as a result

A
  • we can construct approximate MOs between any two bonded atoms e.g. in butane bonding and antibonding MOs for each C-H and C-C
  • the problem is that the AOs which we use to generate these MOs might not point in the correct directions
  • for example the carbons have bonds at 109 degrees around them but the 2p orbitals are all at 90 degrees to each other
  • to overcome this we use hybridization
116
Q

what is hybridization/what does it allow us to do

A
  • hybridization allows us to mix a number of AOs on the same atom to create hybrid atomic orbitals (HAOs)
  • these HAOs point directly to the atoms we want to construct bonds with
117
Q

How would hybridization work in methane

A
  • to form the carbon HAOs we combine the 2s AO with the three 2p AOs (the 1s orbitals are not involved in bonding)
  • this forms 4 hybridized sp3 (because formed from one s and three p) HAOs, they form a tetrahedral shape with 109 degrees between them and look like an asymmetrical dumbbell
  • these HAOs can overlap with the 1s orbitals of a hydrogen and form bonding and antibonding MOs for each C-H bond
118
Q

How do HAOs compare to the true MOs that would occur in a molecule

A
  • the HAOs are all equivalent in energy and shape
  • in reality there are two MOs (in methane or other CH3 groups) where electrons are at different energies
119
Q

what labels can we give the resultant MOs formed in C-C or C-H from HAOs interacting with other HAOs or 1s (in hydrogen) and why

A
  • each MO (bonding and antibonding) are still symmetrical about the internuclear axis so can still be labelled with a sigma or sigma*
  • e.g. in methane 4 HAOs interact with 4 1s orbitals from H’s to form 4 sigma bonding MOs and 4 sigma* antibonding MOs (in this case only bonding MOs are filled)
120
Q

What must we note about the uses of hybridization and when its used

A
  • hybridization cannot be used to predict the shape of molecules
  • but rather the shape of molecules is known and the suitable AOs are hybridized to give the correct HAOs for the shape
121
Q

BeH2 is a linear molecule, what is the hybridization that occur and explain which AOs cannot hybridize to form this shape

A
  • can’t use 1s of Be, because it’s too low in energy
  • can’t use 2py or 2px because they have the wrong symmetry for forming linear HAOs
  • so we use the Be 2s orbital and the Be 2pz orbital to form two HAOs at 180 degrees to each other and two unchanged py and px orbitals
  • these then overlap with the 1 orbitals of Hydrogen
122
Q

explain the hybridization in ethene

A
  • we need three HAOs at 120 degrees and in the same plane
  • this means we cannot use the 2px orbital because it is in/out of the plane where the hydrogens and carbons lie
  • so we combine 2s, 2py and 2pz to form 3 sp2 HAOs at 120 degrees
  • this occurs on both carbons
  • two of the three sp2 HAOs on each carbon bond to hydrogens
  • the final HAO bonds to the other carbon
  • and the px AOs on each carbon overlap to form the pi bond of the double bond
123
Q

what are the HOMO and LUMO in ethene

A
  • the 2pi MO is the HOMO
  • the 2pi* MO is the LUMO
124
Q

list 4 advantages of hybridization

A
  • it simplifies the bonding scheme
  • it gives more directional HAOs that point towards the bonding atoms (or where we think of the lone pairs as pointing to)
  • it gives MOs in which the electrons are not delocalised over many atoms but instead the bonding electrons are localized in between two atoms and the lone pair electrons on one atom
  • more useful when trying to draw mechanisms
125
Q

list 2 disadvantages to hybridization

A
  • simple hybridization does not give the best picture of the different energy levels within the molecule - further refinements necessary
  • encourages a localised view of electrons whereas in reality, they are spread across the molecule
126
Q

how can we use hybridization to explain other bond angles such as H2O

A
  • mixing the s orbital with all three p orbitals gives a tetrahedral shape, this is not the shape of H2O
  • a standard sp3 HAO has 1/4 ‘s character’ and 3/4 ‘p character’
  • so instead we make weird combinations of the s and p orbitals e.g. water has an s:p ratio of 0.19:0.81
127
Q

when can d orbitals be used in hybridization, and what is a common example of this

A
  • d orbitals can be used in hybridization if they are low enough in energy
  • this regularly occurs in transition metal complexes
128
Q

what hybridization occurs in octahedral transition metal complexes

A
  • the s orbital, all three p orbitals and the dz^2 and d(x^2 - y^2) orbitals combine
  • 6 AOs are needed to form 6 HAOs
  • the other d orbitals are not suitable as they point between the axes and not along them
129
Q

what hybridization occurs in a square planar molecule

A
  • to bond to the 4 ligands in a square planar molecule we need 4 HAOs in the same plane
  • as before we cannot use the dxy, dyz, or dxz orbitals because these lie between the axes not on them
  • we also cannot use pz because this does not lie on the xy plane in which the ligands sit

Hence we use EITHER
s, px, py, d(x^2-y^2) = sp2d
OR
px,py,d(x^2-y^2), dz^2 = p2d2

130
Q

give a summary of the different hybridization states (of the central atom) for different shapes of molecule and give an example for each

A
  • linear = sp, e.g. BeH2
  • trigonal planar = sp2, e.g. BF3
  • tetrahedral = sp3, e.g. CF4
  • square planar = sp2d or p2d2, e.g. Pt(Cl)2(NH3)2
  • trigonal bipyramidal = sp3d or spd3, e.g. PF5
  • octahedral = sp3d2, e.g. SF6
131
Q

what is the deciding factor if different hybridization schemes are possible for a metal ion complex

A
  • the energy match between the central atom’s AOs and the ligands’ AOs
132
Q

what is the best way to consider the conjugation in a molecule with alternating single and double bonds e.g. butadiene or benzene

A
  • we can consider the sigma bonding framework using our HAOs as usual but for the double bonds it is best to use MOs
133
Q

what do the HAOs form for a molecule containing alternating single and double bonds

A
  • each carbon is sp2 hybridized, this allows for bonds to 1 or two other carbons and 1 or two other hydrogens, it leaves a p orbital perpendicular to the plane available for pi bonding
134
Q

where we have a system that can delocalize (i.e. alternating single and double bonds) what is the best rule for predicting the form of the n MOs formed from the overlap of n p AOs

A
  • we can use a sinusoidal wave rule
  • mark dots on a line for as many carbon atoms as you have
  • move one further out each side of the terminal carbons and mark some end points
  • draw some standing sinusoidal waves with 1 node, 2 nodes, 3 nodes….., up to n nodes
  • the height under the sine wave at each ‘dot’ or ‘carbon’ predicts the size of the coefficient when adding the orbitals
  • if the value of the sine wave changes then then reverse the in-phase/ out-of-phase
  • from this the size and phase of the p orbitals can be sketched and B/AB interactions can be viewed
135
Q

what do the four pi MOs look like for butadiene, identify the HOMO and LUMO and explain why we put the double bonds where we do

A
  • 1pi = delocalised system over all 4 carbons (one phase above, one phase below carbons), 3 B interactions
  • 2 pi = carbons 1 and 2 have a delocalised pi bond over them and carbons 3 and 4 have the same, 2B and 1AB interactions
  • 3 pi = carbon 1 and 4 have almost unchanged p orbitals, carbons 2 and 3 have a delocalised pi bond over them in antiphase to the orbitals on 1 and 2, 2AB and 1B interactions
  • 4 pi = each carbon has their own ‘p like’ orbital, 3 AB interactions
  • the four pi electrons fill 1pi and 2pi, the HOMO is 2pi, the LUMO is 3pi
  • the although there is pi bonding character about carbons 2 and 3 in 1pi, it is largely cancelled out by the AB interaction in 2pi, however the coefficient in 1pi is greater so its still got more than a single bond
  • carbons 1 to 2 and 3 to 4 definitely have double bonds because both 1pi and 2pi have B interactions between them
136
Q

what are the forms of the pi MOs for a three carbon system, C=C-C (+ or-), allyl anion or cation

A
  • using the sine rule/trick
    thing we get
  • MO with no nodes/1pi = delocalised over all carbons, 2 B interactions
  • MO with 1 node/2pi = end carbons have almost normal p orbitals, middle has 0 electron density as it sits on a node, non-bonding
  • MO with 2 nodes/3pi = each carbon has almost normal p orbitals, end carbons have orbitals slightly tilted out and smaller, pi* MO, 2 AB interactions
137
Q

using the pi MOs for a 3 carbon C=C-C chain, what MOs are filled for an allyl cation, what are the HOMO and LUMO and how does this link to how we can draw it

A
  • as the allyl cation is C3H5+ we only have two pi electrons
  • these two electrons only occupy the lowest energy MO, the 1pi
  • this means it’s a 3 carbon, 2 electron pi system
  • this orbital suggests there’s the highest electron density is on the central carbon so the ends are slightly positive
  • this means the C-C bonds in the molecule are equivalent so technically it’s wrong to draw it with 1 double bond and 1 single bond
  • the HOMO is the 1pi MO and the LUMO is the 2pi MO
138
Q

using the pi MOs for a 3 carbon C=C-C chain, what MOs are filled for an allyl anion, what are the HOMO and LUMO and how does this link to how we can draw it

A
  • we have the exact same orbitals as in the cation
  • the allyl anion is C3H5- so we have 4 pi electrons
  • these two electrons occupy the 1pi and 2pi systems fully
  • this makes it a 3 carbon, 4 electron delocalised pi system
  • the HOMO is the 2pi MO and the LUMO is the 3pi MO
  • because of both the 1pi and 2pi MOs being filled it means we have the greatest electron density on the end carbons
  • the structure is still symmetrical in terms of electron density though so it’s still not right to draw it how we do
139
Q

does the combining of p orbitals need to be from atoms of the same element, what is the difference?

A

no

the difference is that the p orbitals will no longer be at the same energy, hence the form of the MOs will no longer be symmetrical so the electron density will not be shared evenly

140
Q

what do the pi MOs look like for a fomamide molecule look like and what is the HOMO and LUMO, what type are the orbitals, what is the system called

A

1pi = Bonding, delocalised electrons over all 3 atoms (N,C,O)

2pi = non-bonding, MO looks just like 2 AOs on the oxygen and nitrogen, this is the HOMO, sides are opposite phases

3pi = antibonding = largest electron density on middle carbon, antibonding interactions with both the nitrogen and oxygen, this is the LUMO

it is a 3-centre 2-electron bond

141
Q

what are the valence bond picture options and what is the more realistic but less practical bond picture for formamide

A

1) as normal with double bond C=O and single C-N bond, N has lone pair

2) N lone pair delocalises into C-N bond, C=O bond breaks to give O formal charge so now there’s C=N and C-O-

more realistic but less practical = 1 and 1/2 bond over O-C-N with slight -ve charge on O and slight +ve charge on N

142
Q

what is a complicated (but theoretically possible) way to predict the shape of a molecule

A
  • we assume a starting geometry for a molecule, we then work out its energy
  • we then adapt the shape slightly and recalculate the energy
  • this process is repeated until energy is at a minimum
  • it would require a lot of computing power
143
Q

using H3+ as an example how can energies be lowered by adopting a different geometry to those assumed

A
  • a triangular geometry for H3+ is lower in energy than a linear geometry is
  • so the true shape of H3+ is triangular with 2 electrons delocalised over the entire shape
144
Q

what is a very common way in which reactions occur, how does this link to MOs

A
  • the highest energy electrons in one species may be lowered in energy by reacting with a second species
  • this is often a HOMO-LUMO interaction
145
Q

how can we represent the movement of 1 electron or a pair of electrons

A
  • using a curly arrow
  • double headed for pair of electrons
  • fish hook or single headed for 1 electron
  • they start at the point where the electrons are initially and point to where the electrons end up
146
Q

using curly arrows describe the difference between the reaction of H- with H+ and two hydrogen atoms H* (radical)

A
  • in the reaction between H- and H+ there’s a curly arrow pointing from the lone pair/negative charge on H- to the gap between the H+ and H-, this is a double headed curly arrow
  • in the reaction between two H radicals there is a single headed/half/fish-hook arrow from each of the unpaired electrons to the gap between the two radicals
  • we get the same result, an H2 molecule
147
Q

what are some important summary points when considering MOs in reactions

A
  • the MOs for a product may be considered as arising from interactions between the MOs in the reactants
  • Many of the MOs in the reactants remain virtually unchanged
  • one of the most important interactions is that between the HOMO of one species and the LUMO of another, it will lead to the lowering in energy of the most energetic electrons
148
Q

what is a HOMO-LUMO interaction and why is it important, why does the same effect not occur for filled or vacant MOs, sketch their MO diagrams to prove it

A
  • the interaction between a filled MO from one species and a vacant (or partially vacant) MO from a second
  • it will always lead to a net lowering of energy of the electrons
  • this is not the case between two filled MOs or two empty MOs
  • for the filled initial MOs, electrons are both raised and lowered in the resultant MOs and hence there’s no net lowering in energy
  • for the vacant initial MOs no electrons are lowered
  • for the filled initial MO and the vacant initial MO, electrons are only lowered
149
Q

why are there two HOMO-LUMO interactions possible in any reaction and which will be the ‘best’ (have the biggest net lowering in energy)

A
  • if two reactants A and B interact
  • then the HOMO of A can interact with the LUMO of B
  • and the LUMO of A can interact with the HOMO of B
  • the ‘best’ interaction is the one where the two MOs are closest in energy
  • this will be whichever the higher energy HOMO is and the LUMO from the other reactant
150
Q

what is a guide to help us work out which orbitals or electrons are the HOMO and LUMO, what is the energy scale for the different types of orbital

A

in increasing energy the orbitals are

sigma(B), pi(B), non-bonding(e.g. lone pairs, radicals and negative charges), pi(AB), sigma(AB)

151
Q

why are the pi bonding higher in energy than the sigma bonding orbitals

A

the side-on overlap of pi MOs is not as effective as end-on in sigma so it’s higher energy

152
Q

what are the types of electron in NH3 in increasing energy order and which is the HOMO and LUMO

A

in increasing energy order

N-H sigma bonds, N lone pairs (HOMO), N-H sigma* (LUMO)

153
Q

what are the types of electron in BF3 in increasing energy order and which is the HOMO and LUMO

A

in increasing energy

B-F sigma bonds, F lone pairs (HOMO), vacant B pz (LUMO), B-F sigma* (vacant)

154
Q

what are the types of electron in CH2O in increasing energy order and which is the HOMO and LUMO

A

in increasing energy

filled sigma bonding MOs, filled C=O pi bonding MO, O lone pairs (HOMO), vacant C=O pi* antibonding (LUMO), vacant sigma* antibonding

155
Q

what is the general reaction in nucleophilic addition to a carbonyl group

A

Nu(-) + RCOR –> RCO(-)NuR

156
Q

why does the reaction occur how it does for a hydride ion attacking the carbon of a carbonyl

A
  • the carbon has a partial positive charge due to the electronegative oxygen, so there’s an initial electrostatic attraction to the carbon
  • we need to consider the HOMO-LUMO interactions
  • the HOMOs are the sigma bonding electrons in the H- and the lone pairs on the carbonyl oxygen in the carbonyl
  • the sigma bonding electrons in H- are higher in energy (see most important graph in chem) so this is the HOMO which interacts
  • the LUMO it interacts with is the LUMO of the carbonyl group, this is the pi* antibonding MO of the C=O bond
157
Q

if we do isosurface plots of the MOs at different stages of the reaction for the nucleophilic addition of a hydride to a carbonyl, what do we see and why is it significant

A
  • we can look at the H(-) sigma and the C=O pi*, they interact and 2 electrons are added to the antibonding orbital formed, this causes the C=O bond to break forming a C-O bond
  • we see electron density move from the hydrogen to the oxygen
  • we can also look at the C=O pi bond which ends with much less electron density between C and O than it started with

For more see pg59

158
Q

how can the MOs help show which geometry’s are best for the attack of the nucleophile, draw some diagrams

A
  • if the hydride ion attacks the carbonyl from end on then both constructive and destructive overlaps occur with the pi* MO of the carbonyl. Hence there’s no
  • if the hydride ion attacks from directly above then there’s no net interaction for the same reason
  • hence the optimal angle to attack from is 107degrees to the horizontal, on the carbon of the carbonyl
159
Q

why do nucleophiles readily attack the C-O pi* but not the C-C pi*

(3 reasons)

A
  • the carbon on the C=C bond is not delta(-ve) so there’s no initial electrostatic attraction
  • along with this the C-C pi* orbital is higher in energy than the C-O pi* so there’s a worse match between the HOMO of the nucleophile and the pi*
  • one more reason is that if the reaction did occur the C=C double bond would break leaving the negative charge on the carbon, this is not as energetically stable as the negative charge being left on the oxygen because the oxygen orbitals are lower in energy
160
Q

what can we do to understand why the C-C pi* is higher in energy than the C-O pi*

A

The oxygen AOs are lower in energy than the carbon AOs

so when the two atoms interact the antibonding orbitals are raised less from the carbon than they are for two carbons

161
Q

what is the general reaction for nucleophilic substitution

what are the HOMO and LUMO

A

Nu(-) + CR3Nu’ —> Cr3Nu + Nu’(-)

HOMO = lone pair on Nu-
LUMO = sigma* on C-Nu’

it is the C-Nu’ sigma* not C-H sigma* because the Nu’ will be lower in energy than H so the sigma* will be lower in energy

162
Q

where a molecule has sp2 HAOs and no lone pair, what will the LUMO be?

A
  • the empty p orbital

PT IA Chemistry (9 decks)

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