UNICAMP Chem Handout

8/12/2019 UNICAMP Chem Handout

1/99

Instituto de Qumica

Chemical structure and reactivity:

an orbital based approachJames KeelerUniversity of Cambridge

Department of Chemistry

0: Introduction 1

0: Introduction

0: Introduction 2

How the lectures work1.1.1 Section title

we will follow the book pretty closely: what you will see on

the screen is in the text

the title of each frame will (usually) match the section

headings in the book

if a frame is just a figure, then the t itle will give the Figure

number for example:

0: Introduction 3

Fig 2.9 Shaded plot of 1sorbital

0: Introduction 4


2/99

Philosophy

we can understand, or at least rationalize, a great deal of

chemistry by thinking about orbitalsand their interactions

not only structure and shape, but also reactivity

a qualitative understanding of atomic and molecular orbitals is

sufficient we do not need to do any calculations

I hope to show you how this unifying concept can be used to

make sense of a lot of chemistry

0: Introduction 5

Rough outline

thorough revision of atomic orbitals (especially shapes and

energies)

simple molecular orbitals (diatomics)

extending these ideas to larger molecules and in brief solids

how orbitals can be used to understand reactions

if time permits, some other topics (e.g. thermodynamics)

0: Introduction 6

Lets begin

0: Introduction 7

Instituto de Qumica

Chemical structure and reactivity:

an orbital based approachJames KeelerUniversity of Cambridge


2: Electrons in atoms 1


3/99

2: Electrons in atoms


Fig 2.1

ene

rgy

classical quantum

for small objects energy is quantized and can only take specific

values: contrast the classical world where energy varies smoothly


Fig 2.2

classical quantum

in quantum mechanics we can only specify the probabilityof an

object being at a particular place


2.1.2 Probability interpretation

[(x,y,z)]2 is the probability density

formally:

prob. of being in volume Vat position (x,y,z)

=[(x,y,z)]2prob. density

Vvolume

loosely: the probability of finding the electron at a particular

position is proportional to the square of the wavefunction



4/99

2.1.3 Energy

kinetic energy: due to motion

potential energy: e.g. due to interaction between charges

hydrogen atom: negative electron interacting with positive

nucleus (Coulomb potential)

r

r

V(r)

0


2.2 Introducing orbitals

solve the Schrodinger equation for the hydrogen atom

the resulting wavefunctions are called the (hydrogen) atomic

orbitals (AOs)

the mathematical forms and energies can be found exactly

these are the familiar 1s,2s,2petc.

understanding their shapes and energies is crucial


2.2.1 Representing orbitals on paper mathematical form of the 1sorbital

1s(r) = N1sexp (r/a0)

r: electronnucleus distance; a0: Bohr radius(52.9 pm);

N1s is a normalizing factor

plot1s(r)as a function ofr

r/ pm

a00

0 40 80 120

2a0 3a0

1

s(r)

need to understand that this is a three-dimensional function

there are many ways of representing it


Fig 2.7: cubes

(a) (b) (c)

0 0 0 00 0 0

0 00 0 0

00

0 0 11

10

0

00

11

10

0

00

11

10

0

00

00

00

0

00

00

00

0

0 0

1 1 1 0 00 1 3 5 3 1 0

1 3 10 15 10 31

1 5 15 2615

51

1 310

1510

31

01

35

31

0

00

11

10

0

0 0 1 1 1 0 0

0 2 5 7 52 0

1 5 1526 15 5 1

17 26 100 26

71

1 5 1526

155

1

02

57

52

0

00

11

10

0

shading and numbers indicate relative value of the orbital

wavefunction



5/99

Fig 2.8: Contour plot

x

y

(a) (b) (c)

1

1 1

3 3

1010 30

lines connect positions at which the wavefunction has a constant

value (like contours on a map)


Fig 2.9: Shaded plot

density of colour indicates value of wavefunction (ring at maximum

density)


Fig 2.10: Iso-surface

(a) (b) (c)

surface connects positions at which the wavefunction has a

constant value (a three-dimensional contour)

the apparent size depends on value at which the iso-surface

is drawn


2.2.2 Radial distribution function

probability of finding the electron in a thin shell of radius rand

thicknessri.e. summed over all angles

a thin shell

x y

z

defineradial distribution function,P1s(r)

P1s(r) = 4r2 [1s(r)]

2

P1s(r) r is the probability of finding the electron in a shell of

radiusrand thicknessr



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Fig 2.12

r

a00 2a0 3a0 a00 2a0 3a0 a00 2a0 3a0

1s(r) P1s(r)[1s(r)]2

r2

(a) (b) (c)

4r2 ris volume of shell

asrincreases volume of shell increases, but value of

wavefunction decreases

result in a maximum in the RDF


2.3 Hydrogen atomic orbitals

threequantum numberscharacterise each orbital

principalquantum number,n: takes values 1, 2, 3 . . .

orbital angular momentumquantum number,l: takes values

from(n 1)down to0, in integer steps

magneticquantum number,ml: takes values from +lto lin

integer steps

there are(2l + 1)different values of ml


K shell

n = 1

l = 0 only

ml = 0 only

letters represent different values of l

l letter

0 s

1 p

2 d

3 f

only orbital is 1s


L shell

n = 2

ltakes values (n 1)down to0 i.e.l = 1 and l = 0

mltakes values from +lto lin integer steps

forl = 1,ml =1, 0, 1 the three 2pAOs

forl = 0,ml =0 the 2sAO

L shell: three 2p,2s



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M shell

n = 3, hencel = 2, 1, 0

forl = 2,ml = 2, 1, 0, 1, 2 the five 3dAOs

forl = 1,ml = 1, 0, 1 the three 3pAOs

forl = 0,ml = 0 the 3sAO

M shell: five3d, three3p,2s


Orbital energies the energy onlydepends on n

En =Z2RH

n2

Z: nuclear charge (here = 1);RH: Rydberg constant

RH = 2.180 1018 J 1312 kJ mol1 13.6 eV

energies measured downwards from ionization

energy/eV

0

-5

-10

-15

n=1

n=2

n=3

ionization


Orbital energies

the energy onlydepends on n

En =Z2RH

n2

orbitals with the same n have thesameenergy: they are said

to bedegenerate

e.g. the five 3d, the three 3pand the3sall have the same

energy


2.3.2 Shapes of the 2sand 2porbitals; Fig 2.14 cubes

(a) (b) (c)

positive at the centre, crossing to negative at large distances



8/99

Fig 2.15 Iso-surface

x y

z

x y

z(a) (b)

positive at the centre, crossing to negative at large distances

spherical


Fig 2.16 Contour plot/density plot

(a) (b) (c)

-4 0 4 8

-8

-4

0

4

8

-

8

red: positive; blue: negative; green: zero

radial node: value ofrat which the wavefunction crosseszero

(note that at a node the wavefunction must crosszero, not just

be zero)


2p: Fig 2.17 Iso-surface

x y

z

x y

z

x y

z(a) (b) (c)

2pz 2px 2py

red: positive; blue: negative

three degenerate orbitals, pointing along x,yand z

each has a nodal plane(also called an angular nodei.e.

angles at which the wavefunction crosses zero)


Fig 2.18 2pz cubes

(a) (b) (c)

z

x

y

thexyplane is a nodal plane for the2pz AO



9/99

Fig 2.19 2py cubes

(a) (b) (c)

z

x

y

thexzplane is a nodal plane for the2py AO


Fig 2.20 Contour plot/density plot

(a) (b)

-4 0 4 8-8

-4

0

4

8

-8

x

z


nodal plane (xy) appears as a line


Fig 2.21 Hownotto draw aporbital

the positive and negative lobes do not touch

the shape of the lobes is incorrect


2.2.3 Mathematical form of the 1s, 2sand 2porbitals

x

y

r

z

usespherical polar coordinates

wavefunction is a product of a radial partand an angular part

n,l,ml (r, , ) = Rn,l(r)radial part

Yl,ml (, )angular part

note that the wavefunctions are labelled with quantum

numbers2: Electrons in atoms 29


10/99

Recall radial distribution function

probability of finding the electron in a thin shell of radius rand

thicknessri.e. summed over all angles

Pn,l(r) is related simply to the radial part of the wavefunction

Pn,l(r) = r2 [Rn,l(r)]

2

the angular part Yl,ml (, )does not affect the RDF


Fig 2.23 Radial parts and RDFs for 1s, 2sand 2p

r/ a0

r/a0

P(r)R(r)

(a) (b)

2 4 6 8 10

2 4 6 8 10

1s

1s

2p 2p

2s

2s

radial node for2sbut not for1sand 2p;2pgoes to zero at

nucleus (but this is not a node, as wavefunction does not

cross zero)

2sand 2plarger than 1s: principal maximum at larger r

2sand 2phave subtly different distributions of electron

density; note especially the subsidiary maximum, close to the

nucleus, for the 2s2: Electrons in atoms 31

Fig 2.28 Radial parts and RDFs for 3s, 3pand 3d

r/ a0r/a0

(r) P(r)

(a) (b)

2s

3p 3p 3s

3s

3d

3d

5 10 15 20 25 30

5 10 15 20 25 30

two radial nodes for 3s, one for3p, none for 3d

3pand 3dgo to zero at nucleus

larger than2s(and1s)


Fig 2.29 Contour plot/density plot for 3s

(a)

(b) (c)

-10

0

10

20

-20-10 0 10 20-20


spherical; two radial nodes



11/99

Fig 2.30 Iso-surface plots of three 3p

x x xy y y

z z z(a) (b) (c)

3pz 3px 3py

basically like2pwith a nodal plane (angular node)

radial node cuts the lobes


Fig 2.31 Contour plot/density plot for 3pz(a)

(b)

y

orz

x

-10

0

10

20

-20-10 0 10 20-20


radial node appears as a (green) circle

angular node appears as a (green) line


Fig 2.32 Iso-surface plots of three of the 3d

x x xy y y

z z z(a) (b) (c)

3dxz 3dyz 3dxy

two angular nodes

e.g. for3dxz,yzand xyare nodal planes


Fig 2.33 Contour plot/density plot for 3dxz

(a)

(b)

z

x

-10

0

10

20

-20-10 0 10 20-20


angular nodes (two) appear as a (green) line



12/99

Fig 2.34 Iso-surface plots of the other two 3d

x xy y

z z(b)(a)

3dx2-y2 3dz2

still have two angular nodes, but these no longer correspond

to simple planes such as thexyplane


Fig 2.35 Contour plot/density plot for3dx2y2 and3dz2

(a) (b) (c) (d)

y z

xor yx3dx2-y2 3dz2

the two angular nodes appear as a green lines

for3dx2y2 these two angular nodes are planes bisecting the

xzand yzplanes


Fig 2.36 Angular nodes in 3dz2

x y

z

angular nodes are two cones with (latitude) angles of 54.7

and 125.3


2.4 Spin

electrons have intrinsic source of angular momentum, called

spin angular momentum

need to add another quantum number s, the spin angular

momentum quantum number; always takes value of 12

associated quantum numberms takes values +12

and 12

(like

relation betweenl and ml)

ms = +12

, spin up or

ms =12

, spin down or



13/99

2.5 Hydrogen-like atoms

these have just oneelectron, but nuclear charge is larger

e.g. He+ (Z= 2), Li2+ (Z= 3)

energy scales as Z2

En =Z2RH

n2

orbitals shrink

1 2 3 4 5

r/ a0

P(r)

Z=1

Z=2

Z=3


2.6 Multi-electron atoms

hydrogen is particularly simple as there is only one electron

from He onwards we need to take account of

electronelectronrepulsion: this makes things much more

complicated

1

2

+2

we use theorbital approximation: electrons are assumed to

be moving in themean fieldof all the other electrons

as a result, each electron can be assigned to a hydrogen-like

orbital


2.6 Multi-electron atoms: electronic configurations

electrons assigned hydrogen-like orbitals; up to two (spin

paired) electrons per orbital

leads to the familiar electronic configurations:

He: 1s2; Li1s2 2s1; B:1s2 2s2 2p1; Ne1s2 2s2 2p6

the orbitals are similar to, but not the same as, the hydrogen

orbitals; exact form can only be found from computer

calculations

in particular the energies are no longer simply related to the

value ofn


Fig 2.43 Orbital energies: first two periods

1 2 3 4 5 6 7 8 9 10

-60

-50

-40

-30

-20

-10

0

1s

1s2

1s22s1

1s22s2

...2p1

...2p2

...2p3

...2p4

...2p5

...2p6

1s1

2s2p

H He Li Be B C N O F N e

energy/eV

1squickly falls in energy after it is filled

2slower in energy than 2p: this is why 2s is filled first

2sand 2pfall steadily in energy across the period

gap between2sand 2pwidens



14/99

Screening in Li, the 1s2 are closer to the nucleus than the 2s

the1s2 can be thought of a screeningthe 2sfrom the nucleus

if the screening is perfect, the 2ssees a nuclear charge of +1

Z=+32s

1s2

(a)

Zeff=+12s

(b)

the charge experienced by an electron is called the effective

nuclear charge,Zeff2: Electrons in atoms 46

Effective nuclear charge

assume orbital energy follow same form as for hydrogen but

replaceZwithZeff

En =Z2

effRH

n2

rearrange to give Zeff in terms of the orbital energyEn

Zeff =n2En

RH

for Li, 2shas energy 5.34 eV, giving Zeff=1.25

Zeff is much less than the actual nuclear charge (3), so

screening is significant (but not perfect)


Penetration another way of looking at this is say that the 2spenetratesto a

small extent inside the region occupied by the 1s2

useful way of explaining why 2sand 2phave different energies

approximate RDFs for Li (Zefffor 1sis 3, but for 2sand 2pis 1)

2s

1s

2p

2 4 6 8 10

r/ a0

the2spenetrates more into the region occupied by the 1sthan

does the2p

2sexperiences a higherZeffand so is lower in energy than 2p2: Electrons in atoms 48

Fig 2.43 Orbital energies: first two periods

1 2 3 4 5 6 7 8 9 10

-60

-50

-40

-30

-20

-10

0

1s

1s2

1s22s1

1s22s2

...2p1

...2p2

...2p3

...2p4

...2p5

...2p6

1s1

2s2p

H He Li Be B C N O F N e

energy/eV

1squickly falls in energy after it is filled:the 2sand 2p

electrons are not effective at shielding the 1sfrom the

increased nuclear charge

2sand 2pfall steadily in energy across the period: the 2sand

2pelectrons are only partially effective at screening one

another from the increased nuclear charge

gap between2sand 2pwidens: 2spenetrates more to the

nucleus and so experiences more of the increased nuclear

charge 2: Electrons in atoms 49


15/99

Fig 2.42 Orbital energies: first three periods

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

-120

-140

-80

-100

-60

-40

-20

0

1s

1s2

1s22s1

1s22s2

...2p1

...3p1

...3p2

...3p3

...3p4

...3p5

...3p6

...3s1

...3s2

...2p2

...2p3

...2p4

...2p5

...2p6

1s1

2s

2p

3s

3p

H He Li NaBe B C N O F Ne Mg Al S i P S C l Ar

energy/eV

3s/3phigher in energy than 2s/2p:attributed to increase in n

2s/2pquickly fall in energy after after they are filled

3sand 3pfall steadily in energy; 3sis below3p

gap between 3sand 3pwidens


2.6.5 Excited states and empty orbitals recall: orbital energy depends on average repulsion with all

other electrons

thus orbital energy depends on precisely which other orbitals

are occupied

orbital energies are thereforenota fixed ladder of levels

for example:

-25

-20

-15

-10

-5

2s22p2

2p

2s

2s12p3 2p4

energy/eV

energy of an empty orbital is the energy that an electron

would have, were it to occupy that orbital


2.7 Ionization energy

ionization energy (IE) is energy required to remove a

particular electron to infinity

i.e. for the process

A(g)A+(g) + e

energy change is

IE = energy of A+ energy of A

one electron atom: energy of A + is zero and energy of A is

simply the energy of the electron in its orbital, hence

one-electron atom: IE = orbital energy


Ionization of multi-electron atoms

ionization energy (IE) is energy required to remove a

particular electron to infinity


recallenergy of an electron in an orbital depends on which

other orbital are occupied

i.e. orbital energieschangewhen an electron is removed

it can therefore only be anapproximationthat

IE orbital energy



16/99

Fig 2.51 Ionization energies

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 180

5

10

15

20

25

1s2

1s22s1

1s22s2

...2p1

...3p1

...3p2

...3p3

...3p4

...3p5

...3p6

...3s1

...3s2

...2p2

...2p3

...2p4

...2p5

...2p6

1s1

H He Li NaBe B C N O F Ne Mg Al Si P S Cl Ar

energy/eV

ionization energy

-(orbital energy )

IE orbital energy works pretty well

but drop in IE from N to O is a particularly noticeable deviation

explanation for this is due to the effects of quantum

mechanicalexchange energy


2.7.3 Exchange energy

how do we fill up the 2pAOs as we go across the Second

Period?

experimentally, the lowest-energy arrangements are

p6

p1

p2

p3

p4

p5

B

C

N

O

F

Ne

rationalization is that we are maximizing the number of

parallel spins


Fig 2.53

possible arrangements forp3

(a)

(b)

(c)

(d)

(a) has three pairs of parallel spins: 12, 13 & 23

(b) has one pair of parallel spins: 12; (c) has the same, but

this time it is 23

(d) has one pair of parallel spins: 12

each pairof parallelspinslowersthe energy due to a purely

quantum mechanical effect (the exchange energy); (a) is

therefore the lowest energy arrangement


Fig 2.53

possible arrangements forp3

(a)

(b)

(c)

(d)

(a) is the lowest energy arrangement

this is notbecause: electrons repel one another, so placing them in separate

orbitals minimizes the repulsion

electrons with parallel spins avoid one another, so having

parallel spins further lowers the energy

common, but entirely wrong, explanations



17/99

Exchange energy

ionization energy is difference


need to work out the exchange contribution to the total energy

of A and A+

simple model: each pair of parallel spin electrons lowers the

energy byK

e.g.2p2 has an exchange energy of K, but2p3 has an

exchange energy of 3K


Tabulate exchange energy

atom ion

element config. Eexch(atom) config. Eexch(ion) Eexch(ion) Eexch(atom)

B 2p1 0 2p0 0 0

C 2p2 K 2p1 0 K

N 2p3 3K 2p2 K 2K

O 2p4 3K 2p3 3K 0

F 2p5 4K 2p4 3K K

Ne 2p6 6K 2p5 4K 2K

right-hand column gives the exchange contribution to the

ionization energy

exchange contributionincreasesthe ionization energy of N

relative to C, and C relative to B


Fig 2.54 Exchange contributions

p1 p2 p3 p4 p5 p6p0 p1 p2 p3 p4 p5

exchangeenergy

E

exch

(ion)-Eexch

(atom)

atom

ion(a) (b)

(c)

-6K

-4K

-2K

0

2K

-6K

-4K

-2K

0

2K

5

10

15

20

B C N O F Ne

p1 p2 p3 p4 p5 p6p0 p1 p2 p3 p4 p5

atom

ionB C N O F Ne

p1 p2 p3 p4 p5 p6p0 p1 p2 p3 p4 p5

atom

ionB C N O F Ne

ionizationenergy

/eV

(a) gives exchange contribution to each configuration, (b)

gives difference in these contributions, (c) shows effect of

superimposing the exchange effects on a general rise in the IE


Fig 2.54 Exchange contributions

p1 p2 p3 p4 p5 p6p0 p1 p2 p3 p4 p5

exchangeenergy

E

exch

(ion)-Eexch

(atom)

atom

ion(a) (b)

(c)

-6K

-4K

-2K

0

2K

-6K

-4K

-2K

0

2K

5

10

15

20

B C N O F Ne

p1 p2 p3 p4 p5 p6p0 p1 p2 p3 p4 p5

atom

ionB C N O F Ne

p1 p2 p3 p4 p5 p6p0 p1 p2 p3 p4 p5

atom

ion

B C N O F Ne

ionizationenerg

y/eV

the high IE of N relative to O is often said to be due to some

special stability of a half-filled shell: the proper explanation is

more complex, and is due to the change in the exchange

contribution on going from A to A+



18/99

2: Electrons in atoms

the end


Instituto de Qumica

Chemical structure and reactivity:an orbital based approach

James KeelerUniversity of Cambridge


3: Electrons in molecules: diatomics 63

3: Electrons in molecules:

diatomics


Molecular orbitals

assign electrons to molecular orbitals (MOs) in analogy to

atomic orbitals

a powerful framework for understanding structure and bonding

can explain, for example: H2 exists as a stable molecule, He2 is unknown, but He

+

2

has

been detected the bond dissociation energy of Be2 is less than one tenth of

that of either Li2 or B2 N2 has a stronger bond than does O2 the bond in N+

2is weaker than that in N2, but the bond in O

+

2 is

stronger than that in O2 in LiH, the hydrogen has a partial negative charge, whereas

in FH, the hydrogen has a partial positive charge O2 is paramagnetic, meaning that it is drawn into a magnetic

field, but N2 is not



19/99

3.1 Introducing molecular orbitals

+1+1

-1

in H +2

there are several interactions which contribute to the

energy

electronnuclear attraction (blue)

nuclearnuclear repulsion (red)

there is also a contribution to the energy from the kinetic

energy of the electron

the equilibrium bond length (that with the lowest energy) is

determined by the balance between all of these


Fig 3.2 Contributions to the energy

RRe

0

energy

e-n

n-n

ke

tot

H+H+

H2+

en: electronnuclear attraction (blue)

nn: nuclearnuclear repulsion (red)

ke: kinetic energy of the electron (green)

total energy (black) is the sum of allthese, and is a minimum atRe


3.1.1 Linear combination of atomic orbitals

a convenient and intuitive way of constructing MOs

combine1sAOs from the two atoms (A and B)

MO = cA (1sAO on atom A) + cB (1sAO on atom B)

cAand cB are theorbital coefficients; they are just numbers,

whose values determine how much of each AO is present inthe MO

in H +2

the situation is very simple: there are two MOs

+ = [(1sAO on atom A) + (1sAO on atom B)]

= [(1sAO on atom A) (1sAO on atom B)]


Fig 3.3 Bonding and antibonding MOs

quantum mechanics allows us to compute the energies of+andas a function of the bond length

R/ a0

E+

E-

energy

2 4 6 8

energy of + (E+) falls as atoms approach, reaching a

minimum: called the bonding MO

energy of (E) rises as atoms approach: called the

antibonding MO



20/99


x

y z

x

y z

x

y z

x

y z

x

y z

x

y z

R= 6a0 R= 4a0 R= 2.5a0

(a)

(b)

equilibrium bond length is 2.5a0

(a), bonding MO: results from constructiveor in-phaseoverlap

(b), antibonding MO: results from destructiveor out-of-phase

overlap



R= 6a0 R= 4a0 R= 2.5 a0

(a)

(b)

(a), bonding MO: builds up electron density between the

nuclei this is why the energy falls

(b), antibonding MO: electron density pushed away from the

internuclear region3: Electrons in molecules: diatomics 71

3.1.3 Symmetry labels for MOs

assign, or based on what happens to the wavefunction

as you traverse the indicated path

no sign change

cross one nodal plane

cross two nodal planes


3.1.3 Symmetry labels for MOs

(a) (b) (c)

start at any point, move to centre, carry on in same direction

for same distance

if end up at an equivalent point, object possesses a centre of

inversion

(a) and (c) each have a centre of inversion, (b) does not

a homonuclear diatomic has a centre of inversion3: Electrons in molecules: diatomics 73


21/99

Fig 3.9 Symmetry labels for MOs(a) (b)

(a) in bonding, (b) is antibonding MO

both have no sign change as you go round the internuclear

axis

on inversion, bonding MO does not change sign g

on inversion, antibonding MO changes sign u

bonding MO isg; antibonding MO is u


3.1.4 MO diagrams

(a) (b) (c)

R= 6a0 R= 4a0 R= 2.5a0

1g

1u

1s(A)

A B A B A B

1s(B)energy

vertical scale is energy

energies of AOs shown to left and right; energies of MOs

shown in middle

as bond length decreases, interaction increases and energy

shift to MOs increases

antibonding MO goes up in energy more than bonding MO

goes down


3.1.6 Overlap

R/a0

1s(A)

1s

(A)

1s

(B)

1s(B)

overlap,

S(R)

1 2 3 4 5 6

0.2

0.0

0.4

0.6

0.8

1.0

the overlap integral is theareaunder a plot of the productof

the two wavefunctions

S(R)falls off as the internuclear separation increases

S(R) is a guide to how much the bonding MO will be lowered

in energy compared to the AOs

note that at short distances internuclear repulsion increases

and the bonding MO will then rise in energy3: Electrons in molecules: diatomics 76

3.1.7 Summary

MOs are formed from the linear combination of AOs on

different atoms

in H +2

combining two 1sAOs gives two MOs: one bonding and

one antibonding

bonding MO shows a minimum in its energy at a certain

separation; the energy of the antibonding MO simply

increases as the internuclear separation decreases bonding MO arises from an in-phase or constructive

combination of the AOs; this leads to a concentration of

electron density in the internuclear region

antibonding MO arises from an out-of-phase or destructive

combination of the AOs; this leads to electron density being

excluded from the internuclear region and being concentrated

on the periphery



22/99

3.2 H2, He2and their ions

1g

1u

H2

+

1g

1u

H2

1g

1u

He2

+/ H2

-

1s 1s

molecule configuration diss. energy / kJ mol1 bond length / pm

H+2 1

1g 256 106

H2 12g 432 74

He+2 1

2g1

1u 241 108

He2 12g1

2u not observed


3.3 Homonuclear diatomics of the Second Period

moving from H2 to Second Period diatomics brings extra

complications

more AOs to overlap (2sand 2p)

different types of overlap (and )

we are guided by five rules for forming MOs; these arise from

the underlying quantum mechanics


3.3.1 Rules for forming MOs

1. the combination of a certain number of AOs produces the

same number of MOs

2. only AOs of the correct symmetrywill interact to give MOs

3. the closer in energy the AOs, the larger the interaction when

MOs are formed

4. each MO is formed from a different combination of AOs; AOs

which are close in energy to the MO contribute more than

those which are further away in energy

5. the size of the AOs must be compatible for there to be a

strong interaction when MOs are formed


Rule 2: symmetry

consider the overlap of a2swith a2px AO (internuclear axis is

alongz)

positiveoverlap

negativeoverlap

1s

2px

z

positive overlap of the upper lobe is equal and opposite to the

negative overlap of the lower lobe

net overlap is zero: no MOs are formed

2s is symmetric with respect to the yz plane, whereas2pxis

anti-symmetric i.e. they do not have the same symmetry



23/99

Rule 2: symmetry

correct symmetry to overlap

no overlap

1s

1s

1s 1s

2pz 2pz

2pz

2pz

2px 2px

2px 2px

overlap which leads to bonding is shown; there is also an

antibonding MO formed


Rules 3 and 4: energy match

increasing energy gap between AOs

A B A

B

A

B

A

B

**

**

as energy separation of AOs increases, change in energy on

forming the MO decreases

antibonding MO (indicated by ) liesabovethe highest energy

AO, bonding MO liesbelowthe lowest energy AO

if the energy mismatch between the AOs is large, then there is

hardly any change in the energy


Rules 3 and 4: energy match(a) (b) (c)

*

**A B A

B

A

B

unless AOs have the same energy they make an uneven

contribution to the MOs

greatest contribution is from the AO closest in energy to the

MO

if the energy mismatch between the AOs is large, then there is

hardly any contribution from one of the AOs3: Electrons in molecules: diatomics 84

Rules 3 and 4: energy match

As the energy separation between the AOs increases:

the bonding MO lies closer and closer in energy to that of the

lower energy AO

the antibonding MO lies closer and closer in energy to that of

the higher energy AO

the contribution to the bonding MO from the lower energy AO

increases, while that from the higher energy AO decreases

the contribution to the antibonding MO from the higher energy

AO increases, while that from the lower energy AO decreases



24/99

Rule 5: size

(a) (b)

(a) overlap of 2s, (b) overlap of1s(the1sare contracted due

to experiencing a relatively high effective nuclear charge)

although energy match is good, the two 1sare too contracted

to overlap significantly

for the Second Period diatomics, we can simply ignore the 1s

AOs


Types of MOs from 2sand 2pAOs; Fig 3.19

(a) (b)

-4 -2 0 2 4 6-6

-4

-6

-2

0

2

4

6g u

assume only2s2sand 2p2poverlap (energy match)

2s2sgives bonding g and antibondingu

u has nodal plane between the two atoms

only outer part of 2sinvolved in overlap


Fig 3.20 MOs from 2soverlap

(a) (b)x

yz

x

yz

g u

2s2sgives bonding g and antibondingu



Fig 3.21 Cartoon representation of MOs from 2s

overlap

+

+

g

u

grey indicates positive; white indicates negative

in-phase and out-of-phase overlap



25/99

Fig 3.22 MOs from 2poverlap

(a) (b)

g u

x

yz

x

yz

head on overlap of2pz gives bonding g and antibondingu




(a) (b)

g u

-4 -2 0 2 4 6-6

-4

-6

-2

0

2

4

6

head on overlap of2pz gives bonding g and antibondingu



Fig 3.24 Cartoons representation of 2p-type overlap

+ g

+ u


Fig 3.26 MOs from 2poverlap(a) (b)

(c) (d)

gu

x

yz

x

yz

x

yz

x

yz

side ways overlap of 2px(or 2py)gives bonding u and

antibondingg

both have a nodal plane containing the internuclear axis

form a degenerate pair3: Electrons in molecules: diatomics 93


26/99


(a) (b)

u g

-

4 -

2 0 2 4 6-

6

-4

-6

-2

0

2

4

6

z

x

both have a nodal plane containing the internuclear axis

antibonding MO has nodal plane between the two nuclei

second pair of MOs in perpendicular plane


Fig 3.27 Cartoons representation of 2p-type overlap

+ u

+ g


3.3.3 Idealized MO diagram for homonuclear diatomic

only allow 2s2sand 2p2poverlap

ignore1sAOs (too contracted)

sequentially number MOs with same symmetry label e.g. 1g,

2g etc.


Fig 3.28 Idealized MO diagram for homonuclear diatomic

2g

2u

3g

1u

2p

2s2s

2p

1g

3u

AOs on A AOs on BMOs



27/99

Fig 3.29 MO diagram for O2

16 electrons giving configuration

12g12u 2

2g2

2u 3

2g 1

4u1

2g

1g and 1u neither bonding nor

antibonding: ignore

8 bonding electrons, 4 antibonding hence netbonding i.e. stable with respect to

dissociation into atoms

bond order (BO)

BO = 12

(no. ofbondingelectrons

no. ofantibondingelectrons)

=2

2g

2u

3g

1u

1g

3u


Fig 3.30 O2 is paramagnetic

paramagnetism: drawnintoa magnetic field

associated with unpaired electrons

configuration of O2 has two unpaired electrons in g

12g12u 2

2g2

2u 3

2g1

4u1

2g

triumph of MO theory to explain paramagnetism


MO diagram for F2

18 electrons giving configuration

12g12u 2

2g2

2u 3

2g 1

4u1

4g

8 bonding electrons, 6 antibonding hence net

bonding i.e. stable with respect to

dissociation into atoms

BO = 1

dissociation energy of O2 494 kJ mol1, but

for F2 it is 154 kJ mol1

2g

2u

3g

1u

1g

3u


Ions of O2

3g

1u

1g

3u

species bond length / pm diss. energy / kJ mol1 BO paramag?

O+2 112 643 2.5 yes

O2 121 494 2 yes

O2

135 395 1.5 yes

O22

149 204 1 no



28/99

Fig 3.31 Allowing forspmixing

2g

2u

3g

2g

3g

3u

2u

3u

(a)

MO1

MO2

MO3

MO4

MO5

MO6

2p

2s

(b)

1u1u

1g1g

orbitals (including MOs) with same symmetryand which are

reasonably close in energy canmix

result is the higher energy MO goesupin energy, and the

lower energy MO goes down

2g and 3g mix: 3g becomes less bonding and maylie

above1u

spmixing decreases across the period3: Electrons in molecules: diatomics 102

Fig 3.34 Energies of occupied MOs

2u

1u

1g

2g

3g

Li2 Be2 B2 C2 N2 O2 F2

-5

0

-10

-15

-20

-25

-30

-35

orbitalene

rgy/eV


Fig 3.38 Heteronuclear diatomic: LiH

1s

2s

Li H

2

3

only consider 2son Li and 1son H

AO energies do not match resulting in unsymmetrical MOs

2bonding MO mostly on H: H is

note: nogor ulabels on MOs


Fig 3.39 Heteronuclear diatomic: HF

1s

FH3 3

11

4

2px/2py

2pz

H-F

H

-

F

only consider 2pon F since 2stoo low in energy; 1son H

2pxand 2py have no AOs on H with correct symmetry to

overlap nonbonding MOs (1)

3bonding MO mostly on F: F is



29/99

Fig 3.41 NO, NN and CO

2u

3u

1u

1g

1

2

2

2g 3

4

6

51

3

4

5 3g

2s

2s

2p2p

C-ON-NN-O

(c)(b)(a)6

more electronegative atom has lower energy AOs

1/1u are bonding

5/3g weakly bonding

NN and CO are isoelectronic and filled up to 5/3g; NO

has one extra electron in 2


Fig 3.42 NO, NN and CO

3

4

1

5

2

3

4

1

5

2u

3g

2g

1u

C-ON-NN-O

symmetrical in NN, but bonding MOs polarized towards more

electronegative atom in NO and CO


Fig 3.44 Heteronuclear diatomic: LiF

FLi

33

44

5

1

1

2s

2s

2p

Li-F

Li

-

F

AOs on F much lower in energy than those on Li

2pxand 2py have no AOs on Li with correct symmetry to

overlap nonbonding MOs (1)

energy separation of Li and F AOs so large that there is little

mixing

as if an electron transferred from Li to F (ionic)


3.5.3 The HOMO and the LUMO

highest occupied molecular orbital(HOMO)

lowest unoccupied molecular orbital(LUMO)

in CO HOMO is 5and LUMO is 2



30/99


diatomics

the end


Instituto de Qumica




4: Electrons in molecules: polyatomics 111


polyatomics


MOs for larger molecules

as the number of atoms and AOs increases, so does the

number and complexity of the MOs

computer programs available to compute MOs

but a pencil and paper approach is still a useful guide

symmetrycan provide helpful simplification

likewisehybrid atomic orbitals



31/99

4.1 Linear H+3

2

1

1

3

3

2

mirror plane swaps atoms 1 and 3 i.e. they are equivalent

mirror plane does not swap atom 2 with any of the others

recall only orbitals with the same symmetry overlap

we will classify AOs according to their symmetry with respect

to this mirror plane


Fig 4.2 Effect of mirror plane

2

1

3

symmetric

swapped by

mirror plane

consider1sAO on each H, and consider effect of reflection in

mirror plane

AO on atom 2 reflected into itself: symmetric

AO on atom 1 swapped with 3, likewise 3 swapped with 1

AOs on 1 and 3 are neither symmetric nor antisymmetric


Fig 4.3 Symmetry orbitals

1 3

1 3

symmetric: 1s(1) + 1s(3)

antisymmetric: 1s(1) -1s(3)

the combination1s(1) + 1s(3) is reflected onto itself with nosign change: symmetric

the combination1s(1) 1s(3)is reflected onto minusitself:

antisymmetric

combinations of AOs which are either symmetric or symmetric

under a particular symmetry operation are called symmetry

orbitals(SOs)

symmetric: orange; antisymmetric: green4: Electrons in molecules: polyatomics 116

Fig 4.4 MO diagram for H+3

SO1SO2

SO3

MO1

MO3

MO2

atom 2 on left, atoms 1 and 3 on right

note colour coding: symmetric: orange; antisymmetric: green

only orbitals with the same symmetry interact

as usual, form bonding MO from in-phase overlap and

antibonding MO from out-of-phase overlap



32/99

Fig 4.4 MO diagram for H+3

SO1SO2

SO3

MO1

MO3

MO2

MO1 is symmetric and bonding

MO3 is symmetric and antibonding

SO3 is the only antisymmetric orbital: it has nothing to overlap

with

SO3 becomes the nonbonding MO2

SO2 and SO3 have similar energies as little overlap between

non-adjacent atoms4: Electrons in molecules: polyatomics 118

Fig 4.5 Calculated MOs for H+3

MO1 MO2 MO3

these fit closely with our simple MO picture

however, it turns out that H+3

is not linear but an equilateral

triangle

. . . use the same procedure


Fig 4.6 Triangular H+3

: effect of mirror plane

2

1

3

symmetric

swapped by

mirror plane

choose one mirror plane (dashed line)

AO on atom 1 reflected into itself: symmetric

AO on atom 2 swapped with 3, likewise 3 swapped with 2


Fig 4.7 Symmetry orbitals

symmetric

antisymmetric

SO1 SO2

SO3

combine the AOs on atoms 2 and 3 to form SOs

symmetric (orange): SO2; antisymmetric (green): SO3

orbital on atom 1 forms the symmetric SO1



33/99

Fig 4.8 MO diagram for triangular H+3

SO2

SO1

MO1

MO3MO2 SO3

SO2 has adjacent AOs with positive overlap; SO3 has

adjacent AOs with negative overlap

SO2 is lower in energy than SO3

SO3 is only antisymmetric SO; becomes nonbonding MO2

symmetric SOs, SO1 and SO2 overlap to give bonding MO1

and antibonding MO34: Electrons in molecules: polyatomics 122

Fig 4.9 Calculated MOs for triangular H+3

MO1 MO3MO2

1

3 2

these fit closely with our simple MO picture


4.1.3 Optimum geometry for H+3

SO1SO2

SO3

MO1

MO3

MO2

SO2

SO1

MO1

MO3MO2 SO3

there are just two electrons, located in MO1

MO1 is lower in energy/more strongly bonding for the

triangular case than for the linear case . . .

. . . because there arethreefavourable interactions (one along

each edge) as opposed to only two in the linear case


Fig 4.10 Geometry optimization

60 100 140 180

energy

/

lowesttotalenergy with a bond angle of 60

computer programs which calculate MOs will also optimise

the geometry by seeking lowest energy



34/99

MOs of H2O: Fig 4.14 symmetry orbitals

SO1 SO2 SO4 SO5SO3symmetric symmetricsymmetric antisymmetric

2s 2pz 1s(A) 1s(B)2py

swapped by mirror plane

antisymmetric

classify according to mirror plane

oxygen2sand 2pz are symmetric;2py is antisymmetric

oxygen2pxhas node in plane of molecule: cannot overlap

with hydrogen1s

as before, form two SOs from hydrogen 1s


Fig 4.15 MOs of H2O

MO1

MO2

2px

MO3

MO4

MO5

MO6

SO1

SO2

SO4

SO5

SO3

orange: symmetric; green: antisymmetric

SO1, SO2 and SO4 all overlap to give the symmetric MOs

MO1 lies below all the symmetric SOs; MO6 lies above

MO3 somewhere in the middle4: Electrons in molecules: polyatomics 127

Fig 4.15 MOs of H2O

MO1

MO2

2px

MO3

MO4

MO5

MO6

SO1

SO2

SO4

SO5

SO3

SO3 and SO5 all overlap to give the antisymmetric MOs (MO2

& MO5)

the oxygen2px is nonbonding (MO4)

8 electrons fill up to MO4; only onestrictly nonbonding pair of

electrons (in MO4)4: Electrons in molecules: polyatomics 128

Fig 4.16 Computer-calculated MOs of H2O

MO1

MO2

MO3

MO4


Fi 4 17 C l l d MO f CH 4 4 H b id i bi l


35/99

Fig 4.17 Computer-calculated MOs of CH4

MO1 MO2

MO3 MO4

going much further pretty hard .. .

. . . hybrid atomic orbitalsprovide a way forward


4.4 Hybrid atomic orbitals

problem with MOs is that the AOs do not necessarily point in

the right directions e.g. tetrahedral CH4

several AOs overlap to form an MO, which is likely to be

spread over several atoms

a different approach: combine the AOson one atomto formnew orbitalsdesignedto point in the desired directions

these are called hybrid atomic orbitals, HAOs

an HAO can then overlap with just one other orbital to give a

bonding and an antibonding MO


4.4.1sp3 hybrids; Fig 4.18

(a) (b) (c) (d)

four sp3hybrids

x y

z

a b

cd

HAOs formed from2sand the three 2p

designed to point to the corners of a tetrahedron


Fig 4.19 contour plot of one of the sp3 hybrids

-4-6 -2 0 2 4 6

-4

-6

-2

0

2

4

6

note the directional properties


Fi 4 19 D ibi b di i CH i 3 h b id Fi 4 21 CH MO di i 3 h b id


36/99

Fig 4.19 Describing bonding in CH4usingsp3 hybrids

*

sp3 1s

C-H

one HAO overlaps with one hydrogen 1sto give a bonding

MO and a antibonding MO

bonding MO concentrates electron density along the CH

direction

repeat for all four HAOs, giving four bonding MOs

2 electrons is each gives four two-centre two-electronbonds


Fig 4.21 CH4MO diagram using sp3 hybrids

4()

4(*)

4(sp3)2p

2s

4(1s)

note four HAOs giving four distinct and directional bonding

MOs


Describing the bonding in ethane usingsp3 hybrids

in each CH3 fragment, three MOs formed by overlap of an

sp3 HAO with a hydrogen1s

remainingsp3 HAO on each carbon overlaps with the same

orbital on the other carbon to give CC and MOs

contour plots

*

C-C


4.4.2sp2 hybrids

these lie on a plane and point at 120 to one another: ideal for

describing doubly-bonded carbon compounds

ethene

C C

H

HH

H


4 4 2 2 h b id Fi 4 24 Fi 4 25 C t l t f 2 h b id


37/99

4.4.2sp2 hybrids; Fig 4.24

(a)

a b

c(b) (c) (d)

x y

z

three sp2hybrids 2pz

three HAOs formed from2sand two2pAOs

lie in a plane, and pointing at 120 to one another

the remaining 2pAO (here2pz) is notinvolved in forming the

HAOs and points out of the plane of the HAOs


Fig 4.25 Contour plots of sp2 hybrids

(a) (b) (c)

y

x

-4-6 -2 0 2 4 6

-

4-6

-2

0

2

4

6

note directional properties


Fig 4.26 bonding framework in ethene

*

sp21s

*

sp2 sp2

sp2 overlaps with hydrogen 1sto give CH and

sp2 HAOs on different carbons overlap to give CC and

occupation of all the MOs accounts for 10 electrons


Fig 4.26 bonding in ethene

2pz2pz

*

2pz AOs on each carbon overlap to give and MOs

these have a nodal plane in the plane of the molecule (no

overlap between the and orbitals, therefore)

occupation of the MO accounts for the final two electrons


Fig 4 27 Rotation abo t the bond in ethene 4 4 3 sp h brids Fig 4 28


38/99

Fig 4.27 Rotation about the bond in ethene

(a) (b) (c)

rotation about the CC bond reduces overlap between the2pzAOs

if rotate through 90 then no overlap i.e. the bond is broken

origin of high barrier to rotation about a C=C bond


4.4.3sphybrids; Fig 4.28

(a)

a

b

(b) (c) (d)

two sphybrids 2px 2py

z x

y

two HAOs formed from2sand one2pAO

HAOs point at 180 to one another

the remaining two2pAOs (here 2pxand 2py) arenot involved

in forming the HAOs; these 2pAOs point perpendicular to the

line of HAOs


Fig 4.29 Contour plots of sphybrids and unhybridized 2p

(a) (b) (c)

z

x

-4-6 -2 0 2 4 6

-

4-6

-2

0

2

4

62px

note directional properties


Fig 4.30 Bonding in ethyne

*

sp1s

*

sp sp

2px2px

*

2py2py

*

spoverlaps with hydrogen 1sto give CH and

spHAOs on different carbons overlap to give CC and


Fig 4 30 Bonding in ethyne 4 5 Comparing the hybrid and full MO approaches


39/99

Fig 4.30 Bonding in ethyne

*

sp

1s

*

sp sp

2px2px

*

2py2py

*

2pxAOs on each carbon overlap to give and MOs

2pyAOs on each carbon overlap to give and MOs

occupation of all the bonding MOs accounts for all 10

electrons4: Electrons in molecules: polyatomics 146

4.5 Comparing the hybrid and full MO approaches

we have already looked at the MO description of N2: involves

2sand 2pAOs, with significant mixing between 2sand 2p

HAO approach: usesphybrids on the nitrogen

overlap of two spHAOs gives bonding and antibonding MO

otherspHAOs point away from bond and when filled become

lone pairs

overlap of2px and 2pygives two bonding MOs and two

antibonding MOs


Fig 4.31 Comparing the hybrid and full MO approaches

*

*

2g

3g

2u

3u(a) (b)

2p2px,y

spd

spa spb

spc

2s

2p

2s

1u

1g

AO AOHAO HAO

HAO:()2 ()4 (spc)2 (spd)

2

i.e. bond, two bonds and two lone pairs

MO:(2g)2 (2u)

2 (1u)4 (3g)

2

i.e. two bonds and . . .4: Electrons in molecules: polyatomics 148

4.5.1 More about lone pairs

two electrons in out-of-plane 2pxAO in H2O are clearly

non-bonding and hence alone pair

ifboththe bondingandthe corresponding antibonding MOs

are filled, the result is no net bonding arising from the four

electrons

could describe this a two lone pairs

e.g. in N2 (2g)2 (2u)

2 is equivalent to two lone pairs, just as

in the HAO approach

e.g. in F2 (2g)2 (2u)

2 (1u)4 (3g)

2 (1g)4 is equivalent to two

lone pairs from filled MOs and four lone pairs from filled

MOs i.e. six lone pairs in total


4 6 Extending the hybrid concept 4 33 HAOs for H O


40/99

4.6 Extending the hybrid concept

recall HAOs designed to point towards the other atoms

by varying the ratio ofsto pin the hybrids can adjust the

angle between the hybrids

sp

sp2

sp3

%s

ch

aracter

/ degree100 120 140 160 180

10

20

30

40

50

e.g. in H2O to obtain the required bond angle of 104.5 we

need 21% 2sin two of the hybrids (the other two have

proportionately morescharacter)


4.33 HAOs for H2O

HAO1 HAO2 HAO3 HAO4

x

y

z

x x

y

z

x

y

z

y

z

HAO1 and HAO2 point toward H, and form two-centre

two-electron bonds

HAO3 and HAO4 point away from the H atoms; occupied to

give two lone pairs

could describe the hybridization as approximately sp3


4.7 Bonding in organic molecules: saturated systems

CH3F

CH3OH

CH3NH2

can describe carbon assp3 hybridized

can assume the same for the F, O and N

HAOs not involved in bonding become lone pairs if filled

HOMO is lone pair (higher in energy than bonding pairs);

LUMO?4: Electrons in molecules: polyatomics 152

Fig 4.35 Identifying the LUMO

*

sp3 sp3

C-C

*

sp3 1s

C-H

*

sp3

1s

X-H

*

sp3

sp3

C-X

(a) (b) (c) (d)

X is an electronegative atom (lower energy AOs than carbon)

compared to CC, poorer energy match in CX so not

raised in energy so much

hence CX lower than CC ; former is the LUMO

same when comparing CH and XH; latter is the LUMO


4 7 2 Aldehydes ketones and imines 4 7 4 Energy ordering of orbitals


41/99

4.7.2 Aldehydes, ketones and imines

methanal

imine

can describe carbon assp2

hybridized; same for oxygen andnitrogen

CO interaction involves overlap of two sp2 HAOs to give a

bond

and two out-of-plane 2pAOs to give a bond

other HAOs on N and O filled to give lone pairs

HOMO is lone pair; LUMO is (lower in energy than )


4.7.4 Energy ordering of orbitals

as a general rule

highest energy antibonding

antibonding

nonbonding orbitals (including lone pairs)

bonding

lowest energy bonding

in CH3X have already seen we need to choose between

different MOs to identify LUMO


4.8 Delocalized bonding

C

C

C

C

C

C

H

H

H

H

H

H

benzene

R C

OH

O

R C

O

O

R C

O

O

(a)

(b) (c) HC

C

C

C

H

H

H

H

H

butadiene

examples of molecules with evidence for delocalized bonding

benzene is a regular hexagon: all CC bonds same length

CO bond length in carboxylate intermediate between COH

and C=O

in butadiene barrier to rotation about central CC bond is

significantly greater than that for rotation about CC single

bond, but less than that for rotation about C=C double bond


4.8.1 Orbitals in a row for a row of orbitals there is a simple way of finding the MOs

(a) number the atoms (here 14)

1 2 3 4

10 2 3 4 5

10 2 3 4 5

10 2 3 4 5

(a)

(b)

(c)

(d)

(b) add an extra atom at position 0 and position 5

(c) inscribe half sine-wave between 0 and 5

(d) orbital coefficients are proportional to the heights


4 8 1 Orbitals in a row Fig 4 38 MOs for four orbitals in a row ( overlap)


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4.8.1 Orbitals in a row

1 2 3 4

10 2 3 4 5

10 2 3 4 5

10 2 3 4 5

(a)

(b)

(c)

(d)

inscribed half sine wave gives the lowest energy MO

now inscribe two half sine waves between 0 and 5 to give next

highest energy MO

and then three half sine waves and so on


Fig 4.38 MOs for four orbitals in a row (overlap)

1

2

3

4

one, two, three and four half sine wave to give four MOs

recall number of MOs = number of AOs

note increasing number of nodes as energy increases4: Electrons in molecules: polyatomics 159

Butadiene

H

C

C

C

C

H

H

H

H

H

butadiene carbonsp2-carbon sp2

carbonsp2-hydrogen 1s

carbonsp2 hybridized: overlap to give framework

2pAOs out of plane of molecule: overlap to give four

delocalized MOs

four electrons in system: occupy 1and 2MOs


Fig 4.38 MOs of butadiene

1

2

configuration (1)2 (2)2

1bonding between all adjacent atoms

2bonding between 12 and between 34; antibonding

between 23

result is significant bonding between 12 and 34, and

partial bonding between 23


Fig 4 38 Computed MOs of butadiene Allyl cation and allyl anion


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Fig 4.38 Computed MOs of butadiene

HOMO

LUMO LUMO + 1

HOMO -1

very similar to simple prediction

HOMO is 2; LUMO is 3


Allyl cation and allyl anion

HC

CC

H

H

H

H

allyl cation

HC

CC

H

H

H

H

allyl anion

ions thought to be planar and symmetrical

sp2 hybridize carbons: overlap to form framework

three out of plane 2pAOs overlap to give system: three

orbitals in a row


Fig 4.1 MOs for allyl cation and allyl anion

1

2

3

anion cation

inscribe sine waves as before

1bonding across adjacent atoms; 2nonbonding

anion has 4 electrons: (1)2 (2)2

bonding entirely due to 1, but occupation of 2increases

electron density on end atoms


Fig 4.1 MOs for allyl cation and allyl anion

1

2

3

anion cation

cation has 2 electrons: (1)2

bonding entirely due to 1

highest electron density in middle



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4 9 Delocalized structures including heteroatoms 4 9 1 Carboxylate anion


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4.9 Delocalized structures including heteroatoms

e.g. the carboxylate anion

carboxylate anion

R C

O

O

R C

O

O

resonance structures rationalize that fact that CO bond

length is intermediate between typical CO and C=O

alternatively, choosesp2 hybrids for C and O . . .

. . . and form MOs from three out-of-plane 2pAOs


4.9.1 Carboxylate anion

approximate system by three 2pAOs in a row (not in fact

identical)

1 2 3

four electrons, occupy 1and 2: partial bond across all

atoms and highest electron density on end atoms

computed MOs compare well

HOMOHOMO -3 LUMO


4.9.2 Enolates

formed by removing a proton from a carbon adjacent to

carbonyl (the carbon)

H3C

C

H

O

C

C

H

O

H

H

base

C

C

H

O

H

H

C

C

H

O

H

H

A B

enolate

both resonance structures A and B contribute significantly

model bonding in CCO as three atoms in a row


4.9.2 Enolates

approximate system by three 2pAOs in a row (not in fact

identical)

1 2 3

four electrons, occupy 1and 2

computed MOs compare well

HOMOHOMO -2 LUMO

computed charges 0.81 on oxygen, +0.34 on carbonyl

carbon,0.62 on carbon


4.9.3 Amides 4.9.3 Amides


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4.9.3 Amides

e.g. methanamide

H

C

NH2

O

methanamide

X-ray diffraction data indicates a planar structure

taken to imply delocalized bonding over OCN fragment


4.9.3 Amides

computed MOs

HOMOHOMO -2 LUMO

again reminiscent of three orbitals in a row

resonance structures: B must contribute significantly

H

C

N

O

A B

H

H

H

C

N

O

H

H


Summary

can always find the MOs using a computer program, but we

can make some useful predictions about the MOs using

pencil and paper

symmetry can help in forming MOs

hybrid atomic orbitals can be used to give a simple description

of the bonding

the MOs in simple delocalized systems can be constructed

using a geometric argument

for delocalized systems these MOs give a useful alternative

to the use of resonance structures



polyatomics

the end



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Instituto de Qumica




5: Bonding in solids 178

5: Bonding in solids


Bonding in solids

types of bonding in solids:

molecular solids, which contain discrete molecules, held

together by weak interactions, such as hydrogen bonds

giant covalent solids, in which there is a network of covalent

bonds extending throughout the entire structure

metallic solids, in which there is extensive delocalization of the

electrons

ionic solids, in which it is the interactions between discrete

ions which hold the structure together


5.1 Metallic bonding: introducing bands

in a metal, electrons occupy orbitals which are delocalized

through the entire structure

formed in exactly the same way as MOs, but on a much larger

scale

the resulting sets of MOs are know as bands

to start with, think about this in one dimension only


Fig 5.1 Chain of sorbitals Fig 5.2 Density of states


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g 5 C a o s o b ta s

number of atoms in chain

energy

MOs arising from longer and longer chains of sorbitals

NAOs give NMOs

ranging from fully bonding between all adjacent orbitals to

fully antibonding5: Bonding in solids 182

g 5 e s ty o states

energy

density ofstates

in the limit of large N, so many orbitals that the energy is quasi

continuous

density of states number of levels per unit energy greatest

at lowest and highest energies

these MOs called crystal orbitals(COs) form a band


Fig 5.3 Formation of a band

energy

decreasingspacing

increasinginteraction

bonding

COs

antibonding

COs

range of energies as a function of separation

at large separation, no interaction energy same as AO

as separation decreases, interaction increases so separation

between most bonding and most antibonding CO increases:

bandwidthincreases

note half bonding and half antibonding COs


5.1.2 Conduction of electricitypartially filled band

energy

+

+

+

+

+

+

+

+

+

+

-

-

-

-

-

-

-

-

-

-

(a) (b) (c)

key feature of metals is that they conduct electricity

this is due to bands being partially filled

an applied electric field shifts the energies of the COs (up at

negative end, down at positive)

electrons can cascade down in energy from filled to empty

COs: henceconduction5: Bonding in solids 185

5.1.2 Conduction of electricity 5.1.3 Bands in three dimensions / overlapping bands


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y

filled band

energy

+

+

+

+

+

+

+

+

+

+

-

-

-

-

-

-

-

-

-

-

(d) (e) (f)

now consider case of a full band

COs still shift in energy .. .

. . . but no empty lower energy COs for electrons to cascade

down into

henceinsulator


pp g

same basic idea applies in three dimensions, but resulting

bands more complex

overlap between AOs in different directions will be different

bands may overlap: for example bands arising from 2sand

from 2p

s

band

p

ban

d

(a) (b)

rather than filling the top of sband, electrons fill the bottom of

pband as this lowers overall energy: complex pattern of band

occupancy


5.1.3 Overlapping bands

Norbitals overlap to giveNCOs that form a band

it takes 2Nelectrons to fill this band

in Li, band from 2s is half full, but in Be band is completely full

insulator?

Be is a metal, so presumably another band overlaps (e.g. 2p)

sband

p

band

(a) (b)


5.1.5 Band gaps and semiconductors

energy band gap

conduction

band

valence

band

separation between full and empty band is called the band

gap

filled band is called the valence band; empty band is called

theconduction band

if band gap is large, the material is an insulator (full band)


5.1.5 Band gaps and semiconductors Band gaps from Group 14 elements


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g p

if band gap is small enough, electrons can be thermally

excited to the conduction band

now have a partially filled (conduction) band, so material will

conduct . . .

. . . but not very well asemiconductor

conductivityincreaseswith increasing temperature as more

electrons are promoted to conduction band

opposite to a metal, where conductivity decreaseswith

increasing temperature


g p p

measured band gaps

band gap

element / eV / kJ mol1

C (diamond) 6.0 580

Si 1.1 107

Ge 0.67 64.2

Sn 0.08 7.7

Sn 0 0

Pb 0 0

C insulator; Si & Ge semiconductors; Sn and Pb metals


Band gaps from Group 14 elements in solid with tetrahedral coordination can assume sp3

hybridization at each atom

overlap ofsp3 on adjacent atoms gives and MO

these then overlap to give a band (full; the valence band)

and a band (empty; conduction band)

band structure varies down group

*

C Si Ge Sn

band gap depends on energy separation of and MOs,

and interaction between these orbitals as the band is formed

from these MOs: opposing factors 5: Bonding in solids 192

5.2 Ionic solids

ionic solids are held together by electrostatic interaction

between (charged) ions

lattice enthalpy: energy released on forming lattice from

gaseous ions

M+

(g)+

X

(g)MX(s) values can be determined experimentally using BornHaber

cycle

values can be estimated using the simple ionic model


5.2.1 The ionic model for lattice enthalpies 5.2.3 The Kapustinskii equation


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p

assume the ions are hard spheres with an electrostatic

interactionanda short-range repulsive term

resulting expression

Hlattice = NAAz+ze

2

40re

1

1

n

NA Avogadros constantAMadelung constant (depends on crystal structure)

z numerical charges on the ions

nparameter describing repulsion term (n = 9 . . . 12)

re separation of ions in lattice

can assign (and tabulate) radii of individual ions and hence

computere = (r+ + r)


p q

a more approximate, but still useful, expression for the lattice

enthalpy

Hlattice/ kJ mol1

1.07 105 nionsz+z

(r+ + r)/pm

nions is the number of ions in the formula unit e.g. 2 for NaCl,

3 for CaF2

table of ionic radii (subject to considerable variation)

ion r+ / pm ion r+ / pm ion r / pm ion r / pm

Li+ 68 O2 142 F 133

Na+ 100 Mg2+ 68 S2 184 Cl 182

K+ 133 Ca2+ 99 Se2 197 Br 198

Rb+ 147 Sr2+ 116 Te2 217 I 220

Cs+ 168 Ba2+ 134


5.2.4 Validity of the ionic model

0 .0 02 0 0 .0 02 5 0 .0 03 0 0 .0 03 5 0 .0 04 0 0 .0 04 5 0 .0 05 0 0 .0 05 5500

600

700

800

900

1000

1100

Group I halides

Cu(I) halides

Tl(I) halides

Ag(I) halides

1/(r++r-) / pm

-

1

-latticeenergy

/kJmol-1

lattice energy plotted against 1/(sum of ionic radii); dashed

line is prediction of Kapustinskii equation

good agreement for Group 1 halides

much poorer agreement for Tl(I), Ag(I) and Cu(I) halides;

lattice energiesgreaterthan predicted by ionic model

attributed to a covalentcontribution to the lattice energy5: Bonding in solids 196

5: Bonding in solids

the end



52/99

Instituto de Qumica




6: Thermodynamics and the Second Law 198

6: Thermodynamics and

the Second Law


Important questions

what determines whether or not a reaction will go?

what determines the position of equilibrium?

it is the Second Law of Thermodynamicsthat controls these

things

key relationships

rG

=RT ln K rG

= rH TrS

rG: standard Gibbs energy change; rH

: standard

enthalpy change; rS: standard entropy change for the

reaction


6.1 Spontaneous processes

a spontaneous process is something that goes without

intervention from us e.g.

NH3(g) + HCl(g) NH4Cl(s)

the reverse does not happen (but we can intervene to make it

happen)

more subtly, reactions come to a position of equilibriumdefined by a particular value of Ke.g.

CH3COOH + H2O CH3COO

+ H3O+

K= [CH3COO

][H3O+]

[CH3COOH][H2O]

the approach to equilibrium is spontaneous; when we reach

equilibrium, there is no further change


Is energy minimization the criterion Spontaneous endothermic processes


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for a spontaneous process?

reaction goes because the products are more stable than the

reactants

more stable means lower energy . . .

. . . so heat is given out?

energy

reactants

products

+ heat

thus spontaneous reactions must be exothermic

WRONG!


dissolvingNH4NO3(s)in water

the equilibrium2 NO2(g) N2O4(g)

2NO2N2O4

exothermic

low pressure100% NO2

high pressure100% N2O4

atmos. pressure70% NO2

N2O42NO2

endothermic

exothermic

2 NO2(g) N2O4(g)

endothermic

N2O4(g) 2 NO2(g)


6.2 Properties of matter: state functions

the density of a material is independent of how that material is

prepared: it is a property of matter a state function

state defined by temperature, pressure etc.

enthalpy,H, is a state function

the enthalpy change, rH, for

H2(g) + 12

O2(g)H2O(g)

has a fixed value (at given temperature)

we will identify two other important state functions: entropy

andGibbs energy


6.3 Entropy and the Second Law

the Second Law controls whether or not a process will be

spontaneous

Second Law: In a spontaneous process, the entropy of

the Universe increases

but what is entropy? how can we measure it? how can wemeasure its change for the Universe?

to start with we will take a microscopic view of entropy

it is often said that entropy is randomness but what does

this actually mean and how can we use this concept?


6.3.1 A microscopic view of entropy A simple example


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molecules have a large number of energy levelsavailable to

them

e.g. energy levels due to translation,rotationand vibration

in a macroscopic sample there are an extremely large number

of ways that the molecules could be arranged amongst the

energy levels

it is quite impossible to know the details, but we can fall back

on a statistical approach

this leads to a (microscopic) definition of entropy


14 molecules

energy levels 0, 1, 2, .. . in arbitrary units

total energy is 10 units

think about how we can arrange the 14 particles in the energy

levels such that the total energy is 10


A simple example

one possible distributionis

energy,i 0 1 2 3 4 5

population,ni 8 3 2 1 0 0

check number 8 + 3 + 2 + 1 + 0 + 0 = 14

check energy

E = n00 + n11 + n22 + n33 . . .

= 8 0 + 3 1 + 2 2 + 1 3

= 10

how many ways can we arrange the molecules in the energy

levels and achieve this distribution?


A simple example

how many ways can we arrange the molecules in the energy

levels and achieve this distribution?

energy,i 0 1 2 3 4 5


a simple combinatorial problem

W =

N!

n1! n2! n3! . . .

where e.g.5! = 5 4 3 2 1and 0! = 1

in this caseN= 14,n0 = 8,n1 = 3,n2 =2 and n3 = 1, so

W =1.8 10 5

even for just 14 molecules there are very many possible ways

of achieving this distribution


Other distributions Fig 6.2 Possible distributions


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another possible distribution with 14 molecules and 10 units of

energy

energy,i 0 1 2 3 4 5


for which W =91

another possibility

energy,i 0 1 2 3 4 5


for whichW =2.4 104

and on and on . . .


the three we have considered so far

0

1

2

3

4

5

6

7

(a) (b) (c)

(a)W =1.8 105, (b)W =91, (c)W =2.4 10 4


The most probable distribution can be shown that one distribution has the largest value ofW

this is the one for which the populations niobey the

Boltzmann distribution

ni =n0exp

i

kBT

iis the energy,T is the temperature, andkB is the Boltzmann

constant

populations fall off exponentially with the energy, and more

slowly at higher temperatures

energy

population

low T high T


Entropy and the most probable distribution

for large numbers of molecules it turns out that the most

probable distribution is effectively the onlydistribution which

occurs

recall that this is the Boltzmann distribution

Boltzmann hypothesized that

S= kBln Wmax

whereWmaxis the number of ways that the most probable

distribution can be achieved and Sis the entropy

units ofSare J K1 or J K1 mol1


Fig 6.2 Possible distributions Using Boltzmanns definition of entropy


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the three we have considered so far

0

1

2

3

4

5

6

7

(a) (b) (c)

(a) has the largest Wand is the one which obeys the

Boltzmann distribution


recallS= kBln Wmax

this puts in a quantitative and precise form the idea that

entropy is randomness: in fact, entropy is related to the

number of ways that molecules can be arranged amongst

energy levels

can use this to understand how entropy responds to various

changes

e.g. heating, expansion, molecular mass, change of state


Fig 6.4 Heating the sample

heating means putting in energy, so molecules promoted to

higher energy levels

for example

0

1

2

3

4

5

(a) (b)

increaseenergy

W= 1.8X105 W= 2.5X106

Boltzmann distribution in each case

heating increases Wand hence the entropy increases


Fig 6.5 Effect of temperature effect of adding a fixed amount of energy (here 5 units)

0

1

2

3

4

5

(a) cool ( b) warm (c) hot

4.0X103 1.8X105 2.5X106

(a) is cool, (b) is warm and (c) is hot i.e. temperature

reflects the amount of energy

Wincreases by more in going cool warm, than warm hot

when certain amount of energy is absorbed, the increase in

entropy isgreaterthe lower the temperature


Fig 6.6 Expanding a gaseous sample Increasing the molecular mass


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quantum mechanics tells us that when a gas is expanded the

(translational) energy levels get closer together

e.g.

0

1

2

3

4

5

(a) (b)

decreasespacing

W= 1.8X105 W= 1.5X107

decreasing the spacing increases W

expandinga gas increasesthe entropy


quantum mechanics tells us that when the mass of a molecule

increases the (translational) energy levels get closer together

hence, as with expanding a gas,Wincreases

heaviermolecules havegreaterentropy than lighter ones (all

other things being equal)


Changes of state

in a gas molecules are free to move, and hence have many

translational energy levels

in a solid, the molecules are not free to move, and so have

significantly fewer energy levels available

for liquids, the situation is intermediate

hence entropy of gas > liquid> solid


Summary

the entropy increases with temperature i.e. as energy is

supplied to the system, its entropy will increase

absorption of a given amount of energy gives rise to a larger

increase in entropy the lower the initial temperature

the entropy increases as a gas is expanded, and decreases

as a gas is compressed

the entropy of a gas increases as the mass of the

atoms/molecules increases

the entropy of the gaseous state of a substance is greater

than that of the liquid state, which in turn is greater that the

entropy of the solid state


The way forward What is heat?


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you may have come across the idea that entropy is

randomness; but what is randomness?

we have seen that entropy is related to the number of ways

that molecules can be arranged amongst energy levels

the is quantifiable (unlike randomness)

to proceed further we need to use a different, but entirely

equivalent, definition of entropy in terms of heat


a form of energy involved in bringing a hot object in contact

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