CHEMICAL BONDS

A chemical bond is formed between two atoms if theresulting arrangement of the atoms and their electronshas a lower energy than the separate atoms themselves.

Two main ways of creating a chemical bond:

Complete electron transfer IONIC BOND

Electron sharing COVALENT BOND

IONIC BONDS : formed between atoms withlow ionisation energies and those with highELECTRON AFFINITIES, e.g. Na+Cl.

You will hear a lot more about ionic compoundsand their properties in later modules.

We are going to concentrate on

COVALENT BONDING

Characteristic of elements in the middle of the Periodic Table - complete electron transfer is not practicable.

Nature of covalent bonding first established byG.N.Lewis, in 1916, before modern quantum theoryhad been thought of.

Lewis’ idea was that the formation of a covalentbond occurred when one electron from each atombecame shared to form an ELECTRON PAIR.

For hydrogen we would have:

nuclei

electrons

The two atoms are heldtogether by the attractionbetween the pair of electronsand both nuclei

Unequal sharing - polar covalent bonds

Homonuclear bond when the two atoms forming a covalent bond arethe same - the electrons will be equally shared.

Heteronuclear bond if the atoms are different – the sharing will be unequal!

Which atom gets more of the electrons?

Depends on the electronegativities of the atoms

Electronegativity Basic definition:

The ability of an atom in a molecule to attractelectron density to itself

How to make this idea quantitative?

Simplest (Mulliken)

affinity electron E

energy ionisation I

ativityelectroneg where

)EI(

a

a

2

1Ionisation energydefined earlier.Electron affinity:see next slide:

Definition of electron affinity:

anion of energy )X(E

atom X neutral of energy )X(E where

)X(E)X(EEa

Thus - the harder it is to remove an electron (I),and the more energy you get back by attachingan extra electron (Ea) - the higher the

electronegativity.In Periodic Table: high electronegativity at top right,low electronegativity at bottom left.

Very large electronegativity difference ionic bond (X+Y-)No electronegativity difference ‘pure’ covalent bond (XX)Small to moderate electronegativity difference polar covalent bond (X+Y-, where Y is moreelectronegative than X)

Lewis Structures We will first concentrate onH and first row elements.

Basic rules: 1. Every atom tries to achieve anoctet of electrons (2 for H).

2. Each pair of shared electrons one bond - symbolised by a dash ()

How to achieve an octet:F-

F

O2-

O-

O

O+

N3-

N2-

N-

N

N+

C4-

C3-

C2-

C-

C

Note: if an atom loses an electron (giving a positive charge), then it needs to form more bonds to achieve its octet (see extra sheet)

Note : (1) never more than 4 bonds to central atom(max. 8 electrons)

(2) keep formal charges to a minimum

In polyatomic molecules - distinguish between central and terminal atoms. Difficult to give general rule to specify which is which.

Often - atom with lowest electronegativity is central;also chains of same atoms are very rare.

Lewis structures tell us nothing about shapes of molecules.

G. N. Lewis

Some simple examples:

O needs 2 electrons to achieve octetEach H needs 1 electron to achieve its full shell

H O

H

:

:lonepairs (remember - each dash

is a pair of shared electrons)

NH3N needs 3 electrons to achieve octetEach H needs 1 electron to achieve octet

N H

H

H

H2O

There will also be a lone pair on N

Ammoniaplant

Liquid NH3 being sprayed as a fertilizer

CH4C needs 4 electrons to achieve octetEach H needs 1 electron to achieve its full shell

H C

H

H

H No lone pairs - all C electronsinvolved in bonding.

Remember - Lewis structures are2-dimensional - do not give correct structures.

COCl2C needs 4 electrons to achieve octetEach Cl needs 1 electron to achieve its octetO needs 2 electrons to achieve octet

Therefore to fill all octets we have:

C O

Cl

Cl

Note double bond between C and O

Phosgene (COCl2) in WW1

CH3CO2H (ethanoic acid,acetic acid)

The only way to achieve octets for C, O, and full shells for H is

H

CH

H

C

O

O H

NH4+ Also works for ions

N+ needs 4 electrons to achieve octetEach H needs 1 electron to achieve its full shell

H N+

H

H

H

Mendeleev vodka:

old style bottle

There are many more simple examples of organic andinorganic molecules where the drawing of Lewisstructures using the above rules is very straightforward.

But - there are exceptions

1. Some elements (towards left hand side of the Periodic Table) cannot achieve a full octet.

BF3B only has 3 valence electrons - would need5 more to fill octet. But can only form 3 bonds.Thus BF3 is

B F

F

F

2. Elements in second and subsequent rows ofthe Periodic Table can accommodate >8electrons (can ‘expand their octet’)

Such elements possess d-orbitals of suitable energyfor bonding - so >4 bonds can be formed (not restrictedto one s + three p orbitals).

SF6

Because S can expand its octet, it is quitein order to have the Lewis structure:

F

F

F

F

F

F

S

Expansion of octet very common for compounds ofSi, Ge, P, As, Sb, S, Se, Te, Cl, Br etc.

3. It is often not possible to draw a single,unambiguous, Lewis structure for a givenmolecule or ion.

CO32- C is the central atom; O can achieve

its octet by forming 2 bonds (=O), orone bond + picking up an electron (-O)

Also - remember that C cannot form >4 bonds

One possible Lewis structure would therefore be:

C O

-O

-O But this is not the only possibility!

The structure could equally be

C O-

-O

O

or C O-

O

-O

Each of these structures would predict one doublebond and two single bonds, i.e. one shorter and two longer bonds.

Experiment shows that all C-O bonds in the carbonateare the same length.

The correct structure is a blend of all three forms(NOT hopping between them). This phenomenonis called RESONANCE - symbolised by:

C O-

-O

O

C O

-O

-O

C O-

O

-O

The actual structure is called a resonance hybrid,and the pair of electrons comprising the secondcomponent of the double bond is said to bedelocalised.

Another example of resonance:

Acetate (ethanoate) ion, CH3CO2

Here there are TWO possible resonance structures:

H3C C

O

O-

H3C C

O-

O

Both C-O bond lengths are again the same (andintermediate between the lengths of a single and a double bond)

Benzene, C6H6

The bonding in benzene was not properlyunderstood for many years.

When the possibility of resonance became recognised, the problem was largely solved.

Simple ideas would give the structure

C

CC

C

CC

H

H

H

H

H

HBUT the CC bond lengths are all thesame (intermediate between singleand double), and benzene doesnot behave chemically like a moleculecontaining isolated C=C bonds.

Hence we must introduce the second possible form,in resonance with the first:

C

CC

C

CC

H

H

H

H

H

H

C

CC

C

CC

H

H

H

H

H

H

This explains the observed physical and chemical properties of benzene, and it can also be drawn as

with the circle symbolising the delocalised electrons.

Resonance is also possible with moleculescontaining atoms where expansion of the octet has occurred.

Perrier water & Benzene

ClO3

Cl O

O

-O

Cl O

-O

O

Cl O-

O

O

and

ClO4

ClOO

O-

O

ClO-O

O

O

ClOO

O

-O

ClO-O

O

ONote that in all oxy-anions, the negative chargesare formally situated on the oxygen atoms

(chlorate ion)

(perchlorate ion)

oxidizing agent

rocket oxidizer

KClO3 + Sugar

Space ShuttleSolid boosterAl + NH4ClO4

Ammonium perchlorate works, Henderson, Nevada.

All of the examples of resonance shown so farinvolve two or more equivalent structures. It isquite possible to have resonance betweennon-equivalent forms.

Resonance involves the blending of two or more Lewis structures, where the atoms have the samerelative positions, but with different electronicarrangements. Electrons are delocalised, and theenergy of the resonance hybrid is lower than forany single component form.

Non-equivalent resonance hybrids

e.g. thiocyanate ion, (NCS)

N C S- N- C STwo forms the one with the negative charge on Nwill be more important (electronegativity N > S),although both will contribute to the final hybrid.

This introduces the idea of preferred Lewis structures,and this will help us to predict relative stabilities of isomers.

Note - do not confuse the resonance forms shown so farwith isomers, where the spatial arrangements of the atomsdiffer

Thiocyanate[CNS]-

Why is carbon dioxide OCO and not COO?

Lewis structures: O C O

compared to C2- O2+ O

The latter has (i) large formal charges, and (ii) negative charge on less electronegative atom.

Another example:

The ion (NCO) is stable; the ion (CNO)is known,but very unstable. Why is this?

N C O- N- C O

Lewis structures for the two isomers are:

and C- N+ O- respectively.

This again explains the observed properties.

Note that in the former, the negative charge willbe predominantly on O rather than N, from theirrelative electronegativities.

Summarise requirements for a preferred Lewis structure:

1. Keep formal charges to a minimum.

2. Where formal charges are necessary,they should be consistent with relative electonegativities (-ve charge on moreelectronegative atom), and

3. Keep like charges as far apart as possible.

Two final examples (more in tutorial work):

Because F is so electronegative, in the other casewe can have:

F O O F F- O O+ F

F-O+OF

This accounts for the observation that the O-Odistance in H2O2 is much longer than in F2O2

H2O2 and F2O2

For H2O2 the only realistic Lewis structure is

H O O H

O2F2

Dinitrogen tetroxide

N2O4 is an unusual molecule - it has an N-N bond,but this is very long (weak) and the moleculereadily splits into two NO2 fragments. CanLewis structures explain this?

There are several equivalent possible Lewis structures, but they are all of the form

N+ N+

-O

O O-

O

Repulsion between the +ve charges on the N atomsleads to the N-N bond being very weak.

Liquid/GaseousN2O4 ( + NO2)

Before moving on to the actual shapes of molecules, we must recognise that not allcovalent bonds are formed by taking one electronfrom each of two atoms.

Dative (coordinate) bonds

Sometimes a covalent bond is formed by one atom(DONOR) ‘giving’ both electrons to another (ACCEPTOR).

Ammonia (NH3) has a LONE PAIR of electrons onthe N atom.

Boron trifluoride (BF3) only has 6 electrons associated with the B atom. It can complete theoctet by accepting the lone pair from NH3.

N

H

H

H

:

F

BF

F

Now both B and N have effectively completed their octets. The arrow is the standard symbol for such a dative (or coordinate or donor/acceptor) bond.

In abbreviated form this is:NH3F3B

Note that, once formed, this type of bond behavesjust like any other covalent bond.

Another way of writing these bonds is:N+H3F3B-

SHAPES OF COVALENTMOLECULES

Need a method for predicting shapes of molecules.

Here is one which works very well for compoundsof s- and p-block elements (not for d- and f-block - see later lectures)

VValence

SShell

EElectron

PPair

RRepulsion

VSEPR tells us that the preferred structure ofa molecule minimises repulsions between electrons

Repulsions between nuclei are assumed to benegligible.

Core electrons (close to nuclei) are also neglected,i.e. valence electrons only.

Electrons will form pairs (lone or bonding)wherever possible.

Ron Gillespieco-inventor of VSEPR

Standard procedure:1. Determine Lewis structure(s) for

the molecule - as above.2. Count valence electrons according to

well-defined rules - see later.

All structures are based on a few fundamental geometries

Start off with simple examples - only oneCENTRAL ATOM, with ALL of its valence electronsused to form SINGLE BONDS.

General formula: ABn

AB2TWO pairs of valence electrons around A

Repulsions minimised by a LINEAR arrangement

B A BBond angle exactly 180º - any change increasesrepulsions.

E.g. BeCl2: Valence electrons at Be 2two Be-Cl bonds 2Total valence electrons at Be 4

TWO pairs - therefore AB2, linear:

Cl Be Cl Note - equal bond lengths

AB3THREE pairs of valence electrons around A

Repulsions minimised by a TRIGONAL PLANAR arrangement

Bond angles all 120º ; bond lengths equal

E.g. BF3: Valence electrons at B 3three B-F bonds 3Total valence electrons at B 6

THREE pairs - therefore AB3, trigonal planar:

B

A

BB

F

B

FF

BF3

AB4FOUR pairs of valence electrons around A

Repulsions minimised by a TETRAHEDRAL arrangement

Bond angles all 109º28’ ; bond lengths equal

E.g. CH4: Valence electrons at C 4four C-H bonds 4Total valence electrons at C 8

FOUR pairs - therefore AB4, tetrahedral:

B

A

BB

B

H

C

HH

H

AB5

Possible ambiguity - two geometries withsimilar electrostatic repulsions

A

B

B B

B

B

Square-basedpyramid

and B A

B

B

B

B

Trigonal bipyramid

Accurate calculations show that the TRIGONALBIPYRAMID is slightly favoured - hence this is the predicted geometry.

Note: in the trigonal bipyramid not all angles are equal,some 90º, some 120º.

E.g. PF5: Valence electrons at P 5five P-F bonds 5Total valence electrons at P 10

FIVE pairs - therefore AB5, trigonal bipyramidal:

(eq)F P

F(eq)

F(eq)

F(ax)

F(ax)

Note: F(eq)PF(eq) = 120º F(ax)PF(eq) = 90º

AB6

Here we have no ambiguity.

Only one realistic possibility:

OCTAHEDRALA

B

B B

B

B

B

All angles 90º; all bond lengths equal.All B atoms equivalent (cf. AB5)

E.g. SF6: Valence electrons at S 6six S-F bonds 6Total valence electrons at S 12

SIX pairs - therefore AB6, octahedral: S

F

F F

F

F

F

AB7

Largest number of electrons aroundcentral atom found for p-blockcompounds.

Repulsions minimised by a PENTAGONALBIPYRAMID.

BB

B

BB

A

B

B

Distinct axial and equatorial B atoms.

B(eq)AB(eq) = 72º B(ax)AB(eq) = 90º

E.g. IF7:Valence electrons at I 7seven I-F bonds 7Total valence electrons at I 14

SEVEN pairs - therefore AB7, pentagonal bipyramidal:

FF

F

FF

I

F

F

I2 + F2 IF7

Shapes of Ions

Stick to compounds with only single bonds, and withall valence electrons used for bonding.

Ionic charge formally on central atom - simply adjust electron count accordingly.

E.g. PH4+:

Valence electrons at P 5four P-H bonds 4+ve charge on P -1Total valence electrons at P 8

FOUR pairs - therefore AB4, tetrahedral:

H

P+

HH

H

PF6: Valence electrons at P 5

six P-F bonds 6ve charge on P +1Total valence electrons at P 12

SIX pairs - therefore AB6, octahedral:

P

F

F F

F

F

F

Shapes of Molecules and Ions containing Lone Pairs

If some of the central-atom electrons do notparticipate in bonding they are called LONE PAIRS

They still affect the shape, and must be included inthe electron count, but they are not easy to detectexperimentally.

A lone pair will be given the symbol L.

We will work through the ABn systemsdescribed above.

AB2LExample: SnCl2Electron count:

Valence electrons at Sn 4two Sn-Cl bonds 2Total valence electrons at Sn 6

Three pairs, but only2 bonds, thereforeAB2L

Shape based on trigonal plane - with one vertexoccupied by the lone pair.

Appears to be V-shaped:Sn

ClCl

..

Can we predict the bond angle?

SnCl2

The lone pair electrons are closer to the Sn thanare the bond pairs. Therefore:

REPULSIONS FROM LONE PAIRS ARE GREATERTHAN THOSE FROM BOND PAIRS

Hence the ClSnCl angle will be <120º, in fact it is 95º.

AB3LExample: NH3

Electron count:

Valence electrons at N 5three N-H bonds 3Total valence electrons at N 8

Four pairs, but only3 bonds, thereforeAB3L

Structure based on tetrahedron.

N

HH

H

.. Shape is PYRAMIDAL.The bond angle will be <109º28’, infact 107º - due to greater repulsion by lone pair.

AB2L2Example: H2O

Electron count:

Valence electrons at O 6two O-H bonds 2Total valence electrons at O 8

Four pairs, but only2 bonds, thereforeAB2L2

Note that we now have repulsions between twolone pairs.

Order of repulsions:

LP/LP > LP/BP > BP/BP

H2O is V-shaped:

O

HH

The HOH is 104.5º, i..e. less than the HNH in ammonia(107º), because of the greaterrepulsion due to LP/LP.

AB4L Now much more difficult - because basic AB5 shape has two different types of B.

We can place the lone pair either in an axial position (i), or in an equatorial position (ii).

B A

B

B

B

..(i)

A

B

B

B

B

..

(ii)

or

Which will be preferred? Need to count up repulsions.

(i) 3 LP/BP at 90° 1 LP/BP at 180°

(ii) 2 LP/BP at 90° 2 LP/BP at 120°

Fewer repulsions at smaller angle lead to (ii) being preferred. NOTE - LP repulsions mean that B(ax)AB(eq) < 90°, B(eq)AB(eq) < 120°.

e.g. SF4 Valence electrons at S 6 4 S-F bonds 4 total 10Therefore AB4L

S

F(eq)

F(eq)

F(ax)

F(ax)

..

Bond angles found experimentally to be as predicted above, i.e. F(eq)SF(eq) < 120°F(ax)SF(eq) < 90°

SF4

AB3L2Again more than one possibility.

B A

B

B

..

..

A

B

B

B

..

.. AB

B

B..

..or or

(i) (ii) (iii)

We can forget about (ii) - 2 lone pairs at 90°. (i) looks as if it should be best (lone pairs furthest apart) - but add up all repulsions:

(i) LP/LP at 180° + 6LP/BP at 90°

(iii) LP/LP at 120°, + 4LP/BP at 90° + 2 LP/BP at 120°

Detailed calculations show (iii) actually preferred, i.e. T-shaped molecule minimises repulsions.

e.g. ClF3 Valence electrons at Cl 7 3 Cl-F bonds 3 total 10Therefore AB3L2

Cl

F

F

..F

..

The FClF angle is 87°,i.e. <90°, because oflone pair repulsions

AB2L3

Again several possibilities - but onlyone which doesn’t have LP/LPrepulsions at 90°.

A

B

B

..

..

..

This gives a LINEAR MOLECULE- exactly so since the three lone pair repulsions cancel out.

e.g. XeF2 Valence electrons at Xe 8 2 Xe-F bonds 2 total 10Therefore AB2L3

(LINEAR) Xe

F

F

..

..

..

AB5LSimple again - only one possibility:

A

(eq)B

(eq)B

B(eq)

B(ax)

..

B(eq)

Octahedron with onevacant vertex:

SQUARE-BASEDPYRAMID

Bond angle B(ax)AB(Eq) < 90° - due to LP repulsion

e.g. BrF5 Valence electrons at Br 7 5 Br-F bonds 5 total 12Therefore AB5L

Br

F

F

F

F

..

FStructure confirmedexperimentally, withFBrF = 85°

AB4L2

Only one significant possibility(all others have LP/LP at 90º

A

B

B

B

..

..

B

SQUARE PLANAR (exact - as lone pair repulsions cancel)

e.g. BrF4 Valence electrons at Br 7

4 Br-F bonds 4-ve charge 1

total 12Therefore AB4L2

Br

F

F

F..

..F

Note - no AB3L3 or AB2L4 known - all would have LP/LP repulsions at90º

Some trends in bond angles

A series of AB3L molecules - all pyramidal, but they have different bond angles:

NH3 107º PH3 93º AsH3 92º

NF3 101º PF3 99º

All angles less than tetrahedral value (lone pairrepulsion) - but why so different?

Compare NH3 and PH3 - two factors:

(i) SIZE

(ii) ELECTRONEGATIVITY

SIZE - since P > N, then P-H bond longer than N-H,but 2 electrons in each bond, can be symbolised(exaggerated!):

N H

P H

Much easier to push P-H bonds than N-H bondstogether.

ELECTRONEGATIVITY - power of atom to attractelectrons to itself: N > P H

Thus in N-H the electron density will concentrated atN end, in P-H equally distributed:

N H

P H

Preparation of PH3

Now, compare NH3 and NF3

N-F bond longer than N-H - thus similar effect as forN-H and P-H; also electronegativities F > N > H.Thus electron distributions

N H N F

Hence easier to push together N-F than N-H bonds.

Finally compare PH3 and PF3. Expect same effect as for

NH3 and NF3 , BUT FPF angle > HPH angle.

Reason is that there is significant repulsion betweenF atoms at small angles - therefore angle increased.This is a breakdown of VSEPR, which assumesNO internuclear repulsion.

Electronegativity differences also account for thediverse reactivity of organic molecules.

Carbon bonds to many different atoms and if there is asufficient difference in electronegativity the bondsare polarised, giving rise to a dipole moment.

Some common functional groups in organicchemistry with their dipole.

H3C Cl H3C OH C O

There is a list of common organic functional groups atthe front of your book. You will study these later this year.

+ ▬

High purity NF3

Molecules with >1 central atoms

All examples so far - only 1 central atom.

Consider C2H6 (ethane)

Valence electrons at each C 43 C-H + 1 C-C bonds 4Total valence electrons at C 8

FOUR pairs - therefore AB3B’, tetrahedral:

Therefore each CH3C is tetrahedral - but how is the whole molecule put together?

C

H

HH H

C

HH

All angles near, but not exactly 109º28’.

Preferred conformation is staggered - to minimise repulsions, but low barrier to rotation aboutC-C bond.

Another example : methylamine, CH3NH2

Valence electrons at C 43 C-H + 1 C-N bonds 4Total valence electrons at C 8

Valence electrons at N 52 N-H + 1 C-N bonds 3Total valence electrons at N 8

FOUR pairs - therefore AB3B’, tetrahedral

FOUR pairs - therefore AB2B’L, tetrahedral

By analogy with ethane, the preferred structure will be

..C

H

HH

N

HH

Angles at C neartetrahedral values,angles at N near 107º,as in ammonia.

Similar calculations for other organic andinorganic molecules.

Shapes of molecules containing coordinate (“dative”) bonds

Remember from Lewis structures - some bondsformed by one atom (donor) donating BOTH electrons in a bond to the other (acceptor) atom, e.g. in

NH3F3B

Valence electrons at B(acceptor) 33 B-F bonds 31 B-N coordinate bond 2Total valence electrons at B 8

Valence electrons at N(donor) 53 N-H bonds 31 B-N coordinate bond 0Total valence electrons at N 8

FOUR pairs - therefore AB3B’, tetrahedral

FOUR pairs - therefore AB3B’, tetrahedral

Just like ethane, preferred structure is:

B

F

FF H

N

HH

Always count TWO electrons at ACCEPTOR atomand ZERO electrons at DONOR atom.

Shapes of molecules containing multiple bonds

Basic principles:

Only the first bond between two atomssignificantly affects the structure.

Subtract central atom electrons used to form second or third bonds.

Deduce initial shape assuming only single bonds.

Allow for some modification due to the multiple bonds.

Electron count is therefore now:

(a) valence electrons at central atom

(b) +1 for every single bond

(c) -1 for every double bond (-2 for every triple bond)

E.g. phosphorus(V) oxide trichloride (‘phosphorusoxychloride’)

Lewis structure: O=PCl3The electron count is therefore:

Valence electrons at P 53 P-Cl bonds 31 P-O bond 1-1 for one double bond -1Total valence electrons at P 8

i.e. 1 P electron used in formingdouble bond

Therefore essentially tetrahedral stucture:

O

P

ClCl

Cl

>109.5

<109.5

Note effect of double bond onangles. TWO electron pairs indouble bond, therefore morerepulsion and hence greater OPCl angle.Double bond affects detailsof structure.

POCl3

O=SCl2 Valence electrons at S 62 S-Cl bonds 21 S-O bond 1-1 for one double bond -1Total valence electrons at S 8

i.e. 4 electron pairs, 3 bonds (1 double) and one lonepair - pyramidal structure based on tetrahedron:

S

O Cl

Cl

..

Difficult to predict bond angles, but ClSCl <109.5º, OSCl approx. 109.5º,

CoCl3 + SOCl2

Before After

CO2Valence electrons at C 42 C-O bonds 2-1 for each double bond -2Total valence electrons at C 4

Therefore LINEAR CO O

HCN Valence electrons at C 41 C-H bond 11 C-N bond 1-2 for triple bond -2Total valence electrons at C 4

Therefore LINEAR CH N

HCN in outer space

Ethene, C2H4Lewis structure requires a doubleC=C bond, i.e. H2C=CH2

Two central atoms: Valence electrons at each C 42 C-H bonds 21 C-C bond 1-1 for double bond -1Total valence electrons at C 6

Therefore each ‘end’ of the molecule is based ona trigonal plane, with bond angles near 120º

C C

H

H H

H

Note: (i) repulsion due to double bondmeans HCH angle < 120º, HCC angle> 120º.

(ii) reason for planar form being preferred not clear from VSEPR - seelater for explanation.

Propanone (acetone), (CH3)2C=O

Each methyl group will have the tetrahedral geometryshown for ethane (above). For the C2C=O unit we have:

Valence electrons at ‘central’ C 42 C-C bonds 21 C-O bond 1-1 for double bond -1Total valence electrons at C 6

Therefore trigonal planar, with CCC <120º,and OCC >120º,

C O

H3C

H3C

So far, only one sensible Lewis structure each time.What if there are SEVERAL EQUIVALENT Lewis structures?

e.g. ethanoate (acetate) ion, CH3CO2

C

O

O-

H3C C

O-

O

H3C

Valence electrons at ‘central’ C 42 C-O bonds 21 C-C bond 1-1 for double bond -1Total valence electrons at C 6

:both possible

Electron countfor either one:

N.B. negative charge on O’s - agrees withrelative electronegativities.

Electron count suggests structure

C

O

O-

H3C

With two different types of CO bond. BUT both Lewis structures are equally likely,so in fact both CO bondsare equivalent (e.g. lengthbetween single and double).

The double bond is DELOCALISED, and a morerealistic drawing of the structure is:

C

O

O

H3C

CO32- Possible Lewis structures:

C

O-

O-

O C

O

O-

-O C

O-

O

-O

As for previous example:

Valence electrons at C 43 C-O bonds 3-1 for double bond -1Total valence electrons at C 6

Therefore 3 electron pairsgiving a planar arrangement

Again - all Lewis structures equallylikely, therefore structure is:

C

O

O

O

2- Double bond isDELOCALISED,giving structure withall bond lengths thesame, and all angles 120º

Many other examples of oxy-anions withdelocalised double bonds, e.g. sulfite ionO=S(-O)2

Valence electrons at S 63 S-O bonds 3-1 for double bond -1Total valence electrons at S 8

i.e. pyramidal AB3L, withone delocalised doublebond:

S

O O

O

.. 2–

Note all S-O bonds equal, andangles near 109.5º - lone pair,double bond approx. cancel.

General rule: for XOnm, there are (n-m)

double bonds (X=O) and m single (X-O)bonds

One exception: nitrate ion, NO3, where the Lewis

structures are of the form:

N+

O-

O-

O

Valence electrons at N 53 N-O bonds 3-1 for double bond -1-1 for positive charge -1Total valence electrons at N 6

Delocalisation givesregular AB3:

N

O

O

O

-