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CHEM 1000 3.0 Chemical Bonding 1 Chemical Bonding Petrucci, Herring Madura and Bissonnette : Chapters 10 and 11 • Aims: To look at bonding and possible shapes of molecules We will mainly do this through Lewis structures To look at ionic and covalent bonds Use valence shell electronic structure to predict shapes of molecules
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Page 1: 4 Chemical Bonding

CHEM 1000 3.0 Chemical Bonding 1

Chemical Bonding

• Petrucci, Herring Madura and Bissonnette : Chapters 10 and 11

• Aims: – To look at bonding and possible shapes of molecules

• We will mainly do this through Lewis structures– To look at ionic and covalent bonds– Use valence shell electronic structure to predict shapes of

molecules

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Problem set

Chapter 10 questions 3, 4, 11, 17, 18, 21, 22b), 59, 62, 141

Chapter 11 questions 7, 13, 17

Yousif
Yousif
Yousif
Yousif
Yousif
Yousif
Yousif
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Chemical Bonding

• Lewis Theory:– Electrons, particularly valence electrons play a

fundamental role in chemical bonding.– When elements combine to produce compounds

they are attempting to achieve a full valence shell (usually 8 electrons)

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Chemical Bonding

• Lewis Theory: – Electrons can be transferred from one atom to

another to make ions. The atoms are then held together by coulombic forces in an ionic bond

– More often the only way an atom can gain electrons is by sharing. This sharing produces a covalent bond

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Lewis Symbols• This is a way of representing the valence

electrons in an element– It does not include the inner shell electrons– It does not include the spin of an electron

– e.g. Si ([Ne]3s23p2)

– N ([He]2s22p3)

Si

N

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Lewis Structures

• These are the combination of Lewis symbols that represents the sharing or transfer of electrons in a molecule– Ionic bonding examples ( electron transfer)

Na + Cl [Na]+ [ Cl ]-xx

Mg + 2 Cl [Mg]2+ 2 [ Cl ]-xx x

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Ionic Compounds

• We don’t usually see isolated ionic compounds– Normally they are in crystals where one anion

(negative) is attached to several cations (positive) and vice versa. Electrical neutrality means the total number of each ion is the same.

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Ionic Compounds

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Lewis Structures

• These are the combination of Lewis symbols that represents the sharing or transfer of electrons in a molecule– Covalent bonding example (sharing)

H + Cl H Cl x x

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Covalent Compounds

• Here electrons are shared between two atoms. – Why? Because the energy cost of making the ions is too

high– Could be more than just a couple of shared electrons– The electrons are associated with each atom in the

covalent bond– The overall effect is that each atom has “more” (usually

a full shell of) valence electrons.

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Covalent Compounds

• Examples– Single covalent bond HCl

Note that there are 6 electrons around the Cl that are not involved in bonding. It is normal to talk about these as lone pairs, in contrast to bond pairs

Lone pairsBond pair

H Cl H Cl

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Coordinate Covalent Bonds

Bonds do not have to come from equal sharing of electrons

In NH4+ there are 8 electrons around the N, but it has 5

to start with. Hence the hydrogens only contribute 3 of the 8 bonding electrons.

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Multiple Covalent Bonds

• Often the sharing of one electron does not “fill” the valence shell of an atom.

• e.g. N2

N N N N

N N N N

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Multiple Bonds• We use a number of different ways to

describe multiple bonds• For 1 bond pair For 3 bond pair

– Single bond Triple bond– bond order =1 bond order =3

• For 2 bond pairs– Double bond – bond order =2

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Bond Length• Bond Length

– The distance between two atoms joined by a covalent bond

– As the bond order increases the bond length decreases

Bond Length/pm Bond Length/pm

C-C 154 N-N 145C=C 134 N=N 123C C 120 N N 110

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Bond Energy• To separate two atoms that are joined by a

covalent bond, energy must be supplied. This is the Bond Dissociation Energy (D).

• This is equal to the energy released when the bond is formed.H2(g) ĺ2H(g) ǻH = D(H-H) = +435.93 kJ mol-1

• Listed bond energies are usually an average over a number of compounds

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Polarity of Bonds

In most molecules the bonding is not 100% ionic or 100% covalent (equal sharing)– Start from the covalent side– Unequal sharing of electrons in a bond means

one atom is slightly positive (į+) and the other slightly negative (į-).

– This leads to a polar covalent bond.

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Charge density (electrostatic potential)

Ionic

<100% Covalent

100% covalent

This is a map that shows the surface that shows where the charge is located

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Polarity of Bonds

• The ability to attract electrons in a bond appears to be related to electron affinity but we need to use a molecular property to describe it.

• Electronegativity is a quantitative measure of an atom’s ability to compete for electrons with other atoms to which it is bonded.– Actually the difference between an atom’s

ionization energy and its electron affinity

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Polarity of Bonds:Electronegativities

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Polarity of Bonds

• Thus the difference in electronegativities between atoms in a bond gives a measure of the polarity.

• This can also be interpreted as the ionic character of a bond

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Polarity of Bonds:Ionic Character

The larger the electronegativity difference between the atoms, the more ionic the bond.

1.7

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Lewis Structures

• Lewis structures are useful in indicating the bonding in molecules.

• Rules:– All valence electrons must appear– The electrons are usually paired– The valence shells are usually filled (2 electrons for H,

8 for systems with s and p orbitals)– Multiple bonds are often needed, especially in C, N, O,

P, S.

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Lewis Structures1. Check if the compound is ionic. If it is, treat

each ion separately.2. Add up the number of valence electrons (that’s

from all atoms). Add or subtract electrons to give the right charge. This is the available electrons A.

3. Draw a skeletal structure. The central atom will normally have the lowest electronegativity. Carbon is always a central atom. Hydrogen is always terminal.

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Lewis Structures4. Calculate the number of valence electrons

needed to give all the atoms a full shell (2 or 8). This is the number of needed electrons N

5. Determine the number of electrons that must be shared S=N-A

6. Place single bonds in the skeletal structure (2 electrons per). Add extra bonds to satisfy S.

7. Place remaining electrons, in (lone) pairs to complete the octets for each atom.

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Lewis StructuresExample C2N2

Carbon has 4 valence electrons, nitrogen 5. A=18Each atom wants an octet, N=32Shared electrons S= 32-18 =14 (7 bonds)Skeleton: (Electronegativities C 2.5, N 3.0)

N-C-C-N : Still need 4 more bonds. N C C N is the only way to add the 4

bonds without having more than 8 electrons around a carbon.

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Lewis StructuresExample C2N2

This uses 14 of the 18 electrons, so there are only 4 left. The carbons have their octet, each nitrogen is 2 short so we must have one lone pair on each nitrogen.

N C C N

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Formal Charges• The method of obtaining the Lewis structures

does not keep track of where the electrons came from. – Hence we can get multiple results. However in

some structures atoms are not contributing equal numbers of electrons to the bonds. This is less likely to result in a stable configuration.

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Formal Charges• E.g. ONCl

O N Cl O N Cl

Cl donates 3 electrons to the bondO donates 0 electrons to the bond

Cl and N donate one electron eachO and N donate 2 electrons eachThis seems more reasonable.

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Formal Charges

• The formal charge is the apparent charge on an atom that comes from unequal sharing of electrons.

• It can be used to decide between possible structures.

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Formal Charges

• The formal charge is the apparent charge on an atom that comes from unequal sharing of electrons.

• It can be used to decide between possible structures.

• The formal charge on an atom is:FC = # of valence electrons in free atom

í # of lone pair electrons on the atomí½ (# of bonding electrons)

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Formal Charges

Rules for the most plausible Lewis structure:1. The most plausible is the one where no atoms

have a formal charge.2. Where formal charges are required, they should

be as small as possible.3. Negative formal charges should be on the most

electronegative atoms.4. Same sign formal charges on adjacent atoms

are unlikely.

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Formal Charges• E.g. ONCl

O N Cl O N Cl

Formal chargesFC(O) = 6 -6-1 = -1 wrongFC(N) = 5-2-3 = 0FC(Cl) = 7-4-2 = +1 (wrong)2

Formal chargesFC(O) = 6 -4-2 = 0FC(N) = 5-2-3 = 0FC(Cl) = 7-6-1 = 0

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Formal ChargesExample: HCN

H C N H N CRemembering that C has 4 valence electrons and N 5FC(H) = 0 FC(H) = 0FC(C) = 0 FC(C) = í1FC(N) = 0 FC(N) = +1HCN is most likely.

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Formal Charges

• Formal charges allow us to follow electrons in bonding.

• They are not the actual charges on atoms• What is the charge on an atom?

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Charges on atoms

• Oxidation numbers assume that the bonds are 100% ionic– ON(S) = +6, ON(O) = í2

• Formal charges assume bonds are 100% covalent– FC(S) = 0, FC(O) = í0.5

• The real charge is somewhere between and is determined by the polarity of the bonds

S O

O

O

O

2-

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Lee Allen’s Method

• The degree of polarity in a bond involving atom “A” and “B” depends on the electronegativity of “A” and “B”

• IF EN(A) = EN(B) then half the charge is on atom “A” (the bond is covalent)

EN(A)polarity EN(A) EN(B)

v�

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Lee Allen’s Method

• In a molecule the charge on A is

– GA group number (the number of electrons in the atom)– NA Number of lone pair electrons on “A”– NAB is the number of bonding electrons between “A”

and “B”

A A A ABB A

EN(A)q = G -N - N EN(A) EN(B)z �¦

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Lee Allen’s Method

• For a 100% covalent bond EN(A) =EN(B)

• This is the calculation for formal charge

A A A ABB A

A A A AB

EN(A)q = G -N - N EN(A) EN(B)

1q = G -N - N2

becomesz �¦

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Charges on atoms

• Oxidation numbers assume that the bonds are 100% ionic– ON(S) = +6, ON(O) = í2

• Formal charges assume bonds are 100% covalent– FC(S) = 0, FC(O) = í0.5

• Lee Allen’s method (EN(S) = 2.5, EN(O) = 3.5)

– qS = 1.0, qO = í0.75

S O

O

O

O

2-

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Resonance

• Consider ozone O3

These are both reasonable Lewis structures BUT experimental evidence shows the bonds are the same length.

O O O O O O

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Resonance

• The real structure is a hybrid of these two

This is called resonance. The real structure is a resonance hybrid of these two structures.

The real structure has bonds which are between single and double bonds. So what is the bond order of the oxygen-oxygen bond in ozone?

O O O O O O

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Exceptions/Extensions• Odd electron species: eg NO (11 electrons)

– Lewis structures don’t do a good job in this case.

– In general you can produce a Lewis structure without the last electron and this goes on the atom to minimize the formal charge.

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Exceptions/Extensions

• Incomplete octets. – Sometimes when you make a sensible structure you

don’t have enough electrons. – BF3

– All atoms have a zero formal charge– Boron only has 6 electrons

B

F

FF

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Exceptions/Extensions

• BF3– Boron has a formal charge

of -1 The double bonded fluorine +1

– The need to obtain an octet overrides the rule for minimal formal charge AND the –ve formal charge being on the most electronegative atom.

– Note we also have resonance structures.

B

F

F

F

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Exceptions/Extensions• Expanded valence shells.

– To explain the bonding it sometimes seems necessary to have more than 8 electrons in the valence shell. This can be done for third period elements by invoking the “d” orbitals.

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Expanded Octet Example

S O

O

O

O

2-

Sulphate, SO42-

Octets are complete, but the formal charges are horrible.

Sulphur is in the 3rd period so d orbitals may be available for expanding the octet to reduce formal charges.

FC(-O) = -1

FC(S) = +2

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Expanded Octet Example (cont.)

The octet around the S has been expanded to hold 10.

On basis of formal charge this is better.

Would formal charges be further reduced by further expansion?

FC(S) = +1

FC( single bonded O) = -1

FC(double bonded O) = 0

S O

O

O

O

2-

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Expanded Octet Example (cont.)

Is this the best possible structure?

Yes because we cannot lower the formal charges further.

FC(S) = 0

FC( single bonded O) = -1

FC(double bonded O) = 0

Since there are resonance forms FC(O) = 0.5

S O

O

O

O

2-

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Shapes of Molecules

• In this section we will use Lewis structures as an introduction to the shapes of molecules.

• The key concepts are:– Electron pairs repel each other.– Electron pairs assume orientations to minimize

repulsion• This is the Valence Shell Electron Pair Repulsion

Theory (VSEPR)

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VSEPR

• Example: Methane CH4

The 4 bond pairs must orient themselves to minimize their repulsion.

H C H

H

H

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VSEPR Methane

The minimum interaction occurs when the electron pairs point towards the vertices of a tetrahedron.The carbon is in the centre and the hydrogen are at the vertices. The molecule is tetrahedral.

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VSEPR Ammonia

• Example: Ammonia NH3

The 4 electron pairs must still orient themselves to minimize their repulsion.

N H

H

H

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VSEPR Ammonia

The minimum interaction still occurs when the electron pairs point towards the vertices of a tetrahedron.The nitrogen is in the centre and the hydrogens are at three of the vertices. The lone pair points to the fourth. The molecule is trigonal pyramidal.

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VSEPR Water

• Example: Water H2O

The 4 electron pairs must orient themselves to minimize their repulsion.

O

H

H

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VSEPR Water

The minimum interaction still occurs when the electron pairs point towards the vertices of a tetrahedron.The oxygen is in the centre, the hydrogens are at two of the vertices. The lone pairs point to the other two. The molecule is bent.

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Bond Angles• This analysis suggests that all three molecules

should have bond angles of 109.5o.

– The methane bond angle is 109.5o but for ammonia it is 107o and in water it is 104.5o

– The lone pair electrons are not constrained as much as the bonding pairs. They spread out thus the repulsive forces are:

Lone pair-lone pair>Lone pair-bond pair>bond pair-bond pair

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Molecular Geometry

• VSEPR theory can be used to describe the shapes of most molecules.

• Warnings– When you describe the shape, don’t include the

lone pairs. (water is bent, not tetrahedral)– Molecules have three dimensions (methane is a

tetrahedron not a square)• Be familiar with table 10.1 pg 399-400

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Going Beyond Lewis Structures

• Lewis structures are very useful in explaining the bonding in simple molecules and in predicting molecular shapes.

• They do not explain why electrons in bond pairs bring nuclei together.

• They can not be used to estimate bond lengths or bond strengths.

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Going Beyond Lewis Structures

• There are two currently used bonding theories, valence bond theory and molecular orbital theory.

• Valence bond theory envisions bonding as resulting from the overlap of atomic orbitals.

• Molecular orbital theory moves past atomic orbitals and derives orbitals that belong to the molecule as a whole.

• We will concentrate on valence bond theory as that is sufficient to rationalize the structures we see.

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Valence Bond Method• This method considers what happens to the

valence orbitals (the outer shell).• If the atoms start a long way apart, there is no

interaction.• As the atoms move closer together the orbitals

may overlap.• If there is an electron in the orbital then there is a

high probability of finding an electron between the nuclei.

• This is the region where there is a covalent bond.

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Valence Bond Method

• We cannot ignore the previously developed rules, so there are only two electrons per bond– Overlap can only occur for 2 half filled orbitals

or one empty and one filled orbital.

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Valence Bond MethodH2S

Not shown

This suggests the bond angle should be 90o. It is 92o.

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Valence Bond method for methane

• Chemically methane is stable• We know from VSEPR that methane is

tetrahedral • BUT

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Valence Bond method for methane• The valence bond theory (so far) would

suggest that the 1s electrons from H will be donated to the p orbitals to “pair-up” the unpaired electrons.

• In this case we would have CH2

• That’s not what is seen. We need CH4.

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Valence Bond method for methane

• Somehow the carbon needs to have 4 unpaired electrons

• This gives 4 unpaired electrons

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Valence Bond method for methane• This costs energy, as it makes an excited

state C atom.• Also if we try to use these orbitals in

bonding, the orientation is wrong. – If hydrogens attached to the “p” orbitals, they

would be orthogonal. – Where would the overlap with the “s” be? – We know the molecule is tetrahedral.

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Valence Bond Rethinking• We can no longer assume that the orbitals in a

bonded atom are the same as those of an isolated atom.

• Remember that the atomic orbitals come from solving an equation assuming a single nucleus. In a molecule the electron has to respond to more nuclei.

• We should solve the “new” problem, but we usually say we can combine the atomic orbitals in some way.

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Valence Bond method for methane• Since the orbitals are wave functions, we can take

the 2s and 2p wave functions and combine them to give 4 equivalent wave functions. These can be made to have the same shape and energy.

• This is called hybridization and the resulting orbitals are called hydrid orbitals.

• The combination of one “s” and three “p” orbitals gives four “sp3” hybrid orbitals.

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Generation of sp3

hybrid orbitals

Same shape but different orientation to minimize interaction

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Methane

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Energy and Hybrid Orbitals

• What about the energy?• The energy of the orbitals is conserved because

the sp3 orbitals have an energy between the s and p orbitals.

C

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Energy and Hybrid Orbitals

• However there will be an energy cost in making the sp hybrids with one electron in each. – The orbitals are generated assuming only one electron

• So why would the molecule do something that will cost energy?

• Once the molecule is formed you release the bond energy. If this is greater than the “cost” then it is worthwhile.

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Ammonia

• Now we have a pair of electrons in one of the sp3 hybrid orbitals, so it can’t bond with hydrogen.

N

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Ammonia

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NH3 versus PH3

• NH3 has sp3 bonding and a bond angle of ~109q.

• PH3 has a bond angle of 93 q so it looks like it does not hybridize.

• It still hybridizes but the angle is 93q– The lone pair on the P is larger than that on the N– Thus the bond pair-bond pair angle is even

smaller in PH3

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• The sp3 hybrid orbitals don’t explain the bonding for Boron (group 13) or for Beryllium (group 2)

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• The problem appears to be that there are insufficient electrons to even half fill the 4 sp3 orbitals.

• In these case we need to consider different hybridization schemes.

• We define hybrids such that the number of hybrids is the same as the number of valence electrons.

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sp2 and sp hybrid orbitals

Boron

Beryllium

The number of orbitals is conserved.

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sp2 hybrid orbitals (BCl3)

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sp hybrid orbitals (BeCl2)

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Multiple bonds

• How do we deal with the Lewis structures that have multiple bonds?

• Consider ethylene C2H4

C C

H

H H

H

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Multiple bonds: ethylene (C2H4)

• The geometry suggests sp2 hybrid orbitals are involved.

There is still an electron in the remaining 2p orbital.

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Multiple bonds: ethylene (C2H4)The sp2 hybrid orbitals are in purple. The “p” orbital is blue/red

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Multiple bonds: ethylene (C2H4)

The orbitals that overlap along the axis of the nuclei are called sigma bonds - ı bonds.

Those where the overlap is side-to-side are pi bonds - ʌ bonds.

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Multiple bonds: ethylene

The ı bonds determine the shape of the molecule, the ʌ bonds restrict the rotation about the C-C axis.

The orbital overlap is more extensive for the ı bonds so the ʌ bond is weaker than the ı bond.

Bond strengthsC-C 347 kJ mol-1 C=C 611 kJ mol-1

difference = 264 kJ mol-1

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Multiple bonds: acetylene (ethyne)

• The geometry suggests sp hybrid orbitals are involved.

There is still an electron in each of the remaining 2p orbitals.

C CH H

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Multiple bonds: acetylene (ethyne)

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Multiple bonds: d orbitals

S O

O

O

O

2-

• In sulphate we needed pi bonding and to use the “d” orbitals on the sulphur.

• Without worrying about resonance structures, can we explain the bonding?

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Multiple bonds: d orbitals sulphate

S O

O

O

O

2-

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Multiple bonds: d orbitals sulphate

S O

O

O

O

2-

8 electrons6 orbitals

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Multiple bonds: d orbitals sulphate

S O

O

O

O

2-

8 electrons6 orbitals

Why 6 orbitals?For the resonant structures

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Delocalization

• This is an alternative method of describing molecules that have resonant structures.

• Eg Ozone

• We can describe the pi bonding as being delocalised– The pi electrons are delocalised among 3 atoms

O O O O O O

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Delocalization

• Eg Ozone

O O O O O O

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Multiple Bonds: Benzene

This structure is planar and suggests sp2.

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Multiple Bonds: Benzene

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Multiple Bonds: Benzene

• In this case the 6 pi electrons are delocalized over all 6 atoms of the molecule.