RESONANCE STRUCTURES

For certain molecules or molecular ions, two or more EQUIVALENT Lewis structures can be drawn.

The equivalent structures have the same formal charge and are equal in energy.

2-Example: CO3 , carbonate ion

C

O

1

2

3

O O

-2

Three equivalent structures for the carbonate ion – all have the same formal charge.

The double bond electron pair is DELOCALIZED over all three C-O bonds and hence all three C-O bonds are between pure-single and pure-double bonds.

Note: This differs from the two structures for N3- which had

different bonding arrangements for the atoms, and hence not equivalent.

CO O

O

1

2

3

CO O

O

1

2

3

CO O

O

1

2

3

0-1

0

-10 0

-1

-1

-1

-1 00

These equivalent structures are called RESONANCE STRUCTURES.

The carbonate ion is NOT a mixture of the three structures, nor does any one of the three exist independently.

The real structure is an average of the three structures, and properties of these bonds are such that they are intermediate between single and double bonds.

All three C-O bonds identical, the bond lengths are between a single bond between C and O and a double bond between C and O

Other examples of molecules that have more than one equivalent resonance structures are NO3

-, SO3

EXCEPTIONS TO THE OCTET RULE

Although the octet rule is very useful, it does not apply in all cases. There are three classes of exceptions

1) Compounds with odd number of electrons

2) Compounds with a shortage of electrons, less

than the octet.

3) Compounds with an abundance of electrons,

exceeding the octet

1) Molecules with ODD number of valence electrons

An example is nitrogen monoxide, NO.

Molecules with an odd number of electrons cannot satisfy the octet rule for all its atoms.

In this case the octet rule must be given up for one of the atoms by leaving an unpaired lone electron.

ALL BONDING ELECTRONS MUST BE PAIRED.

NO is stable molecule with 11 valence electrons

In (i) N has an octet, but not O; in (ii) O has an octet but not N

Calculate the formal charge of each atom for (i) and (ii) to determine which is the more favored structure.

N=O N=O

(i) (ii)

0 0-1 +1

Structure (ii) is favored since the formal charges on both atoms is zero; in (i) N is -1 and O is +1, with the more electronegative atom, O, possessing a +1 formal charge

N=O N=O

(i) (ii)

Two Lewis structures can be drawn for NO

2) Electron Deficient Molecules

Example BF3

Lewis model predicts the following structure for BF3

B

F

FF+1-1

However, experimental evidence suggests that the structure of BF3 is:

B

F

FF00

(i) assigns a +1 formal charge to a very electronegative atom which is undesirable

(ii) is the observed structure, even though B does not have an octet, the formal charges on all atoms are zero.

B

F

FF+1

-1

B

F

FF0

0

(i) (ii)

3) Valence-Shell Expansion

The elements in Groups IIIA to VII A in the third period and beyond show a tendency to surround themselves with more than 8 electrons.

It is possible for these elements through “valence shell expansion” for the central atom to have more than 8 electrons

In cases like this, after accounting for bonding electrons, assign lone pairs to the outer atoms to give them octets.

If any electrons still remain, assign them to the central

atoms as lone pairs.

Example: SF6

Number of valence electrons = 6 + (6x7) = 48

Total number of electrons needed for each atom to satisfy an octet = 7x8 = 56

Number of shared electrons = 56-48 = 8

8 electrons, 4 bonds, is not enough bonds for SF6

Hence, assign one bond pair to each bond, which for SF6 is 6.

Then assign lone pairs to the outer atoms to give them an octet. If any electrons remain, assign them to the central atom.

SF

F

FF

F

F

Count all electrons

(6x6) + (6x2) = 48 electrons

SF

F

FF

F

F

Formal charge on all atoms: 0

POCl3 - phosphoryl chloride

P Cl

Cl

Cl

O-1

+1 0

0

0P Cl

Cl

Cl

O0

0 0

0

0

(i) (ii)

i) non zero formal charges

ii) valence shell expansion for P

Molecules have definite shape i.e. definite geometries.

For example, H2O is bent and CO2 is linear.

Shapes of Molecules

OHH 104.5o

CO O==180o

Lewis stuctures do not provide information on a molecule’s geometry, i.e. shape.

VALENCE SHELL ELECTRON-PAIR REPULSION (VSEPR) THEORY: predicts molecular shapes based on Lewis stuctures

VSEPR derives from the realization that bonding and non-bonding electron pairs will locate themselves in space so as to minimize repulsive interactions.

The geometry of the electron pairs, and the covalent bonds that form, can be predicted using VSEPR

Molecular geometry depends on the total number of electron pairs

To use the VSEPR theory to predict molecular geometry, need to know how electron pairs can be arranged around a central atom so as to minimize electron-electron repulsion.

An electron pairs arranges itself in three-dimensional space such that it is equally far apart from all the other electron pairs.

The resulting arrangement of the electron pairs in a molecule depends on the NUMBER of electron pairs.

Note: For VSEPR, the electron pairs we refer to are the valence electrons

For a central atom surrounded by TWO electron pairs, the optimum geometry which minimizes repulsion between the electrons pairs is LINEAR

180o

LINEAR

120o

TRIGONAL PLANAR

For a central atom surrounded by THREE electron pairs, the optimum geometry which minimizes repulsion between the electrons pairs is TRIGONAL PLANAR

109.47o

TETRAHEDRAL

For a central atom surrounded by FIVE electron pairs, the optimum geometry which minimizes repulsion between the electrons pairs TRIGONAL BIPYRAMIDAL

For a central atom surrounded by FOUR electron pairs, the optimum geometry which minimizes repulsion between the electrons pairs is TETRAHEDRAL

TRIGONAL BIPYRAMIDAL90o

120o

For a central atom surrounded by SIX electron pairs, the optimum geometry which minimizes repulsion between the electrons pairs OCTAHEDRAL

90o

OCTAHEDRAL

90o

We have been talking about the location of the electron pairs about the central atom.

This maybe different from molecular geometries i.e. the structure or shape of a molecule

Water has a “BENT” structure with the H-O-H angle measured to be 104.5o.

OHH 104.5o

If O is bonded to 2 H atoms, why is the structure of water not linear?

If there are non-bonding electrons on the central atom then the molecular geometry differs from the electron geometry but is obtained from knowledge of the electron geometry about the central atom.

MOLECULAR GEOMETRY

Molecular geometry is the arrangement of ATOMS about the central atom.

If there are no non-bonding electrons, then the electron geometry and the molecular geometry is the same.

To determine the molecular geometry about the central atom of a molecule:

1) Write the best Lewis structure for the molecule or ion

2) Count the total number of ATOMS attached to the central atom plus the number of NONBONDING ELECTRON PAIRS on the central atom .

3) Using the VSEPR geometries establish the ELECTRON geometry about the central atom.

4) Finally examine the placement of atoms and identify the molecular geometry.

Step 1: Write the best Lewis structure for the molecule

Total number of valence electrons = 1+1+6 = 8

Number of electrons needed for each to complete an octet (remember for H need two electrons) = 2 + 2 + 8 = 12

Shared electrons = 12- 8=4 => two bonds

Number of unpaired electrons = 4 => 2 lone pairs on O

Lewis structure of H2O is:

HH O

2) Count the total number of atoms attached to the central atom plus the number of non-bonding electron pairs on the central atom

HH O

For water,

Total number of electron pairs = 2 pairs of bonding electrons (bp) + 2 pairs of non-bonding electrons (np) = 4 pairs

3) Then using the VSEPR geometries establish the electron geometry about the central atom.

According to VSEPR theory: For a central atom surrounded by four electron pairs, the optimum geometry which minimizes repulsion between the electrons pairs is tetrahedral

Arrange the two nonbonding electron pairs and the two H atoms at the apices of the tetrahedron, with O at the center

OH

H

To determine the molecular geometry of water, look at the arrangement of just the atoms (in this case, H,O,H)

4) Finally examine the placement of atoms and identify the molecular geometry.

Looking at the arrangement of the atoms in H2O, we see that the molecule is bent.

OH

H

For a tetrahedral geometry, expect that the H-O-H angle is 109.5o

The measured H-O-H angle is 104.5o, slightly smaller than predicted.

Non-bonding electron pairs are more diffuse than bonding electron pairs, taking up more space.

Model

OH

H 104.5o

Draw the Lewis structure to determine the number of bonding electron pairs and number of nonbonding pairs on the central atom (a double or triple bond counts as a single electron pair when using VSEPR theory).

Arrange the bonding and nonbonding electron pairs around the central atom in a geometry predicted by VSEPR.

Position the atoms bonded to the central atom where the bonding electrons are positioned.

Determine the molecular geometry based on the arrangement of the atoms

Two electron pairs

Example: CO2

Number of electron pairs around C = 2 (count the double bond as a single electron pair).

Both are bonding pairs

VSEPR predicts a linear molecular geometry for CO2

Lewis structure: O=C=O

Model

Three electron pairs

The three electron pairs around the central B atom are all bonding pairs.

According to VSEPR the three electron pairs around the central atom are positioned at the vertices of an equilateral triangle

In the case of BF3, since the three electron pairs are bonding pairs, the three F atoms are positioned at the vertices of the equilateral triangle.

B

F

FF0

0BF3 – Lewis structure

Model

For VSEPR count two bonding electron pairs and one nonbonding electron pair.

Hence the three electron pairs are situated at the vertices of an equilateral triangle.

B

F

F F

120o

trigonal planar

NO2- Lewis structure: O=N-O

The two O atoms are situated at two of the vertices and the nonbonding electron pair at the third.

N

O O

>120o

bent

-1

<120o

The molecular geometry if NO2- is not trigonal planar, but

bent.

The O-N-O angle is less than 120o due to the presence of the nonbonding pair on N

Four electron pairs

For molecules with four electron pairs, the electron pairs surround the central atom in a TETRAHEDRAL geometry.

The molecular geometry depends on the number of bonding and nonbonding electron pairs on the central atom.

Examples:

CH

H

H

H

methane, CH4

tetrahedral

OH

H

water, H2Obent

HN

H

H

ammonia, NH3

pyramidalModel

Five Electron Pairs

Electronic geometry is trigonal bipyramidal.

However, unlike the 2, 3 and 4 electron geometries all vertices of a trigonal bipyramid are NOT equivalent.

There are two AXIAL positions and three EQUATORIAL positions.

90o

120o

axial

equatorial

axial

equatorial

equatorial

Angle between

axial-central atom-axial = 180o

axial-central atom-equatorial = 90o

equatorial -central atom- equatorial = 120o

90o

120o

The positions of the nonbonding electron pairs relative to each other and the bonding pairs have to account for the non-equivalence of the axial and equatorial positions.

To position the nonbonding and bonding electrons: nb-nb repulsion > nb-b > b-b

Hence, position nonbonding pairs first, then bonding pairs accounting for the repulsion between the electron pairs.

PCl5

P

Cl

Cl

ClCl

Cl

ClCl

ClCl

Cl

P

trigonal

bipyramidal

Model

SF4

S

F

F F

F

Position the nonbonding electron pair on S at the equatorial position since this minimizes nb-b repulsions

FF

F

F

S

“see-saw”

Angles in SF4 are different from the predicted 180o, 120o and 90o due to the non-bonding electron pair

FF

F

F

S187o

87o

102o

Other examples

ClF3 - distorted T

XeF2 - linear

Model

Six electron pairs

Octahedral geometry

All vertices are the equivalent, hence bonding and nonbonding electron pairs can be positioned at any vertex

OCTAHEDRAL

Molecular geometry determined by number and position at atoms and nonbonding electron pairs

Model

Examples

SF

F

FF

FF

SF6

octahedral

BrF

F

FF

F

BrF5

square pyramidal

XeF

F

F

F

XeF4

square planar

Molecular Properties

Bond Lengths

The bond distance is the distance between the two nuclei of the two atoms bonded together.

Bond lengths depend on the elements and whether the bond between the two elements is single, double or triple

Typically:

single bond length > double bond length > triple bond length

Bond Energies

If two atoms in a diatomic molecule are pulled far enough apart the bond between the atoms breaks.

The energy needed to break the bond is called the bond energy or bond dissociation energy.

Bond energies depend on the nature of the atoms bonded together and whether the bond is single, double or triple.

Typically single bond energies < double bond energy < triple bond energy

Dipole Moments

The bonding electron pair in HCl is not shared equally between the H and Cl atoms.

This is because Cl is more electronegative than H, and tends to attract the electron pair closer towards it.

Hence the electron distribution for the bonding pair of electrons is not evenly distributed over both the H and Cl atoms, but is more concentrated over the Cl atom.

The more electronegative atom develops a slight negative charge and the more electropositive atom a slight positive charge

This charge separation results in a DIPOLE MOMENT, and a POLAR HCl bond.

The dipole moment is defined as the product of the magnitude of the separated charges and the distance between the centers of the charges.

The dipole moment is a vector

H Cl

The larger the electronegative difference between the bonded atoms, larger is the magnitude of the dipole moment and hence more polar is the bond.

All heteronuclear diatomic molecules (A-B. where A≠B) are polar molecules with the degree of polarity depending on the electronegativity differences between the two atoms.

Heteronuclear diatomics have non-zero dipole moments

All homonuclear diatomic molecules (A2) are non-polar, with zero dipole moments

Polyatomic molecules

Bonds between atoms of different elements will each have a dipole moment associated with them.

The magnitude of the dipole moment of a bond depends on the electronegativity differences between the bonded atoms.

However, whether the molecule as a whole has a dipole moment depends on the geometry of the molecule.

The dipole moment is a vector and hence the dipole moment of the molecule is the vector sum of the dipole moments of each bond in the molecule.

Methane, CH4

The net dipole moment for CH4 is zero

CH

H

H

H

Water, H2O

OHH

OHH

Model