Molecular Structure

transcript

Chemistry 120

Molecular Structure

Both atoms and molecules are quantum systems

We need a method of describing molecules in a quantum mechanical way so that we can predict structure and properties

The method we use is the

Linear Combination of Atomic Orbitals

where we can use the properties of atoms to predict the properties of molecules.

Chemistry 120

Molecular Structure

We combine atoms to form molecules by considering the phase of the atomic orbitals we are using

We represent the phase via the shading we give the orbital.

The phase represents the sign of the wavefunction

Chemistry 120

Molecular Structure

We combine atoms to form molecules by considering the phase of the atomic orbitals we are using

The phase represents the sign of the wavefunction

We represent the phase via the shading we give the orbital.

Chemistry 120

Molecular Structure

For an s orbital, the orbital has the same phase everywhere:

For a p orbital, there is a

change in the sign of the

wavefunction across the

nodal plane:2p orbital, n = 2, l = 1, ml = -1

1s orbital, n = 1, l = 0

Chemistry 120

Molecular Structure

Consider two H atoms (1s1) coming together from infinite separation.

There are two possibilities:

1 The wavefunctions are in phase

2 The wavefunctions are not in phase

Chemistry 120

Molecular Structure

Case 1: The wavefunctions are in phase

The atoms move together and the electron waves overlap with the same phase, producing constructive interference and a build up of electron density between the nuclei

The energy of the system drops and we form a bond

Chemistry 120

r = 8 Å

r = 7 Å

r = 6 Å

r = 5 År = 0.75 Å

r = 1 Å

r = 2 Å

r = 3 Å

Chemistry 120

Molecular Structure

Case 2: The wavefunctions are out of phase

The atoms move together and the electron waves have opposite phase.

The electron waves overlap producing destructive interference and electron density between the nuclei is reduced.

The energy of the system rises and we have an antibonding situation

Chemistry 120

r = 8 Å

r = 7 Å

r = 5 Å

r = 4 Å r = 0.75 Å

r = 1 Å

r = 2 Å

r = 3 Å

Chemistry 120

Bonding Antibonding

Two atoms with wavefunctions in phase overlap with constructive interference. Electron density increases between the nuclei and the overall energy decreases.

When the wavefunctions are of opposite phase, the electron density between the nuclei decreases due to destructive interference. The energy of the system rises and we have an antibonding situation

Chemistry 120

Bonding

Antibonding

Chemistry 120

Organic Structure and Bonding

Review of diatomic bonding

There are two types of bond that are important in this part of the Periodic Table

bonds and bonds

Chemistry 120

bonds and bonds

bonds are in general stronger than bonds and can be formed from either s or p orbitals:

Chemistry 120

bonds and bonds

bonds have no nodal plane that contains the two nuclei.

The * antibonding orbital has a nodal plane between the two nuclei

Chemistry 120

bonds and bonds

bonds have a nodal plane that contains both nuclei,

The * antibonding orbital also has a plane between the nuclei

Chemistry 120

bonds and bonds

These , bonding orbitals and *, * antibonding orbitals are the orbitals that are used to bind all simple organic molecules together.

We can also describe the bonding in diatomic molecules

important models for larger organic systems

Chemistry 120

bonds and bonds

To describe the bonding in the diatomic molecules such as O2, N2 and X2 (X = F, Cl, Br and I), we use both the s orbitals and the p orbitals on the two atoms as a basis set - the palette of atomic orbitals from which we will build the molecular orbitals.

The energies of the two different l states, s and p, are slightly different in polyelectronic atoms.

Chemistry 120

bonds and bonds

The s orbitals and the p orbitals appear as follows

Chemistry 120

Organic Structure and Bonding bonds and bonds

We arrange the atoms along one of the axes for convenience and so the first pair of orbitals we construct are the s and s* orbitals from the s orbitals on the atoms.

Chemistry 120

We now us the higher energy p orbitals to construct p and porbitals

Chemistry 120

The complete molecular orbital diagram for all the diatomic molecules from Li2 to N2

Chemistry 120

The complete molecular orbital diagram for all the diatomic molecules from Li2 to N2

As each molecule has a different number of electrons,

Li2 2 Be2 4 B2 6 C2 8

N2 10 O2 12 F2 14 Ne2 16

Chemistry 120

Li2 2 Be2 4 B2 6 C2 8

N2 10 O2 12 F2 14 Ne2 16

We can write the electronic structure of each molecule by placing electron pairs into the orbitals.

Chemistry 120

Li2 2 Be2 4 B2 6 C2 8

N2 10 O2 12 F2 14 Ne2 16

Something peculiar happens after N2

Recall that as the charge on the nucleus increases, the orbitals become more stabilized and the electrons become more strongly bound.

Chemistry 120

Li2 2 Be2 4 B2 6 C2 8

N2 10 O2 12 F2 14 Ne2 16

This happens by different amounts, depending on the orbital.

After N2 (10 e-), the ordering of the orbitals derived from p change their order in the molecule

Chemistry 120

For N2 (10 e-), the ordering is this

For O2 (12 e-), the ordering is this

Chemistry 120

This is an example of configurational interaction

Each electron moves in the field of the other electrons. If the energies of the two molecular orbitals are sufficiently close and the nodal properties are correct, molecular orbitals will interact and shuffle their energies in the molecule.

This causes the orbitals to change their energetic ordering but only when the nuclear charge is high enough to force the electrons close in energy.

Chemistry 120

Configurational interaction

Each electron moves in the field of the other electrons. If the energies of the two molecular orbitals are sufficiently close and the nodal properties are correct, molecular orbitals will interact and shuffle their energies in the molecule.

This causes the orbitals to change their energetic ordering but only when the nuclear charge is high enough to force the electrons close in energy.

Chemistry 120

Molecular Structure

A full description of the structure of a molecule requires the solution of the Schrödinger equation for the entire molecule.

The potential term is far too complicated to be solved analytically and so we need an empirical approach to molecular structure.

,,ˆˆ

,,ˆ rV

Chemistry 120

Molecular Structure

There are two common approaches

- Lewis description

- Valence Shell Electron Pair Repulsion (VSEPR)

theory

and both are based on the electron count at the central atom of the molecule or fragment of the molecule.

Chemistry 120

Molecular Structure

Lewis description

The covalent chemical bond can be thought of as a pair of electrons shared between atoms;By considering the number of electrons in the valence shell and the number of electrons in the outer atoms, we can explain the presence of lone pairs and the gross structure of the molecule.

G. N. Lewis

Chemistry 120

Molecular Structure

Lewis description

The covalent chemical bond can be thought of as a pair of electrons shared between atoms;

By considering the number of electrons in the valence shell and the number of electrons in the outer atoms, we can explain the presence of lone pairs and the gross structure of the molecule in simple cases.

G. N. Lewis

Chemistry 120

Molecular Structure

Lewis description

The Lewis description arose form an attempt to cram the observed properties of atoms in combination into a mechanically classical picture of the physical world then prevalent; in fact even classically, the structure of the atom was not explicable.

G. N. Lewis

http://www.chem.yale.edu/~chem125/125/history99/7BondTheory/LewisOctet/

cubicoctet.html

Chemistry 120

Molecular Structure

Lewis description

The Lewis description is based on the observed requirement that the atom achieves the valence shell octet associated with the noble gases - a noble gas configuration.

Consider the formation of MgCl2

Mg: 1s22s22p63s2 or [Ne]3s2

Cl: 1s22s22p63s23p5 or [Ne]3s23p5

Chemistry 120

Molecular Structure

Lewis description

We know that MgCl2 is ionic and so the changes in the valence shell configurations are

Mg: 1s22s22p63s2 or [Ne]3s2

Mg2+: 1s22s22p6 or [Ne]

Cl: 1s22s22p63s23p5 or [Ne]3s23p5

Cl-: 1s22s22p63s23p6 or [Ne]3s23p6 (i.e. [Ar])

Chemistry 120

Molecular Structure

Lewis description

We therefore account for the stability of MgCl2 through the formation of closed shell ions with noble gas configurations, namely

Mg2+: 1s22s22p63s2 or [Ne]

Cl-: 1s22s22p63s23p5 or [Ne]3s23p6 (i.e. [Ar])

Chemistry 120

Molecular Structure

Lewis description

In this respect, the Lewis description of bonding is accurate, but there are major failures with molecules.

Lewis described molecular structure through the idea that the atom had some inherent tetrahedral quality and that the electrons were distributed in static manner at the vertices of the tetrahedron

Chemistry 120

Molecular Structure

Lewis description

Lewis described molecular structure through the idea that the atom had some inherent tetrahedral structure and that the electrons were distributed in static manner at the vertices of the tetrahedron.

Chemistry 120

Molecular Structure

Lewis description

Molecular species therefore take structures via sharing electrons through the vertices of the tetrahedron. This naturally implies that all molecules are tetrahedral, which causes major problems for those that are not……….

Chemistry 120

Molecular Structure

Lewis description

Examples: BH3, CH4, NH3, OH2 and FH

All these structures are based on the tetrahedron and the sharing of electrons in bonds or the presence of lone pairs at the corner of the tetrahedron.

Chemistry 120

Molecular Structure

Lewis description

We can depict the valence shell (i.e. the shell with the highest principle quantum number) as

B C N O F

1s22s22p1 1s22s22p2 1s22s22p3 1s22s22p4 1s22s22p5

Chemistry 120

Molecular Structure

Lewis description

We satisfy the open valences of these atoms with H atoms :

B C N O F

1s22s22p1 1s22s22p2 1s22s22p3 1s22s22p4 1s22s22p5

Chemistry 120

Molecular Structure

1s22s22p1

1s22s22p4

1s22s22p2

1s22s22p6

1s22s22p3

1s22s22p6

1s22s22p4

1s22s22p6

1s22s22p5

1s22s22p6

Chemistry 120

Molecular Structure

Chemistry 120

Molecular Structure

Chemistry 120

Molecular Structure

The structures of the first row hydrides are not accurately predicted by the Lewis Theory of structure and bonding.

Chemistry 120

Molecular Structure

The other model for molecular structure is VSEPR.

We consider a closed shell atom and we also assume that it is spherical. The structure is then determined by the number of “stereochemically active units” present in the outer shell.

These stereochemically active units are the ‘lone pairs’ and the bond pairs that are formally assumed to exist in a molecule from a Lewis picture of structure and bonding.

Chemistry 120

Molecular Structure

Once we assume that bond and lone pairs exist, we introduce some other assumptions, one about structure and one about energies of interactions.

The structural types that we use are based on the distribution of points on the surface of a sphere such that the distance between them is a maximum.

Chemistry 120

Molecular Structure

For two stereochemically active units, the obvious geometry is linear:

Chemistry 120

Molecular Structure

For three stereochemically active units, we form a triangular arrangement of atoms around the central atom:

This geometry is termed

Trigonal Planar

Chemistry 120

Molecular Structure

Four stereochemically active units are arranged in the form of a tetrahedron

Chemistry 120

Molecular Structure

For five stereochemically active units, there are two choices. The one most commonly encountered is the trigonal bipyramid

Chemistry 120

Molecular Structure

Six stereochemically active units have only one choice for the base geometry – the octahedron

Chemistry 120

Molecular Structure

In each of these geometries, the sites that are predicted are occupied either by an atom or by an ‘electron pair’

The final requirement is to detail the interaction between these various pairs of electrons – bond and lone – in the atom.

Chemistry 120

Molecular Structure

The interaction energies that we are interested in are the repulsions between these pairs of electrons.

As bond pairs are more tightly confined, the reulsions due to bond pairs are less. Lone pairs, assumed to be more diffuse, suffer from higher repulsions and thus the energy ordering is:

B.P.-B.P. < B.P.-L.P. < L.P.-L.P

Chemistry 120

Molecular Structure

With these rules in hand, and a knowledge of the possible geometries, we can now predict with some certainty the expected molecular geometry of any main group species.

Chemistry 120

Molecular Structure

Example: What are the structures of SF4, CF4 and XeF4?

Step 1: Count the electrons on each central metal atom.

S 6 C 4 Xe 8

Chemistry 120

Molecular Structure

Step 1: Count the electrons on each central metal atom.

S 6 C 4 Xe 8

Step 2: Determine the number of bond pairs that each atom has.

In this case it is 4 each.

Chemistry 120

Molecular Structure

Step 3: Determine the number of ‘lone pairs’ that each atom has. As each bond has 1 e- from the central atom, then the number of electrons in lone pairs is just

S 6 - 4 = 2 C 4 - 4 = 0 Xe 8 - 4 = 4

Chemistry 120

Molecular Structure

Step 4: Determine the number of sterochemically active species by type

SF4 4 bond pairs, 1 lone pair

CF4 4 bond pairs, 0 lone pair

XeF4 4 bond pairs, 2 lone pairs

Chemistry 120

Molecular Structure

Step 5: Apply the energy rules to each structure type and determine the structure type

SF4 trigonal bipyramid

CF4 tetrahedron

XeF4 octahedron

Chemistry 120

Molecular Structure

And so the structures are……..

Chemistry 120

Molecular Structure Review

• Lewis electron dot structures

• Valence Shell Electron Pair Repulsion (VSEPR)

theory

Both are based on the electron count at the central atom, A, of the molecule or fragment of the molecule AXn.

Chemistry 120

We observe that the atoms achieve a noble gas configuration in the valence shell, often (but not always) an octet.

The first three noble gas configurations:

[He]: 1s2 2 valence electrons

[Ne]: 1s22s22p6 8 valence electrons

[Ar]: 1s22s22p63s23p6 8 valence electrons

Chemistry 120

Be B C N O F Ne

Na Mg Al Si P S Cl Ar

K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

Rb Sr Y Zr Nb Mo Tc Os Rh Pd Ag Cd In Sn Sb Te I Xe

Cs Ba La Hf Ta W Re Ru Ir Pt Au Hg Tl Pb Bi Po At Rn

Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu

Th Pa U Np Pu Bk Cf EsCmAm Fm Md No Lr

Rf Db Bh MtAc Sg Hs

Chemistry 120

AHn Examples:

BH3, CH4, NH3, OH2 and FH

We satisfy the open valences of these atoms with H atoms :

B C N O F

[He] 2s22p1 [He] 2s22p2 [He] 2s22p3 [He] 2s22p4 [He] 2s22p5

Chemistry 120

Be B C N O F Ne

Na Mg Al Si P S Cl Ar

K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

Rb Sr Y Zr Nb Mo Tc Os Rh Pd Ag Cd In Sn Sb Te I Xe

Cs Ba La Hf Ta W Re Ru Ir Pt Au Hg Tl Pb Bi Po At Rn

Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu

Th Pa U Np Pu Bk Cf EsCmAm Fm Md No Lr

Rf Db Bh MtAc Sg Hs

Chemistry 120

AHn: BH3, CH4, NH3, OH2 and FH

[He]2s22p1

[He]2s22p4

[He]2s22p2

[He]2s22p6

[He]2s22p3

[He]2s22p4

[He]2s22p6

[He]2s22p5

[He]2s22p6

Chemistry 120

VSEPR: What are the structures of BH3, CH4, NH3, OH2, and FH?

Step 1: Count the electrons on each central atom.

B 3 [He]2s22p1

C 4 [He]2s22p2

N 5 [He]2s22p3

O 6 [He]2s22p4

F 7 [He]2s22p5

Chemistry 120

Step 2: Determine the number of bond pairs for each central atom.

3 4 3 2 1Number of bond pairs:

Chemistry 120

Step 3: Determine the number of ‘lone pairs’ remaining on the central atom.

3 4 3 2 1

0 0 1 2 3

Number of bond pairs:

Number of lone pairs:

Chemistry 120

Step 4: Determine the total number of stereochemically active units (bond pairs and lone pairs of electrons).

BH3 3 bond pairs + 0 lone pairs = 3

CH4 4 bond pairs + 0 lone pairs = 4

NH3 3 bond pairs + 1 lone pair = 4

OH2 2 bond pairs + 2 lone pairs = 4

FH 1 bond pairs + 3 lone pairs = 4

Chemistry 120

Step 5: Determine the structure type by applying the energy rules to each structure type.

The most favorable structure minimizes steric interactions among the stereochemically active units.

For a molecule with a central atom (B), imagine the stereochemically active units (A) on the surface of a sphere as far apart in space as possible.

Chemistry 120

BA2 BA3 BA4

The number of stereochemically active units determines the geometry around the central atom.

linear trigonal planar tetrahedral

Chemistry 120

BA5 BA6

trigonal bipyramid octahedral

Chemistry 120

Which VSEPR geometry is appropriate for these compounds?

CH4 4 bond pairs + 0 lone pairs = 4

NH3 3 bond pairs + 1 lone pair = 4

OH2 2 bond pairs + 2 lone pairs = 4

FH 1 bond pairs + 3 lone pairs = 4

Chemistry 120

trigonal planar

Chemistry 120

CH4, NH3, OH2, FH: All have 4 stereochemically active units.

Chemistry 120

CH4, NH3, OH2, FH: All have 4 stereochemically active units.

tetrahedral

Chemistry 120

CH4, NH3, OH2 and FH

All have a tetrahedral geometry at the central atom.

Chemistry 120

Q: If there are a mix of bonding pairs and lone pairs around the central atom, what determines their positions?

Chemistry 120

Molecular Structure

A: As bond pairs are more tightly confined, the repulsions due to bond pairs are less. Lone pairs, assumed to be more diffuse, suffer from higher repulsions and thus the energy ordering is:

B.P.-B.P. < B.P.-L.P. < L.P.-L.P

Q: If there are a mix of bonding pairs and lone pairs around the central atom, what determines their positions?

Chemistry 120

Molecular Structure

B.P.-B.P. < B.P.-L.P. < L.P.-L.P

Examples :

PCl5 PCl3

Chemistry 120

Molecular Structure

B.P.-B.P. < B.P.-L.P. < L.P.-L.P

Examples :

PCl5 PCl3

Cl PCl

Chemistry 120

Molecular Structure

B.P.-B.P. < B.P.-L.P. < L.P.-L.P

Examples :

PCl5 PCl3

Cl PCl

AB5 AB4

Chemistry 120

Molecular Structure

B.P.-B.P. < B.P.-L.P. < L.P.-L.P

Examples :

SF2 SF4

Chemistry 120

Molecular Structure

B.P.-B.P. < B.P.-L.P. < L.P.-L.P

Examples :

SF2 SF4

Chemistry 120

Trigonal Bipyramid

Chemistry 120

Molecular Structure

B.P.-B.P. < B.P.-L.P. < L.P.-L.P

Examples :

SF2 SF4

Chemistry 120

Q: What if the species is charged, e.g. PH4+ ?

Chemistry 120

A: Take the charge into account in the valence electrons of the central atom.

P0 5 valence electrons

P+ 4 valence electrons

Chemistry 120

A: Take the charge into account in the valence electrons of the central atom.

P0 5 valence electrons

P+ 4 valence electrons

Chemistry 120

Q: If there are three different atoms in a molecule, which one is the central one?

Chemistry 120

A: The least electronegative atom (most electropositive atom) is in the center.

Chemistry 120

Example:

Chemistry 120

Example:

Chemistry 120

Organic chemistry is the chemistry of the top right corner of the Periodic table.

It is the branch of chemistry that is most closely connected to biology and health-related science:

Medicine and Pharmaceuticals

The key elements are carbon, hydrogen, oxygen, nitrogen and phosphorous.

Chemistry 120

Organic Structure and Bonding and review

bonds: No nodal plane that contains both nuclei

bonds: 1 nodal plane that contains both nuclei

Chemistry 120

Bonding in organic molecule almost always only contains these two types of bond.

In organic molecules, neutral carbon atoms always have 4 and only 4 bonds:

They can be single, double or triple

Chemistry 120

Alkyne: 2-butyne single and triple bonds

Alkene: 2-butene single and double bonds

Alkane: butane all single bonds

Bonding in organic molecule almost always only contains these two types of bond.

In organic molecules, neutral carbon atoms always have 4 and only 4 bonds:

They can be single, double or triple

Chemistry 120

In methane, all the bonds are identical and methane is tetrahedral.

How do we describe the bonding in methane?

Two methods:

1. Full molecular orbital theory

needs group theory and quantum mechanics

2. Hybridization of atomic orbitals

an approximation that works well in organic chemistry

Chemistry 120

This is a method for describing the bonding in organic molecules by adding all of the wavefunctions together on a single carbon atom, using the three 2p orbitals and the 2s orbital.

We can form three different combinations:

sp3 sp2 sp

where the superscripts show the number of p orbitals that we are adding to the s orbital

Hybridization

Chemistry 120

In terms of energy, sp3 hybridization looks like:

sp3 hybridization

Chemistry 120

The four sp3 hybrid orbitals point naturally at the corners of the tetrahedron.

Carbon is tetrahedrally coordinated

All are equal in length and the angle between the orbitals and therefore the bonds is ~ 109.5°

Carbon atoms with four single bonds are sp3 hybridized

sp3 hybridization

Chemistry 120

We can also add two p orbitals to the s orbital to form an sp2 hybrid, leaving one p orbital unused

sp2 hybridization

Hybridize

three sp2 hybrids

Atom Hybridized atom

Chemistry 120

The three sp2 hybrid orbitals point naturally at the corners of a triangle – the coordination at carbon is trigonal planar. All are equal in length and the angle between the orbitals and therefore the bonds is 120°

Carbon atoms with two single bonds and one double bond are sp2 hybridized

sp2 hybridization

Chemistry 120

The double bond can occur due to the p orbital that we have not used on the carbon atom:

sp2 hybridization

Chemistry 120

Imagine two sp2 hybridized carbon atoms forming a bond using one sp2 hybrid:

The ‘spare’ p orbitals can then form the bond.

sp2 hybridization

Chemistry 120

Alkenes, ketones, aldehydes, and any double bonded atom are all sp2 hybridized. The bonds are formed from the hybrids and the bond from the p orbital left over on each atom.

sp2 hybridization

Chemistry 120

The last possible hybrid is the sp hybrid. We use one p orbital and 1 s orbital:

sp hybridization

Chemistry 120

The two sp hybrids point at 180° to each other.

The two p orbitals can form two bonds

sp hybridization

Chemistry 120

Any triple bonded atom is sp hybridized

Alkynes, CO and CN- are all triply bonded.

sp hybridization

Chemistry 120

In a alkene, or other double bonded structure, the sp2 hybrids from the framework

Double and Triple bond structures

The bond between carbon atoms contains 4 electrons, two in an sp2 bond and two in the p bond

Chemistry 120

In a alkyne, or other triple bonded structure, the sp hybrids from the framework

Double and Triple bond structures

The bond between carbon atoms contains 6 electrons, two in an sp bond and four in the two p bonds

Chemistry 120

So far, we have considered bonds between identical atoms – homoatomic bonds.

The electron distributions are equal as the orbitals on the atoms have identical energies and sizes.

In a heteroatomic system, this is not true and the atomic orbitals that make up the molecular orbitals have different energies.

Polar and non-polar bonds

Chemistry 120

The orbital energies on an electronegative atom are lower in energy and therefore stabilize an electron more effectively.

In a heteroatomic bond, this causes a small change in the distribution of the bonding electron density and thus a small, permanent charge difference

Chemistry 120

For a non-polar molecule, the molecular orbital diagram is the standard diagram for a diatomic, shown here for O2-F2

Polar molecules are skewed in energy.

Chemistry 120

Polar molecules are skewed in energy.

The bonding molecular orbitals are more similar to the lower energy atomic orbitals – those on the electronegative element.

The opposite is true for the antibonding orbitals.

Chemistry 120

Phase Changes

Matter exists primarily in three phases:

liquid

Both elements and compounds are found in

these three phases, and are denoted by

subscripts, e.g. N2 (g), H2O(l), and Au(s).

Chemistry 120

Phase Changes

The three phases interconvert with one another.

solidliquid

Chemistry 120

Phase Changes

solidliquid

condensationvaporization

Interconversions of gases and liquids are called:

Chemistry 120

Phase Changes

Interconversions of liquids and solids are called:

solidliquid

melting

freezing

Chemistry 120

Phase Changes

Interconversions of gases and solids are called:

solidliquid

sublimation deposition

Chemistry 120

Phase Changes

AB – sublimation/deposition AD – melting/freezing

AC – vaporization/condensation

Chemistry 120

Phase Changes

Changing from a less dense phase to a more dense phase (e.g. condensation) is exothermic. Changing from a more dense phase to a less dense (e.g. vaporization) one is endothermic.

For any two phases, the energy changes in both directions are equal in magnitude, but opposite in sign.

Hvap = -Hcon

Recall that enthalpy is a state function.

Chemistry 120

Phase Changes

H2O enthalpy of fusion (melting)

Chemistry 120

Phase Changes

Liquids and their vapors are in equilibrium.

Pressure of a vapors (gas phase)

= vapor pressure of liquid

Pressure and temperature are directly proportional.

Recall PV = nRT

Chemistry 120

Phase Changes

SCF = supercritical fluid

Chemistry 120

Phase Changes

AB – sublimation/deposition AD – melting/freezing

AC – vaporization/condensation

Chemistry 120

Phase Changes

The temperature at which a liquid boils is called its boiling point (bp). Boiling point is a function atmospheric pressure, or the pressure above the solution.

Normal boiling point is the boiling temperature of a liquid at 1 atmosphere (atm) pressure.

Chemistry 120

Phase Changes

Critical temperature, Tc, is the highest temperature at which liquid and vapor exist in equilibrium.

Critical pressure, Pc, is the vapor pressure at the critical temperature.

Critical point is reached at Tc and Pc.

Triple point is the temperature and pressure at which all three phases coexist.

Chemistry 120

Phase Changes

H2O phase diagram

C = Tc, Pc

A = triple point

Chemistry 120

Phase Changes

Phase diagram for HgI2 (mercuric iodide)

HgI2 () and HgI2 () are both solids but different phases.

Chemistry 120

The phase changes from solid to liquid to gas are governed by intermolecular forces.

Intramolecular forces are the chemical bonding forces discussed previously.

These intermolecular forces have both attractive and repulsive components. Collectively they are called van der Waals forces after the Dutch Nobel laureate (physics) who described them.

Intermolecular Forces

Chemistry 120

Johannes Diderik van der Waals

Van der Waals forces describe the behavior of a non-ideal gas, which includes both attractive and repulsive components.

[P + a(n/V)2](V-bn) = nRT

Chemistry 120

All molecules exert weak attractions on one another due to the mutual attraction of nuclei and electrons. These attractive forces are only significant at very short distances.

At such small distances the intermolecular repulsion of the electrons on different atoms is also significant.

Chemistry 120

The electrons orbiting all atoms and molecules can be perturbed by an electric field, with greater or lesser ease. This property is called polarizability.

The electron cloud around an atom or molecule can give an instantaneous dipole any time that the electrons are not distributed perfectly symmetrically.

Such a dipole can induce dipoles in other species nearby.

Chemistry 120

The attractive forces between an instantaneous dipole and an induced dipole are called London dispersion forces after the physicist Fritz London.

These forces are stronger for more polarizable species.

Chemistry 120

Some molecules have a permanent dipole because of differences in electronegativities among the atoms.

Such molecules experience dipole-dipole forces.

All molecules experience dispersion forces and induced dipoles, and polar molecules also experience dipole-dipole forces.

Chemistry 120

Dipole-dipole interactions.

Chemistry 120

The intermolecular forces that we have seen

London dispersion forces

induced dipole interactions

dipole-dipole interactions

have a strong effect on the boiling points of liquids, along with the molecular weight of a compound, and hydrogen bonding.

Chemistry 120

(a) CS2

(b) CH3OH

(c) CH3CH2OH

(d) H2O

(e) C6H5NH2

Chemistry 120

Hydrogen bonding is an additional type of bonding interaction that requires a hydrogen atom on one molecule and a source of electron density on another molecule, usually a lone pair.

Hydrogen bonding can be intramolecular as well as intermolecular.

Chemistry 120

Boiling points of covalent hydrides

Chemistry 120

Boiling points of covalent hydrides

Chemistry 120

The second-row hydrides NH3, H2O, and HF exhibit much higher boiling points that would be expected based on their molecular weights. Strong hydrogen bonding between the molecules is responsible for the large liquid phase range of these compounds.

CH4 has a low boiling point because it has no lone pairs to form strong hydrogen bonds.

Chemistry 120

Hydrogen bonding is crucial for the double-helical structure of DNA.

An understanding of hydrogen bonding between base pairs made the structural solution possible for James Watson and Francis Crick.

Chemistry 120

DNA base pairs hydrogen bond between

G-C and

Chemistry 120

Structure and Bonding

bonds are in general stronger than bonds and can be formed from either s or p orbitals:

bonds have no nodal planes along the line containing the two nuclei.

Chemistry 120

Structure and Bonding

The * antibonding orbital has one nodal plane between the two nuclei

Chemistry 120

Structure and Bonding bonds have one nodal plane that contains both nuclei.

The * antibonding orbital also has one nodal plane between the nuclei.

Chemistry 120

Hybridization Review

Recall:

sp3 hybridization 4 bonds

sp2 hybridization 3 bonds, 1 bond

sp hybridization 2 bonds, 2 bonds H

HC C CC HHH

sp3 sp2 sp

Chemistry 120

methanol ethanol

Chemistry 120

methanol ethanol

Chemistry 120

methanol ethanol

sp3 sp3 sp3

Chemistry 120

methanol ethanol

sp3 sp3 sp3 sp3

Chemistry 120

carbon disulfide

aniline

Chemistry 120

carbon disulfide

aniline

Chemistry 120

carbon disulfide

aniline

Chemistry 120

carbon disulfide

aniline

Molecular Structure

Documents