## Intermolecular Forces

### Dipole Moment

Remember that Fluorine is has the highest value of electronegativity, χ? Due to electronegativity difference, Δχ, b/w two atoms, bond would polarize.

If the Δχ = 0, then the bond is said to be non-polar.

If the Δχ is non-zero, the bond is polar.

So for molecule N=O, we have

As it is seen, arrow indicates the dipole moment of the bonds between
N and O, and the arrow points toward more negative side, indicated with
δ-, and the other end is more positive side,
δ+. This is so because oxygen is closer to
F in the periodic table than nitrogen.

Adding all Polarities of bonds **→**
molecular dipole moment, μ

The bond dipole moments of O-H bonds (red arrow) in water points toward oxygen atom. When these two bond dipoles are added together, molecular dipole moment (blue arrow) is produced. As in the bond dipole, the direction the arrow is pointing has partially negative charge, and the opposite end has partially positive charge.

**Draw in polar bonds with arrows to show the bond dipole moments and molecular dipole moment of dimethyl ether (CH**

*Example:*_{3}O CH

_{3}), tetrafluoromethane (CF

_{4}) and HCOCH

_{2}CF

_{3}.

In order to deduce dipole moments of molecule is to know the 3-D structure of the molecule. Then, draw in the bond dipoles (red arrows). We consider C-H and C-C bonds to be non-polar bonds, therefore we don't draw in the arrow for these bonds. After bond dipoles are drawn in, all bond dipole vectors are added, and the resulting vector is the molecular dipole moment (blue arrow ).

For CF_{4}, two of the red bond dipoles form a resultant
green dipole in
the plane of the computer monitor. For other two bond dipoles, with
orange color, form another green resultant pointing in the opposite direction.
And, they cancel each other, therefore the moleuclar dipole moment of
CF_{4} is zero.

*Very Important Concept*!! You'd have easier time knowing bond dipole and moleucular dipole moments in organic chemistry,

__Yes, in__

*organic chemistry!*

### Intermolecular Forces

Forces b/w molecules

Forces responsible for condensing gas into liquid and liquid into solid

Forces responsible for making different compounds to have different melting point and boiling point.

Four types:

**Ion-Dipole Interaction**

Water surrounds Na^{+} ion in a) and F^{-} ion surrounded by water in b). The red sphere in water is the oxygen atom. Notice that the oxygen is pointing toward Na ion, and the hydrogen atoms point toward F ion.

**Dipole-Dipole Interaction**

So, if you have a bunch of N=O molecules, you'd see the packing of molecules as shown on the left diagram in the figure below. On the right diagram negative-negative interactions and positive-positive interactions, both represented by green arrows, are the repulsive interaction. So the structure is unstable. It'd eventually become the same packing as on the left.

**Dipolar vs non-dipolar interactions**

N_{2} | NO | O_{2} | |

Dipole(debye) | 0 | 0.153 | 0 |

Mass (g/mol) | 28 | 30 | 32 |

Boiling pt(K) | 77 | 121 | 90 |

**Dispersion Interaction**

As they are close to each other, B sees the electrons on A, and the electrons on the left-side of B starts to avoid the approaching A (lower diagram on left). This induces polarity on molecule B, which in turn can polarize A as well. This is the source for the attractive force between non-polar non-ionic molecules.

If the molecule is easy to deform → said to be *Polarizable*.

Halogen | mp (K) | bp (K) |
---|---|---|

F_{2} | 53.5 | 85.0 |

Cl_{2} | 171.6 | 239.1 |

Br_{2} | 265.9 | 331.9 |

I_{2} | 386.8 | 457.5 |

**Hydrogen Bonding**

Sharing of hydrogen between two molecules

Each molecule must act as both the donor and acceptor.

The picture on the right is a dimer of water. The molecule on left is the H-donor and the one on right is the acceptor.

The shared hydrogen is co-linear with the lone pair-H-donor.

The binding energy for water dimer is quite small (3.4 kcal/mol). But, this is the force holding two strands of DNA.

**Comparison of the strength of intermolecular forces**

Force | Strength in kJ/mol | Strength in kcal/mol |
---|---|---|

Ion-Dipole | 10 - 50 kJ/mol | 2 - 12 kcal/mol |

Dipole-Dipole | 3 - 4 kJ/mol | 0.5 - 1 kcal/mol |

Dispersion | 1 - 10 kJ/mol | 0.2 - 2 kcal/mol |

H-bond | 10 - 40 kJ/mol | 2 - 10 kcal/mol |

**What is the most important intermolecular force exit in each of the following liquid? Can you draw two molecules for each and show how they are aligned? 1) Liquid ammonia (NH**

*Example:*_{3}, 2) C

_{6}H

_{14}, and 3) Formaldehyde (COH

_{2})

1) Liquid ammonia, NH

_{3}, has lone pair electrons and H, therefore it can H-bond to each other. The two molecules can interact in the following way.

2) C_{6}H_{14} (hexane) is non-polar and elongated molecule. So,
it only has dispersion interaction. In order to liquify, two molecules must be
aligned sideways as shown below.

3) Formaldehyde is a dipolar molecule. The dipole-dipole interaction entails lining up the two dipole moments as follows:

## Liquid Properties

**Surface Tension**Energy needed to increase surface area (Cohesive energy)

The red dashed line is the meniscus you read for measurements, like measuring the volume of water with graduated cylinder.

Mercury side = Convex | H_{2}O side = Concave |

Hg Intxn with glass is weak | H_{2}O Intxn with glass is strong |

Cohesive Energy of Hg is smaller than intxn b/w Hg and glass | Cohesive Energy is larger than intxn b/w H_{2}O and glass |

## Crystal Structure

Two types of solids

**Types of Crystaline Solids**

Types | Intermolecular Forces | Properties | Examples |
---|---|---|---|

Ionic | Ion-ion force | Brittle, hard, high-melting | NaCl, KBr, MgCl_{2} |

Molecular | Dispersion, dipole-dipole, H-bond | Soft, low-melting, non-conducting | H_{2}O, Br_{2}, CO_{2}, CH_{4} |

Covalent network | Covalent bonds | Hard, high-melting | diamond, SiO_{2} |

Metallic | Metallic bond | Variable hardness and melting points, conducting | Na, Zn, Cu, Fe |

## X-Ray Diffraction

X-ray is used to determine the structure of crystal -- very important tool in Biochemistry -- Protein structures are determined this way. The structure of DNA was discovered by this technique.

Above diagram shows how the experiment is carried out. X-ray is irradiated onto a crystal. Then the diffraction pattern is photographed. | This is the diffraction pattern of one form of DNA. As can be seen, the structural parameters are obtained from the diffraction pattern. |

The red arrows from left to fall on atoms indicates incident X-ray, and the red arrows going away from atoms are the diffracted (outgoing) X-ray. The incident rays with the wave length, λ, on the 2^{nd} layer must travel extra distances covered by the lengths, C-B and B-D. The diffraction pattern only appears when the extra distances, C-B and B-D are integral multiples of λ

By using trigonometry, one can show the extra distnaces traveled is:

$$CB=BD=d\mathrm{sin}\theta $$

Then, according to Bragg, the distance C-B and B-D is related to integral multiple of λ, thus one obtians the relationship between incident X-ray wave length and the structural parameter, *d* is given
as

$$d=\frac{n\lambda}{2\mathrm{sin}\theta}$$

This is called Bragg equation.

## Crystal Types

Crystalline solids have repeating unit →**Unit Cell**.

The repeating pattern of **unit cells** determine the structure of crystal.

*Crystal packing*

Crystal packing shows how the atoms are arranged in a crystal. There are
four types:

Simple Cubic | ||

Body-Centered Cubic | ||

Hexagonal Closest-Packing | ||

Cubic Closest-Packing |

*Unit Cells*

Within the packing pattern, one can consider a unit of repeating cells, called
*unit cell*, which contains fewest possible atoms. For example, simple cubic packing pattern can be considered to have a unit cell with the edge length

*a*and contains 1/8 of atoms at each corner, which means there are 1 atom in the unit cell, as shown below. There are two other types of unit cell within the so-called simple cubic unit cell.

3 types:

Primitive Cubic | |

Body-centered Cubic | |

Face-centered Cubic |

In the following diagram, the relationship between the edge length and atomic radius of monoatomic solid that packs in a simple cubic cells.

**What is the density of tungsten (W) in g/cm**

*Example:*^{3}which assumes body-centered cubic packing? The atomic radius of W is 1.39 Å.

Using the above diagram, we can get the edge length,

*a*, from the atomic radius,

*r*, by using the relationship shown in the diagram above, and it is $$a=\frac{4r}{\sqrt{3}}$$ Therefore,

*a*= 4(1.39 Å)/√3 = 3.210 Å. So the volume of the unit cell is (3.210 Å)

^{3}.

We need the mass of W atoms in the unit cell. We can obtain it by counting the number of atoms in the cell and convert them into g, using Avogadro's number.

There are two W atoms in the cell. Then,
$$?{g}_{W}={2}_{W}\left(\frac{1mo{l}_{W}}{6.022\times {10}_{W}^{23}}\right)\left(\frac{183.84{g}_{W}}{1mo{l}_{W}}\right)=61.0\times {10}_{{g}_{W}}^{-23}$$
The volume is (3.210 Å)^{3}, but it needs to be converted to
cm^{3}.
$$?c{m}^{3}={\left(3.210\stackrel{\circ}{A}\right)}^{3}{\left(\frac{1m}{{10}^{10}\stackrel{\circ}{A}}\right)}^{3}{\left(\frac{100cm}{1m}\right)}^{3}=33.08\times {10}^{-24}c{m}^{3}$$
The density, then is,
$$d=\frac{m}{V}=\frac{61.0\times {10}_{{g}_{W}}^{-23}}{33.08\times {10}^{-24}c{m}^{3}}=18.4g/c{m}^{3}$$
The measured density is 19.25 g/cm^{3}.

*Allotropes* = same formula with different structural forms

Carbon has 40 allotropes! Graphite (graphene → Nobel Prize for synthesizing), Bucky ball (Nobel for determining structure), Bucky tubes, diamond, etc.

## Phase Changes

### Phase Change

A few names associated with phase change:Fusion (melting) | solid → liquid | Δ H >0 |
---|---|---|

Freezing | liquid → solid | Δ H < 0 |

Vaporization | liquid → gas | Δ H > 0 |

Condensation | gas → liquid | Δ H < 0 |

Sublimation | solid → gas | Δ H > 0 |

Deposition | gas → solid | Δ H < 0 |

This is a plot of heating curve of water. Plotted is the temperature of water by adding more heat (energy) to the water. Notice there are two plateaus at melting point and at boiling point.

In order for you to make 100° liquid H_{2}O to 100° gas H_{2}O (steam), it needs about 40 kJ/mol of energy.

This is the heat of vaporization, ΔH_{vap} = 40.67 kJ/mol.

Similarly, you see less than 10 kJ/mol is needed to melt ice into liquid → heat of fusion, ΔH_{fus} = 6.01 kJ/mol

### Liquid Vapor Equilibrium

Vapor Pressure ← Dyanmic equilibrium b/w liquid and gas

Over the surface of liuquid bromine, you see vapor of gaseous bromine. Liquid and gas establish a so-called dynamic equilibrium.

*Rate _{in} = Rate_{out}*

_{2}(g) goes in to the liquid is equal to the Rate of Br

_{2}

*(l)*becoming gas.

Higher the temperature → Higher the vapor pressure

Higher T → Larger E_{K} → Easier to overcome intermolecular intxns.

T > 0 K, there is always vapor exists over solid or liquid.

*Clausius-Clapeyron Equation*

If you plot P

_{vap}as a fxn of T, you get a picture on the left

But if you plot ln(P) vs. 1/T, instead, the curve becomes linear!

Linear equation is easier to work with!!!

The relationship b/w ln(P) and 1/T is,

$$\mathrm{ln}\left(\frac{{P}_{2}}{{P}_{1}}\right)=\frac{-\Delta {H}_{vap}}{R}\left(\frac{1}{{T}_{2}}-\frac{1}{{T}_{1}}\right)$$
In this equation, *R* = 8.314 J/molK is used. It means that for T, we should be using Kelvin unit. For pressure, you can use whatever the unit as long as P_{1} and P_{2} have the same unit. Δ H_{vap} is the heat of vaporization (for water, it is 40.67 kJ/mol as we discussed above).

**Denver is known as Mile High City. If the barometric pressure is measured to be 0.837 atm, what is the normal boiling point of water in Brooklyn (sea level)?**

*Example:*

We know water boils at 100°C at sea level at which barometric pressure is 1.00 atm. Let's assign P_{1} = 1.00 atm and T_{1} = 100°. So, we are solving for T_{2}. Then,

$$\begin{array}{l}\frac{-R}{\Delta {H}_{vap}}\mathrm{ln}\left(\frac{{P}_{2}}{{P}_{1}}\right)=\frac{1}{{T}_{2}}-\frac{1}{{T}_{1}}\\ \frac{1}{{T}_{2}}=\frac{R}{\Delta {H}_{vap}}\mathrm{ln}\left(\frac{{P}_{1}}{{P}_{2}}\right)+\frac{1}{{T}_{1}}\\ {T}_{2}=\frac{1}{\frac{R}{\Delta {H}_{vap}}\mathrm{ln}\left(\frac{{P}_{1}}{{P}_{2}}\right)+\frac{1}{{T}_{1}}}\end{array}$$

Substitute #'s in. $${T}_{2}=\frac{1}{\frac{8.314\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$molK$}\right.}{40.67\times {10}^{3}\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.}\mathrm{ln}\left(\frac{1.00atm}{0.837atm}\right)+\frac{1}{373.15K}}=\frac{1}{3.6374\times {10}^{-5}/K+2.6799\times {10}^{-3}/K}=\frac{1}{2.7163\times {10}^{-3}/K}$$

*T _{2}* = 368.15K, or 95.0°C.

## Phase Diagram

Phase diagram shows phases of substance as a function of temperature and pressure.

Each solid lines represent phase boundary. For example, the line in the diagram with an arrow is a phase boudnary b/w solid and liquid.

We know:

So, the left-most (green region) is solid, middle part (pink region) is liquid, and bottom part (yellow region) is gas.

**Important points and processes on phase diagram**

Few important Points on the Phase Diagrams:

*Supercritical Fluid* is used to decaffinate coffee beans.

Important Processes: