Liquids and Solids
1.
Are liquids closer in physical properties to solids or gases? Why?
Liquids are more similar to solids. There are many intermolecular forces
experienced by solids and liquids and very few by gases.
2.
Define
a. Intramolecular Interactions
Interactions (bonds) within a molecule.
b. Intermolecular Interactions
Interactions between two or more particles.
3.
How do intermolecular Interactions affect the boiling and freezing point of a
substance?
Increased interactions results in a higher boiling point and higher freezing
point.
4.
Why is ∆Hvap much greater than ∆Hfus ?
Because liquid gas transition breaks virtually all of the intermolecular
bonds.
5.
Types of Intramolecular Bonds and an example of each
a. Non-polar (O2)
b. Polar covalent (HF)
c. Ionic (NaCl)
d. Atomic ( Cgraphite )
6.
Types of Intermolecular Interactions
a. London Dispersion (aka Van Der Waal’s forces or LDF)
i. What type of substances experience this type of interaction?
All polar and non-polar substances.
ii. What occurs to make this interaction happen?
Shift of e- to one side.
iii. What affects the strength of this interaction?
The size of the atom/molecule.
increased mass = increased strength
iv. What evidence suggests this type of interaction exists?
The fact that non-polar substances can be condensed to liquids
and solids.
b. Dipole-Dipole Interactions
i. What type of substances experience this type of interaction?
Polar covalent substances.
ii. What occurs to make this interaction happen?
Dipoles line up due to attraction of partial charges.
iii. What affects the strength of this interaction?
The greater the polarity the greater the strength.
c.
Hydrogen bonding ( a special type of dipole-dipole interaction)
i. What type of substances experience this type of interaction?
Substances with polar N-H, O-H or H-F bonds.
ii. What occurs to make this interaction happen?
Dipoles line up due to attraction. This type of dipole is stronger
than the typical dipole-dipole interaction because F, O and N are
particularly electronegative.
iii. What affects the strength of this interaction?
The more polar the bond and greater the number of N-H, H-F or
O-H bonds, the stronger the interaction.
d. Ionic
i. What type of substances experience this type of interaction?
Ionically bonded substances.
ii. What occurs to make this interaction happen?
Positively and negatively charges line up.
Na+Cl-Na+Cl-Na+Cl- etc….
7.
What is the typical ordering of strength for these interactions? Are there
situations where this ordering may not apply?
Ionic > Hydrogen Bond>Dipole-Dipole>London Dispersion
There are occasions when this order may not apply. A very large non-polar
substance compared to a mildly polar compound may have an LDF that is
stronger than the dipole-dipole interaction.
8.
Identify the most important types of interparticle forces present in the solids
of each of the following:
a. NH4Cl
Ionic compund – ionic interactions
b. Ar
Atom – LDF only
c. HF
Polar – Hydrogen bonding and LDF
d. BF3
Non-polar - LDF
e. CHCl3
Polar – dipole-dipole and LDF
9.
For which molecule in each of the following pairs would you expect the
stronger intermolecular forces?
a. CH3CH2CH2NH2 or H2NCH2CH2NH2
The answer has 2 areas of hydrogen bonding capability.
b. CH3CH3 or H2CO
The answer is polar so it has dipole-dipole interactions.
c. CH3OH or H2CO
Though both are polar, the answer is polar with the ability to hydrogen
bond.
d. HF or HBr
Once again, both options are polar, but HF has the ability to hydrogen
bond.
10.
Characteristics of liquids
a. Surface Tension – The resistance to increase surface area.
i. How do the intermolecular interactions affect this characteristic?
An increase in interactions leads to an increase in surface tension.
b. Capillary Action – The spontaneous rising of a liquid in a narrow tube.
i. What forces are responsible for this characteristic?
1. Adhesive – Interactions of molecule with the container.
2. Cohesive – Interactions between molecules with each
other.
ii. How do intermolecular interactions affect these forces?
Increased interactions between molecules increases cohesiven
interactions.
Increased interactions between the molecule and container
increases adhesive interactions.
c.
Viscosity – Resistance to flow.
i. How do intermolecular forces affect this property?
Increased interactions increased cohesive forces increased
viscosity
d. Vapor Pressure – Pressure above a liquid/solid due to evaporation.
i. How do intermolecular forces affect this property?
Increased interactions decreased vapor pressure
ii. How does temperature affect this property?
Increased temperature increased vapor pressure.
11.
In each of the following groups of substances, pick the one that has the given
property
a. Highest boiling point: HBr, Kr, Cl2
The stronger the interactions the higher the boiling point.
Kr and Cl2 are nonpolar – this means that they only experience LDFs. HBr
is a polar molecule, which means that it experiences LDF and dipoledipole interactions. These forces are stronger.
b. Highest freezing point: H2O, NaCl, HF
The stronger the interactions the higher the freezing point.
H2O and HF are both polar molecules that would experience LDF and
hydrogen bonding intermolecular forces. NaCl, however, is an ionic
compound. Thus it would experience, much stringer, ionic interactions.
c. Lowest Vapor Pressure: Cl2, Br2, I2
The stronger the interactions the lower the vapor pressure. All of these
compounds are nonpolar – meaning they experience only LDF
interactions. Remember, the strength of the LDF very much depends on
the size of the particle. Because I2 is the largest molecule, it would have
the greatest LDF.
d. Lowest freezing point: N2, CO, CO2
The weaker the interactions the lower the freezing point. CO is a polar
molecule and CO2 and N2 are nonpolar. Nonpolar substances experience
the weakest forces. Because N2 is smaller, it would experience the
smallest LDF and, therefore, have the lowest freezing point.
e. Greatest viscosity: H2S, HF, H2O2
The greater the intermolecular forces the greater the viscosity. H2S
experiences dipole-dipole interactions and LDFs. Both HF and H2O2
experience H-bonding and LDFs. Because H2O2 has two areas for Hbonding it has the stringer intermolecular forces.
f. Greatest heat of vaporization: CH3OCH3, CH3CH2OH, CH3CH2CH3
The greatest heat of vaporization would be associated with the
compound that has the greatest intermolecular forces. CH3CH2OH is the
only to experience H-bonding, thus it would have the highest heat of
vaporization.
g. Smallest enthalpy of fusion: I2, CsBr, CaO
Both CsBr and CaO are ionic compounds and I2 is a non-polar compound.
The enthalpy of fusion is amount of heat required to melt a substance
(sl). The smallest enthalpy of fusion would be associated with the
weakest intermolecular forces. Because I2 is non-polar it would only
experience LDFs – the weakest intermolecular force and would therefore
have the smallest ∆Hfusion.
12.
Rationalize the difference in boiling point for:
CH3CH2CH2CH2CH3
(36.2°C)
(9.5°C)
More surface area leads to more effective LDFs. Additionally, long molecules
have the potential for tangling.
13.
Which of the following substance would have a boiling point closest to
argon?
Cl2, HCL, F2, NaF, or HF
Argon is non-polar and therefore has only LDF forces. That means we are
looking for non-polar compounds from the list – only F2 and Cl2 fit the bill.
Remember that LDFs are very much dependent upon the size/mass of the
compound – so we need to determine whether Cl2 or F2 has a mass that is
closer to Ar. The molar mass of Ar is 39 g/mol. F2 has a molar mass = 38
g/mol. These two would therefore have relatively similar LDFs and thus the
closest boiling points.
14.
Match the following boiling points with the correct structure; -42.1°C, -23°C
and 78.5°C
a. CH3CH2OH 78.5oC (strongest forces – H-bonding)
b. CH3OCH3 -23oC (dipole-dipole interactions)
c. CH3CH2CH3 -42.1oC (Weakest forces – LDFs)
15.
Why does water create a concave meniscus and mercury a convex meniscus?
Water’s adhesive forces are greater than its cohesive forces.
Mercury’s cohesive forces are greater than its adhesive forces.
16.
Why does water form into beads on a waxed car?
Surface tension is used to minimize surface area. A bead, or sphere, is the
shape with the lowest possible surface area.
17.
What are two broad categories of solids?
a.
Amorphous (short range order, e.g. glass)
b. Crystalline (long range order)
18.
Types of crystalline solids
a. Ionic
b.
Molecular
c. Atomic
d. Covalent Network
e. Metallix
19.
What is X-ray diffraction used for?
X-ray diffraction is used to determine the structures of crystalline solids.
20.
What is the Bragg equation?
nλ = 2d sin θ
n = integer (order)
λ = wavelength used
d = distance between atoms
θ = angle of incidence/reflection
This equation can be used to interpret the results of an diffraction experiment
and determine the structure of a crystalline solid.
21.
A topaz crystal has an interplanar spacing (d) of 1.36 x 10-10 m. Calculate the
wavelength of the X-ray that should be used if θ = 15.0o (assume n = 1).
These problems are typically just straight plug-ins. We know that we need to
use the Bragg equation because we are dealing with X-ray diffraction.
We will start by organizing our data:
Plugging in:
22.
What are 3 ways metals are typically arranged can be arranged?
a.
Face Centered Cubic (fcc)
i. Illustrations
ii. Facts
1.
Volume = e3
r = radius of atom
2. # of nearest neighbors – 12 nearest neighbors
3 above, 3 below and 6 on the
same level.
3. type of packing – Cubic Closest Packing
4. Total atoms within unit cell – 6(1/2) + 8(1/8) = 4 atoms
5. % of space used – 74%
b.
Body Centered Cubic
i. Illustrations
ii. Facts
1. Volume = e3
r = radius of atom
2.
# of nearest neighbors – 8 nearest neighbors
4 above and 4 below
3. type of packing – You just need to know that it is not cubic
closest packing
4. Total atoms within unit cell – 8(1/8) + 1 = 2 atoms
5. % of space used – 68%
c.
Simple Cubic (aka Primitive)
i. Illustrations
ii. Facts
1. Volume – e3
r = radius of atom
2. # of nearest neighbors – 6 nearest neighbors
1 above, 1 below and 4 on the
same level.
3. type of packing - You just need to know that it is not cubic
closest packing
4. Total atoms within unit cell – 8(1/8) = 1 atom
5. % of space used – 52.4%
23.
A helpful formula for dealing with cubic structures and density:
24.
A certain form of lead has a cubic closest packed structure with an edge
length of 492 pm. Calculate the value of the atomic radius and density of the
lead.
Because this is a cubic closest packed structure (meaning face centered cubic):
Plugging in:
To solve for the density we can just plug into the formula from problem 23.
Because we are talking about a face-centered arrangement, we know that there
are four total atoms per unit cell.
Plugging in:
25.
You are given a small bar of an unknown metal X. You find the density of the
metal to be 10.5 g/cm3. An X-ray diffraction experiment measures the edge of
the face centered cubic unit cell as 4.09 x 10-10 m. Identify X.
The identity of X can be determined by calculating the molar mass. We will
do this once again by using the formula established in question 23.
Once again, this is a face-centered cubic so we know that there are 4 total
atoms per unit cell.
26.
Theories for bonding in metals
a. Electron Sea Model – cations regularly arranged e- move freely around.
This theory explains metals ability to conduct because the electrons are
described as free moving charged particles – the basic requirement for to
conduct.
b. Band Model – uses molecular orbitals to explain properties.
In this theory we are looking at the distance separating filled molecular
orbitals and the unoccupied molecular orbitals to explain the
conducting/insulating properties of various solids.
27.
Using Band Theory, explain the difference between conductors, semiconductors and insulators.
In this model:
A substance is described as a conductor when the filled MOs and unfilled
MOs are in close proximity to one another. Because there is a very small
energy difference, electrons are able to jump into the next unoccupied MO
with relative ease. Thus the electrons are “free moving” charged particles.
Semiconductors are substances who filled MOs and empty MOs have a
greater separation than conductors but are close enough in proximity that
electrons are still able to shift to the empty MOs. Because there is a greater
separation it would take more energy (and thus be “harder”) to get the shift to
occur, but it is still viable.
Insulators are substances whose filled MOs and empty MOs have a significant
difference in energy, and whose electrons, therefore, cannot move very freely
between the two molecular orbitals
28.
What are p-type and n-type semiconductors?
P-type and N-type semiconductors are said to be “doped”. Meaning that they
have has some impurity added to them to enhance their conducting abilities.
Remember that the ability to conduct hinges on the movement of electrons.
Silicon (Si) is a commonly used semi-conductor. If a silicon sample is taken
and some of the silicon atoms are replaced with gallium a p-type semiconductor would be created.
The “p” stands for positive. It has this namesake because while Si has 4
valence electrons, Ga only has 3 – thus there is one less electron leaving the
sample “positive” by comparison.
The reason that this enhances the conducting abilities is that with less
electrons in the filled MOs there are “open” spaces created for remaining
electrons to move into leading to a greater movement of charged particles
(electrons) which leads to better conducting abilities.
If, instead, a silicon sample is taken and some of the silicon atoms are
replaced with arsenic, an n-type semi-conductor would be created.
The “n” stands for negative. It has this naming because Aresenic has one
additional valence electron compared to Si – thus there is one more electron
making this sample more “negative” by comparison.
The reason this enhances the conducting abilities is that with more electrons
present, there are more charged particles to transition over the band separating
the filled MOs and the unfilled MOs. More moving charged particles means
more conductive ability.
29.
What is a network atomic solid? What are examples?
A solid that is made with covalent bonds throughout like Cdiamond
30.
In molecular solids the intramolecular forces are very strong whereas the
intermolecular forces are very weak.
31.
The bonding in ionic solids will be primarily explained by the closest packing
model.
32.
What is typically larger, cation or anion? How does this affect the typical
arrangement within a salt?
Anions are typically larger due to e-/e- repulsion. Because of the
consequences of sizing, typically the larger anion will be packed and the
smaller cations will be in the holes (or spaces between packed anions).
33.
What two types of holes in a closest packed structure?
a. Tetrahedral
b. Octahedral
34.
Arrange these holes in order of increasing size.
tetrahedral < octahedral
35.
Guidelines for deciding which type of hole is used
0.225 R- < r+ <0.414 R0.414R- < r+ < 0.732 R-
use tetrahedral
use octahedral
Where:
“R – “ refers to the radius of the anion
“r + “ refers to the radius of the cation
36.
Why are holes that are slightly smaller than the cation used?
Because when the cation is squeezed into the hold it pushes apart the anions a
bit and that minimizes the anion/anion repulsion.
37.
What is the ration of packed spheres to holes for
a. Tetrahedral
1 : 2 (there are twice as many holes as spheres).
b. Octahedral
1:1 (same number of spheres as holes)
38.
Equation for vapor pressure change with temperature change
39.
Label the following heating curve for water:
40.
Label the following phase diagrams – how can you identify which is water?
Water has a negative slope between S/L phases because ice is less dense than
water. That means that if the pressure is increased (thereby forcing an
increased density) water would convert to the liquid phase, as it is the most
dense of all phases.