APSC 131 Final Exam Workshop Solutions

APSC 131 Final Exam Workshop Solutions

References [1] M. Mombourquette, APSC 131 Final Examination, Kingston: Department of Chemistry, Queen’s University, 2012

[2] M. Mombourquette, APSC 131 Final Examination, Kingston: Department of Chemistry, Queen’s University, 2013



Quantum Mechanical Models 1.1 Atomic Model and Quantum Numbers Electrons are described by a wave function (𝝍), which gives the probability of finding it inside a 3D

region.

Prob(r) = 𝑟2𝜓(𝑟, 𝜃, 𝜙, 𝑛)2

Where (𝑟, 𝜃, 𝜙) is the spherical position and n is an integer representing the energy level of the electron

and the number of nodes (points of 0 probability) in the wave. The Rydberg Equation describes the

energy an electron needs to go from one energy level to another.

∆𝐸 = ℎ𝑓 = 𝑍2𝑅𝐻 |1

𝑛12 −

1

𝑛22| , 𝑅𝐻 =

𝑚𝑒 𝑒4

8 휀02 ℎ3 𝑐 (1 +

𝑚𝑒𝑚𝑝)= 2.179 × 10−18 𝐽

Where Z is the nuclear core charge, me and mp are the mass of an electron and proton respectively, ε0 is

the standard permeability of free space, and c is the speed of light.

Electrons in atoms are arranged in orbitals. Pauli’s Exclusion Principle states that no two electrons within

an atom can be in the same location, specified by their quantum numbers, at the same time:

Name Principle Angular Magnetic Spin Letter n l ml ms

Description Energy Level Orbital Type Degenerate Orbitals

Electron within orbital

Values 1, 2, 3, 4… 0 to n-1 ±l ½ or -½

1.1 Example (2014 final) [3]:

The equation below represents the radial-part of a hydrogen-like orbital wavefunction. What kind of

orbital does it represent and why? ao is the Bohr Radius – the distance from the nucleus to the first bohr

orbital

𝑅(𝑟) =1

9√3(𝑍

𝑎0)

32(6 − 6𝜎 + 𝜎2)𝑒−

𝜎2 , where σ =

2𝑍𝑟

𝑛𝑎0

Solution Additional Explanation

1

3s Start by looking at each part of the function:

1

9√3(𝑍

𝑎0)

32

Is a constant and can be ignored.

𝑒−𝜎2

Continues to decrease as radius

increases. This is a decay function

which only depends on radius. This would

be part of an s orbital (spherical). (6 − 6𝜎 + 𝜎2)

Has 2 zeros (corresponding to radial

nodes), hence the orbital must be in n =

3.

1.1 Example (2015 Quiz):

An electron in the hydrogen atom’s n=6 drops to n=2. (c = 3.0 x108 m/s)

a) What is the wavelength of the light emitted?

∆𝐸 = ℎ𝑣 = 𝑅𝐻 |1

𝑛12 −

1

𝑛22|

(6.626 × 10−34 𝐽 ∙ 𝑠)𝑣

= 2.179 × 10−18 |1

62−1

22|

𝑣 = 7.308 × 1014 𝐻𝑧

𝜆 = 𝑐/𝑣

𝜆 =(3.0×

108𝑚

𝑠)

7.308×1014 𝐻𝑧= 4.105 × 10−7 𝑚

First solve and rearrange for v. Solve for wavelength.

1.2 Aufbau and Hund’s Rule The Aufbau Principle is a method of filling orbitals in order from the lowest energy upwards. Hund’s

Rule states that one electron is placed into each degenerate orbital before a second is added. An

exception exists for nearly full or half-full d orbitals such as s2 d4 or s2 d9, which become s1 d5 or s1 d10

instead.

1.2 Example (Practice Final):

According to Hund’s Rule, which of the following atoms has exactly three unpaired electrons in its ground

state?

A) B

B) C

2

C) N

D) O

F) F

C) Nitrogen N has 3 electrons in the p orbital. Each orbital gets an electron before any gets a second, so it has 3 unpaired.

1.2 Example:

What is the electronic configuration of As? How many electrons are in each of its valence p orbitals and

provide the quantum numbers of one of them. Given As5+, what element would it be isoelectric with?

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3 As is element 33 in the 4th row in column 15. Use Aufbau to fill the electrons (33 of them).

1 electron is in each of its valence p orbitals From Aufbau, there will be 3 electrons left to fill into the 3 valence p orbitals. Hund`s rule states that each p orbital will receive 1 before any will get a 2nd.

n = 4, l = 1, ml can be -1, 0, or 1 and ms is 1/2 4th row (n = 4), p ortbial (l = 1). It doesn’t specify which p orbital, therefore any ml can be used. ms must be ½ since it is the only electron in the orbital.

The electronic configuration for As5+ is: 1s2 2s2 2p6 3s2 3p6 4s2 3d8

This will be isoelectric with Ni

Find the corresponding element on the periodic table with that electron config.

1.3 Heisenberg and De Broglie De Broglie extended the wave equation, f = vλ, to describe the wavelength of a moving particle:

𝜆 =ℎ

𝑚v , ℎ = 6.626 × 10−34 J s

Heisenberg’s Uncertainty Principle describes how the position and momentum of a particle cannot

both be exactly known at the same time:

∆(𝑚v)∆𝑥 ≥ℎ

4𝜋

1.3 Example (Adapted from 2012 final) [1]:

What is the ‘wavelength’ of a 0.45g bullet moving at 2000 m/s? How accurately can you predict its

position if you know its momentum is accurate to ±1 x 10-5 kg m s-1?

3

𝜆 =ℎ

𝑚v

=6.626 × 10−34

0.00045 × 2000

= 7.36 × 10−34 m

Use De Broglie’s equation to solve for wavelength

∆(𝑚v)∆𝑥 ≥ℎ

4𝜋

(1 × 10−5)(∆𝑥) ≥6.626 × 10−34

4𝜋

∆𝑥 ≥ 5.27 × 10−30 m

Then use heisenberg’s uncertainty principle to solve for the error in position, ∆𝑥

1.4 Periodic Table Trends

Trends in the Periodic Table can explain reactivity of atoms and bonds present based on

electronegativity. Trends are related, such as electronegativity increases as atomic radius decreases, as

the attraction between the positively charged nucleus and negatively charged electrons is stronger with

a smaller radius.

Ionization Energy: Energy Required to remove the outermost electron (valence electron) from a

gaseous atom to form a cation.

Atomic Radius: Typical distance from the center of the nucleus to the outermost point of the

surrounding electron cloud (considers: nuclear charge increases as we move left to right, orbital size

increases as we move down).

1.4 Example (2014 Final):

4

Answer: C Cs is lower on the periodic table, so atomic radius is larger. There is less of an attraction between the nucleus and electron so it requires less energy to leave.

Bonding 2.1 Lewis Structures, Charges, and Resonance We use Lewis dot structures for simple molecules to show bonding electrons. If many structures can be

drawn, there is resonance and electron delocalization: the electrons shift between all the structures.

The resonant structure with the fewest and smallest formal charges is the most stable. The following

table can be used to determine the formal charge and Lewis structure for any molecule:

Valence State

Group

1 2 3 4 5 6 7 8

Normal 1, 0 2, 0 3, 0 4, 0 3, 1 2, 2 1, 3 0, 4

Hypervalence 5, 0 4, 1 3, 2 2, 3

6, 0 5, 1 4, 2

7, 0 6, 1 8, 0

The first number represents the number of bonds the atom can form. The second number represents the

number of lone pairs on an atom. The table won’t be provided on your exam but it’s easy to fill-in. Add 2

to the first number and subtract 1 from the second number as you move down the table. Stop when the

second number is zero.

2.1 Example :

Draw the Lewis structure for SO42-.

5

2.2 Valence Shell Electron Pair Repulsion (VSEPR) Theory VSEPR describes the 3D geometric shape of a molecule around an atom. The AXE naming convention is

used to describe the ligands on a central atom and its lone pairs. The central atom is given the letter ‘A’,

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while any attached ligands are given ‘X’ and lone pairs ‘E’. The steric number of an atom is the number

of attached ligands plus the number of electron pairs around it.

2.3 Hybridization and Valence Bond (VB) Theory Hybridization combines atomic orbitals to form identical mixed orbitals to make molecules symmetrical.

The number of atomic orbitals which combine will form the same amount of hybrid orbitals. Hybrid

orbitals are named after their components, (i.e. five sp3d hybrid orbitals are formed by one s, three p and

a d orbital).

Chemical bonds form when two orbitals overlap. Sigma bonds are head-on overlaps, while pi bonds are

side-on. All single bonds are sigma bonds, while all subsequent bonds (2nd and 3rd bonds) are pi.

2.3 Example:

Draw the Lewis structure(s) of HNO3 and describe its VSEPR shape. What hybridization does the central

N have? Describe its bonding with the oxygen atoms.

This offers resonance stability between the non-acidic oxygens, has the fewest and smallest charges and is a symmetrical arrangement.

Trigonal planar N is the central atom, with 3 bonds and no lone pairs (AX3).

Sp2 hybridization, with 3 sigma bonds and 1 pi bond

3 ligands means 3 identical bonds are needed (sp2), with the remaining p orbital in the pi bond.

Bonding MC Examples (2015 Final)

Answer: B First draw structure to answer question.

7

18. B NOT A - Double bonds have 1 sigma and 1 pi bond NOT C – sigma bonds are endwise overlap, pi bonds form from orbitals side by side NOT D – pi bonds can exist not in hybrid orbitals NOT E – sigma bonds can only form in s orbitals

2.4 Molecular Orbital (MO) Theory MO Theory assumes no hybridization occurs. Whenever two atomic orbitals overlap, they form a

bonding molecular orbital which is lower in energy and an antibonding (*) orbital which is higher.

Electrons are still filled into these new orbitals via Aufbau. σ and π antibonds look similar to their bonding

orbitals, however they contain a node between the atoms which destabilizes them.

Bond Order =# e− in Bonds − # 𝑒− in Antibonds

2

Paramagnetic compounds have unpaired electrons in their MOs while diamagnetic compounds have all

paired electrons.

8

2.4 Example:

Draw the MO diagram for OF. What is its bond order? Is it paramagnetic?

Bond order = 4bonds – 2.5 anti-bonds = 1.5 How is 1.5 possible? OF has a free radical and accounts for the ambiguous bond order This molecule is paramagnetic due to the lone electron in the pi anti-bond.

When more than two atoms are bonded together, nonbonding (n) molecular orbitals can form which

are at the average energy of all participating atomic orbitals and don’t contribute towards the bond

order.

2.5 Band Theory Conductors allow electrons to move freely on their surface while semiconductors allow only some

motion of electrons. Insulators have limited electron motion across them.

Bands occur when atomic orbitals are packed so closely in energy they merge to form a “continuous”

orbital allowing electrons to transition between atoms. The difference in energy between two bands is

its band gap. The valence band is the highest energy band which has electrons, while the lowest energy

empty band is the conduction band.

Semiconductors and insulators require heat to promote electrons across the band gap in order to begin

conduction. Doping adds impurities to reduce this gap: n-doping uses atoms with extra electrons, raising

the valence band while p-doping uses atoms with fewer electrons, lowering the conduction band.

Band theory can be understood by looking at the energy versus orbital diagram shown below.

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2.5 Example:

Si is a common semiconductor. What elements could Si be doped with to form a p-type or n-type

semiconductor? If the bandgap of an n-type Si semiconductor is 100 kJ/mol e-, what wavelength of light

would you expect it to give off in an LED?

Phosphorus will form an n-type Aluminum will form a p-type

Si has 4 valence electrons. To form an n-type, you need an element with 5; to form a p-type you need an element with 3. The atoms should also be similar in size (similar atomic number)

𝜆 =ℎ𝑐

𝐸

=(6.626 × 10−34)(3 × 108)

100000× (6.02 × 1023)

= 119 nm

Start with 𝐸 = ℎ𝑓 and 𝑓 =𝑐

𝜆 to rearrange for λ.

We scale by 6.02x1023 since the energy is given as per-mole, not per-photon This would be in the high-energy UV

2.5 MC Example (2014 Final):

Answer: C: n-type Phosphorus has more electrons.

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2.6 Bond Classification, Energies and Dipole moments Using the average energy needed to break/form a bond, the enthalpy of a reaction can be approximated:

∆𝐻𝑟𝑥𝑛 =∑∆𝐻(Bonds Broken) −∑∆𝐻(Bonds Formed)

In a bond, one atom can be more electronegative than the other, causing it to become more negative.

In polar bonds, this results in the two atoms having partial charges (𝛿). A dipole forms from the charge

separation in polar and ionic bonds, shown as an arrow pointing from positive to negative. The electric

dipole moment (μ) takes into account the size of the charge and the distance:

∆EN >2 2> ∆EN >0.4 <0.4

Bond Type Ionic Polar Covalent Non-Polar Covalent

H𝛿+ �⃑⃑� → Cl𝛿− 𝜇 = 𝛿 ∙ �⃑�

For more complex molecules, the overall μ is the sum of the moments across each bond. Therefore, it’s

possible to have a molecule with polar bonds but is overall non-polar.


Given the bond enthalpies C-C (348), C=O (707), O=O (498), H-O (464), C-H (414) in kJ/mol, compute ∆Ho

kJ/mol for the complete combustion reaction: 𝐶7𝐻16 + 11𝑂2 → 7𝐶𝑂2 + 8𝐻2𝑂. Does CO2 or H2O have a

dipole?

Reactants: 11 O = O 6 C − C 16 C − HProducts: 14 C = O 16 O − H

∆𝐻 = 𝐵𝑜𝑛𝑑𝑠 𝐵𝑟𝑜𝑘𝑒𝑛 − 𝐵𝑜𝑛𝑑𝑠 𝐹𝑜𝑟𝑚𝑒𝑑

Count up all the bonds broken and bonds formed, then use the equation for enthalpy

∆𝐻 = (11(498) + 6(348) + 16(414))− (14(707) + 16(464))

= −3132 kj/mol

No dipole for CO2, there is one for H2O

CO2 does not have a dipole even with its polar bonds because it has a linear shape (cancels); while H2O is bent, allowing the dipoles to add

2.6 MC Examples (2015 Final)

Answer: D

11

NaCl is a salt (comprised of ionic bonds) and can be torn apart by a polar solvent

because the dipoles present in the solvent attract the cation and anion away from one

another causing the molecule to dissolve. Can be summed-up as “like dissolves

like”.

States of Matter 3.1 Solids Solids are characterized by their regularity – either crystalline or amorphous; bonding as ionic, covalent,

metallic, or intermolecular; and their dimensionality – lines (i.e. oil), sheets (i.e. graphene), or networks

(i.e. diamond). A lattice is a series of points in space with a repeatable unit. Their coordination number

is the number of neighbors each atom has.

Crystal Structure Simple Cubic Body Centered Cubic

Face Centered Cubic

Hexagonal

Atoms/Cell 1 2 4 2

Coordination # 6 8 12 12

Simple Cubic Body Centered Cubic Face Centered Cubic Hexagonal

Density =#Atoms per Cell

Volume of Cell∙

𝑀𝑎𝑡𝑜𝑚6.02 × 1023

Face-centered and hexagonal structures are closest packing (i.e. most dense). In ionic lattices, typically

one ion will be larger than the other and form the structure while the smaller ion will fill the holes.


Vanadium has a body centered cubic cell structure with a unit cell dimension of 305pm. What is the

atomic radius of Vanadium? What is its density?

(large diagonal) = 4𝑟

= √3(side length)

𝑟 =√3

4(305)

= 132.07 pm

For a BCC cell, the large diagonal (i.e. from lower left to upper right) contains 4 of the atomic radii Since it is a cube, the large diagonal can be found:

𝑐2 = 3𝑎2

The density is expressed as g/cm3

The density is found using the molar mass

12

𝐷 =2 ∗ 50.94

(6.022 ∗ 1023)(305 ∗ 10−10)3

= 5.96g

cm3

2 atoms of V fit into 1 unit BCC cell; therefore, 2 moles of V fit into 2 moles of the volume

3.2 Solid-Liquid-Gas Equilibria and Phase Changes The vapor pressure of a substance is the equilibrium pressure between its liquid/solid and gas phases

related via Clausius Clapeyron:

ln (𝑃2𝑃1) = −

∆𝐻𝑣𝑎𝑝0

𝑅(1

𝑇2−1

𝑇1)

At the boiling point of a liquid, its vapor pressure equals atmospheric pressure. During a phase change,

energy added goes to breaking IMFs rather than increasing temperature.

Phase Diagrams are graphs between two state variables that describe the phase of a compound at each

combination. Crossing a line in a process indicates a phase change.

The triple point occurs where all three phases coexist in equilibrium. The critical point occurs where any

higher T or P causes the substance to become a supercritical fluid – a gas/liquid hybrid where phase

changes between the two can occur without a distinct shift in properties.

3.2 Example:

The normal boiling point of acetone is 56oC and the molar heat of vaporization is 32.0 kJ/mol. What

temperature will acetone boil at under a pressure of 0.38atm? If 1 mol of liquid acetone inside a 1L tank

at 31oC is depressurized from 1atm to 0.38atm isothermally, how much work was performed? Assume

the volume of the liquid and the vapor pressure of acetone at 31oC and 1atm is negligible.

ln (𝑃2𝑃1) = −

∆𝐻𝑣𝑎𝑝0

𝑅(1

𝑇2−1

𝑇1)

ln (0.38

1) = −

32000

8.3145(1

𝑇2−

1

56 + 273)

𝑇2 = 303.9 K

Rearranging clausius clapeyron to solve for T2

13

𝑤1 = −𝑃𝑒𝑥𝑡∆𝑉

𝑃1𝑉1 = 0.38𝑉2 𝑉2 = 2.63

𝑤1 = −0.38(2.63 − 1) = −0.62 J

The container expands isothermally irreversibly to 0.38atm (step 1), then maintains that pressure as the acetone evaporates (step 2). We find the ending volume of step 1 using Ideal gas law

𝑤2 = −𝑞2 = −∆𝐻𝑣𝑎𝑝

= −32000 J

𝑤𝑡𝑜𝑡𝑎𝑙 = −32000.62 J

Since the phase change is isothermal and isobaric, q = -w = ∆H In this case, Cp∆T is not applicable since it is a phase change, but the ∆Hvap is given

3.3 Intermolecular Forces (IMFs) London Dispersion Forces are the weakest IMF. They are caused by spontaneous temporary dipoles

formed as electrons move. Dipole-Dipole Forces are stronger but occur only between molecules with

permanent dipoles. Hydrogen Bonds are the strongest IMF and only occur between compounds with

hydrogens bonded to N, O, or F.

4. Gases Gases fill their containers and exert a pressure (P). The partial pressure (ρ) of one component is the

fraction of the total pressure contributed by that component.

𝑃𝑡𝑜𝑡 =∑𝜌𝑖 𝜌𝑖 =𝑛𝑖𝑛𝑡𝑜𝑡𝑎𝑙

𝑅𝑇

𝑉=

𝑛𝑖𝑛𝑡𝑜𝑡𝑎𝑙

𝑃𝑡𝑜𝑡

The Ideal gas law assumes the size of gas molecules is negligible compared to the distance between

molecules, travel in constant random motion and do not have any intermolecular forces between them.

𝑃𝑉 = 𝑛𝑅𝑇𝑃1𝑉1𝑃2𝑉2

=𝑛1𝑇1𝑛2𝑇2

The Van Der Waals Equation corrects for the gas molecule size assumption by introducing an ‘a’ and ‘b’

factor unique to each gas:

(𝑃 +𝑎𝑛2

𝑉2) (𝑉 − 𝑏𝑛) = 𝑛𝑅𝑇

4.1 Example:

A sealed container at 298K initially has 5 mol O2 and 1 mol S8. If the S8 is completely burned to form SO2,

but the pressure inside the container is constant at P, how much did the temperature change by? What

is the partial pressure of SO2 in the container?

The reaction equation is:

𝑂2 +1

8𝑆8 → 𝑆𝑂2

Therefore, O2 is limiting

To consume 1mol S8 you need at least 8mol O2. Since you have less, O2 is limiting

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At completion, products are 3/8 S8 + 5 SO2

𝑃1𝑉1𝑃2𝑉2

= 1 =𝑛1𝑇1𝑛2𝑇2

1 =6(298)

5.375(𝑇2)

𝑇2 = 332.65 K

Using the ideal gas law as a ratio, solve for T2

It equals 1 since PV doesn’t change

𝜌𝑆𝑂2 =𝑛𝑆𝑂2𝑛𝑡𝑜𝑡

𝑃

=5

5.375𝑃

= 0.93𝑃

Solve for partial pressures using the mole fraction

4.2 Example:

If you have a mole of a real gas in which there are substantial intermolecular forces between the gas

molecules in a 22.4L container at 273K will the pressure be?

1. 1 atomsphere

2. Greater than 1 atmosphere

3. Less than 1 atmosphere

According to the Van der Waals theory of gases, the pressure would be greater than 1 atmosphere due

to the substantial IMFs even though the ideal gas law says it would be 1 atm.

APSC 131 Final Exam Workshop Solutions

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