CHEM 120 — Test 2 Crash Course

Everything on the test, explained from scratch with worked examples.

Wed Nov 11 · 45 min · 17 multiple-choice questions · bubble sheet.
Module 2 → 6 Q · Module 3 → 6 Q · Module 4 → 5 Q (Lewis & resonance only).
The periodic table + the key equations are printed on your data sheet — you don't memorize those.
Read this first. You don't need to be a genius — this test is really about 8 repeating question types. Learn how to do each one with the worked examples below and you've basically got it. Do them with a pen, not just your eyes. You have time. 💪

Module 2 · Quantum & the H Atom

6 questions · focus is Bohr onward (§2.4+)

The one big idea

Electrons in atoms can only have certain allowed energies — not just any value. This is called quantization. Everything in this module is a consequence of that. Light comes in packets (photons) of energy E = hν, and when an electron jumps between allowed energy levels, it absorbs or emits exactly one photon.

The 3 experiments that proved it (know what each showed)

  • Blackbody radiation (Planck): energy comes in packets, E = n·hν.
  • Photoelectric effect (Einstein): light is photons; electrons only pop off a metal if the light's frequency is high enough (not if it's just brighter).
  • Line spectra: atoms give off only specific colours → electron energies are quantized.

1. Electron transitions (the Bohr calculation)

When an electron moves between levels n, the energy change is ΔE = −R_H(1/n_f² − 1/n_i²) (this formula is on your data sheet; R_H = 2.179×10⁻¹⁸ J).

  • Electron drops to a lower level → emits a photon → ΔE is negative.
  • Electron jumps up → absorbs a photon → ΔE is positive.
Worked example

An electron in H falls from n = 4 to n = 2. Absorbed or emitted, and how much energy?

  1. Plug in: ΔE = −2.179×10⁻¹⁸ × (1/2² − 1/4²) = −2.179×10⁻¹⁸ × (0.25 − 0.0625)
  2. = −2.179×10⁻¹⁸ × 0.1875 = −4.09×10⁻¹⁹ J
  3. It's negative → the electron dropped → a photon is emitted.
Answer: emitted, 4.09×10⁻¹⁹ J. (For wavelength you'd use λ = hc/|ΔE|.)
On the test it looks like: "An electron falls from n=5 to n=2… is light absorbed or emitted / what is ΔE or λ?" Same steps every time.

2. Quantum numbers (the address of an electron)

Every electron has four quantum numbers — think of them as an address:

SymbolNameWhat it setsAllowed values
nprincipalsize / energy level1, 2, 3, …
ℓangularshape (0=s, 1=p, 2=d, 3=f)0 up to n−1
mₗmagneticorientation−ℓ … 0 … +ℓ
mₛspinspin up/down+½ or −½

The key rule: ℓ must be less than n, and mₗ can't be bigger than ℓ.

Worked example

Which set (n, ℓ, mₗ, mₛ) is not allowed?  (2, 2, 0, +½)

  1. Here n = 2, so ℓ can only be 0 or 1 (must be ≤ n−1 = 1).
  2. But ℓ = 2 is written → impossible.
Answer: (2, 2, 0, +½) is invalid — ℓ can't equal n.

3. Orbitals & nodes (counting)

An orbital is a region where an electron is likely to be. Orbitals have nodes — spots where the electron is never found. Two kinds, each with a formula:

CountFormula
# orbitals in level nn²
angular nodesℓ
radial nodesn − ℓ − 1
total nodesn − 1
Worked example

How many radial and angular nodes does a 4d orbital have?

  1. 4d means n = 4 and ℓ = 2 (d = 2).
  2. Angular nodes = ℓ = 2.
  3. Radial nodes = n − ℓ − 1 = 4 − 2 − 1 = 1.
Answer: 1 radial, 2 angular.
The #1 trap: people swap the two rules. Lock it in: Angular = ℓ (the letter). Radial = n − ℓ − 1 (the longer formula).

4. Reading orbital graphs

They show you a plot of R(r) vs r (wave amplitude vs distance from the nucleus) and ask which orbital it is. Two clues:

  • Where it starts at r = 0: starts at a peak → it's an s orbital. Starts at zero → it's p, d, or f.
  • How many times it crosses the x-axis = the number of radial nodes.
Worked example

A plot starts at a maximum at r = 0 and crosses the axis twice. Which orbital?

  1. Starts at a peak → s orbital → ℓ = 0.
  2. Crosses twice → 2 radial nodes. Use radial = n − ℓ − 1 → 2 = n − 0 − 1 → n = 3.
Answer: 3s.

Also know: the "+" and "−" (or two colours) on orbital lobes show the phase of the wavefunction — not electric charge. And for hydrogen's 1s, the most likely distance of the electron is a₀ = 52.9 pm (memorize this number — it's not on the data sheet).

Module 3 · Electron Configs & Periodic Trends

6 questions · all of it is fair game

5. Writing electron configurations

You fill orbitals from lowest energy up. Memorize this filling order:

1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s …

Three rules: Pauli (max 2 electrons per orbital, opposite spins), Aufbau (fill lowest first), Hund (in a set like the three p orbitals, put one electron in each — all same spin — before doubling up).

How many electrons per sublevel: s = 2, p = 6, d = 10, f = 14. Valence electrons = group number (groups 1–2) or group number − 10 (groups 13–18).

Worked example

Write the configuration of iron (Fe, 26 electrons) and count unpaired electrons.

  1. Fill to 26: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶ → short form [Ar] 4s² 3d⁶.
  2. The 3d⁶ fills by Hund: ↑↓ ↑ ↑ ↑ ↑ (one orbital doubled, four singles).
  3. That's 4 unpaired electrons.
Answer: [Ar] 4s² 3d⁶, 4 unpaired.
Two exceptions to memorize: Cr = [Ar] 4s¹ 3d⁵ and Cu = [Ar] 4s¹ 3d¹⁰ (they "borrow" one electron from 4s). Don't overthink why — just know them.

6. Ions (the biggest trap in the course)

For a + ion (cation): write the neutral atom first, then remove electrons from the highest n first. For transition metals that means the 4s electrons leave before the 3d — even though 4s filled first. For a − ion (anion): just keep filling.

Worked example

Write the configuration of Mn²⁺ (manganese has 25 electrons).

  1. Neutral Mn = [Ar] 4s² 3d⁵.
  2. It's 2+, so remove 2 electrons — from the highest n first, which is 4s (n=4), not 3d (n=3).
  3. Remove both 4s electrons → [Ar] 3d⁵.
Answer: [Ar] 3d⁵.
The trap: if you "just refill" for 23 electrons you'd get [Ar] 4s² 3d³ — that's wrong (it's actually neutral vanadium). Always build the neutral atom, then strip electrons off the top (highest n).

7. Magnetism (it's just counting)

Paramagnetic = has at least one unpaired electron. Diamagnetic = every electron is paired. So the whole question is: write the config, count unpaired electrons. Zero unpaired → diamagnetic.

Worked example

Is Zn paramagnetic or diamagnetic? Zn = [Ar] 4s² 3d¹⁰.

  1. Every sublevel is completely full → all electrons paired.
Answer: diamagnetic.

8. Periodic trends

Four properties, and they all point the same way except atomic size:

PropertyAcross a period →Down a group ↓
Atomic radius (size)gets smallergets bigger
Ionization energy (IE)goes up*goes down
Electron affinitygoes up (bigger magnitude)goes down
Electronegativity (EN)goes upgoes down

Memory hook: IE, EA, and EN all get biggest at the top-right (fluorine, EN = 4.0). Size is the opposite — biggest bottom-left.

* The two IE "dips" (a favourite trick question)

Ionization energy generally rises across a period, but it dips twice:

  • Group 2 → 13 (e.g. Be → B): boron's outer electron is a higher-energy 2p, so it's easier to pull off.
  • Group 15 → 16 (e.g. N → O): nitrogen's 2p is exactly half-full (very stable, all unpaired); oxygen has one paired 2p electron that repels its partner, so it's easier to remove.
Worked example

Rank Na, Mg, Al, Si by ionization energy, lowest to highest.

  1. General rule says Na < Mg < Al < Si… but the Mg→Al dip flips Mg and Al.
  2. Real order: Na < Al < Mg < Si.
Answer: Na < Al < Mg < Si.

Module 4 · Lewis Structures & Resonance

5 questions · pages 1–14 only (no VSEPR, no shapes, no polarity)

9. Count the valence electrons

Before drawing anything, add up all the valence electrons: sum each atom's valence electrons, then subtract the charge (a − charge means you add electrons).

Worked example

How many valence electrons in carbonate, CO₃²⁻?

  1. C is group 14 → 4. Each O is group 16 → 6, and there are 3 → 18.
  2. The 2− charge adds 2 more.
  3. Total = 4 + 18 + 2 = 24.
Answer: 24 valence electrons.

10. Draw the Lewis structure

Steps: (1) count valence electrons, (2) put the least-electronegative atom in the centre (never H), (3) single-bond everything, (4) fill the outer atoms' octets, (5) put leftovers on the centre, (6) if the centre is short, make double/triple bonds.

Rules to remember:

  • H only ever has 2 electrons (one bond), always on the outside.
  • C, N, O, F (period 2) never go past 8 electrons (no expanded octet).
  • Period 3 and below (S, P, Cl, Br, I, Xe) can hold more than 8 — called an expanded octet.
Worked example — "which atom has an expanded octet?"

In XeF₄, xenon is bonded to 4 F and still has 2 lone pairs = 12 electrons around it. Xe is in period 5, so it's allowed.

Answer: Xe in XeF₄ has an expanded octet. (C and N never can; B is electron-deficient, a different thing.)

11. Formal charge

Formal charge tells you if your structure is good. For an atom:

FC = (valence electrons) − (lone-pair electrons + ½ of bonding electrons)

Shortcut: FC = valence − (dots + bonds), counting each line as 1 bond. The best structure has the smallest formal charges, with any negative charge on the most electronegative atom.

Worked example

Formal charge on the central N in nitrate (NO₃⁻), drawn with one N=O and two N–O.

  1. N has 4 bonds total (one double + two single) and no lone pairs.
  2. FC = 5 − (0 lone-pair e⁻ + ½·8 bonding e⁻) = 5 − 4 = +1.
Answer: +1.

12. Resonance

Resonance = when you can draw the same molecule more than one way, differing only in where the double bond / lone pairs sit. The real molecule is the average of all of them (so, e.g., all three bonds in carbonate are identical).

  • You need at least one double or triple bond for resonance to exist (you move those π electrons; you never break a single σ bond).
  • If two versions are equally good, they contribute equally. If not, the major one puts the negative charge on the more electronegative atom.
Worked example

In cyanate OCN⁻, why is the structure with the −1 charge on O the major one (not the one with −1 on N)?

  1. Both structures have the same size of formal charge (−1), so that can't decide it.
  2. Oxygen is more electronegative than nitrogen, so it holds the negative charge better.
Answer: because O is more electronegative. (Watch the trap answer "smaller formal charge" — they're equal.)

Foundation you lean on

quick recap — not questions on their own, but you need them

Reading the periodic table (it's given to you)

Columns = groups (1–18), rows = periods. Blocks: s, p, d, f. Metals on the left, nonmetals upper-right. Atomic number = number of protons = number of electrons (for a neutral atom). Molar masses and atomic numbers are printed on your data sheet.

Light & energy (Module 2 setup)

c = λν (wave speed) and E = hν = hc/λ (photon energy) — both on the data sheet. Higher frequency → higher energy → shorter wavelength.

Electronegativity

How hard an atom pulls on shared electrons. Rises toward the top-right: F > O > N ≈ Cl. You use this to decide where a negative charge goes (Module 4).

Ionic vs molecular & polyatomic ions

Metal + nonmetal = ionic; two nonmetals = molecular (covalent). Know the common polyatomic ions by name/formula: nitrate NO₃⁻, sulfate SO₄²⁻, carbonate CO₃²⁻, phosphate PO₄³⁻, ammonium NH₄⁺, hydroxide OH⁻, acetate C₂H₃O₂⁻.

The night-before game plan

  1. Read each worked example above with a pen — redo the arithmetic yourself, don't just nod along.
  2. Do the practice test (the 17-question one) closed-book. Mark it. For every miss, come back to the matching section here.
  3. Do the professor's own practice on LEARN — "Review-Material for Test 2" has a solutions file. That's the closest thing to the real questions.
  4. Tomorrow morning: glance at the Last-Look sheet, then stop. Walk in knowing the 8 question types.
  5. Sleep. Seriously — a rested brain beats one more hour of cramming on a test like this.

You've got the whole blueprint and every question type is a repeatable recipe. You can absolutely do this. 🧪

Your other Test 2 materials