Midterm Exam Review

Overview

John Flood reviews the midterm’s cumulative chemistry topics, noting that the in-person class average rose from 74% to 78% while the online section averaged 69%, and that more than half the exam should cover Units 3 to 5. The review connects dimensional analysis and solution concentrations with electron structure, molecular geometry, bonding, nomenclature, and photon energy, using worked examples such as 300 mL of 2.1 M iron(III) sulfite and an electron transition from n = 2 to n = 6.

Key takeaways

Treating molarity as moles per liter makes the dimensional-analysis path explicit: volume gives moles of compound, then the formula’s subscripts give moles of a specific ion.
A 1.5 M NaNO₃ solution is 1.5 M in nitrate ions but 3.0 M in total ions, because each formula unit separates into one sodium ion and one nitrate ion.
For dilute aqueous solutions, 150 ppm is approximately 150 mg/L, so a 4.6 L sample contains 690 mg of solute.
Count electron regions on an atom to predict hybridization and electron geometry: four regions correspond to sp³ and tetrahedral geometry, while three correspond to sp² and trigonal planar geometry.
Use Lewis structures to count bonds: every bonded atom pair has one sigma bond, and additional bonds in a multiple bond are pi bonds formed by unhybridized p orbitals.
For empirical formulas from percent composition, assume a 100 g sample, convert each element’s mass to moles, and reduce the mole amounts to the smallest whole-number ratio.

Chapters

0:00 Midterm Preparation and Unit-Test Results

John Flood reports that the in-person class average increased from 74% on the previous unit test to 78% on the latest one.
The online section averaged 69%, and Flood suggests that keeping pace and studying with classmates may help address gaps.
Students are encouraged to use the study center, STEM center, and discussion boards to study together.

3:08 Building the Review List from Cumulative Exam Topics

Students request review on ion counts, hybridization, photon energy, electron configurations, mole ratios, sigma and pi bonds, percent conversions, subshells, and electron transitions.
Flood emphasizes that earlier nomenclature matters because later questions depend on recognizing compound and geometry names.
The exam review list highlights that material from different units increasingly overlaps.

15:24 Midterm Format, Timing, and Unit Emphasis

Flood says the midterm resembles the multiple-choice unit tests in question style and format, but allows three hours instead of two.
The exam is expected to emphasize Units 3 to 5, which cover more modules and contribute more connected material.
Unit 2 nomenclature remains relevant across later questions, while Unit 1 is expected to contribute fewer broad questions.

17:27 Molarity as a Road Map from Volume to Moles

Flood starts the Unit 5 review by treating molarity as moles per liter, not as a standalone number.
To find moles from molarity, begin with solution volume and use molarity as a conversion factor.
A final answer about a particular ion may require an additional mole-to-mole ratio from the compound formula.

19:00 Counting Nitrate and Total Ions in Sodium Nitrate

A 1.5 M NaNO₃ solution contains 1.5 M nitrate because each formula unit supplies one nitrate ion.
Each NaNO₃ formula unit produces two ions, one Na⁺ and one NO₃⁻, so total ion concentration is 3.0 M.
Flood distinguishes a polyatomic ion such as nitrate, which remains one ion in solution, from the individual ions produced by an ionic compound.

25:24 Iron(III) Sulfite: Specific-Ion Moles and Total-Ion Counts

Flood corrects a worksheet problem by adding the missing volume: 300 mL of 2.1 M Fe₂(SO₃)₃.
The formula gives a 3:1 ratio of sulfite ions to iron(III) sulfite formula units; dimensional analysis yields 1.89 mol SO₃²⁻.
For total ions, each formula unit produces two Fe³⁺ ions and three sulfite ions, or five ions per formula unit.

34:20 Solving Dilution Problems with M₁V₁ = M₂V₂

Flood explains that molarity multiplied by volume represents moles, which remain constant when a solution is diluted.
To prepare 200 mL of 0.1 M HCl from 1.0 M HCl, use 20 mL of the original solution.
Diluting 50 mL to 150 mL is a threefold dilution; a final concentration of 0.15 M means the initial concentration was 0.45 M.

40:57 Converting Aqueous ppm to Milligrams of Solute

For dilute aqueous solutions, Flood uses the practical conversion 1 ppm ≈ 1 mg/L.
A 4.6 L sample at 150 ppm contains 690 mg of pollutant, found by multiplying 150 mg/L by 4.6 L.
The mass-fraction definition of ppm requires dividing by 10⁶; the mg/L shortcut avoids unnecessary density conversions for water.

50:23 Drawing Water’s Lewis Structure Before Predicting Shape

H₂O has eight valence electrons: six from oxygen and one from each hydrogen.
Oxygen is the central atom, forms two single bonds, and holds two lone pairs; hydrogen follows a duet rather than an octet.
The Lewis structure provides the electron regions needed to determine oxygen’s hybridization and molecular geometry.

56:00 Why Four Electron Regions Correspond to sp³ Hybridization

Oxygen in water has four electron regions, two bonding pairs and two lone pairs, which correspond to four hybrid orbitals.
Flood explains hybridization as a model for arranging electron regions farther apart than the 90-degree orientations of unhybridized p orbitals.
Four regions map to sp³ orbitals and a tetrahedral electron geometry; the ideal angle is 109.5 degrees, while water’s bond angle is about 105 degrees.

1:02:40 Water’s Bent Molecular Shape and Sigma Bonds

Water has tetrahedral electron geometry but a bent molecular geometry because its two lone pairs are excluded when naming the molecular shape.
Flood cautions that a flat Lewis-structure drawing can misrepresent a molecule’s three-dimensional arrangement.
Each O–H single bond is one sigma bond, so H₂O contains two sigma bonds and no pi bonds.

1:07:35 Carbonic Acid Geometry and Carbon Hybridization

In H₂CO₃, carbon has three electron regions, giving it trigonal planar electron and molecular geometry.
Three regions correspond to sp² hybridization; the notation counts one s orbital and two p orbitals.
Flood connects hybridization names to geometry: sp² is associated with trigonal planar arrangements, while sp³ is associated with tetrahedral arrangements.

1:11:50 Counting Sigma and Pi Bonds in Carbonic Acid

Each bonding region around carbon contributes a sigma bond, and the double bond adds one pi bond from overlapping unhybridized p orbitals.
The complete carbonic acid structure contains five sigma bonds and one pi bond.
In the course-level model, an sp² atom retains one unhybridized p orbital, while an sp atom can retain two for pi bonding.

1:17:54 Subshells, Orbitals, and the First Two Energy Levels

The first shell, n = 1, has only an s subshell and one orbital.
The second shell has s and p subshells; the p subshell contains three orbitals with different spatial orientations.
Each orbital can hold an electron pair, and the s and p orbitals differ in shape and nodal structure.

1:19:30 Electron Configurations, Sulfide, and Chromium’s Exception

Flood reviews the Aufbau principle, the Pauli exclusion principle, and Hund’s rule as guides for filling orbital diagrams.
S²⁻ has 18 electrons and the configuration 1s² 2s² 2p⁶ 3s² 3p⁶, matching argon.
Chromium is an exception to the simplest filling pattern: its configuration is [Ar] 3d⁵ 4s¹, associated with the stability of a half-filled d subshell.

1:25:45 Quantum Numbers n, l, mₗ, and mₛ

The principal quantum number n is a positive integer; the angular momentum number l ranges from 0 to n − 1.
Subshell labels correspond to l values: s = 0, p = 1, d = 2, and f = 3.
The magnetic quantum number mₗ ranges from −l to +l, while mₛ is +½ or −½ for the two electron spin states.

1:34:55 Acid Nomenclature Review Begins

Flood shifts to naming acids and asks students to distinguish oxyacids from binary acids.
The distinction depends on whether the acid contains oxygen, which changes the naming pattern.
Acid names are revisited later with examples including sulfurous, nitric, and hydroiodic acid.

1:36:24 Photon Energy from a 424-Nanometer Wavelength

Flood presents the photon equations E = hν and E = hc/λ, using a wavelength of 424 nm.
The speed of light must be used with compatible units; convert nanometers to meters or convert meters per second to nanometers per second.
The constant h has units of joule-seconds, and carrying units through the calculation helps verify that the answer is in joules.

1:43:32 Connecting Electron-Transition Energy to Photon Wavelength

An electron transition’s energy change is related to the energy of the photon absorbed or emitted.
Once the transition energy is known, rearrange E = hc/λ to obtain λ = hc/E.
Flood notes that transition questions may ask for energy, wavelength, or a principal quantum number, so the equation may need algebraic rearrangement.

1:46:42 Why Percent Composition Can Be Treated as Grams

For an empirical-formula problem with percentages but no sample mass, assume a 100 g sample so each percentage becomes the same number of grams.
Convert those masses to moles, then divide by the smallest mole amount to find the lowest whole-number ratio.
A different assumed sample mass, such as 89 g, gives the same composition ratios but requires more arithmetic.

1:50:17 Naming Oxyacids and Binary Acids

Examples include H₂SO₃ as sulfurous acid, HNO₃ as nitric acid, and H₂CrO₄ as chromic acid.
Binary acids such as HI and H₂S use the hydro- prefix, giving hydroiodic acid and hydrosulfuric acid.
Flood distinguishes oxyacid naming from binary-acid naming as a key nomenclature rule.

1:51:44 Worked n = 2 to n = 6 Transition and Photon Count Scaling

Flood substitutes initial and final energy levels into the hydrogen transition equation, then uses λ = hc/|ΔE| to find a wavelength.
The classroom calculation reports 489 nm, though the displayed arithmetic is inconsistent with the stated transition values.
Energy scales with photon count: multiply energy per photon by the number of photons, or by Avogadro’s number to obtain energy per mole.

1:57:27 Water, Acids, and the Limits of Simple Naming Rules

Flood notes that whether water acts as an acid depends on the acid-base definition and reaction context.
Most introductory inorganic acids begin with H, but acetic acid is an exception in how its formula is written.
The review ends by cautioning students not to treat introductory acid-pattern rules as universal.

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, John Flood Chemistry.