CHEM 105 Comprehensive Exam Review & Ionic Compound Drill

⚡ Interactive Ionic Compounds Drill Engine

Master charges, IUPAC names, and formulas across 76 presentation compounds.

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Compound 1 of 76

GIVEN CHEMICAL FORMULA:

Fe(HSO₃)₂

Category: Type II Polyatomic • Slide 40

Cation & Charge

Anion & Charge

IUPAC Systematic Name

Chemical Formula

💡 Hint: Naming & Charge Balancing Approach

Cations: Group 1 (+1), Group 2 (+2), Group 13 Al (+3). For transition metals, calculate the Roman numeral from the total negative charge of the anions so the sum equals 0!
Exceptions: Silver ($Ag^+$), Zinc ($Zn^{2+}$), and Cadmium ($Cd^{2+}$) have invariable charges; Roman numerals are not strictly required for them.
Polyatomic Ions: Look out for oxygen count suffixes: -ate has more oxygens than -ite. The bi- prefix indicates an added $H^+$ (e.g. $HCO_3^-$ bicarbonate, $HSO_3^-$ bisulfite).

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📖 Browse All 76 Presentation Compounds (Master Reference Table)

#	Formula	Cation	Anion	Systematic Name	Type

SPECIAL REFERENCE • SLIDE 39 Polyatomic Study List

Polyatomic Ions Mastery Drill

Slide 39 of the presentation lists 27 essential polyatomic ions. Test your recall of charges, chemical formulas, and names below:

Nitrate

Formula

Net Charge

📝 Full Polyatomic Ion Master List (Slide 39)

Name	Formula	Charge	Structural & Chemical Notes

1. Chemical Bonding & Electronegativity (ΔEN)

Bond Dissociation Energy, ΔEN Scale, Bond Polarity • Slides 5–16

QUESTION 1.1 Bond Classification (ΔEN)

According to Slide 12 of the presentation, bonds are classified by difference in electronegativity (ΔEN):
• Nonpolar Covalent: ΔEN = 0.0 to 0.4
• Polar Covalent: ΔEN = 0.4 to 1.8
• Ionic: ΔEN = 1.8 to 3.3

Slide 15 asks you to predict whether the following bond is ionic or covalent: K – O (Potassium with Oxygen; EN values: K = 0.8, O = 3.5).

Calculate ΔEN

Bond Classification

💡 Hint: Formulas & Approach

$$\Delta EN = |EN_O - EN_K|$$ Compare the resulting difference against the cutoff of 1.8. Metal + Non-metal with large ΔEN results in electron transfer rather than sharing.

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1. Calculate difference in electronegativity: $$\Delta EN = 3.5 - 0.8 = 2.7$$ 2. Evaluate against Slide 12 scale: Since $\Delta EN = 2.7 > 1.8$, the electron density is transferred almost completely to oxygen.

Classification: Ionic bond (K⁺ and O²⁻ in lattice arrangement).

QUESTION 1.2 Increasing Polarity Ranking

Slide 16 presents the exercise: "Arrange these bonds in order of increasing polarity: O–H, C–H, N–H, S–H, C–N, C–O."
(Paulings: H = 2.1, C = 2.5, S = 2.5, N = 3.0, O = 3.5).
Which of the following represents the correct sequence from lowest ΔEN to highest ΔEN?

A) S–H < C–H < C–N < N–H < C–O < O–H B) C–H < S–H < C–N < N–H < C–O < O–H C) C–N < C–H < S–H < C–O < N–H < O–H D) O–H < C–O < N–H < C–N < C–H < S–H

💡 Hint: Formulas & Approach

Calculate $\Delta EN$ for each individual bond: $$\Delta EN(\text{C-H}) = |2.5 - 2.1| = 0.4$$ $$\Delta EN(\text{S-H}) = |2.5 - 2.1| = 0.4$$ $$\Delta EN(\text{C-N}) = |2.5 - 3.0| = 0.5$$ $$\Delta EN(\text{N-H}) = |3.0 - 2.1| = 0.9$$ $$\Delta EN(\text{C-O}) = |2.5 - 3.5| = 1.0$$ $$\Delta EN(\text{O-H}) = |3.5 - 2.1| = 1.4$$

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Comparing all calculated $\Delta EN$ values:
• $\text{C-H} = 0.4$ and $\text{S-H} = 0.4$ (tied as least polar)
• $\text{C-N} = 0.5$
• $\text{N-H} = 0.9$
• $\text{C-O} = 1.0$
• $\text{O-H} = 1.4$ (most polar)

Correct sequence: Option A or B (C–H / S–H < C–N < N–H < C–O < O–H).

2. Covalent Compounds, Hydrates & Acids/Bases

Binary Covalent Prefixes, Hydrate Nomenclature, Acid Rules • Slides 18–22, 42–46

QUESTION 2.1 Covalent Nomenclature

Apply the binary covalent nomenclature rules from Slides 19–22:
1. What is the systematic name for P₄O₆?
2. What is the chemical formula for Dinitrogen tetroxide?

Name for P₄O₆

Formula for Dinitrogen tetroxide

💡 Hint: Formulas & Approach

Greek prefixes: 1 = mono, 2 = di, 3 = tri, 4 = tetra, 5 = penta, 6 = hexa, 7 = hepta, 8 = octa, 9 = nona, 10 = deca. Do not use mono- on the first element. When tetra- or hexa- precedes oxide, the 'a' is typically dropped (hexoxide).

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1. P₄O₆: 4 phosphorus atoms = tetraphosphorus; 6 oxygen atoms = hexoxide (or hexaoxide).
→ Tetraphosphorus hexoxide

2. Dinitrogen tetroxide: di- indicates 2 nitrogens ($N_2$); tetr- indicates 4 oxygens ($O_4$).
→ N₂O₄

QUESTION 2.2 Hydrates & Acids Naming

From Slides 43–45:
1. Name the hydrate with formula MgSO₄ • 7H₂O (commonly known as Epsom salt).
2. Name the binary acid HCl(aq) versus the oxyacid HNO₃(aq).

Name for MgSO₄ • 7H₂O

Name for HNO₃

💡 Hint: Formulas & Approach

Hydrates: Name ionic compound first, then prefix + hydrate (7 = hepta). Acids: If anion ends in -ate (nitrate), acid name is (root)-ic acid. Binary acids without oxygen take hydro-(root)-ic acid.

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1. MgSO₄ • 7H₂O: Magnesium sulfate + 7 water molecules (heptahydrate).
→ Magnesium sulfate heptahydrate

2. HNO₃: Contains polyatomic ion nitrate ($NO_3^-$). Rule: -ate becomes -ic acid.
→ Nitric acid (whereas $HCl$ is hydrochloric acid).

3. Lewis Structures, Octet Rule & Formal Charge

Valence Counting, Octet Exceptions, Resonance, Formal Charge • Deck: Lewis Structures

QUESTION 3.1 Valence Counting & Octet Exceptions

Slides 10–14 discuss elements that deviate from the standard octet rule:
1. How many total valence electrons are in the electron-deficient molecule BF₃ (Boron trifluoride)? Does Boron obey the octet rule in this stable compound?
2. How many total valence electrons are in SF₆ (Sulfur hexafluoride)?

Valence e⁻ in BF₃

Valence e⁻ in SF₆

💡 Hint: Formulas & Approach

Boron is Group 13 (3 valence e⁻); Halogens are Group 17 (7 valence e⁻). Sulfur is Group 16 (6 valence e⁻). Sum: $V = V_{\text{central}} + \sum V_{\text{ligands}}$.

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1. BF₃: $$V = 1(3) + 3(7) = 3 + 21 = \mathbf{24\text{ valence electrons}}$$ Boron forms 3 single bonds with the 3 fluorines, using 6 electrons. It has an incomplete octet (6 electrons) and does not obey the octet rule.

2. SF₆: $$V = 1(6) + 6(7) = 6 + 42 = \mathbf{48\text{ valence electrons}}$$ Sulfur uses empty 3d orbitals to accommodate 6 bonds (12 shared electrons), forming an expanded octet.

QUESTION 3.2 Formal Charge Calculation

Slide 18 gives the governing formal charge equation: $$\text{Formal Charge} = (\text{Valence } e^-) - (\text{Non-bonding } e^-) - \frac{1}{2}(\text{Shared bonding } e^-)$$ Slide 20 evaluates the Lewis structure of POCl₃ where Phosphorus has 1 double bond to Oxygen and 3 single bonds to Chlorine:
• Oxygen has 1 double bond (4 bonding e⁻) and 2 lone pairs (4 non-bonding e⁻).
• Phosphorus has 5 bonds total (10 bonding e⁻) and 0 lone pairs.
Calculate the formal charge on Phosphorus ($P$) and Oxygen ($O$) in this structure.

Formal Charge on P

Formal Charge on O

💡 Hint: Formulas & Approach

Valence electrons: P = 5, O = 6. $$FC(P) = 5 - 0 - \frac{1}{2}(10) = 5 - 5$$ $$FC(O) = 6 - 4 - \frac{1}{2}(4) = 6 - 4 - 2$$

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1. Phosphorus (P): $$FC(P) = 5 - 0 - \frac{1}{2}(10) = 5 - 5 = \mathbf{0}$$ 2. Oxygen (O): $$FC(O) = 6 - 4 - \frac{1}{2}(4) = 6 - 6 = \mathbf{0}$$ 3. Chlorine (Cl): Each Cl has 1 single bond and 3 lone pairs: $$FC(Cl) = 7 - 6 - \frac{1}{2}(2) = 0$$
Because all atoms achieve a formal charge of 0, the structure with a $P=O$ double bond (expanded octet) is preferred over the single-bonded structure ($P^+ - O^-$).

4. VSEPR Theory: Geometries & Bond Angles

Electron Domains, Molecular Shapes, Lone Pair Repulsion • Deck: VSEPR.pptx

QUESTION 4.1 Lone Pair Compression

Slides 11–16 compare three molecules with 4 electron domains (steric number = 4):
• CH₄ (Methane): 4 bonding pairs, 0 lone pairs
• NH₃ (Ammonia): 3 bonding pairs, 1 lone pair
• H₂O (Water): 2 bonding pairs, 2 lone pairs

According to VSEPR theory, lone pairs occupy more space and exert stronger repulsions than bonding pairs. What are the experimentally observed bond angles for CH₄, NH₃, and H₂O?

CH₄ Bond Angle (°)

NH₃ Bond Angle (°)

H₂O Bond Angle (°)

💡 Hint: Formulas & Approach

Standard ideal tetrahedral angle is $109.5^\circ$. Each additional lone pair compresses the remaining bonding angles by approximately $2^\circ$ to $2.5^\circ$.

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• CH₄ (Tetrahedral): 0 lone pairs → Ideal tetrahedral angle = 109.5°.
• NH₃ (Trigonal pyramidal): 1 lone pair compresses bond angles to 107° (Slide 15).
• H₂O (Bent): 2 lone pairs exert strong LP-LP and LP-BP repulsions, compressing bond angles to 104.5° (Slide 16).

QUESTION 4.2 Expanded Octet Shapes

Slides 18–24 cover 5- and 6-electron domain molecules:
1. XeF₄ has 6 electron domains (4 single bonds, 2 lone pairs). What is its molecular geometry?
2. PCl₅ has 5 electron domains (5 single bonds, 0 lone pairs). What are its two characteristic bond angles?

XeF₄ Molecular Geometry

PCl₅ Molecular Geometry

💡 Hint: Formulas & Approach

For 6 domains with 2 lone pairs: To minimize repulsion, the 2 lone pairs position themselves $180^\circ$ apart (trans/axial), leaving the 4 bonding pairs in a flat plane.

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1. XeF₄: • Electron geometry: Octahedral (6 domains). • 2 lone pairs sit opposite each other (axial positions) at $180^\circ$. • Molecular geometry: Square planar (Slide 23).

2. PCl₅: • 5 electron domains, 0 lone pairs. • Molecular geometry: Trigonal bipyramid (Slide 18). • Angles: 120° (equatorial-equatorial) and 90° (axial-equatorial).

5. Valence Bond Theory & Orbital Hybridization

sp, sp², sp³, sp³d, sp³d², Sigma (σ) and Pi (π) Bonds • Deck: Hybridization.pptx

QUESTION 5.1 σ and π Bond Counting

Slides 15, 30, and 55 examine Ethylene (C₂H₄) and Acetylene (C₂H₂):
• A single bond consists of 1 σ bond.
• A double bond consists of 1 σ bond + 1 π bond.
• A triple bond consists of 1 σ bond + 2 π bonds.

In a molecule of Ethylene (H₂C=CH₂), calculate the total number of σ bonds and π bonds, and state the hybridization of each Carbon atom.

Total σ (sigma) bonds in C₂H₄

Total π (pi) bonds in C₂H₄

Carbon Hybridization

💡 Hint: Formulas & Approach

Count all single C–H bonds plus the single component of the C=C double bond. Steric number on each Carbon: 3 electron domains (2 single bonds to H + 1 double bond to C). Steric number 3 corresponds to mixing one s and two p orbitals ($sp^2$).

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1. σ bonds: Four C–H single bonds + one σ component of C=C double bond = 5 σ bonds.

2. π bonds: Formed by sideways overlap of unhybridized 2p orbitals on the carbons = 1 π bond.

3. Hybridization: Each Carbon has 3 effective electron domains → sp² hybridization with trigonal planar local geometry (~120° bond angles).

6. Molecular Polarity & Intermolecular Forces (IMFs)

Net Dipoles, Symmetry, London Dispersion, Dipole-Dipole, H-Bonding • Deck: Molecular Polarity.pptx

QUESTION 6.1 Molecular Symmetry & Net Dipole

Slides 6–9 explain why molecules with polar bonds can be nonpolar overall.
Compare Carbon dioxide (CO₂) and Water (H₂O):
• Both molecules have two polar bonds ($C=O$ ΔEN = 1.0; $O-H$ ΔEN = 1.4).
Explain why CO₂ is nonpolar while H₂O is polar.

A) CO₂ is linear ($180^\circ$); its two equal bond dipole vectors point in opposite directions and cancel out ($\mu = 0$). H₂O is bent ($104.5^\circ$), so its bond dipoles do not cancel ($\mu > 0$). B) Carbon is less electronegative than Hydrogen, making the C=O bond nonpolar. C) Water is nonpolar because oxygen has two lone pairs that neutralize the charge.

💡 Hint: Formulas & Approach

Net molecular polarity is the vector sum of individual bond dipole moments: $$\vec{\mu}_{\text{net}} = \sum \vec{\mu}_i$$ In linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral geometries with identical surrounding atoms, vector sum is zero.

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Correct Answer: Option A.
• CO₂: Linear geometry ($180^\circ$). The two polar $C^{\delta+} = O^{\delta-}$ bond dipoles pull with equal magnitude in exactly opposite directions, cancelling out completely (μ = 0).
• H₂O: Bent geometry ($104.5^\circ$) due to 2 lone pairs on oxygen. The two $O^{\delta-} - H^{\delta+}$ dipoles add vectorially to produce a strong upward net dipole moment pointing toward oxygen.

QUESTION 6.2 IMF Types & Boiling Point Trends

Slides 14–19 describe the four types of Intermolecular Forces (IMFs):
1. London Dispersion Forces (LDF): Temporary fluctuating dipoles present in all molecules; increases with polarizability/molecular size.
2. Dipole-Dipole: Permanent electrostatic attraction between polar molecules.
3. Hydrogen Bonding: Exceptionally strong dipole-dipole force occurring when Hydrogen is directly bonded to N, O, or F.
4. Ion-Dipole: Interaction between an ionic charge and polar solvent molecules.

Which of the following substances exhibits Hydrogen Bonding as its dominant intermolecular force and therefore has an unusually high boiling point?

A) CH₄ (Methane) B) H₂S (Hydrogen sulfide) C) NH₃ (Ammonia) D) CH₃Cl (Chloromethane)

💡 Hint: Formulas & Approach

Hydrogen bonding requires H bonded directly to Nitrogen, Oxygen, or Fluorine. Check whether H is bonded to N, O, or F in each candidate molecule.

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Correct Answer: C) NH₃.
• CH₄: Nonpolar → London dispersion forces only.
• H₂S: Polar, but S is not N, O, or F → Standard dipole-dipole forces.
• NH₃: H is directly bonded to highly electronegative Nitrogen with a lone pair → Hydrogen Bonding.
• CH₃Cl: Dipole-dipole forces (no H bonded to Cl).

7. Molecular Orbital (MO) Theory & Bond Order

Bonding vs Antibonding, Period 2 Diatomics, Bond Order, Paramagnetism • Deck: MO Theory

QUESTION 7.1 Bond Order & Stability

Slides 8–10 define Bond Order in Molecular Orbital Theory: $$\text{Bond Order} = \frac{1}{2}\left(N_{\text{bonding electrons}} - N_{\text{antibonding electrons}}\right)$$ Calculate the bond order for:
1. H₂ ($\sigma_{1s}^2$)
2. He₂ ($\sigma_{1s}^2\ \sigma_{1s}^{*2}$)
3. N₂ (Valence MOs: $\sigma_{2s}^2\ \sigma_{2s}^{*2}\ \pi_{2p}^4\ \sigma_{2p}^2$)

Bond Order for H₂

Bond Order for He₂

Bond Order for N₂

💡 Hint: Formulas & Approach

Apply $BO = \frac{1}{2}(N_b - N_a)$. Orbitals with asterisks (*) are antibonding. Orbitals without asterisks are bonding. A bond order of 0 means the molecule does not stably form under ordinary conditions.

📝 Show Full Worked Solution

1. H₂: 2 bonding electrons in $\sigma_{1s}$, 0 antibonding electrons: $$BO(H_2) = \frac{2 - 0}{2} = \mathbf{1}\quad\text{(Single stable covalent bond)}$$ 2. He₂: 2 bonding electrons in $\sigma_{1s}$, 2 antibonding electrons in $\sigma_{1s}^*$: $$BO(He_2) = \frac{2 - 2}{2} = \mathbf{0}\quad\text{(Unstable; does not exist as a molecule)}$$ 3. N₂: 8 bonding valence electrons ($\sigma_{2s}^2, \pi_{2p}^4, \sigma_{2p}^2$) and 2 antibonding electrons ($\sigma_{2s}^{*2}$): $$BO(N_2) = \frac{8 - 2}{2} = \mathbf{3}\quad\text{(Very strong triple bond; 945 kJ/mol)}$$

QUESTION 7.2 Paramagnetism & Oxygen (O₂)

Slides 2, 3, 17, and 18 address the experimental observation that liquid oxygen is paramagnetic (strongly attracted into a magnetic field):
1. Why does Lewis structure fail to explain why O₂ is paramagnetic?
2. How does Molecular Orbital Theory correctly explain this?

A) Lewis structure shows all paired electrons (:Ö=Ö:), predicting diamagnetism. In MO theory, Hund's rule places the last 2 valence electrons with parallel spins into two separate degenerate antibonding orbitals ($\pi_{2p}^{*1}, \pi_{2p}^{*1}$), creating 2 unpaired electrons (paramagnetic). B) Lewis structures only apply to solids; liquid oxygen decomposes into free radicals. C) Oxygen is diamagnetic, and the experiment was mistaken.

💡 Hint: Formulas & Approach

Paramagnetic = contains at least one unpaired electron. Diamagnetic = all electrons paired. O₂ has 12 valence electrons. Valence configuration: $\sigma_{2s}^2\ \sigma_{2s}^{*2}\ \sigma_{2p}^2\ \pi_{2p}^4\ \pi_{2p}^{*2}$. By Hund's rule, the 2 electrons in the degenerate $\pi_{2p}^*$ level occupy separate orbitals with parallel spins.

📝 Show Full Worked Solution

Correct Answer: Option A.
• In the standard Lewis structure $:\ddot{O}=\ddot{O}:$, every electron appears paired up as either a lone pair or a shared double bond. A student relying strictly on Lewis structures would incorrectly deduce that $O_2$ is diamagnetic.
• In Molecular Orbital theory, the 12 valence electrons of $O_2$ fill: $$\sigma_{2s}^2\ \sigma_{2s}^{*2}\ \sigma_{2p}^2\ \pi_{2p_y}^2\ \pi_{2p_z}^2\ \pi_{2p_y}^{*1}\ \pi_{2p_z}^{*1}$$ Because the two $\pi_{2p}^*$ antibonding orbitals have identical energy, Hund's rule dictates that they each receive one electron with parallel spins. These two unpaired electrons confer paramagnetism, perfectly matching experiment!