Chemistry 121 Spring 2025 Practice Exam 3 Notes

Chemistry 121 Spring 2025 Practice Exam 3 Notes

Part I: Electron Configuration and Lewis Structures

• Electron Configuration (9 points)

  • A) He: $1s^2$

  • B) B: $1s^22s^22p^1$

  • C) N: $1s^22s^22p^3$

• Condensed Electron Configuration (16 points)

  • Uses noble gas core abbreviation.

  • A) Sr: $[Kr]5s^2$

  • B) Cl: $[Ne]3s^23p^5$

  • C) P^{3-}: $[Ne]3s^23p^6$

  • D) Ga^{+}: $[Ar]3d^{10}4s^2$

• Alanine Lewis Structure Analysis (25 points)

  • A) Electron-domain Geometry

    • N: Tetrahedral

    • C: Tetrahedral

    • O: Tetrahedral

  • B) Bond Angles

    • a) Approximately $109.5^{\circ}$

    • b) Approximately $109.5^{\circ}$

    • c) Approximately $109.5^{\circ}$

  • C) Hybridization

    • N: $sp^3$

    • C: $sp^3$

    • O: $sp^3$

  • D) Molecular Geometry of O: Bent

Part II: Multiple Choice Questions

• (3 points each = 75 points total)

• Only one correct answer per question.

Atomic Properties and Trends

• Question 1: Largest Atomic Radius

  • Answer: B) S

  • Atomic radius increases down a group.

• Question 2: Largest First Ionization Energy

  • Answer: B) N

  • Ionization energy generally increases across a period and up a group. Nitrogen has a special stability due to its half-filled p subshell.

• Question 3: Second Ionization of Phosphorous

  • Answer: C) $P^+(g) \rightarrow P^{2+}(g) + e^-$

  • Second ionization refers to removing an electron from the +1 ion.

• Question 4: Double Bond

  • Answer: B) 2

  • A double bond consists of two pairs of shared electrons.

• Question 5: Not Eight Valence Electrons

  • Answer: B) $Ca^+$

  • $Ca^+$ has lost one electron from its $s^2$ configuration, leaving it with only one valence electron after losing one of it's two valence electrons. The others have noble gas configurations.

• Question 6: Longest Covalent Bond

  • Answer: A) single

  • Single bonds are longer than double or triple bonds because there is less electron density between the atoms, leading to weaker attraction and larger bond length.

• Question 7: Least Electronegative

  • Answer: D) Rb

  • Electronegativity decreases down a group and increases across a period. Rubidium is an alkali metal found at the bottom of the group.

• Question 8: Least Polar Bond

  • Answer: B) P - S

  • The smaller the electronegativity difference between two atoms, the less polar the bond.

• Question 9: Formal Charge in ClF

  • Answer: C) 0, 0

  • In ClF, both chlorine and fluorine have a formal charge of 0 when they both have three lone pairs.

• Question 10: Ionic Bonds

  • Answer: C) CaO

  • Ionic bonds typically occur between metals and nonmetals with a large electronegativity difference.

• Question 11: Resonance Structures of SO3

  • Answer: C) 3

  • Sulfur can accommodate more than an octet. There are three resonance structures for $SO_3$

• Question 12: Central Atom Obeying Octet Rule

  • Answer: A) CF4

  • $CF_4$ obeys the octet rule. The others violate it.

• Question 13: Lattice Energy

  • Answer: D) decreases, increases

  • Lattice energy increases with decreasing ionic radius and increasing ionic charge (charge is more important).

• Question 14: Enthalpy Change (∆H) Calculation

  • Reaction: $2HCl (g) + F<em>2 (g) \rightarrow 2HF (g) + Cl</em>2 (g)$

  • Given bond dissociation enthalpies (D in kJ/mol): H-Cl = 431, F-F = 155, H-F = 567, Cl-Cl = 242

  • Calculation: $\Delta H = [2(431) + 155] - [2(567) + 242] = -223$

  • Answer: B) -223

• Question 15: Nonpolar Molecule

  • Answer: D) CO2

  • $CO_2$ is linear and symmetrical, making it nonpolar overall.

• Question 16: Electron-Domain Geometry of Oxygen in OF2

  • Answer: D) tetrahedral

  • Oxygen in $OF_2$ has two bonding pairs and two lone pairs, giving a tetrahedral electron-domain geometry.

• Question 17: Geometry of Ammonia (NH3)

  • Answer: A) tetrahedral, trigonal pyramidal

  • $NH_3$ has a tetrahedral electron-domain geometry and a trigonal pyramidal molecular geometry due to the lone pair.

• Question 18: F-N-F Bond Angle in NF3

  • Answer: B) 109.5 ˚

  • In $NF_3$, the F-N-F bond angle is slightly less than $109.5^{\circ}$ because the lone pair repels the bonding pairs more strongly.

• Question 19: Angles Between sp2 Orbitals

  • Answer: C) 120˚

  • The angles between $sp^2$ orbitals are $120^{\circ}$.

• Question 20: Hybrid Orbitals

  • Answer: B) three sp2 hybrid orbitals

  • The blending of one s atomic orbital and two p atomic orbitals produces three $sp^2$ hybrid orbitals.

• Question 21: π Bond in Ethylene (CH2CH2)

  • Answer: D) p atomic orbitals

  • The $\pi$ bond in ethylene results from the overlap of p atomic orbitals.

• Question 22: Pressure Conversion

  • Given: 1.20 atm

  • Conversion: $1 atm = 101.325 kPa$

  • Calculation: $1.20 atm * 101.325 kPa/atm = 121.59 kPa$

  • Answer: A) 122

• Question 23: Temperature Calculation

  • Given: 0.444 mol CO, 11.8 L, 889 torr

  • Conversion: $889 torr * (1 atm / 760 torr) = 1.1697 atm$

  • Ideal Gas Law: $PV = nRT \rightarrow T = PV/nR$

  • $T = (1.1697 atm * 11.8 L) / (0.444 mol * 0.0821 L \cdot atm / (mol \cdot K)) = 378.8 K$

  • $T(^{\circ}C) = 378.8 - 273.15 = 105.65 ^{\circ}C$

  • Answer: D) 106

• Question 24: Amount of Gas Calculation

  • Given: 60.82 L, 31˚C, 367 mm Hg

  • Conversion: $31 ^{\circ}C + 273.15 = 304.15 K$ and $367 mm Hg * (1 atm / 760 mm Hg) = 0.4829 atm$

  • Ideal Gas Law: $PV = nRT \rightarrow n = PV/RT$

  • $n = (0.4829 atm * 60.82 L) / (0.0821 L \cdot atm / (mol \cdot K) * 304.15 K) = 1.18 mol$

  • Answer: A) 1.18

• Question 25: Similar Properties to Fluorine

  • Answer: C) Cl

  • Chlorine is in the same group (halogens) as fluorine and therefore has similar chemical and physical properties. They have the same number of valence electrons and undergo similar chemical reactions.

Exam Scoring

• Part I: __/50

• Part II: __/75

• Total: __/125