Exercise 1
Which of the following statements best describes London Dispersion Forces (LDF)?
Options:
- (A) LDF are the strongest type of intermolecular forces present in all polar molecules.
- (B) LDF arise due to permanent dipoles in polar molecules.
- (C) LDF are induced dipole interactions present between nonpolar molecules.
- (D) LDF involve the sharing of electrons between molecules.
View Answer
Answer Explanation:
London Dispersion Forces are induced dipole interactions that occur between nonpolar molecules due to temporary dipoles formed by the random movement of electrons.
Correct Answer: (C)
Exercise 2
In the process of solution formation, which step is exothermic?
Options:
- (A) Separating solute molecules.
- (B) Separating solvent molecules.
- (C) Mixing solute and solvent molecules.
- (D) Both separating solute and solvent molecules.
View Answer
Answer Explanation:
Mixing solute and solvent molecules is exothermic because energy is released when solute-solvent interactions are formed.
Correct Answer: (C)
Exercise 3
Consider the following set of compounds: Methanol (CH₃OH), Dimethyl Ether (CH₃OCH₃), Ethanol (C₂H₅OH), and Diethyl Ether (C₂H₅OC₂H₅). Arrange these compounds in order of increasing boiling points and explain the reasoning based on their intermolecular forces.
Options:
- (A) Methanol < Ethanol < Dimethyl Ether < Diethyl Ether
- (B) Diethyl Ether < Dimethyl Ether < Methanol < Ethanol
- (C) Dimethyl Ether < Diethyl Ether < Methanol < Ethanol
- (D) Methanol < Dimethyl Ether < Diethyl Ether < Ethanol
View Answer
Answer Explanation:
The boiling points of these compounds are influenced by the types and strengths of intermolecular forces present:
- Methanol (CH₃OH): Capable of hydrogen bonding due to the -OH group, leading to strong intermolecular attractions.
- Ethanol (C₂H₅OH): Also capable of hydrogen bonding with a larger molecular size compared to methanol, resulting in higher boiling points.
- Dimethyl Ether (CH₃OCH₃): Exhibits dipole-dipole interactions due to the C-O-C linkage but lacks hydrogen bonding, resulting in lower boiling points than alcohols.
- Diethyl Ether (C₂H₅OC₂H₅): Similar to dimethyl ether but with a larger molecular size and surface area, leading to stronger London Dispersion Forces and slightly higher boiling points than dimethyl ether.
Therefore, the order of increasing boiling points is:
Dimethyl Ether < Diethyl Ether < Methanol < Ethanol
Correct Answer: (C)
Exercise 4
Which of the following statements about resonance structures is FALSE?
Options:
- (A) Resonance structures contribute to the actual hybrid structure of a molecule.
- (B) All resonance structures must have the same arrangement of atoms.
- (C) Resonance structures indicate delocalization of electrons.
- (D) Resonance structures represent rapidly interchanging Lewis structures in dynamic equilibrium.
View Answer
Answer Explanation:
- Option A: True. Resonance structures are different Lewis structures that contribute to the actual electronic structure of a molecule.
- Option B: True. All resonance structures must have the same arrangement of atoms; only the placement of electrons changes.
- Option C: True. Resonance structures show the delocalization of electrons within molecules.
- Option D: False. Resonance structures do not represent rapidly interchanging structures or a dynamic equilibrium. Instead, they represent a hybrid or average of all possible valid Lewis structures.
Correct Answer: (D)
Exercise 5
Consider the ozone molecule (O₃) with the following Lewis structure: the central oxygen atom is bonded to two terminal oxygen atoms with one double bond and one single bond, respectively. Each terminal oxygen has two lone pairs, and the central oxygen has one lone pair. Calculate the formal charge on the central oxygen atom.
Options:
- (A) +1
- (B) 0
- (C) -1
- (D) +2
View Answer
Answer Explanation:
To calculate the formal charge on the central oxygen atom in ozone (O₃), use the formal charge formula:
$$ \text{Formal charge} = (\text{Valence electrons}) – (\text{Non-bonding electrons}) – \frac{1}{2} (\text{Bonding electrons}) $$
For the central oxygen atom:
- Valence electrons: Oxygen is in group 16, so it has 6 valence electrons.
- Non-bonding electrons: The central oxygen has one lone pair, which equals 2 non-bonding electrons.
- Bonding electrons: It forms one double bond (4 electrons) and one single bond (2 electrons), totaling 6 bonding electrons.
Plugging these values into the formula:
$$ \text{Formal charge} = 6 – 2 – \frac{6}{2} = 6 – 2 – 3 = +1 $$
Correct Answer: (A)
Exercise 6
Phosphorus pentachloride (PCl₅) is a molecule where the central phosphorus atom is bonded to five chlorine atoms. Based on the number of valence electrons and the resulting hybridization, determine the hybridization state and the molecular geometry of the phosphorus atom in PCl₅.
Options:
- (A) sp³ hybridization; tetrahedral geometry
- (B) sp³d hybridization; trigonal bipyramidal geometry
- (C) sp² hybridization; trigonal planar geometry
- (D) sp³d² hybridization; octahedral geometry
View Answer
Answer Explanation:
To determine the hybridization and molecular geometry of phosphorus in PCl₅, follow these steps:
- Determine the number of valence electrons on phosphorus: Phosphorus is in group 15, so it has 5 valence electrons.
- Count the regions of electron density around phosphorus: In PCl₅, phosphorus forms five bonds with chlorine atoms and has no lone pairs, resulting in five regions of electron density.
- Determine the hybridization:
- 2 regions: sp hybridization
- 3 regions: sp² hybridization
- 4 regions: sp³ hybridization
- 5 regions: sp³d hybridization
- 6 regions: sp³d² hybridization
- Determine the molecular geometry:
- 2 regions: Linear
- 3 regions: Trigonal planar
- 4 regions: Tetrahedral
- 5 regions: Trigonal bipyramidal
- 6 regions: Octahedral
Therefore:
Hybridization: sp³d
Molecular Geometry: Trigonal bipyramidal
Correct Answer: (B)
Exercise 7
Which of the following compounds has the lowest boiling point?
Options:
- (A) H₂O
- (B) NH₃
- (C) CH₄
- (D) CO₂
View Answer
Answer Explanation:
CH₄ has the lowest boiling point due to weaker London dispersion forces compared to the other compounds.
Correct Answer: (C)
Exercise 8
In semiconductor physics, doping with a group V element such as phosphorus in silicon introduces extra electrons. Explain how this affects the band structure of silicon and its electrical conductivity. Additionally, describe what type of charge carriers are predominant in this n-type semiconductor.
Options:
- (A) The conduction band is lowered, increasing the band gap; holes are the predominant charge carriers.
- (B) Extra electrons occupy the valence band, decreasing electrical conductivity; electrons are the predominant charge carriers.
- (C) Extra electrons occupy the conduction band, bridging the gap between the valence and conduction bands; electrons are the predominant charge carriers.
- (D) The band structure remains unchanged, but the mobility of charge carriers decreases; both electrons and holes are equally predominant.
View Answer
Answer Explanation:
Doping silicon with a group V element like phosphorus adds extra valence electrons that occupy energy levels slightly below the conduction band. These electrons can easily move into the conduction band with minimal energy input, thereby increasing electrical conductivity.
- Option A: Incorrect. The conduction band is not lowered, and holes are not the predominant charge carriers in n-type semiconductors.
- Option B: Incorrect. Extra electrons do not occupy the valence band; they contribute to the conduction band.
- Option C: Correct. Extra electrons introduced by doping occupy energy levels near the conduction band, allowing electrons to move freely, thereby increasing conductivity. Electrons are the predominant charge carriers in n-type semiconductors.
- Option D: Incorrect. The band structure is altered by doping, but electrons are the primary charge carriers, not both electrons and holes equally.
Correct Answer: (C)
Exercise 9
What is the formal charge on the oxygen atom in the hydroxide ion (OH⁻)?
Options:
- (A) 0
- (B) -1
- (C) +1
- (D) -2
View Answer
Answer Explanation:
In OH⁻, oxygen has a formal charge of -1.
Correct Answer: (B)
Exercise 10
What type of doping is achieved by adding boron to silicon?
Options:
- (A) n-type doping
- (B) p-type doping
- (C) Intrinsic doping
- (D) Compensated doping
View Answer
Answer Explanation:
Adding boron, which has one less valence electron than silicon, results in p-type doping.
Correct Answer: (B)
Exercise 11
Which of the following molecules can form hydrogen bonds?
Options:
- (A) CH₃CH₂CH₃
- (B) H₂S
- (C) NH₃
- (D) CCl₄
View Answer
Answer Explanation:
NH₃ can form hydrogen bonds because it has hydrogen atoms bonded to nitrogen, a highly electronegative atom.
Correct Answer: (C)
Exercise 12
Which of the following pairs of molecules exhibit dipole-dipole interactions?
Options:
- (A) CO and CO₂
- (B) H₂ and N₂
- (C) HCl and HBr
- (D) CH₄ and C₂H₆
View Answer
Answer Explanation:
HCl and HBr are polar molecules that exhibit dipole-dipole interactions.
Correct Answer: (C)
Exercise 13
Ion-dipole forces are most significant in which of the following scenarios?
Options:
- (A) Dissolving nonpolar molecules in hexane.
- (B) Dissolving NaCl in water.
- (C) Dissolving NH₃ in ethanol.
- (D) Dissolving CH₄ in water.
View Answer
Answer Explanation:
Ion-dipole forces are significant when dissolving ionic compounds like NaCl in polar solvents like water.
Correct Answer: (B)
Exercise 14
Which of the following factors increases the strength of London Dispersion Forces (LDF) in a molecule?
Options:
- (A) Decreasing molecular mass.
- (B) Increasing molecular symmetry.
- (C) Decreasing surface area.
- (D) Increasing the number of electrons.
View Answer
Answer Explanation:
Increasing the number of electrons increases polarizability, thereby strengthening LDF.
Correct Answer: (D)
Exercise 15
Which of the following solutes is least likely to dissolve in water?
Options:
- (A) NaCl
- (B) CH₄
- (C) NH₃
- (D) C₂H₅OH
View Answer
Answer Explanation:
CH₄ is nonpolar and has weak intermolecular forces, making it least likely to dissolve in polar water.
Correct Answer: (B)
Exercise 16
Consider the following molecules: water (H₂O), hydrogen fluoride (HF), and ammonia (NH₃). Which of the following statements correctly describes their hydrogen bonding abilities and how these bonds influence their physical properties?
Options:
- (A) Ammonia (NH₃) forms stronger hydrogen bonds than water (H₂O) due to the higher electronegativity of nitrogen compared to oxygen.
- (B) Hydrogen fluoride (HF) forms the strongest hydrogen bonds among the three, leading to its higher boiling point despite having the smallest molecular size.
- (C) Water (H₂O) exhibits hydrogen bonding that is less extensive than ammonia, resulting in a lower boiling point.
- (D) Hydrogen bonding in ammonia (NH₃) occurs because hydrogen atoms are bonded to highly electropositive nitrogen, which attracts electrons away from hydrogen.
View Answer
Answer Explanation:
Hydrogen fluoride (HF) forms the strongest hydrogen bonds due to the high electronegativity of fluorine, leading to a higher boiling point. While water (H₂O) can form extensive hydrogen bonding networks, ammonia (NH₃) forms weaker hydrogen bonds because nitrogen is less electronegative than oxygen or fluorine.
Correct Answer: (B)
Exercise 17
Consider the following molecules: water (H₂O), hydrogen sulfide (H₂S), methanol (CH₃OH), and dimethyl ether (CH₃OCH₃). These molecules exhibit different intermolecular forces based on their molecular structures. Identify the correct sequence of these substances in increasing order of their boiling points and justify the order based on their molecular interactions.
Options:
- (A) CH₃OCH₃ < H₂S < CH₃OH < H₂O
- (B) H₂S < CH₃OCH₃ < H₂O < CH₃OH
- (C) H₂S < CH₃OCH₃ < CH₃OH < H₂O
- (D) CH₃OH < H₂O < H₂S < CH₃OCH₃
View Answer
Answer Explanation:
The boiling points of these molecules are determined by the type and strength of intermolecular forces present:
- Dimethyl ether (CH₃OCH₃): Weak dipole-dipole interactions + London forces, no hydrogen bonding → relatively low boiling point.
- Hydrogen sulfide (H₂S): Dipole-dipole interactions, but sulfur is less electronegative → slightly higher boiling point than CH₃OCH₃.
- Methanol (CH₃OH): Hydrogen bonding due to -OH → much higher boiling point.
- Water (H₂O): Strong hydrogen bonding (two -OH bonds per molecule) → highest boiling point among these substances.
The correct order of increasing boiling points is:
CH₃OCH₃ < H₂S < CH₃OH < H₂O
Correct Answer: (A)
Exercise 18
Predict the solubility of silver chloride (AgCl) in water and explain the reasoning based on ion-dipole interactions and lattice energy.
Options:
- (A) Highly soluble due to strong ion-dipole interactions overcoming lattice energy.
- (B) Slightly soluble because ion-dipole interactions are weaker than the lattice energy of AgCl.
- (C) Insoluble because there are no significant intermolecular forces to facilitate dissolution.
- (D) Moderately soluble due to a balance between ion-dipole interactions and lattice energy.
View Answer
Answer Explanation:
Solubility of ionic compounds in water depends on the balance between the lattice energy of the solid and the ion-dipole interactions formed between the ions and water molecules.
- Lattice Energy of AgCl: AgCl has a high lattice energy due to the strong electrostatic attraction between Ag⁺ and Cl⁻ ions.
- Ion-Dipole Interactions: Water can form ion-dipole interactions with Ag⁺ and Cl⁻, but not enough to fully overcome the lattice energy of AgCl.
Thus, AgCl is slightly soluble in water.
Correct Answer: (B)
Exercise 19
Determine the formal charge on each oxygen atom in the permanganate ion (MnO₄⁻) assuming the central manganese atom is bonded to four oxygen atoms with equivalent bonds in the resonance structure.
Options:
- (A) Each oxygen has a formal charge of -1
- (B) Two oxygens have a formal charge of -1 each, and two oxygens have a formal charge of 0
- (C) All oxygens have a formal charge of 0
- (D) One oxygen has a formal charge of -2, and three oxygens have a formal charge of 0
View Answer
Answer Explanation:
When manganese forms four equivalent bonds with four oxygens (in resonance), the typical formal representation is that each O can be considered to have a -1 formal charge in one resonance form, averaged out in the resonance hybrid. Conventionally, we often say each oxygen is -1 for simple counting, though in reality the negative charge is delocalized.
Correct Answer: (A)
Exercise 20
The benzene molecule (C₆H₆) is often represented by two Kekulé structures. Explain why these resonance structures contribute to the stability of benzene and describe how resonance affects the bond lengths within the benzene ring.
Options:
- (A) Resonance structures increase the overall energy, making benzene less stable; bond lengths are unequal.
- (B) Resonance structures distribute electron density evenly, enhancing stability; bond lengths are equivalent.
- (C) Resonance structures have no impact on stability; bond lengths vary randomly.
- (D) Resonance structures create localized pi bonds, increasing reactivity; bond lengths are shortened.
View Answer
Answer Explanation:
Benzene’s resonance structures depict alternating single and double bonds, but experimentally, all C-C bonds in benzene are of equal length, intermediate between single and double bonds. Resonance allows for the delocalization of pi electrons across the entire ring, distributing electron density evenly.
- Option A: Incorrect. Resonance lowers overall energy → more stable; bond lengths are equal.
- Option B: Correct. Resonance distributes electron density uniformly, resulting in equivalent bond lengths and enhanced stability.
- Option C: Incorrect. Resonance significantly impacts stability and bond lengths.
- Option D: Incorrect. Resonance delocalizes electrons, not localizes them.
Correct Answer: (B)
Exercise 21
Determine the hybridization state and molecular geometry of the central sulfur atom in sulfur hexafluoride (SF₆).
Options:
- (A) sp³ hybridization; octahedral geometry
- (B) sp³d hybridization; trigonal bipyramidal geometry
- (C) sp³d² hybridization; octahedral geometry
- (D) sp² hybridization; trigonal planar geometry
View Answer
Answer Explanation:
In SF₆, sulfur forms six bonds with fluorine, no lone pairs → 6 electron domains → sp³d² → octahedral geometry.
Correct Answer: (C)
Exercise 22
The benzene molecule (C₆H₆) is often represented by two Kekulé structures. Which statement best explains how resonance contributes to the stability and bond characteristics of benzene?
Options:
- (A) Resonance allows for localized double bonds, increasing bond strength and stability.
- (B) Resonance structures create a dynamic equilibrium with rapidly shifting electrons.
- (C) Resonance delocalizes the pi electrons over all six carbon atoms, resulting in equivalent bond lengths and enhanced stability.
- (D) Resonance has no significant effect on bond lengths or stability in benzene.
View Answer
Exercise 23
What is the formal charge on the nitrogen atom in the nitrate ion (NO₃⁻)?
Options:
- (A) 0
- (B) -1
- (C) -2
- (D) +1
View Answer
Answer Explanation:
Nitrogen in NO₃⁻ typically forms 3 bonds (1 double, 2 single) with no lone pairs. Formal charge calculation → +1.
Correct Answer: (D)
Exercise 24
What type of semiconductor is formed when boron is doped into silicon?
Options:
- (A) n-type doping
- (B) p-type doping
- (C) Intrinsic doping
- (D) Compensated doping
View Answer
Answer Explanation:
Boron (group 13) has fewer valence electrons than silicon, so doping with boron creates electron “holes”: p-type doping.
Correct Answer: (B)
Exercise 25
Determine the formal charge on each oxygen atom in the sulfate ion (SO₄²⁻) using the most stable resonance structure.
Options:
- (A) Each oxygen has a formal charge of -1
- (B) Two oxygens have a formal charge of -1 each, and two oxygens have a formal charge of 0
- (C) All oxygens have a formal charge of 0
- (D) One oxygen has a formal charge of -2, and three oxygens have a formal charge of 0
View Answer
Answer Explanation:
In SO₄²⁻ (resonance), typically two O’s have single bonds (formal charge -1 each) and two O’s have double bonds (formal charge 0), summing to -2 overall.
Correct Answer: (B)
Exercise 26
Determine the hybridization state and molecular geometry of the central boron atom in boron trifluoride (BF₃).
Options:
- (A) sp³ hybridization; tetrahedral geometry
- (B) sp² hybridization; trigonal planar geometry
- (C) sp hybridization; linear geometry
- (D) dsp³ hybridization; trigonal bipyramidal geometry
View Answer
Answer Explanation:
Boron trifluoride has 3 electron domains (3 bonds to F, no lone pairs) → sp² → trigonal planar.
Correct Answer: (B)
Exercise 27
Explain how resonance contributes to the stability and bond length uniformity in the nitrite ion (NO₂⁻).
Options:
- (A) Resonance localizes double bonds on one oxygen, creating unequal bond lengths and decreasing stability.
- (B) Resonance structures indicate a dynamic equilibrium with rapid electron shifting.
- (C) Resonance delocalizes the pi electrons over both oxygen atoms, resulting in equivalent bond lengths and enhanced stability.
- (D) Resonance has no effect on NO₂⁻ stability or bond lengths.
View Answer
Answer Explanation:
In NO₂⁻, resonance shares the double bond among the two O atoms, giving identical N–O bond lengths & more stability.
Correct Answer: (C)
Exercise 28
In the permethylate ion (C₅H₁₁O⁻), determine the formal charge distribution assuming the negative charge is localized on the oxygen atom. Which statement is true regarding the formal charges?
Options:
- (A) The oxygen atom has a formal charge of -1, and all carbon atoms have a formal charge of +1.
- (B) The oxygen atom has a formal charge of -1, and all carbon atoms have a formal charge of 0.
- (C) Both oxygen and one carbon atom have a formal charge of -1 each.
- (D) The oxygen atom has a formal charge of 0, and one carbon atom has a formal charge of -1.
View Answer
Answer Explanation:
O has 6 valence e⁻, 3 lone pairs (6 e⁻), 1 bond (2 e⁻). Formal charge = 6 – 6 -1 = -1. Carbons = 0.
Correct Answer: (B)
Exercise 29
Dinitrogen tetroxide (N₂O₄) exhibits resonance. Which statement best describes the effect of resonance on N₂O₄?
Options:
- (A) Resonance causes the N–N bond to be a perfect single bond with localized electrons.
- (B) Resonance delocalizes electrons between the two nitrogens, resulting in equal N–O bond lengths.
- (C) Resonance leads to unequal N–O bond lengths due to alternating single/double bonds.
- (D) Resonance has no effect; all bonds are static.
View Answer
Answer Explanation:
In N₂O₄, resonance equalizes the N–O bonds via electron delocalization, making them all the same length.
Correct Answer: (B)
Exercise 30
Determine the hybridization state and molecular geometry of the central phosphorus atom in phosphorus trichloride (PCl₃).
Options:
- (A) sp³ hybridization; tetrahedral geometry
- (B) sp² hybridization; trigonal planar geometry
- (C) sp³ hybridization; trigonal pyramidal geometry
- (D) sp hybridization; linear geometry
View Answer
Answer Explanation:
P has 3 bonds + 1 lone pair → 4 electron domains → sp³ → trigonal pyramidal geometry.
Correct Answer: (C)
Exercise 31
Determine the formal charge on each atom in the cyanate ion (OCN⁻) using the most stable resonance structure.
Options:
- (A) O: -1; C: +1; N: 0
- (B) O: 0; C: -1; N: +1
- (C) O: -1; C: 0; N: 0
- (D) O: 0; C: +1; N: -1
View Answer
Answer Explanation:
Most stable structure: Oxygen single-bonded (FC = -1), carbon triple-bonded to nitrogen (FC = 0 each).
Correct Answer: (C)
Exercise 32
What is the electron domain geometry around the central phosphorus atom in phosphorus trifluoride (PF₃), and how does this geometry influence the bond angles?
Options:
- (A) Tetrahedral; lone pair → bond angles < 109.5°
- (B) Trigonal planar; no lone pairs → 120° angles
- (C) Trigonal bipyramidal; lone pairs occupy axial positions
- (D) Linear; 180° bond angles around P
View Answer
Answer Explanation:
PF₃ has 3 bonds + 1 lone pair → 4 e⁻ domains → tetrahedral electron geometry, but molecular shape is trigonal pyramidal. Bond angles < 109.5°.
Correct Answer: (A)
Exercise 33
Determine the hybridization state and molecular geometry of the central xenon atom in xenon tetrafluoride (XeF₄).
Options:
- (A) sp³ hybridization; tetrahedral geometry
- (B) sp³d hybridization; trigonal bipyramidal geometry
- (C) sp³d² hybridization; octahedral geometry
- (D) sp² hybridization; trigonal planar geometry
View Answer
Answer Explanation:
XeF₄ has 4 bonds + 2 lone pairs → 6 domains → sp³d² → electron geometry = octahedral, but with 2 lone pairs arranged to give a square planar shape.
Correct Answer: (C)
Exercise 34
Determine the molecular geometry and hybridization state of the central nitrogen atom in nitrogen trichloride (NCl₃).
Options:
- (A) sp³; tetrahedral geometry
- (B) sp³; trigonal pyramidal geometry
- (C) sp²; trigonal planar geometry
- (D) sp; linear geometry
View Answer
Answer Explanation:
NCl₃: 3 bonds + 1 lone pair → 4 domains → sp³ → trigonal pyramidal molecular geometry.
Correct Answer: (B)
Exercise 35
What is the electron domain geometry around the central sulfur atom in sulfur dioxide (SO₂), and how does this geometry influence the bond angles?
Options:
- (A) Trigonal planar; the lone pair reduces bond angles < 120°
- (B) Tetrahedral; the lone pair → ~109.5°
- (C) Linear; 180° bond angles
- (D) Octahedral; 90° angles
View Answer
Answer Explanation:
SO₂ has 3 domains (2 bonds + 1 lone pair) → trigonal planar electron geometry; molecular shape is bent; angle < 120°.
Correct Answer: (A)
Exercise 36
Which of the following molecules does NOT exhibit resonance?
Options:
- (A) CO₃²⁻
- (B) NO₃⁻
- (C) O₂
- (D) SO₄²⁻
View Answer
Answer Explanation:
O₂ does not exhibit resonance. CO₃²⁻, NO₃⁻, and SO₄²⁻ do have resonance forms.
Correct Answer: (C)
Exercise 37
Determine the hybridization state and molecular geometry of the central xenon atom in xenon difluoride (XeF₂).
Options:
- (A) sp³; trigonal bipyramidal geometry
- (B) sp³d; seesaw geometry
- (C) sp³d²; linear geometry
- (D) sp; linear geometry
View Answer
Answer Explanation:
XeF₂ → 2 bonds, 3 lone pairs = 5 electron domains → sp³d. Electron geometry = trigonal bipyramidal, lone pairs occupy equatorial positions → linear shape.
Correct Answer: Actually, the conventional label is sp³d for 5 domains. The final shape is linear. (Beware that some texts label it differently, but typically 5 domains is sp³d.)
Exercise 38
Calculate the formal charge on each nitrogen atom in dinitrogen pentoxide (N₂O₅), assuming the most stable resonance structure.
Options:
- (A) Both nitrogen atoms have +1.
- (B) One nitrogen has +1, the other 0.
- (C) Both nitrogen atoms have 0 each.
- (D) One nitrogen has +2, the other -1.
View Answer
Answer Explanation:
In N₂O₅, each nitrogen typically ends up with a +1 formal charge in stable resonance forms. This balances with oxygens (some at -1) to maintain neutrality.
Correct Answer: (A)
Exercise 39
The benzyl carbocation (C₆H₅CH₂⁺) exhibits resonance stabilization. Which statement best describes this effect?
Options:
- (A) The positive charge is localized on the benzylic carbon, destabilizing it.
- (B) The positive charge is delocalized over the ring, increasing stability.
- (C) Resonance does not affect the benzyl carbocation.
- (D) Resonance forms a double bond to remove the positive charge.
View Answer
Answer Explanation:
In the benzyl carbocation, the cationic center’s positive charge can be delocalized onto the ring carbons via resonance, stabilizing it.
Correct Answer: (B)
Exercise 40
Using VSEPR theory, predict the molecular geometry of phosphorus pentachloride (PCl₅) and explain the electron pair arrangement.
Options:
- (A) Trigonal bipyramidal; 3 equatorial + 2 axial positions.
- (B) Octahedral; 6 equivalent positions.
- (C) Tetrahedral; 4 equivalent positions.
- (D) Square pyramidal; 5 positions but 1 apex.
View Answer
Answer Explanation:
PCl₅ → 5 bonds, no lone pairs = 5 electron domains → trigonal bipyramidal. 3 chlorines in equatorial plane, 2 axial.
Correct Answer: (A)
Exercise 42
Determine the formal charge on the oxygen atom in the hydronium ion (H₃O⁺).
Options:
- (A) +1
- (B) 0
- (C) -1
- (D) +2
View Answer
Answer Explanation:
Oxygen in H₃O⁺: 6 valence e⁻ – 2 nonbonding e⁻ – (6 bonding e⁻)/2 = +1.
Correct Answer: (A)
Exercise 43
Determine the hybridization state and molecular geometry of the central sulfur atom in sulfur dioxide (SO₂).
Options:
- (A) sp³; bent geometry
- (B) sp²; bent geometry
- (C) sp; linear geometry
- (D) sp²; trigonal planar geometry
View Answer
Answer Explanation:
SO₂ has 3 electron domains (2 double bonds + 1 lone pair) → sp². Molecular shape = bent, bond angle ~119°.
Correct Answer: (B)
Exercise 44
Explain how resonance contributes to the stability and bond length uniformity in the acetate ion (CH₃COO⁻).
Options:
- (A) Resonance localizes negative charge on one oxygen → unequal bonds.
- (B) Resonance delocalizes negative charge over both oxygens → equal bond lengths, stable.
- (C) Resonance forms a permanent double bond to one O, single to the other.
- (D) Resonance has no effect on stability or bond lengths.
View Answer
Answer Explanation:
Acetate (CH₃COO⁻) shares the negative charge equally between two oxygens, giving identical C–O bond lengths & increased stability.
Correct Answer: (B)
Exercise 45
Using VSEPR theory, predict the molecular geometry of phosphorus trifluoride (PF₃) and explain its electron pair arrangement.
Options:
- (A) Trigonal planar; 3 bonds, no lone pairs.
- (B) Tetrahedral; 3 bonds + 1 lone pair.
- (C) Trigonal pyramidal; 3 bonds + 1 lone pair.
- (D) Linear; 2 bonds, no lone pairs.
View Answer
Answer Explanation:
PF₃ has 4 electron domains (3 bonding + 1 lone pair) → sp³. Molecular shape = trigonal pyramidal.
Correct Answer: (C)
Exercise 46
Compare the intermolecular forces in hydrogen peroxide (H₂O₂) and water (H₂O). Which statement is TRUE regarding their boiling points?
Options:
- (A) H₂O₂ has a lower boiling point than H₂O due to weaker H-bonding.
- (B) H₂O has a higher boiling point than H₂O₂ because it forms stronger H-bonds.
- (C) Both have identical boiling points as hydrogen compounds.
- (D) H₂O₂ has a higher boiling point than H₂O due to additional H-bonding sites.
View Answer
Answer Explanation:
Hydrogen peroxide can form more extensive hydrogen bonding networks (two -OH groups) than water, giving it a higher boiling point.
Correct Answer: (D)
Exercise 47
The allyl carbocation (C₃H₅⁺) exhibits resonance stabilization. Which statement best describes the effect of resonance on its stability?
Options:
- (A) Positive charge is localized, reducing stability.
- (B) Positive charge is delocalized over three carbons, increasing stability.
- (C) Resonance has no effect on allyl cation stability.
- (D) Resonance forms a double bond that removes the positive charge entirely.
View Answer
Answer Explanation:
Allyl cation resonance spreads the charge across multiple carbons → more stable than if it were localized.
Correct Answer: (B)
Exercise 48
Determine the hybridization state and molecular geometry of the central phosphorus atom in phosphorus pentachloride (PCl₅).
Options:
- (A) sp³; trigonal bipyramidal
- (B) sp³d; trigonal bipyramidal
- (C) sp³d²; octahedral
- (D) sp²; trigonal planar
View Answer
Answer Explanation:
PCl₅ → 5 electron domains → sp³d → trigonal bipyramidal geometry.
Correct Answer: (B)
Exercise 49
Determine the formal charge on each atom in the hydrosulfite ion (HSO₃⁻) using the most stable resonance structure.
Options:
- (A) H: +1; S: +1; each O: -1
- (B) H: 0; S: 0; two O: 0 each; one O: -1
- (C) H: 0; S: +1; each O: -1
- (D) H: +1; S: 0; three O: -1 each
View Answer
Answer Explanation:
Typically H = 0, S = 0, one O is -1, the other two are 0, net -1 charge for HSO₃⁻.
Correct Answer: (B)
Exercise 50
Compare the intermolecular forces in ethylene (C₂H₄) and ethanol (C₂H₅OH). Which of the following statements is TRUE regarding their boiling points?
Options:
- (A) Ethylene has a higher boiling point than ethanol due to stronger London dispersion forces.
- (B) Ethanol has a higher boiling point than ethylene because it can form hydrogen bonds.
- (C) Both ethylene and ethanol have similar boiling points as they have the same number of carbon atoms.
- (D) Ethylene has a higher boiling point than ethanol due to the presence of a double bond.
View Answer
Answer Explanation:
Ethylene (C₂H₄) relies mainly on London forces. Ethanol (C₂H₅OH) can form hydrogen bonds, a much stronger interaction → higher boiling point.
Correct Answer: (B)