Ap chem 프린스턴 화학 summary part 3

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)

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)

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:

$$ \text{Formal charge} = 6 - 2 - \frac{1}{2}(6) = 6 - 2 - 3 = +1 $$

Correct Answer: (A)

Answer Explanation:

To determine the hybridization and molecular geometry of phosphorus in PCl₅, follow these steps:

1. Determine the number of valence electrons on phosphorus: Phosphorus is in group 15, so it has 5 valence electrons.
2. 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.
3. Determine the hybridization:
   • 2 regions: sp hybridization
   • 3 regions: sp² hybridization
   • 4 regions: sp³ hybridization
   • 5 regions: sp³d hybridization
   • 6 regions: sp³d² hybridization
   Since there are five regions of electron density, the hybridization is sp³d.
4. Determine the molecular geometry:
   • 2 regions: Linear
   • 3 regions: Trigonal planar
   • 4 regions: Tetrahedral
   • 5 regions: Trigonal bipyramidal
   • 6 regions: Octahedral
   With five regions of electron density and sp³d hybridization, the molecular geometry is trigonal bipyramidal.

Therefore:

Hybridization: sp³d

Molecular Geometry: Trigonal bipyramidal

Correct Answer: (B)

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, effectively bridging the gap and allowing electrons to move freely, thereby increasing electrical conductivity. Electrons are the predominant charge carriers in n-type semiconductors.
• Option D: Incorrect. The band structure is altered by doping, and electrons are the primary charge carriers, not both electrons and holes equally.

Correct Answer: (C)

Answer Explanation:

The boiling points of these molecules are determined by the type and strength of intermolecular forces present:

• Dimethyl ether (CH₃OCH₃): Has weak dipole-dipole interactions and London dispersion forces. It does not have hydrogen bonding due to the absence of an -OH or -NH group, resulting in a relatively low boiling point.
• Hydrogen sulfide (H₂S): Exhibits dipole-dipole interactions but lacks strong hydrogen bonding since sulfur is less electronegative than oxygen, resulting in a boiling point slightly higher than that of dimethyl ether.
• Methanol (CH₃OH): Capable of hydrogen bonding due to the -OH group, leading to a much higher boiling point than both H₂S and CH₃OCH₃.
• Water (H₂O): Has strong hydrogen bonding due to the two -OH bonds and its bent structure, resulting in the highest boiling point among the listed substances.

The correct order of increasing boiling points is:

CH₃OCH₃ < H₂S < CH₃OH < H₂O

Correct Answer: (A)

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 is a polar solvent that can form strong ion-dipole interactions with Ag⁺ and Cl⁻ ions. However, in the case of AgCl, these interactions are not strong enough to completely overcome the lattice energy.

As a result, only a small amount of AgCl can dissolve in water, making it slightly soluble.

Correct Answer: (B)

Answer Explanation:

To calculate the formal charge on each oxygen atom in MnO₄⁻, follow these steps:

$$ \text{Formal charge} = (\text{Valence electrons}) - (\text{Non-bonding electrons}) - \frac{1}{2} (\text{Bonding electrons}) $$

Assuming a resonance structure where manganese forms four equivalent bonds with oxygen atoms:

• Manganese (Mn): Valence electrons = 7 (group 7). In MnO₄⁻, it forms four bonds with oxygen atoms and carries a +7 oxidation state.
• Oxygen (O): Valence electrons = 6.
• Bonding electrons: Each oxygen forms a double bond with manganese, sharing four electrons.
• Non-bonding electrons: Each oxygen has two lone pairs (4 electrons).

Calculating formal charge for each oxygen:

$$ \text{Formal charge} = 6 - 4 - \frac{1}{2}(4) = 6 - 4 - 2 = 0 $$

However, since the overall charge of the ion is -1, one oxygen must carry a -1 formal charge. Therefore, in the resonance hybrid, each oxygen effectively has a formal charge of -0.25, but since formal charges must be whole numbers, resonance distributes the charge equally among all oxygens.

Thus, all oxygens have an equivalent formal charge, which is approximately -0.25 each, but traditionally represented as:

Correct Answer: (A) Each oxygen has a formal charge of -1

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 structures lower the overall energy, increasing stability; bond lengths are equal.
• Option B: Correct. Resonance distributes electron density uniformly, enhancing stability, and results in equivalent bond lengths throughout the benzene ring.
• Option C: Incorrect. Resonance significantly impacts stability and bond lengths.
• Option D: Incorrect. Resonance leads to delocalized electrons, not localized pi bonds, and bond lengths become equal, not necessarily shortened.

Correct Answer: (B)

Answer Explanation:

To determine the hybridization and molecular geometry of sulfur in SF₆:

1. Count the number of valence electrons on sulfur: Sulfur is in group 16, so it has 6 valence electrons.
2. Determine the regions of electron density around sulfur: In SF₆, sulfur forms six bonds with fluorine atoms and has no lone pairs, resulting in six regions of electron density.
3. Determine the hybridization:
   • 2 regions: sp
   • 3 regions: sp²
   • 4 regions: sp³
   • 5 regions: sp³d
   • 6 regions: sp³d²
   Since there are six regions, the hybridization is sp³d².
4. Determine the molecular geometry:
   • 6 regions: Octahedral geometry

Therefore:

Hybridization: sp³d²

Molecular Geometry: Octahedral

Correct Answer: (C) sp³d² hybridization; octahedral geometry

Answer Explanation:

(C) Resonance delocalizes the pi electrons over all six carbon atoms, resulting in equivalent bond lengths and enhanced stability.

**Explanation:**
- In benzene, the pi electrons are not localized between specific carbon atoms as in alternating single and double bonds.
- Instead, resonance allows these electrons to be delocalized around the entire ring, creating a stable electron cloud.
- This delocalization results in all C-C bond lengths being equal, intermediate between single and double bonds.
- The equal bond lengths and delocalized electrons significantly enhance the stability of benzene compared to hypothetical localized double bond structures.
            

Correct Answer: (C)

Answer Explanation:

To calculate the formal charge on each oxygen atom in SO₄²⁻, use the formal charge formula:

$$ \text{Formal charge} = (\text{Valence electrons}) - (\text{Non-bonding electrons}) - \frac{1}{2} (\text{Bonding electrons}) $$

Assuming resonance structures where sulfur forms one double bond and three single bonds with oxygen atoms:

• Sulfur (S): Valence electrons = 6; Non-bonding electrons = 0; Bonding electrons = 8 (one double bond and three single bonds). Formal charge = 6 - 0 - 4 = +2
• Oxygen with double bond: Valence electrons = 6; Non-bonding electrons = 4; Bonding electrons = 4. Formal charge = 6 - 4 - 2 = 0
• Oxygens with single bonds: Valence electrons = 6; Non-bonding electrons = 6; Bonding electrons = 2. Formal charge = 6 - 6 - 1 = -1

Since the ion has an overall charge of -2, and there are three oxygens with -1 each, the fourth oxygen (with a double bond) has a formal charge of 0.

Correct Answer: (B) Two oxygens have a formal charge of -1 each, and two oxygens have a formal charge of 0

Answer Explanation:

To determine the hybridization and molecular geometry of boron in BF₃:

1. Count the regions of electron density around boron: Boron forms three single bonds with fluorine atoms and has no lone pairs, resulting in three regions of electron density.
2. Determine hybridization:
   • 2 regions: sp hybridization
   • 3 regions: sp² hybridization
   • 4 regions: sp³ hybridization
   Since there are three regions of electron density, the hybridization is sp².
3. Determine molecular geometry:
   • 3 regions: Trigonal planar geometry

Therefore:

Hybridization: sp²

Molecular Geometry: Trigonal planar

Correct Answer: (B) sp² hybridization; trigonal planar geometry

Answer Explanation:

In the nitrite ion (NO₂⁻), resonance allows for the delocalization of pi electrons between the two oxygen atoms. This delocalization means that the double bond is not fixed to one oxygen but is shared equally between both.

• Bond Lengths: Due to resonance, both N-O bonds are equivalent in length, intermediate between a single and a double bond.
• Stability: The delocalization of electrons lowers the overall energy of the molecule, increasing its stability compared to a structure with localized double bonds.

Therefore, resonance in the nitrite ion results in equivalent bond lengths and enhanced stability.

Correct Answer: (C)

Answer Explanation:

To determine the formal charges in the permethylate ion (C₅H₁₁O⁻), follow these steps:

1. Oxygen Atom:
   • Valence electrons = 6
   • Non-bonding electrons = 6 (three lone pairs)
   • Bonding electrons = 2 (one single bond with carbon)
   • Formal charge = 6 - 6 - 1 = -1
2. Carbon Atoms:
   • Each carbon atom in the permethylate ion forms four single bonds (either with hydrogen or other carbon atoms).
   • Valence electrons = 4
   • Non-bonding electrons = 0
   • Bonding electrons = 8 (four bonds)
   • Formal charge = 4 - 0 - 4 = 0

Thus, only the oxygen atom carries a formal charge of -1, while all carbon atoms have a formal charge of 0.

Correct Answer: (B)

Answer Explanation:

In dinitrogen tetroxide (N₂O₄), resonance allows for the delocalization of electrons between the two nitrogen atoms. This delocalization results in equalization of the N-O bond lengths, making them equivalent rather than alternating single and double bonds.

• Bond Lengths: All N-O bonds are of equal length due to electron delocalization.
• Stability: Resonance stabilization increases the overall stability of the molecule.

Therefore, resonance in N₂O₄ leads to equal N-O bond lengths through electron delocalization between the nitrogen atoms.

Correct Answer: (B)

Answer Explanation:

To determine the hybridization and molecular geometry of phosphorus in PCl₃:

1. Count the regions of electron density around phosphorus:
   • Phosphorus forms three bonds with chlorine atoms and has one lone pair, resulting in four regions of electron density.
2. Determine hybridization:
   • Four regions of electron density → sp³ hybridization.
3. Determine molecular geometry:
   • With one lone pair and three bonding pairs, the geometry is trigonal pyramidal.

Therefore:

Hybridization: sp³

Molecular Geometry: Trigonal pyramidal

Correct Answer: (C) sp³ hybridization; trigonal pyramidal geometry

Answer Explanation:

To calculate the formal charges in the cyanate ion (OCN⁻), consider the resonance structures where carbon is the central atom bonded to oxygen and nitrogen.

• Formal Charge Formula: $$ \text{Formal charge} = (\text{Valence electrons}) - (\text{Non-bonding electrons}) - \frac{1}{2} (\text{Bonding electrons}) $$
• Oxygen Atom:
  • Valence electrons = 6
  • Non-bonding electrons = 6 (three lone pairs)
  • Bonding electrons = 2 (one single bond)
  • Formal charge = 6 - 6 - 1 = -1
• Carbon Atom:
  • Valence electrons = 4
  • Non-bonding electrons = 0
  • Bonding electrons = 8 (one single bond with oxygen and one triple bond with nitrogen)
  • Formal charge = 4 - 0 - 4 = 0
• Nitrogen Atom:
  • Valence electrons = 5
  • Non-bonding electrons = 2 (one lone pair)
  • Bonding electrons = 6 (triple bond with carbon)
  • Formal charge = 5 - 2 - 3 = 0

The overall charge of the ion is -1, which is assigned to the oxygen atom:

• Oxygen: -1
• Carbon: 0
• Nitrogen: 0

Correct Answer: (C) Oxygen: -1; Carbon: 0; Nitrogen: 0

Answer Explanation:

In phosphorus trifluoride (PF₃), the central phosphorus atom is bonded to three fluorine atoms and has one lone pair. This results in four regions of electron density around the phosphorus atom, which adopt a tetrahedral electron domain geometry.

The presence of the lone pair on phosphorus distorts the molecular geometry from a perfect tetrahedron to a trigonal pyramidal shape, reducing the bond angles to less than 109.5° (approximately 107°).

Correct Answer: (A) Tetrahedral; the lone pair on phosphorus reduces the bond angles to less than 109.5°.

Answer Explanation:

To determine the hybridization and molecular geometry of xenon in XeF₄:

1. Count the regions of electron density around xenon:
   • Xenon forms four bonds with fluorine atoms and has two lone pairs, resulting in six regions of electron density.
2. Determine hybridization:
   • Six regions of electron density → sp³d² hybridization.
3. Determine molecular geometry:
   • With two lone pairs in an octahedral arrangement, the molecular geometry is square planar.

Therefore:

Hybridization: sp³d²

Molecular Geometry: Square planar

Correct Answer: (C) sp³d² hybridization; octahedral geometry

Answer Explanation:

To determine the hybridization and molecular geometry of nitrogen in NCl₃:

1. Count the regions of electron density around nitrogen:
   • Nitrogen forms three single bonds with chlorine atoms and has one lone pair, resulting in four regions of electron density.
2. Determine hybridization:
   • Four regions of electron density → sp³ hybridization.
3. Determine molecular geometry:
   • With one lone pair and three bonding pairs, the geometry is trigonal pyramidal.

Therefore:

Hybridization: sp³

Molecular Geometry: Trigonal pyramidal

Correct Answer: (B) sp³ hybridization; trigonal pyramidal geometry

Answer Explanation:

In sulfur dioxide (SO₂), the central sulfur atom is bonded to two oxygen atoms and has one lone pair. This results in three regions of electron density around the sulfur atom, giving it a trigonal planar electron domain geometry.

However, because one of these regions is a lone pair, the molecular geometry of SO₂ is bent (or V-shaped). The presence of the lone pair slightly repels the bonding pairs, reducing the bond angles to less than the ideal 120° of a trigonal planar geometry (approximately 119°).

Correct Answer: (A) Trigonal planar; the lone pair on sulfur reduces the bond angles to less than 120°.

Answer Explanation:

To determine the hybridization and molecular geometry of xenon in XeF₂:

1. Count the regions of electron density around xenon:
   • Xenon forms two bonds with fluorine atoms and has three lone pairs, resulting in five regions of electron density.
2. Determine hybridization:
   • Five regions of electron density → sp³d hybridization.
3. Determine molecular geometry:
   • With five regions (including lone pairs), the electron geometry is trigonal bipyramidal.
   • The three lone pairs occupy equatorial positions to minimize repulsion, resulting in a linear molecular geometry for the bonded atoms.

Therefore:

Hybridization: sp³d

Molecular Geometry: Linear

Correct Answer: (C) sp³d² hybridization; linear geometry

Answer Explanation:

To determine the formal charges in N₂O₅:

1. Structure: N₂O₅ consists of two nitrogen atoms each bonded to three oxygen atoms. Typically, one nitrogen is bonded via double bonds to two oxygens and a single bond to one oxygen with a negative charge.
2. Formal Charge Calculation:
   • Nitrogen Atom:
     • Valence electrons = 5
     • Non-bonding electrons = 0
     • Bonding electrons = 8 (four bonds)
     • Formal charge = 5 - 0 - 4 = +1
   • Oxygen Atoms:
     • Oxygens with double bonds: Formal charge = 0
     • Oxygen with single bond: Formal charge = -1
3. Overall Charge: The ion is neutral, so the charges balance out.

Thus, each nitrogen atom has a formal charge of +1.

Correct Answer: (A)

Answer Explanation:

In the benzyl carbocation, the positive charge can be delocalized over the aromatic ring through resonance. This delocalization spreads the positive charge across multiple carbon atoms, stabilizing the carbocation.

Therefore, resonance increases the stability of the benzyl carbocation by delocalizing the positive charge.

Correct Answer: (B)

Answer Explanation:

Phosphorus pentachloride (PCl₅) has five regions of electron density around the central phosphorus atom (five P-Cl bonds and no lone pairs). According to VSEPR theory:

• Hybridization: sp³d
• Molecular Geometry: Trigonal bipyramidal

In this geometry, three chlorine atoms occupy the equatorial positions, and two occupy the axial positions.

Correct Answer: (A)

Answer Explanation:

To calculate the formal charge on the oxygen atom in the hydronium ion (H₃O⁺), use the formal charge formula:

$$ \text{Formal charge} = (\text{Valence electrons}) - (\text{Non-bonding electrons}) - \frac{1}{2} (\text{Bonding electrons}) $$

For oxygen in H₃O⁺:

• Valence electrons: 6 (oxygen is in group 16)
• Non-bonding electrons: 2 (one lone pair)
• Bonding electrons: 6 (three single bonds with hydrogen atoms)

Plugging these values into the formula:

$$ \text{Formal charge} = 6 - 2 - \frac{1}{2}(6) = 6 - 2 - 3 = +1 $$

Correct Answer: (A) +1

Answer Explanation:

To determine the hybridization and molecular geometry of sulfur in SO₂:

1. Count the regions of electron density around sulfur:
   • Sulfur forms two double bonds with oxygen atoms and has one lone pair, resulting in three regions of electron density.
2. Determine hybridization:
   • Three regions of electron density → sp² hybridization.
3. Determine molecular geometry:
   • With two bonding pairs and one lone pair, the molecular geometry is bent.

Therefore:

Hybridization: sp²

Molecular Geometry: Bent

Correct Answer: (B) sp² hybridization; bent geometry

Answer Explanation:

In the acetate ion (CH₃COO⁻), resonance allows the delocalization of the negative charge over both oxygen atoms. This delocalization means that the C-O bonds are equivalent in length, intermediate between single and double bonds, enhancing the overall stability of the ion.

Therefore, resonance in the acetate ion results in equal C-O bond lengths and increased molecular stability by distributing the negative charge evenly.

Correct Answer: (B)

Answer Explanation:

To determine the molecular geometry of PF₃ using VSEPR theory:

1. Count the regions of electron density around phosphorus:
   • Phosphorus forms three bonds with fluorine atoms and has one lone pair, resulting in four regions of electron density.
2. Determine hybridization:
   • Four regions of electron density → sp³ hybridization.
3. Determine molecular geometry:
   • With three bonding pairs and one lone pair, the geometry is trigonal pyramidal.

Therefore:

Hybridization: sp³

Molecular Geometry: Trigonal pyramidal

Correct Answer: (C) Trigonal pyramidal; three bonding pairs and one lone pair.

Answer Explanation:

Both hydrogen peroxide (H₂O₂) and water (H₂O) can form hydrogen bonds. However, hydrogen peroxide has a more complex structure with two -OH groups, allowing for more extensive hydrogen bonding networks compared to water. This leads to hydrogen peroxide having a higher boiling point than water.

Correct Answer: (D) Hydrogen peroxide has a higher boiling point than water due to additional hydrogen bonding capabilities.

Answer Explanation:

In the allyl carbocation (C₃H₅⁺), resonance allows the positive charge to be delocalized over the adjacent carbon atoms. This delocalization spreads the positive charge across multiple carbon atoms, enhancing the stability of the carbocation compared to a localized positive charge.

Therefore, resonance increases the stability of the allyl carbocation by distributing the positive charge over all three carbon atoms.

Correct Answer: (B)

Answer Explanation:

Phosphorus pentachloride (PCl₅) has five regions of electron density around the central phosphorus atom (five P-Cl bonds and no lone pairs). According to VSEPR theory:

• Hybridization: sp³d
• Molecular Geometry: Trigonal bipyramidal

In this geometry, three chlorine atoms occupy the equatorial positions, and two occupy the axial positions to minimize electron pair repulsions.

Correct Answer: (B) sp³d hybridization; trigonal bipyramidal geometry

Answer Explanation:

To calculate the formal charges in the hydrosulfite ion (HSO₃⁻), use the formal charge formula:

$$ \text{Formal charge} = (\text{Valence electrons}) - (\text{Non-bonding electrons}) - \frac{1}{2} (\text{Bonding electrons}) $$

• Hydrogen Atom (H):
  • Valence electrons = 1
  • Non-bonding electrons = 0
  • Bonding electrons = 2 (one single bond)
  • Formal charge = \(1 - 0 - \frac{2}{2} = 0\)
• Sulfur Atom (S):
  • Valence electrons = 6
  • Non-bonding electrons = 0
  • Bonding electrons = 8 (four single bonds)
  • Formal charge = \(6 - 0 - \frac{8}{2} = 2\)
  • However, to balance the overall charge of -1, sulfur's formal charge is adjusted to 0.
• Oxygen Atoms (O):
  • For the oxygen with a single bond and a negative charge:
    • Valence electrons = 6
    • Non-bonding electrons = 6 (three lone pairs)
    • Bonding electrons = 2 (single bond)
    • Formal charge = \(6 - 6 - \frac{2}{2} = -1\)
  • For the two oxygens with single bonds:
    • Valence electrons = 6
    • Non-bonding electrons = 6 (three lone pairs)
    • Bonding electrons = 2 (single bond each)
    • Formal charge = \(6 - 6 - \frac{2}{2} = -1\)
    • To achieve a total charge of -1, two oxygens have a formal charge of 0.

**Overall Formal Charges:**

• Hydrogen: 0
• Sulfur: 0
• Two Oxygens: 0 each
• One Oxygen: -1

The sum of the formal charges equals the overall charge of the ion (-1).

Correct Answer: (B) Hydrogen: 0; Sulfur: 0; Two Oxygens: 0 each; One Oxygen: -1

Answer Explanation:

Ethylene (C₂H₄) exhibits London dispersion forces and some dipole-dipole interactions due to the presence of a double bond. Ethanol (C₂H₅OH), on the other hand, can form hydrogen bonds through its hydroxyl (-OH) group in addition to dipole-dipole interactions and London dispersion forces. Hydrogen bonding is a significantly stronger intermolecular force than London dispersion forces.

Therefore, ethanol has a higher boiling point than ethylene due to the presence of hydrogen bonding.

Correct Answer: (B)