Chapter 2. Molecular and Ionic Compound Structure and Properties
2.1 Types of Chemical Bonds
2.1.1
2.1.2
2.1.3
2.1.4
Key Concepts:
Dipole Moment (D): A measure of the separation of positive and negative charge in a bond, with higher values indicating a larger separation of charge.
Electronegativity: The ability of an atom to attract electrons in a chemical bond. The greater the difference in electronegativity between two atoms, the more polar the bond, which typically increases the dipole moment.
Periodic Trends: Electronegativity increases across a period (left to right) and decreases down a group (top to bottom) in the periodic table.
Solution:
Compare the Electronegativity of Cl and Br:
Chlorine (Cl): Has a higher electronegativity than bromine (Br) because it is located above Br in Group 17 of the periodic table.
Bromine (Br): Has a lower electronegativity compared to chlorine.
Predict the Dipole Moment for H-Cl: Since chlorine is more electronegative than bromine, the electron density will be more strongly pulled toward chlorine in the H-Cl bond than it is toward bromine in the H-Br bond. This stronger pull increases the separation of charge, leading to a larger dipole moment in the H-Cl bond compared to the H-Br bond.
Answer:
The dipole moment for the H-Cl bond should be greater than 0.82 D because chlorine is more electronegative than bromine, leading to a stronger bond polarity and thus a higher dipole moment.
Detailed explanation here.
We need to determine the polarity of each bond based on the difference in electronegativity between the atoms involved. The greater the difference in electronegativity, the more polar the bond.
Identify the Electronegativity Values
Here are the electronegativity values (based on the Pauling scale) for the relevant elements:
Carbon (C): 2.55
Phosphorus (P): 2.19
Fluorine (F): 3.98
Chlorine (Cl): 3.16
Calculate the Electronegativity Differences
C—P bond: $\text{Electronegativity difference} = |2.55 - 2.19| = 0.36$
P—F bond: $\text{Electronegativity difference} = |3.98 - 2.19| = 1.79$
C—Cl bond: $\text{Electronegativity difference} = |3.16 - 2.55| = 0.61$
Arrange the Bonds in Order of Increasing Polarity
Now, we arrange the bonds based on their electronegativity differences. The bond with the smallest difference is the least polar, and the bond with the largest difference is the most polar.
C—P: Electronegativity difference = 0.36 (least polar)
C—Cl: Electronegativity difference = 0.61
P—F: Electronegativity difference = 1.79 (most polar)
The bonds arranged in order of increasing polarity are: C—P < C—Cl < P—F
Detailed explanation here.
PART A.
Answer: Metal
Explanation: Tin (Sn) is a metal because it is located in Group 14 of the periodic table and exhibits metallic properties like electrical conductivity and malleability.
Detailed explanation here.
PART B.
Answer: Nonmetal
Explanation: Chlorine (Cl) is a nonmetal. It belongs to Group 17 (the halogens) of the periodic table and is known for its high electronegativity and ability to form covalent and ionic bonds.
Detailed explanation here.
PART C.
Answer: Ionic bonding
Explanation: Tin (Sn), being a metal, tends to lose electrons and form positive ions (cations), while chlorine (Cl), being a nonmetal, tends to gain electrons and form negative ions (anions). The electrostatic attraction between these oppositely charged ions leads to the formation of an ionic bond.
Detailed explanation here.
PART D.
SnCl₂:
Appearance: White crystalline solid
Melting Point: 247°C
Boiling Point: 623°C
Type of Bonding: Ionic
Explanation: The high melting and boiling points, as well as the solid crystalline structure of SnCl₂, suggest that it has ionic bonding. Ionic compounds typically have high melting and boiling points due to the strong electrostatic forces between the ions.
SnCl₄:
Appearance: Colorless liquid
Melting Point: −34°C
Boiling Point: 114°C
Type of Bonding: Covalent
Explanation: SnCl₄, a colorless liquid at room temperature with relatively low melting and boiling points, suggests covalent bonding. Covalent compounds generally have lower melting and boiling points because the intermolecular forces are weaker compared to the ionic bonds in SnCl₂.
Detailed explanation here.
PART A. Nature of Bonding
In copper metal, the valence electrons are delocalized and form a "sea of electrons" that move freely throughout the solid. These electrons are not associated with any specific atom, which allows the metal cations to remain in a fixed position while the electrons move freely. This delocalization contributes to properties like electrical conductivity, as the free electrons can move under the influence of an electric field, and malleability, as the metal can be deformed without breaking the bonds.
Detailed explanation here.
PART B. Melting Point
The delocalization of electrons in a metallic bond leads to strong attractions between the positively charged metal ions and the sea of electrons, resulting in a high melting point for copper. In contrast, molecular solids like ice have weaker intermolecular forces (hydrogen bonds) holding the molecules together, leading to a much lower melting point.
Detailed explanation here.
PART C. Electrical Conductivity
Copper is a good conductor of electricity because the delocalized electrons can move freely and carry an electric current. As temperature increases, the ions in the metal lattice vibrate more, which can scatter the electrons and slightly reduce conductivity. However, copper still remains a good conductor at elevated temperatures.
Detailed explanation here.
2.2 Intramolecular Forces
2.2.1
2.2.2
2.2.3
PART A
Bond length of the Br–Br bond:
The bond length is the internuclear distance where the potential energy is at a minimum.
From the graph, the minimum point for the Br–Br bond (solid line) occurs around 230 pm.
Bond energy for the Br–Br bond:
The bond energy is the depth of the potential energy well (the lowest point on the curve).
The minimum value for the Br–Br bond energy is approximately −200 kJ/mol (read from the vertical axis at the deepest part of the Br–Br curve).
Detailed explanation here.
PART B
Prediction:
The bond length of the Cl–Cl bond should be shorter than that of the Br–Br bond.
Justification:
Atomic radius trends: As you move up a group in the periodic table, atomic radii decrease. Chlorine (Cl) is above bromine (Br) in Group 17 (the halogens). Chlorine atoms are smaller than bromine atoms.
Bond length and atomic size: Since chlorine atoms are smaller, the internuclear distance between two Cl atoms will be shorter than between two Br atoms. Therefore, the bond length of the Cl–Cl bond should be shorter than that of the Br–Br bond.
Detailed explanation here.
PART C
Error in the student's sketch:
The dotted curve for Cl–Cl shows a longer bond length than the Br–Br curve (solid line), which contradicts periodic trends.
Chlorine atoms are smaller than bromine atoms, so the bond length for Cl–Cl should be shorter, not longer, than Br–Br.
Correction:
The minimum point of the Cl–Cl curve (the dotted line) should be shifted to the left (toward a shorter internuclear distance), reflecting a shorter bond length.
The Cl–Cl bond energy should also be higher than Br–Br because Cl–Cl bonds are generally weaker than Br–Br bonds. The minimum depth of the curve should be less negative than −200 kJ/mol. Thus, the curve should have a higher minimum point than the Br–Br curve, indicating a bond energy closer to −150 kJ/mol.
Detailed explanation here.
When comparing bonds between the same atom, we can make the following general statements.
A double bond is ( pick one: shorter OR longer ) and ( pick one: weaker OR stronger ) than a single bond.
A triple bond is ( pick one: shorter OR longer ) and ( pick one: weaker OR stronger ) than a double bond.
As the bond order increases, the bond length increases and the bond energy increases .
Detailed explanation here.
PART A
Comparison of NaF and NaCl:
Both sodium fluoride (NaF) and sodium chloride (NaCl) involve the same cation, Na+, so the difference in lattice energy will depend on the anion size.
Fluorine (F−) is smaller than chlorine (Cl−) in the periodic table (since atomic radius increases down a group). The smaller the ion, the shorter the distance between the ions, resulting in a stronger electrostatic attraction.
Coulomb's Law Application:
According to Coulomb’s law, the lattice energy is inversely proportional to the distance between the ions. Since fluorine is smaller than chlorine, the distance between N+ and F− in NaF is shorter than that between Na+ and Cl− in NaCl.
Answer: Therefore, the lattice energy of NaCl will be less than that of NaF because of the larger ionic radius of Cl− compared to F−.
Detailed explanation here.
PART B
Comparison of NaF and MgO:
The cation in MgO is Mg2+, and the anion is O2−, whereas in NaF, the cation is Na+ and the anion is F−.
The charges on MgO ions are greater than those in NaF. Mg2+ and O2− both have charges of 2, while N+ and F− only have charges of 1. Since lattice energy increases with the product of the ion charges, the lattice energy for MgO should be higher than that for NaF.
Coulomb's Law Application:
Ionic radii: While magnesium and oxygen ions are slightly smaller than sodium and fluoride ions, the significant factor here is the higher charge on both Mg2+ and O2−. Coulomb's law predicts that a greater product of charges (4 instead of 1) leads to a much higher lattice energy, even if the distance between ions is comparable or slightly different.
Answer: The lattice energy of MgO should be greater than 930 kJ/mol due to the higher ionic charges and the stronger electrostatic attraction between the Mg2+ and O2− ions.
Detailed explanation here.
2.3 Structure of Ionic
2.3.1
Key Concepts:
Magnesium sulfide (MgS) is an ionic compound, meaning it is made of Mg²⁺ cations and S²⁻ anions.
In a solid ionic structure, the ions are arranged in a repeating lattice, with cations and anions alternating to minimize repulsion between like charges and maximize attraction between opposite charges.
Errors in the Diagram:
Incorrect Charges on Ions:
Magnesium ions (Mg) are depicted as Mg⁺ instead of Mg²⁺.
Sulfide ions (S) are depicted as S⁻ instead of S²⁻.
Correcting this, each Mg ion should have a 2+ charge, and each S ion should have a 2− charge.
Charge Balance Issue:
In a neutral ionic compound, the total positive and negative charges must balance. Since MgS consists of Mg²⁺ and S²⁻ ions, there should be an equal number of these ions. However, the diagram shows 14 Mg⁺ ions and 6 S⁻ ions, which does not follow the 1:1 ratio required by the formula MgS.
The correct ratio should be one Mg²⁺ for every S²⁻.
Clustering of Like Charges:
The diagram shows several instances of Mg⁺ ions clustered together without alternating with S⁻ ions. This is incorrect because in an ionic solid, oppositely charged ions should be adjacent to each other to minimize repulsion and maximize electrostatic attraction. The diagram should alternate Mg²⁺ and S²⁻ ions in a regular pattern, not group them by charge.
Corrected Description:
In a correct particle diagram for MgS:
The Mg ions should be labeled Mg²⁺.
The S ions should be labeled S²⁻.
The number of Mg²⁺ ions should equal the number of S²⁻ ions.
Mg²⁺ and S²⁻ ions should alternate in the lattice structure, with no clustering of like-charged ions.
Detailed explanation here.
2.4 Structure of Metals and Alloys
2.4.1
A metallic solid can conduct electricity due to the presence of delocalized electrons within its structure. In metals, the atoms are arranged in a lattice of positively charged metal ions (cations), but the outermost electrons (valence electrons) are not bound to any specific atom. Instead, they move freely throughout the entire structure, forming what is often referred to as a "sea of electrons."
The key factors that enable this conductivity are:
Delocalization of Electrons: The electrons are not tied to any specific atom and can flow through the entire metal structure.
Closely Packed Metal Ions: The tightly packed ions in the lattice create an environment where electrons can move freely between the ions without being impeded.
Detailed explanation here.
2.6 Resonance and Formal Charge
2.6.1
2.6.2
2.6.3
PART A
Resonance Structure 1:
Carbon is double-bonded to one oxygen and single-bonded to the other oxygen.
The single-bonded oxygen has three lone pairs and carries a -1 formal charge, while the double-bonded oxygen has two lone pairs.
Resonance Structure 2:
The positions of the double bond and single bond between carbon and the oxygen atoms are swapped compared to the first structure.
Again, the oxygen atom with the single bond has three lone pairs and a -1 formal charge.
Detailed explanation here.
PART B
The two carbon-oxygen bonds in the HCO₂⁻ ion have the same length. This is because the two resonance structures are equivalent, leading to a situation where the electron density is delocalized equally over both C-O bonds.
This delocalization results in both bonds having a bond order that is intermediate between a single bond and a double bond, rather than distinct single and double bonds. This averaging effect causes the bond lengths to be identical and intermediate between typical C-O single and double bond lengths.
Detailed explanation here.
PART C
The best way to describe the bond order of the carbon-oxygen bonds in the HCO₂⁻ ion is as 1.5 for each C-O bond. This bond order is calculated as the average of the resonance structures, where each structure has one single bond and one double bond. Since there are two equivalent resonance structures:
Resonance Structure 1: C-O bond order = 1 for one bond, 2 for the other.
Resonance Structure 2: C-O bond order = 2 for one bond, 1 for the other.
Therefore, the average bond order is:
Average bond order = (1 + 2) / 2 = 1.5
Thus, each carbon-oxygen bond in the methanoate ion has a bond order of 1.5, which accounts for the equal length and partial double-bond character of both bonds due to resonance.
Detailed explanation here.
Formal Charge Calculation Formula
Formal charge = (Valence electrons) - (Non-bonding electrons) - ½(Bonding electrons)
Structure #1
Carbon (C):
Valence electrons = 4
Non-bonding electrons = 0
Bonding electrons = 8 (four single bonds)
Formal charge = 4 - 0 - ½(8) = 0Oxygen (O):
Valence electrons = 6
Non-bonding electrons = 4
Bonding electrons = 4 (one double bond)
Formal charge = 6 - 4 - ½(4) = 0Chlorine (Cl) (both atoms are the same):
Valence electrons = 7
Non-bonding electrons = 6
Bonding electrons = 2 (one single bond)
Formal charge = 7 - 6 - ½(2) = 0
Overall Formal Charges for Structure #1: All atoms have a formal charge of 0.
Detailed explanation here.
Formal Charge Calculation Formula
Formal charge = (Valence electrons) - (Non-bonding electrons) - ½(Bonding electrons)
Structure #2
Carbon (C):
Valence electrons = 4
Non-bonding electrons = 0
Bonding electrons = 8 (two double bonds)
Formal charge = 4 - 0 - ½(8) = 0
Oxygen (O):
Valence electrons = 6
Non-bonding electrons = 6
Bonding electrons = 2 (one single bond)
Formal charge = 6 - 6 - ½(2) = -1Chlorine (Cl) (both atoms are the same):
Valence electrons = 7
Non-bonding electrons = 6
Bonding electrons = 2 (one single bond)
Formal charge = 7 - 6 - ½(2) = 0
Overall Formal Charges for Structure #2: Oxygen has a formal charge of -1, carbon and chlorine have 0.
Detailed explanation here.
Formal Charge Calculation Formula
Formal charge = (Valence electrons) - (Non-bonding electrons) - ½(Bonding electrons)
Structure #3
Carbon (C):
Valence electrons = 4
Non-bonding electrons = 2
Bonding electrons = 6 (one double bond, one single bond)
Formal charge = 4 - 2 - ½(6) = -1Oxygen (O):
Valence electrons = 6
Non-bonding electrons = 4
Bonding electrons = 4 (one double bond)
Formal charge = 6 - 4 - ½(4) = 0Chlorine (Cl) (both atoms are the same):
Valence electrons = 7
Non-bonding electrons = 6
Bonding electrons = 2 (one single bond)
Formal charge = 7 - 6 - ½(2) = 0
Overall Formal Charges for Structure #3: Carbon has a formal charge of -1, while oxygen and chlorine have 0.
Detailed explanation here.
Determining the Most Preferred Structure
The most preferred Lewis structure is typically the one where:
The formal charges are closest to zero for all atoms.
Any negative formal charges reside on the more electronegative atoms.
Structure #1 is the most dominant or preferred structure because all atoms have a formal charge of 0, which is the most stable configuration. This structure minimizes the formal charges and places the electrons in a way that fulfills the octet rule for all atoms involved without any unnecessary charges.
Detailed explanation here.
PART A: Lewis Structure for NO₂:
Total valence electrons: 17 (5 from N + 6 from each O).
Structure: N is the central atom with one single bond to one oxygen, a double bond to the other oxygen, and one unpaired electron on nitrogen.
Detailed explanation here.
PART B: Limitations:
The Lewis structure model struggles with odd-electron species, as it cannot represent an unpaired electron pair perfectly.
This leads to difficulties in predicting the actual electronic distribution and the molecule's reactivity.
Detailed explanation here.
2.7 VSEPR Theory and Bond Hydridization
2.7.1
2.7.2
2.7.3
PART A: NO₂⁺ (Nitronium Ion)
Draw the Lewis Structure
Valence Electrons Calculation:
Nitrogen (N): 5 valence electrons; Each Oxygen (O): 6 valence electrons × 2 = 12; Positive charge (-1 electron): -1
Total: 5 + 12 - 1 = 16 valence electrons
Lewis Structure:
Nitrogen is the central atom.
Two double bonds between nitrogen and each oxygen atom, and no lone pairs on nitrogen.
Predict the Molecular Geometry
With 2 bonding pairs and no lone pairs around the nitrogen, the geometry is linear. (Consult this chart or memorize)
Bond Angles
For a linear molecule, the bond angle is 180°. (Consult this chart or memorize.)
Dipole Moment
Since NO₂⁺ is linear and the molecule is symmetric with identical O atoms on both sides, the dipole moments cancel each other out.
Dipole Moment: No net dipole moment (nonpolar).
Detailed explanation here.
PART B: PF₅ (Phosphorus Pentafluoride)
Draw the Lewis Structure
Valence Electrons Calculation:
Phosphorus (P): 5 valence electrons; Each Fluorine (F): 7 valence electrons × 5 = 35;
Total: 5 + 35 = 40 valence electrons
Lewis Structure:
Phosphorus is the central atom with five fluorine atoms bonded to it.
Phosphorus has an expanded octet with no lone pairs.
Predict the Molecular Geometry
Phosphorus has 5 bonding pairs around it, leading to a trigonal bipyramidal geometry.
Bond Angles
In a trigonal bipyramidal structure, bond angles are:
90° between axial and equatorial positions.
120° between equatorial positions.
Dipole Moment
PF₅ is symmetric with equal electronegative fluorine atoms surrounding phosphorus, resulting in even charge distribution.
Dipole Moment: No net dipole moment (nonpolar).
Detailed explanation here.
PART A: BeF₂ (Beryllium Fluoride)
Lewis Structure:
Beryllium (Be) is the central atom with two fluorine (F) atoms bonded to it.
Be has no lone pairs.
Structure: F—Be—F
Electron Domains:
2 bonding pairs (2 single bonds with F).
Total = 2 domains.
Hybridization:
sp (2 domains).
Detailed explanation here.
PART B: CO₂ (Carbon Dioxide)
Lewis Structure:
Carbon (C) is the central atom with two oxygen (O) atoms double bonded to it.
No lone pairs on the carbon atom.
Structure: O=C=O
Electron Domains:
2 bonding pairs (2 double bonds with O).
Total = 2 domains.
Hybridization:
sp (2 domains).
Detailed explanation here.
PART C: NH₃ (Ammonia)
Lewis Structure:
Nitrogen (N) is the central atom with three hydrogen (H) atoms bonded to it and one lone pair.
Electron Domains:
3 bonding pairs (N-H bonds) + 1 lone pair.
Total = 4 domains.
Hybridization:
sp³ (4 domains).
Detailed explanation here.
PART D: NO₂⁺ (Nitronium Ion)
Lewis Structure:
Nitrogen (N) is the central atom with two oxygen (O) atoms double bonded to it and no lone pairs.
Structure: O=N=O
Electron Domains:
2 bonding pairs (2 double bonds with O).
Total = 2 domains.
Hybridization:
sp (2 domains).
Detailed explanation here.
PART E: BF₃ (Boron Trifluoride)
Lewis Structure:
Boron (B) is the central atom with three fluorine (F) atoms bonded to it.
No lone pairs on boron.
Electron Domains:
3 bonding pairs (3 B-F bonds).
Total = 3 domains.
Hybridization:
sp² (3 domains).
Detailed explanation here.
PART F: SO₂ (Sulfur Dioxide)
Lewis Structure:
Sulfur (S) is the central atom with two oxygen (O) atoms bonded to it, one with a double bond, the other with a double bond, and one lone pair.
Electron Domains:
2 bonding pairs (S-O double bonds) + 1 lone pair.
Total = 3 domains.
Hybridization:
sp² (3 domains).
Detailed explanation here.
PART G: NH₄⁺ (Ammonium Ion)
Lewis Structure:
Nitrogen (N) is the central atom with four hydrogen (H) atoms bonded to it.
No lone pairs on nitrogen.
Electron Domains:
4 bonding pairs (4 N-H bonds).
Total = 4 domains.
Hybridization:
sp³ (4 domains).
Detailed explanation here.
PART H: NO₂⁻ (Nitrite Ion)
Lewis Structure:
Nitrogen (N) is the central atom with one oxygen (O) atom double bonded and another oxygen single bonded with a lone pair.
Electron Domains:
2 bonding pairs (1 double bond, 1 single bond with lone pair on nitrogen) + 1 lone pair.
Total = 3 domains.
Hybridization:
sp² (3 domains).
Detailed explanation here.
| Molecule | Total Number of Sigma (σ) Bonds | Total Number of Pi (π) Bonds |
|---|---|---|
| Ethane (C₂H₆) | 7 | 0 |
| Ethene (C₂H₄) | 5 | 1 |
| Ethyne (C₂H₂) | 3 | 2 |