Chapter 1 Summary

Question 1

Electrostatic potential plots

Answer 1

Electrostatic potential plots are color-coded maps of electron density, indicating electron rich and electron deficient regions

Question 2

Curved arrow notation shows the movement of an electron pair. The tail of the arrow always begins at an electron pair, either in a bond or a lone pair. The head points to where the electron pair “moves”

Answer 2

Question 3

Curved arrow notation shows the movement of an electron pair. The tail of the arrow always begins at an electron pair, either in a bond or a lone pair. The head points to where the electron pair “moves”

Answer 3

Question 4

Formal charge

Answer 4

Formal charge (FC) is the difference between the number of valence electrons of an atom and the number of electrons it “owns”

Question 5

Formal charge

Answer 5

Formal charge (FC) is the difference between the number of valence electrons of an atom and the number of electrons it “owns”

Question 6

• The general rule of bonding:

Answer 6

Atoms strive to attain a complete outer shell of valence electrons (Section 1.2). H “wants” 2 electrons. Second-row elements “want” 8 electrons.

Question 7

Resonance

Answer 7

The basic principles:
• Resonance occurs when a compound cannot be represented by a single Lewis structure.
• Two resonance structures differ only in the position of nonbonded electrons and bonds.
• The resonance hybrid is the only accurate representation for a resonance-stabilized compound. A
hybrid is more stable than any single resonance structure because electron density is delocalized.

Question 8

Resonance

Answer 8

The basic principles:
• Resonance occurs when a compound cannot be represented by a single Lewis structure.
• Two resonance structures differ only in the position of nonbonded electrons and bonds.
• The resonance hybrid is the only accurate representation for a resonance-stabilized compound. A
hybrid is more stable than any single resonance structure because electron density is delocalized.

Question 9

Resonance

Answer 9

The basic principles:
• Resonance occurs when a compound cannot be represented by a single Lewis structure.
• Two resonance structures differ only in the position of nonbonded electrons and bonds.
• The resonance hybrid is the only accurate representation for a resonance-stabilized compound. A
hybrid is more stable than any single resonance structure because electron density is delocalized.

Question 10

Resonance

Answer 10

The basic principles:
• Resonance occurs when a compound cannot be represented by a single Lewis structure.
• Two resonance structures differ only in the position of nonbonded electrons and bonds.
• The resonance hybrid is the only accurate representation for a resonance-stabilized compound. A
hybrid is more stable than any single resonance structure because electron density is delocalized.

Question 11

What is the general rule of bonding for atoms in terms of their outer shell of valence electrons?

Answer 11

Atoms strive to attain a full outer shell of valence electrons.

Question 12

How many electrons do hydrogen atoms “want” to have in their outer shell to be stable?

Answer 12

2

Question 13

What is the usual number of bonds and nonbonded electron pairs for a neutral nitrogen atom?

Answer 13

Nitrogen in a neutral atom usually forms 3 bonds and has 1 lone pair.

Question 14

According to the chart, which second-row element typically forms one bond and has three nonbonded electron pairs?

Answer 14

Halogens (X = F, Cl, Br, I)

Question 15

What is the formula for calculating the formal charge (FC) of an atom in a molecule?

Answer 15

Formal Charge Formula: Formal Charge = Number of Valence Electrons - (Number of Electrons Owned by the Atom).

Question 16

Calculate the formal charge on a carbon atom that has three bonds and one lone pair.

Answer 16

Formal Charge = 4 (valence electrons for carbon) - (3 bonds * 1 + 2 lone electrons) = 4 - 5 = -1.

Question 17

In curved arrow notation, what does the tail of the arrow represent?

Answer 17

The tail of the arrow shows the origin of an electron pair, either in a bond or a lone pair.

Question 18

What does an arrow pointing from an electron pair on one atom to another atom signify in curved arrow notation?

Answer 18

The arrow indicates the movement of an electron pair from one atom to form a bond or to complete an octet on another atom.

Question 19

What are “electrostatic potential plots,” and what do they show in terms of electron density?

Answer 19

Electrostatic potential plots are color-coded maps that show areas of electron density, indicating electron-rich and electron-deficient regions in a molecule.

Question 20

Here’s what “electrons an atom owns” means:

Answer 20

• An atom owns all its lone pair electrons.
• It owns half of its bonding electrons (i.e., one electron per bond).

Question 21

Curved Arrow Notation:

Answer 21

• Curved arrows are used to show the movement of electrons in reaction mechanisms.
• The tail of the arrow shows the starting point of the electron pair (where the electrons currently reside), which can be a lone pair or a bond.
• The head of the arrow points to where the electron pair will move. If the arrow points from a lone pair to an atom, it typically signifies bond formation. If it points from a bond to an atom, it may indicate bond breaking or electron redistribution.

Question 22

Electrostatic Potential Plots:

Answer 22

• Electrostatic potential plots are visual representations that map electron density in molecules.
• These plots are often color-coded:
• Electron-rich regions are usually shown in colors representing higher electron density (often red or blue).
• Electron-deficient regions are shown in colors representing lower electron density.
• These plots help illustrate the distribution of electron density across a molecule, which is useful in understanding regions of relative positive and negative charge.

Question 23

Electrostatic Potential Plots:

Answer 23

• Electrostatic potential plots are visual representations that map electron density in molecules.
• These plots are often color-coded:
• Electron-rich regions are usually shown in colors representing higher electron density (often red or blue).
• Electron-deficient regions are shown in colors representing lower electron density.
• These plots help illustrate the distribution of electron density across a molecule, which is useful in understanding regions of relative positive and negative charge.

Question 24

What is the importance of a properly drawn Lewis structure in molecular studies?

Answer 24

A properly drawn Lewis structure is important as it determines geometry, hybridization, and bond types for each atom.

Question 25

Which geometries are commonly determined from Lewis structures?

Answer 25

Common geometries are linear, trigonal planar, and tetrahedral.

Question 26

Identify the hybridization associated with a molecule that has a trigonal planar geometry.

Answer 26

A trigonal planar geometry has sp² hybridization.

Question 27

What occurs during resonance in molecular structures?

Answer 27

Resonance involves the delocalization of electrons, which contributes to molecular stability.

Question 28

How many groups around an atom will result in a linear geometry, and what is the bond angle?

Answer 28

2 groups around an atom result in a linear geometry with a bond angle of 180°.

Question 29

How many groups around an atom will result in a linear geometry, and what is the bond angle?

Answer 29

2 groups around an atom result in a linear geometry with a bond angle of 180°.

Question 30

If a molecule has four groups around a central atom, what is the geometry and hybridization?

Answer 30

- Four groups around a central atom result in a tetrahedral geometry and sp³ hybridization.

Question 31

What does the resonance hybrid represent in a molecule with resonance structures?

Answer 31

The resonance hybrid is an average of all resonance forms, showing delocalized electron density.

Question 32

In the table of geometries and bond angles, which geometry corresponds to a bond angle of 120°?

Answer 32

A trigonal planar geometry corresponds to a bond angle of 120°.

Question 33

Provide an example of a molecule that demonstrates sp hybridization and linear geometry.

Answer 33

Beryllium hydride (BeH₂) demonstrates sp hybridization and linear geometry.

Question 34

Importance of Lewis Structures:

Answer 34

• Lewis structures are essential in representing molecules. They show how atoms are connected, the number of bonds, lone pairs, and the electron distribution in a molecule. • From a Lewis structure, you can determine: • Geometry (such as linear, trigonal planar, tetrahedral) based on electron groups around atoms. • Hybridization (sp, sp², sp³) which depends on the number and types of bonds. • Types of Bonds (single, double, triple), which indicate the strength and electron-sharing characteristics of each bond.

Question 35

Resonance:

Answer 35

• Resonance occurs when a molecule cannot be accurately represented by a single Lewis structure. Instead, multiple forms (resonance structures) illustrate the possible locations of electrons. • Key points: • Resonance structures differ only in the position of nonbonded electrons and π bonds. • The resonance hybrid is the true representation, showing a delocalization of electrons across the molecule, which stabilizes it. • Unlike isomers, which have different atomic arrangements, resonance structures involve only the rearrangement of electrons.

Question 36

Difference between Resonance Structures and Isomers:

Answer 36

• Isomers: Different compounds with the same molecular formula but different atom arrangements. • Resonance Structures: Different electron configurations for the same molecular structure, differing only in electron placement.

Question 37

Difference between Resonance Structures and Isomers:

Answer 37

• Isomers: Different compounds with the same molecular formula but different atom arrangements. • Resonance Structures: Different electron configurations for the same molecular structure, differing only in electron placement.

Question 38

Geometry and Hybridization:

Answer 38

Geometry and Hybridization: • The number of electron groups around a central atom dictates the molecule’s geometry and hybridization: • 2 groups: Linear, 180° bond angle, sp hybridization. • 3 groups: Trigonal planar, 120° bond angle, sp² hybridization. • 4 groups: Tetrahedral, 109.5° bond angle, sp³ hybridization. • Examples include: • BeH₂ (sp, linear) • BF₃ (sp², trigonal planar) • CH₄ (sp³, tetrahedral)

Question 39

Geometry and Hybridization:

Answer 39

Geometry and Hybridization: • The number of electron groups around a central atom dictates the molecule’s geometry and hybridization: • 2 groups: Linear, 180° bond angle, sp hybridization. • 3 groups: Trigonal planar, 120° bond angle, sp² hybridization. • 4 groups: Tetrahedral, 109.5° bond angle, sp³ hybridization. • Examples include: • BeH₂ (sp, linear) • BF₃ (sp², trigonal planar) • CH₄ (sp³, tetrahedral)

Question 40

Drawing Organic Molecules

Answer 40

Drawing Organic Molecules (Section 1.7): • Organic molecules are often drawn in simplified forms to represent their structure. Three main ways to represent these structures are: • Skeletal Structure: Lines represent bonds between carbons, with carbon and hydrogen atoms implied. • Condensed Structure: Groups are written linearly without showing individual bonds, e.g., (CH₃)₂CHCH₂CH₃ for isooctane. • Expanded Structure: Shows each atom and bond explicitly. • For a carbon bonded to four other atoms (as in CH₄), the geometry is tetrahedral. To best represent this three-dimensional structure, chemists use: • Two solid lines in the plane of the paper. • One wedge (indicating a bond coming out toward the viewer). • One dashed line (indicating a bond going away from the viewer).

Question 41

Bond length

Answer 41

Bond Length: • Bond length trends vary across the periodic table and depend on the bond order (single, double, or triple) and the hybridization of the bonded atoms.

Question 42

Bond length

Answer 42

Bond Length: • Bond length trends vary across the periodic table and depend on the bond order (single, double, or triple) and the hybridization of the bonded atoms.

Question 43

Bond order

Answer 43

Bond Order: Bond length decreases as the number of shared electrons (bond order) between two nuclei increases. Example trend for C-C bonds:

Question 44

Hybridization:

Answer 44

• Bond length also increases as the percentage of s-character in the orbital decreases. Orbitals with higher s-character (like sp) hold electrons closer to the nucleus, making the bond shorter.

Question 45

Hybridization:

Answer 45

• Bond length also increases as the percentage of s-character in the orbital decreases. Orbitals with higher s-character (like sp) hold electrons closer to the nucleus, making the bond shorter.

Question 46

Bond Length and Bond Strength:

Answer 46

• There is an inverse relationship between bond length and bond strength: shorter bonds are generally stronger bonds. For example, the C≡C triple bond is shorter and stronger than the C=C double bond, which is, in turn, shorter and stronger than the C-C single bond.

Question 47

Bond Length and Bond Strength:

Answer 47

• There is an inverse relationship between bond length and bond strength: shorter bonds are generally stronger bonds. For example, the C≡C triple bond is shorter and stronger than the C=C double bond, which is, in turn, shorter and stronger than the C-C single bond.

Question 48

Which representation is the simplest shorthand method for drawing organic molecules?

Answer 48

Skeletal structure is the simplest shorthand for organic molecules.

Question 49

What type of bond representation is commonly used to indicate a bond coming out toward the viewer?

Answer 49

A wedge indicates a bond coming out toward the viewer.

Question 50

How does bond length change across a period from left to right on the periodic table?

Answer 50

Bond length decreases across a period.

Question 51

Which trend correctly describes bond length down a group on the periodic table?

Answer 51

Bond length increases down a group.

Question 52

Which trend correctly describes bond length down a group on the periodic table?

Answer 52

Bond length increases down a group.

Question 53

What effect does bond order have on bond length?

Answer 53

Higher bond order results in shorter bonds.

Question 54

Arrange the following in order of increasing bond length: C-C, C=C, C≡C.

Answer 54

The correct order is C≡C < C=C < C-C.

Question 55

How does the percent s-character in hybridized orbitals affect bond length?

Answer 55

Higher s-character results in shorter bonds.

Question 56

How does the percent s-character in hybridized orbitals affect bond length?

Answer 56

Higher s-character results in shorter bonds.

Question 57

In terms of bond strength, which type of C-C bond is the strongest?

Answer 57

The triple bond (C≡C) is the strongest.

Question 58

What is the relationship between bond length and bond strength?

Answer 58

Shorter bonds are generally stronger.

Question 59

What is the relationship between bond length and bond strength?

Answer 59

Shorter bonds are generally stronger.

Question 60

Give an example of a molecule with the shortest C-H bond length due to high s-character in hybridization.

Answer 60

Example: An sp-hybridized carbon in a C-H bond, like in ethyne (C₂H₂), has the shortest bond due to high s-character.

Question 61

Give an example of a molecule with the shortest C-H bond length due to high s-character in hybridization.

Answer 61

Example: An sp-hybridized carbon in a C-H bond, like in ethyne (C₂H₂), has the shortest bond due to high s-character.

Question 62

Bond Length Across a Period:

Answer 62

• As you move across a period from left to right, atoms generally become smaller. This is because of increasing nuclear charge (more protons in the nucleus) with each successive element, which pulls the electron cloud closer to the nucleus, reducing atomic radius. • Consequently, the smaller atomic radius results in shorter bond lengths for bonds formed between elements further right in a period. For example, the bond length for O-H is shorter than for C-H.

Question 63

Bond Length Across a Period:

Answer 63

• As you move across a period from left to right, atoms generally become smaller. This is because of increasing nuclear charge (more protons in the nucleus) with each successive element, which pulls the electron cloud closer to the nucleus, reducing atomic radius. • Consequently, the smaller atomic radius results in shorter bond lengths for bonds formed between elements further right in a period. For example, the bond length for O-H is shorter than for C-H.

Question 64

Bond Length Down a Group:

Answer 64

Bond Length Down a Group: • Moving down a group in the periodic table, atoms become larger because they have more electron shells. Each additional shell increases the atomic radius, which makes the atom “bigger” overall. • Larger atoms will form longer bonds because the electron cloud is farther from the nucleus, meaning that the bonding electrons are also farther from the nuclei of the bonded atoms. For example, the bond length for H-F is shorter than for H-I.

Question 65

Bond Order:

Answer 65

• Bond order is the number of shared electron pairs between two atoms. Higher bond order (like a double or triple bond) means more electrons are shared between the two nuclei. • More shared electrons increase the electrostatic attraction between the two atoms, pulling them closer together and resulting in shorter bond lengths. For instance, C≡C (triple bond) is shorter than C=C (double bond), which is shorter than C-C (single bond).

Question 66

Hybridization and s-Character:

Answer 66

• The hybridization of an atom affects the bond length due to the s-character of the hybrid orbitals: • sp hybridized orbitals (as in triple bonds) have 50% s-character, sp² hybridized orbitals (as in double bonds) have 33% s-character, and sp³ hybridized orbitals (as in single bonds) have 25% s-character. • The higher the s-character, the closer the electrons are held to the nucleus because s-orbitals are closer to the nucleus than p-orbitals. This results in shorter bonds for bonds involving sp hybridized atoms compared to sp² or sp³ hybridized atoms. • For example, C-H bonds in sp-hybridized carbons (like in acetylene, C₂H₂) are shorter than those in sp³-hybridized carbons (like in methane, CH₄).

Question 67

Bond Length and Bond Strength:

Answer 67

• There’s an inverse relationship between bond length and bond strength. Shorter bonds are stronger because the bonded nuclei are closer together, which increases the overlap of the bonding orbitals and enhances the attractive forces between the nuclei and the shared electrons. • For example, the C≡C bond in acetylene is shorter and stronger than the C=C bond in ethene, which in turn is shorter and stronger than the C-C bond in ethane.

Question 68

Bond Length and Bond Strength:

Answer 68

• There’s an inverse relationship between bond length and bond strength. Shorter bonds are stronger because the bonded nuclei are closer together, which increases the overlap of the bonding orbitals and enhances the attractive forces between the nuclei and the shared electrons. • For example, the C≡C bond in acetylene is shorter and stronger than the C=C bond in ethene, which in turn is shorter and stronger than the C-C bond in ethane.

Question 69

The number of groups around an atom determines both its geometry (Section 1.6) and hybridization

Answer 69

Question 70

The number of groups around an atom determines both its geometry (Section 1.6) and hybridization

Answer 70

Question 71

Drawing organic molecules

Answer 71

Question 72

Drawing organic molecules

Answer 72

Question 73

Bond length

Answer 73

Question 74

Bond length

Answer 74

Question 75

Sigma (σ) and Pi (π) Bonds:

Answer 75

• Sigma bonds (σ bonds) are generally stronger than pi bonds (π bonds). Sigma bonds involve head-on overlap between atomic orbitals, which allows for a more direct and stronger connection between atoms. Pi bonds, on the other hand, involve side-by-side overlap and are generally weaker. • In multiple bonds (double or triple bonds), the first bond is always a sigma bond, while additional bonds are pi bonds. For example: • A double bond (C=C) consists of one sigma bond and one pi bond. • A triple bond (C≡C) consists of one sigma bond and two pi bonds.

Question 76

Sigma (σ) and Pi (π) Bonds:

Answer 76

• Sigma bonds (σ bonds) are generally stronger than pi bonds (π bonds). Sigma bonds involve head-on overlap between atomic orbitals, which allows for a more direct and stronger connection between atoms. Pi bonds, on the other hand, involve side-by-side overlap and are generally weaker. • In multiple bonds (double or triple bonds), the first bond is always a sigma bond, while additional bonds are pi bonds. For example: • A double bond (C=C) consists of one sigma bond and one pi bond. • A triple bond (C≡C) consists of one sigma bond and two pi bonds.

Question 77

Electronegativity and Polarity:

Answer 77

• Electronegativity is the tendency of an atom to attract electrons in a bond. • Trend: Electronegativity increases across a period (left to right) and decreases down a group in the periodic table. • Polarity occurs when two atoms with different electronegativities are bonded together. • If C or H is bonded to a more electronegative atom like N, O, or a halogen, the bond is polar. • A molecule is polar if it has: • At least one polar bond, or • Two or more polar bonds (or dipoles) that reinforce each other.

Question 78

Shortcut for Drawing Lewis Structures:

Answer 78

Shortcut for Drawing Lewis Structures: • Steps to Determine the Number of Bonds: 1. Count the valence electrons for all atoms in the molecule. 2. Calculate how many electrons would be needed if every atom (except hydrogen) had a full octet (8 electrons). 3. Subtract the actual number of valence electrons from the total electrons needed. This gives the number of electrons that need to be shared, which determines the number of bonds. Divide this number by two to find the number of bonds. • Steps to Draw the Lewis Structure: 1. Arrange the atoms in the usual way. 2. Count up the total number of valence electrons. 3. Use the shortcut to determine the required number of bonds. 4. Draw two-electron bonds to all hydrogens first, then draw remaining bonds between other atoms while ensuring that second-row elements do not exceed eight electrons. 5. Add lone pairs to complete octets for all atoms, where needed, and calculate formal charges to verify the structure.

Question 79

Shortcut for Drawing Lewis Structures:

Answer 79

Shortcut for Drawing Lewis Structures: • Steps to Determine the Number of Bonds: 1. Count the valence electrons for all atoms in the molecule. 2. Calculate how many electrons would be needed if every atom (except hydrogen) had a full octet (8 electrons). 3. Subtract the actual number of valence electrons from the total electrons needed. This gives the number of electrons that need to be shared, which determines the number of bonds. Divide this number by two to find the number of bonds. • Steps to Draw the Lewis Structure: 1. Arrange the atoms in the usual way. 2. Count up the total number of valence electrons. 3. Use the shortcut to determine the required number of bonds. 4. Draw two-electron bonds to all hydrogens first, then draw remaining bonds between other atoms while ensuring that second-row elements do not exceed eight electrons. 5. Add lone pairs to complete octets for all atoms, where needed, and calculate formal charges to verify the structure.

Question 80

Example - Drawing CH₃NCO (Methyl Isocyanate)

Answer 80

1. Arrange the atoms in the suggested order: H₃C-N-C=O. 2. Count the total number of valence electrons. 3. Use the shortcut to determine how many bonds are required. 4. Follow the steps above to add bonds and lone pairs.

Question 81

Which type of bond is stronger, sigma (σ) or pi (π)?

Answer 81

Sigma bonds are stronger than pi bonds.

Question 82

How many sigma and pi bonds are present in a double bond?

Answer 82

A double bond has 1 sigma and 1 pi bond.

Question 83

What happens to electronegativity as you move from left to right across a period?

Answer 83

Electronegativity increases across a period.

Question 84

In which direction does electronegativity decrease on the periodic table?

Answer 84

Electronegativity decreases down a group.

Question 85

Which of the following bonds would be polar? • a) C-H • b) C-O • c) C-C

Answer 85

The C-O bond is polar due to different electronegativities.

Question 86

Which of the following bonds would be polar? • a) C-H • b) C-O • c) C-C

Answer 86

The C-O bond is polar due to different electronegativities.

Question 87

What determines if a molecule is polar? • a) Presence of one polar bond • b) Presence of two or more reinforcing dipoles • c) Both a and b

Answer 87

Both a and b determine if a molecule is polar.

Question 88

In the shortcut for drawing Lewis structures, what does subtracting the actual number of valence electrons from the required number of electrons determine?

Answer 88

Subtracting valence electrons from needed electrons gives the number of bonds.

Question 89

How many bonds would you expect in CH₄ if you calculated with the shortcut method?

Answer 89

b) - CH₄ would need 4 bonds.

Question 90

What is the first step in drawing a Lewis structure using the shortcut method? • a) Draw bonds to hydrogen atoms • b) Count up the total number of valence electrons c) Arrange the atoms in the expected order

Answer 90

Arrange the atoms in the expected structure.

Question 91

What is the first step in drawing a Lewis structure using the shortcut method? • a) Draw bonds to hydrogen atoms • b) Count up the total number of valence electrons c) Arrange the atoms in the expected order

Answer 91

Arrange the atoms in the expected structure.

Question 92

Why are formal charges calculated after drawing the structure? • a) To check that all atoms have a full octet • b) To verify the arrangement and electron distribution are stable • c) To ensure that there are no polar bonds

Answer 92

(b) - Calculating formal charges verifies stability and distribution.