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Chapter8CHEM101FA20PreLesson.pdf
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Chapter 8: Chemical Bonding and Lewis Structures
CHEM 101 Fall 2020
Dr. Lauren Genova
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Chapter 8
By the end of this chapter, you will be able to: • Write Lewis dot structures to depict bonding in simple molecules
• Compute formal charges for atoms an a Lewis structure
• Use formal charges to identify the most reasonable Lewis structure for a given molecule
• Explain the concept of resonance and draw Lewis structures representing resonance for a given molecule
• Assess the polarity of covalent bonds • Describe how three properties of chemical bonds (bond order,
bond length, and bond dissociation energy) relate to one another
• Calculate enthalpies of reaction using average covalent bond energies (ΔHrxn)
Chapter 8: Learning Objectives
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• Valence electrons are all the electrons after the noble gas in the abbreviation with the highest n (e.g., Br has seven valence electrons [Ar]4s2 3d104p5) – Hint: group number
tells the number of valence electrons present in the atom!
• Lewis dot structures show # of valence electronsà Each valence electron is represented by a dot around the element symbol
Periodicity and Lewis Dot Structures
Chapter 8
1 2 3 4 5 6 7 8
(3) (4) (5) (6) (7)
(8)
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• Octet rule: An octet of 8 electrons (4 pairs of e−) represents a fully filled, stable configuration
– Exceptions: H and He, for which the stable configuration is 2 electrons (called the “duet rule”)
• Elements tend to gain, lose, or share electrons with other atoms, in order to complete their octet.
• Most main group elements (groups 1-2 and 13-18) will not allow more than 8 electrons to be assigned to them (we’ll discuss exceptions)
Loss of 2 electrons makes Be similar to He (Be2+ = He)
Gain of 1 electron makes F similar to Ne (F− = Ne)
Sharing 4 electrons allows C to have a full octet
Octet Rule
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Chapter 8
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• Chemical bond: a net attractive force between two atoms which occurs due to the interactions of their valence electrons
• There are two types of chemical bonds:
1.) Ionic bond: Charge-charge attraction between ions. One or more valence electrons are transferred from one atom to another.
2.) Covalent bond: Sharing of valence electrons.
Chemical Bonding
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Chapter 8
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• Compounds composed of ions are called ionic compounds (or salts), and their constituent ions are held together by ionic bonds
• Ionic bonds = electrostatic forces of attraction between oppositely charged cations and anions – Strongest types of bonds – Formed by transfer of electrons between a metal and nonmetal
Ions are held together by electrostatic forces à
– Though you can represent ionic compounds with Lewis structures, we will focus on covalent compounds
Ionic Compounds
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Chapter 8
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• Covalent (molecular) compounds – Nonmetals bound by sharing one or more pairs of electrons
• A valid Lewis structure should have an octet for each atom except hydrogen (exceptions exist)
H2 H H+ or H HH H
or Cl Cl
Bonding electrons belong to both atoms
Nonbonding electrons (or lone pairs) don’t participate in bonding; associated with only one atom 7
Lewis Structures for Covalent Compounds
Cl2 Cl ClClCl +
→
→
A line (dash) represents a shared pair of electrons
Each dot = one valence e−
Chapter 8
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• Lewis structures express connectivity of atoms, distribution of electrons, and bonding patterns
• Lewis structures can show several arrangements of electrons: – Single bonds: one pair of electrons shared between two atoms.
Example: H2
– Double bonds: Two electron pairs (4 e−) shared between two atoms. Example: O2
– Triple bonds: Three electron pairs (6 e−) shared between two atoms. Example: N2
– Lone pairs (non-bonding pairs): a pair of e− not shared between two atoms. Example: (lone pairs in blue)
Lewis Structures for Covalent Compounds
+ or H HH H→H H
+ or O=OO O→O O
+ or N≡NN N→N N
or N≡NN N
Chapter 8
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• If we’re given a formula (e.g., H2O), how do we turn it into a structure?
• H2O has a fairly straightforward Lewis structure, but what about something more complicated? How and when do we know if we need to make double or triple bonds? 9
Lewis Structures of Water
(condensed structure)
Both of these versions are acceptable Lewis structures!
Chapter 8
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Example 1: Chloroform (CHCl3) 1. Draw the molecular “skeleton” (arrangement)
by arranging the atoms around a central atom • The least electronegative atom is usually central;
C, N, P and S are often central. H is never central. – Central atom for CHCl3 = C
2. Count the total # of valence electrons in the molecule or ion • For a neutral molecule, add the valence electrons from each atom • For a cation, subtract the ion’s charge from the sum of the valence e− • For an anion, add the ion’s charge to the sum of the valence e−
– Sum of valence electrons = C (4 valence e−) + H (1 valence e−) + 3*Cl (3*7 valence e−) = total of 26 valence electrons
– CHCl3 is neutral (not an ion), so no need to subtract/add anything10
Cl Cl Cl
H C
1.) Steps for Drawing Lewis Structures
Chapter 8
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\
3. Place single bonds between the central atom and each terminal atom
• Keep track of number of valence electrons used so far (out of the total number calculated in Step 2)
4. Place remaining electron pairs on terminal atoms (except H) so they each have an octet.
• If any electrons are left over, assign them to the central atom – Hydrogen follows duet rule
(maximum of 2 valence e−) – No electrons left over here
Cl Cl Cl
H C
# of valence electrons used (out of 26 total) = 8
Cl Cl Cl
H C
# of valence electrons used (out of 26 total) = 8 + 18 = 26
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Example 1: Chloroform (CHCl3)
1.) Steps for Drawing Lewis Structures
Chapter 8
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5. Count the electrons on the central atom. • If it has fewer than 8 electrons, rearrange the electrons of the outer
atoms to form multiple bonds with the central atom to satisfy the octet rule whenever possible. – This is a trial-and-error approach: if bonding all atoms with single
bonds doesn’t work, try one double bond! – If double bonds don’t work, then try one triple bond!
• Note: H can never form a double or triple bond (and fluorine usually doesn’t either) – Carbon has 8 electrons; we don’t need
to worry about doing anything else. We are finished!
Example 1: Chloroform (CHCl3)
Cl Cl Cl
H C
1.) Steps for Drawing Lewis Structures
Chapter 8
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Cl Cl Cl
H C
Writing Lewis Structures: Covalent Molecules
Cl Cl Cl
H C Cl Cl
Cl
H C
• There are several ways to write out your final structure! Example: Chloroform (CHCl3)
• Unless I tell you otherwise, use the “partially” condensed structure to express your final answers
Full Lewis dot structure (all dots shown)
“Partially” condensed structure
(where lines are used to represent shared pairs of e−; lone pairs are shown)
Condensed structure (where lines are used to represent shared pairs of e−; no lone pairs are shown)
= =
Chapter 8
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Your turn! Let’s practice.
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2.) Practice: Drawing Lewis Structures
1. Decide on the central atom and draw the molecular “skeleton” (arrangement)
• The least electronegative atom is usually central; C, N, P and S are often central. H is never central. ** You can have more than one central atom! **
Here, both C’s are central!
2. Count the total # of valence electrons in the molecule or ion – Sum of valence electrons = 2*C (2*4 valence e−) + 4*H
(4*1 valence e−) = total of 12 valence electrons
HC H
H C H
Example 2: Ethylene (C2H4)
Chapter 8
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3. Place single bonds between the central atom and each terminal atom
• Keep track of number of valence electrons used so far (out of the total number calculated in Step 2)
4. Place remaining electron pairs on terminal atoms (except H) so they each have an octet.
• If any electrons are left over, assign them to the central atom – All terminal atoms (H)’s follow the
duet rule as is; 2 electrons left over
# of valence electrons used (out of 12 total) = 10
# of valence electrons used (out of 12 total) = 10 + 2 = 12
HC H
H C H
HC H
H C H
Example 2: Ethylene (C2H4)
2.) Practice: Drawing Lewis Structures
Chapter 8
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5. Count the electrons on the central atom. • If it has fewer than 8 electrons, rearrange
the electrons of the outer atoms to form multiple bonds with the central atom to satisfy the octet rule whenever possible. – Right now, each carbon has 7 electrons.
How can we use the electrons we have to satisfy the octet rule?
– Form a double bond and recount the number of valence electrons!
– The math works out (12 valence e− total); octet rules are satisfied. We are finished!
HC H
H C H
HC H
H C H
HC
H
H C =
H
or
Example 2: Ethylene (C2H4)
2.) Practice: Drawing Lewis Structures
Chapter 8
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Let’s try one more…
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3.) Practice: Drawing Lewis Structures
Example 3: Nitrosonium ion (NO+) 1. Decide on the central atom and
draw the molecular “skeleton” (arrangement) – ** New tip: to denote an ion, always include brackets around the
structure and indicate the charge outside the brackets with an exponents **
2. Count the total # of valence electrons in the molecule or ion Sum of valence electrons: N (5 valence e− × 1 atom) = 5 O (6 valence e− × 1 atom) = 6 −1 electron (positive charge) = −1
Total = 10 valence electrons
ON +
+
Chapter 8
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Example 3: Nitrosonium ion (NO+) 3. Place single bonds between the
central atom and each terminal atom • Keep track of number of valence
electrons used so far (out of the total number calculated in Step 2)
4. Place remaining electron pairs on terminal atoms (except H) so they each have an octet.
• If any electrons are left over, assign them to the central atom – 8 electrons (4 pairs) remaining in step 3;
4 electrons (2 pairs) each placed on N and O 18
# of valence electrons used (out of 10 total) = 2
ON +
# of valence electrons used (out of 10 total) = 2 + 8 = 10
ON +
3.) Practice: Drawing Lewis Structures
Chapter 8
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Example 3: Nitrosonium ion (NO+) 5. Count the electrons on the central atom. • If it has fewer than 8 electrons, rearrange
the electrons of the outer atoms to form multiple bonds with the central atom to satisfy the octet rule whenever possible. – Right now, N and O each have 6 electrons. Let’s try making a
double bond and seeing if that helps. – Oxygen has 8 electrons now, but N
still only has 6. Let’s try a triple bond! – Both atoms have
8 electrons now. Octet rule is satisfied; we are finished!
ON +
ON +
ON +
ON ≡ +
or
3.) Practice: Drawing Lewis Structures
Chapter 8
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• The octet rule has several notable (and common) exceptions: – Elements in the 3rd row (n=3) or below on the periodic table
may have more than 8 electrons called an expanded octet • Extra electrons may go into the empty d orbitals
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Exceptions to the Octet Rule
http://guweb2.gonzaga.edu/faculty/cronk/CHEM101pub/Lewis_structures_II.html
Chapter 8
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• The octet rule has several notable (and common) exceptions: – Incomplete octets. Examples: beryllium, boron, aluminum
– Free radicals (often nitrogen-containing compounds) have an odd number of electrons. Example: NO2
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Exceptions to the Octet Rule
Advice: When drawing Lewis structures, just count the number of valence e−, use the steps outlined previously, and remember these exceptions exist if you get stuck!
Chapter 8
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• In some structures, it is possible to place lone pairs or multiple bonds on different atoms.
• Structures with the same connectivity of atoms but different arrangement of valence electrons are called resonance structures.
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Resonance Structures
This double arrow indicates resonance
Chapter 8
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• The nitrate ion has three resonance structures as shown:
4.) Draw three resonance structures for the sulfate ion. Could it have more?
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Resonance in Polyatomic Ions
Chapter 8
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5.) Which is the better Lewis structure for HCOOH? Is there any way to predict?
A B
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Comparing Lewis Structures
Chapter 8
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• The predominant resonance structure of a molecule is the one with charges as close to ZERO as possible.
• Two criteria should be considered when selecting the predominant (better) resonance structure:
1.) Formal charge: The resonance structure with the lowest overall formal charge on each atom is the most stable.
Formal charge = [# valence electrons of atom] – [# of bonds to atom + # of non-bonded electrons]
2. Electronegativity: If there is a negative charge assigned, it should be placed on the MOST electronegative element if possible.
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Resonance Structures
Chapter 8
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• Formal charge (F.C.) is determined for each atom in a molecule as follows:
Formal charge = [# valence e− of atom] – [# of bonds to atom + # of non-bonded e−]
• Structure with lowest formal charge values is “best” (most stable!)
H NC All atoms in HCN have a F.C. of 0
F.C. = [1] – [1] = 0
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Formal Charge
F.C. = [4] – [4] = 0 F.C. = [5] – [3+2] = 0
Example: HCN
Chapter 8
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5.) Which is a better Lewis structure for HCOOH? Is there any way to predict?
A B
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Comparing Lewis Structures
Chapter 8
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5.) Which is a better Lewis structure for HCOOH? Is there any way to predict? à Formal charge! (The resonance structure with the lowest overall formal charge on each atom is the most stable.)
A B
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Comparing Lewis Structures
F.C. = [1] – [1] = 0
F.C. = [4] – [4] = 0
F.C. = [6] – [1+6] = −1
F.C. = [6] – [3+2] = +1
F.C. = [1] – [1] = 0
Formal charge = [# valence e− of atom] – [# of bonds to atom + # of non-bonded e−]
Chapter 8
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B
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Comparing Lewis Structures
F.C. = [1] – [1] = 0F.C. = [1] – [1] = 0
F.C. = [4] – [4] = 0
F.C. = [6] – [2+4] = 0
F.C. = [6] – [2+4] = 0
Formal charge = [# valence e− of atom] – [# of bonds to atom + # of non-bonded e−]
A − +
5.) Which is a better Lewis structure for HCOOH? Is there any way to predict? à Formal charge! (The resonance structure with the lowest overall formal charge on each atom is the most stable.)
Chapter 8
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A B
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Comparing Lewis Structures
−
+
5.) Which is a better Lewis structure for HCOOH? Is there any way to predict? à Formal charge! (The resonance structure with the lowest overall formal charge on each atom is the most stable.)
B is the “better” Lewis structure because it represents the more stable molecule! (lowest overall F.C. on each atom)
Chapter 8
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• With covalently bound ions (polyatomics), we add or subtract electrons to account for overall charge.
• The sum of the formal charges should equal the charge of the ion, so not every atom will have a formal charge of zero. This is still ok!
• Example: Carbonate ion (CO32−)
Total valence electrons: 3(6)+4+2 = 24 electrons
Formal charges: −1 on each oxygen = −2 overall (same as the overall charge on the ion).
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Polyatomic Ion Formal Charges
−
−
or
F.C. = [6] – [1+6] = −1
F.C. = [6] – [1+6] = −1
Chapter 8
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• Electronegativity: Relative ability of an atom to attract electrons in a bond to itself; follows same trend as electron affinity
• If non-zero formal charge is necessary, the more electronegative element should get the more negative charge
Electronegativity and Formal Charge
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Chapter 8
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6.) Try drawing a third resonance structure for the cyanate ion, NCO– (with C as the central atom). Label as structure C. Which structure for NCO– is best?
Chapter 7
Resonance Structures
Chapter 6Chapter 2Chapter 2Poll Question #1
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C
?
Formal charge = [# valence e− of atom] – [# of bonds to atom + # of non-bonded e−]
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• The polarity of a bond expresses how equally electrons are shared between two atoms.
• The absolute value of the difference in electronegativity (ΔEN) of two bonded atoms provides a rough measure of the polarity to be expected in the bond and, thus, the bond type
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Electronegativity and Bond Polarity
Chapter 8
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Electronegativity difference (ΔEN)
Bond type
ΔEN ≈ 0 (0.0 to 0.3) Nonpolar covalent
ΔEN between 0.4 and 1.8 Polar covalent
ΔEN > 1.8 Ionic
Using Electronegativity Values to Predict Bond Type
Bond types: 1.) Nonpolar covalent bonds: electrons shared equally between atoms
• Mostly diatomic molecules and C-H 2.) Polar covalent bonds: unequal sharing of electrons
• Different non-metals 3.) Ionic bonds: Metal gives electrons to a nonmetal
• Metal and non-metal
Chapter 8
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• Polar covalent bonds: form between atoms whose electronegativities differ, resulting in an unequal sharing of electrons
• Electrons lie closer to the more electronegative element (unequal distribution of electrons)
• Resulting bond is “polar”, and the two atoms are said to have “dipoles” or partial charges – Dipoles can be notated with a “δ+” (partial positive charge) and “δ−”
(partial negative charge) or an arrow (dipole moment) • δ+ signifies the atom that is being deprived of e−
• δ− signifies the atom that is hogging the e−
• The cross end of the arrow represents the positive end of the dipole (δ+);
• The arrowhead points toward the the negative end of the dipole (δ−).
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Polar Covalent Bonds
https://www.dlt.ncssm.edu/tiger/chem3.htm
Chapter 8
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7.) In HI, iodine is more electronegative than hydrogen, so the dipole points from H to I
• In HF, the dipole is similar, but HF is more polar. Why?
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Polar Covalent Bonds
Chapter 8
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8.) Identify the most polar covalent bond.
A) O−N B) O−F C) O−O D) O−H
Chapter 7
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Bond Polarity
Chapter 6Chapter 2Chapter 2Poll Question #2
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• Lewis theory gives insight into several properties of chemical bonds
1.) Bond Order: The number of bonding electron pairs shared by two atoms
2.) Bond Length: The distance between the nuclei of two bonded atoms
3.) Bond Dissociation Energy: The energy required to break a bond
• These three properties are inter-related.
Fig. 9.19, p.419
Fig. 9.19, p.419
Fig. 9.19, p.419
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More Bond Properties
Chapter 8
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1.) Bond Order: The number of chemical bonds between a pair of atoms • Typically a value of 1, 2, or 3
– Bond order of 1 = single bond (e.g., H–F) – Bond order of 2 = double bond (e.g., O=O) – Bond order of 3 = triple bond (e.g., N≡N)
• Fractional bond orders occur in molecules with resonance structures For example, what do you think the N–O bond order is in NO2− ?
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Bond Order
Chapter 8
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1.) Bond Order: The number of chemical bonds between a pair of atoms • Typically a value of 1, 2, or 3
– Bond order of 1 = single bond (e.g., H–F) – Bond order of 2 = double bond (e.g., O=O) – Bond order of 3 = triple bond (e.g., N≡N)
Bond order increases: Single bond < double bond < triple bond
1 < 2 < 3
9.) In general, the greater the bond order (i.e., the more bonds holding two atoms together),
– The shorter the bond length – The stronger the bond and the larger the bond dissociation energy
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Bond Order
Chapter 8
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2.) Bond Length: The distance between the nuclei of two bonded atoms • Bond length depends on:
– Atomic size(s): Larger atom = longer bond – Bond order: Higher order = shorter bond
• Relationship between bond length and atomic size:
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Bond Length
Chapter 8
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The GREATER the number of bonds (bond order), the HIGHER the bond strength and the SHORTER the bond.
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Bond Energy 3.) Bond Energy (aka Bond Dissociation Energy): The energy required to break 1 mole of a chemical bond in the gas phase • The strength of a bond is measured by the bond energy • Examples:
BOND STRENGTH (kJ/mol) H—H 436 C—C 346 C=C 602 C≡C 835 N≡N 945
Chapter 8
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Identify the weakest covalent bond:
A.) C−C B.) C=C C.) C≡C
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Chapter 7
Bond Strength
Chapter 6Chapter 2Chapter 2Poll Question #3
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• Bond energies can be used to calculate reaction enthalpy (ΔHrxn):
• Breaking bonds REQUIRES energy (use the + value of bond energies).
• Making bonds RELEASES energy (use the – value of bond energies). 45
Using Bond Energy to Calculate Reaction Enthalpy (ΔHrxn)
sum of sum of
Chapter 8
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Visualization of ΔHrxn
Chapter 8
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10.) Find the ΔHrxn for the reaction:
CH4(g) + Cl2(g) à HCl(g) + CH3Cl(g)
Average Bond Energies: (I will always give you this information) H – H 436 kJ/mol Cl – Cl 242 kJ/mol H – Cl 432 kJ/mol H – C 414 kJ/mol C – Cl 339 kJ/mol 47
Reaction Enthalpy (ΔHrxn) Example
Chapter 8
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