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Chapter 8

Covalent Bonding and Molecular Structure

Chapter 8: Covalent
Molecular Structure

Bonding

and

Chapter In Context
In this chapter and the next, we examine chemical bonding in detail. We examined ionic bonding briefly in Chapter 2 and will do so in more detail in Chapter 11. We will also examine intermolecular forces in detail in Chapter 11. Here we will apply what you have learned about atomic structure (Chapter 6), electron configurations, and periodic trends
(Chapter 7) to the chemical bonds formed between atoms and ions and the shapes of molecules and ions that contain covalent bonds.


8-1

Biology: Molecular shape of enzymes specifically allow only certain reactions to occur. Drugs are developed that specifically fit into active sites in the enzyme to affect or even stop its action.

Chapter 8
8.1 Interactions Between
Particles: Coulomb’s
Law
8.2 Covalent Bonding
Basics
8.3 Lewis Structures
8.4 Bond Properties
8.5 Electron Distribution in Molecules
8.6 Valence-Shell
Electron-Pair
Repulsion Theory and Molecular Shape
8.7 Molecular Polarity

Chapter Goals
• Apply Coulomb’s Law.
• Understand forces involved in covalent bonding. • Write Lewis symbols and
Lewis structures.
• Predict bond properties.
• Understand charge distribution in molecules.
• Use VSEPR theory.
• Identify polar and nopolar species.

Chapter 8

Covalent Bonding and Molecular Structure

8-2

8.1 Interactions Between Particles: Coulomb’s Law
OWL Opening Exploration
8.1 Coulomb’s Law

Matter is made up of atoms and ions that experience both attractive and repulsive forces.
The strength of the force holding oppositely charged particles together in any material is described by Coulomb’s Law (equation 8.1). According to this law, force of attraction or repulsion between two charged species is directly proportional to the magnitude of the charge on the particles (qA and qB in equation 8.1) and inversely proportional to the square of the distance between the two particles (r in equation 8.1).
(q )(q )
(8.1)
Force ! A 2 B r For electrons, q = –e, and for nuclei, q = +Ze where Chapter Goals Revisited
• Apply Coulomb’s Law
Understand
relationship between charge, distance, and force of attraction or repulsion. e = magnitude of electron charge (1.6022 x 10–19 C)
Z = nuclear charge (number of protons) r = distance between particles A and B

All chemical attractive forces involve opposite charges, such as those between protons in a nucleus and the electrons surrounding that nucleus, and between positive and negative ions. The type and strength of these attractive forces allow us to categorize the different types of bonding found in matter (Table 8.1). In ionic bonding, found in ionic solids and liquids such as NaCl and CaCO3, there are strong attractive forces between positively and negatively charged ions. Covalent bonding, the attractive forces between electrons and nuclei on adjacent atoms within a molecule, occurs in compounds such as H2O and NH3.
Another type of attraction takes place in pure metals. Metallic bonding, the attractive forces between electrons and nuclei in metals, occurs in metals such as Cu and Fe and will be discussed in more detail in Chapter 9. Finally, there are forces that exist between molecules, called intermolecular forces, which will be discussed in more detail in
Chapter XX.
Table 8.1 Types of Chemical Bonding
Type of Interaction
Source of Attractive Forces
Ionic Bonding

Oppositely charged ions

Covalent Bonding

Nuclei and valence electrons

Metallic Bonding

Nuclei and electrons

Intermolecular Forces

Partial charges on individual, separate molecules

8.2 Covalent Bonding Basics
OWL Opening Exploration
8.2

A covalent bond is characterized by the sharing of valence electrons by two adjacent atoms. This happens most often between nonmetal elements such as carbon, hydrogen, oxygen and nitrogen.
For example, consider a simple covalently bonded molecule, H2. When two isolated H atoms are at a great distance from one another, they feel no attractive or repulsive forces.
However, as the atoms approach more closely, the attractive and repulsive forces between the two atoms become important.

Chapter Goals Revisited
• Understand forces involved in covalent bonding. Identify attractive and repulsive forces between atoms.

Chapter 8

Covalent Bonding and Molecular Structure

8-3

H

represents valence electron represents nucleus

There are two types of repulsive forces between the two atoms. First, the nuclei repel because they are both positively charged. Second, the electrons repel because they are both negatively charged. The attractive forces between the two atoms result from the interaction between the positively charged nucleus on one hydrogen atom and the negatively charged electron on the other hydrogen atom. In summary, for two hydrogen atoms, HA and HB,
Repulsive Forces:
Attractive Forces:

electron A – electron B nucleus A – nucleus B electron A – nucleus B electron B – nucleus A

In general, when two atoms approach each other, these repulsive and attractive forces always occur. When the attractive forces are greater than the repulsive forces, a covalent bond forms. When the repulsive forces are greater than the attractive forces, a covalent bond does not form and the atoms remain isolated.

The balance between the attractive and repulsive forces in H2 is related to the distance between H atoms, as shown in Figure 8.1. At large distances, neither attractive nor repulsive forces are important and no bond forms between H atoms. At short distances, repulsive forces are stronger than attractive forces and no bond forms.

Figure 8.1 Energy of H2 as a function of internuclear distance

At an internuclear distance where the attractive forces are stronger than the repulsive forces, a bond forms between the H atoms. The two valence electrons “pair up” and are shared between the two hydrogen nuclei in a covalent bond that is represented by a single line connecting the H atoms.
H· + H· → H—H
The single line between H atoms in H2 is a very useful representation of a chemical bond, but it is does not give an accurate picture of the distribution of bonding electrons in the

Flashforward
9.6 Molecular Orbital Theory

Chapter 8

Covalent Bonding and Molecular Structure

8-4

H2 molecule. More sophisticated descriptions of chemical bonding will be discussed in
Chapter 9.

8.3 Lewis Structures
OWL Opening Exploration
8.X

One of the most important tools chemists use to predict the properties of a chemical species is its Lewis structure. A Lewis structure (also called a Lewis dot structure or
Lewis diagram) shows the arrangement of valence electrons (both bonding and nonbonding) and nuclei in covalently bonded molecules and ions. The simplest Lewis structure is the Lewis symbol for an element, where the element symbol represents the nucleus and core (non-valence) electrons, and dots represent valence electrons. As shown in the Lewis symbols in Figure 8.2, valence electrons are traditionally arranged around the four sides of the element symbol. The electrons are single or paired to reflect the electron configuration of the element and the Pauli exclusion principle.

Figure 8.2. Lewis symbols for the 2

nd

Chapter Goals Revisited
• Write Lewis symbols and
Lewis structures.
Use the periodic table and valence electrons to write Lewis symbols. period elements.

nd

The Lewis symbols for the elements in the 2 period (Figure 8.2) show the relationship between valence electrons and group number for the A group (main group) elements.
Recall that the number of valence electrons for any main group element is equal to the group number for that element.

Flashback
7.3 Electron Configuration of Elements

EXAMPLE PROBLEM: Lewis Symbols for Atoms
(a) Draw the Lewis symbol for Te, tellurium.
(b) The following Lewis diagram represents the valence electron configuration of a main-group element.
Identify the element in period 5 that has this valence electron configuration.
SOLUTION:
Te

(a)
Tellurium is a Group 6A element and has six valence electrons.
(b) Rb The element has one valence electron, so it is a Group 1A element. Rubidium is the Group 1A element in the fifth period. OWL Example Problems
8.3 Lewis Symbols for Atoms

The representation of H2 shown in the previous section where the bonding electron are represented with a line, H—H, is the Lewis structure of H2. In a Lewis structure, pairs of bonding electrons are represented as lines connecting atom symbols, and nonbonding electrons are shown as dots (Figure 8.3). Nuclei and core electrons are represented by element symbols. Notice in Figure 8.3 that there can be more than one bonding pair between atoms, and that nonbonding electrons usually appear in pairs (called lone electron pairs or lone pairs). A bond consisting of two electrons (one line) is called a single bond, a bond made up of 4 electrons (two pairs, two lines) is called a double bond, and a bond with 6 electrons (three pairs, three lines) is called a triple bond.

Chapter 8

Covalent Bonding and Molecular Structure

8-5

Figure 8.3 Lewis structures for C2H4 and ICl.

Lewis structures are very useful in visualizing the physical and chemical properties of compounds made up of nonmetal elements. We will begin by learning how to create
Lewis dot structures, and then in the following sections use Lewis structures to explore bond properties (bond order, bond length, bond energy, bond polarity), the shapes of molecules, molecular polarity, and how the shape and polarity of molecules influence chemical properties.
Drawing Lewis Structures
The most important guideline to follow when drawing Lewis structures is the octet rule.
The octet rule states that most atoms in a Lewis structures are surrounded by no more than 8 electrons (shared bonding electrons and unshared nonbonding electrons). The octet rule is related to the fact that valence shells contain a single s orbital and three p orbitals that can accommodate up to 8 electrons, and it is these orbitals that are most often involved in forming covalent bonds between nonmetals in covalent compounds.
There are, however, some exceptions to the octet rule.
• Hydrogen has a single valence electron in a 1s orbital and therefore only accommodates two electrons when it forms covalent bonds. It typically forms only one chemical bond to another atom and does not have lone pairs in Lewis structures. • Beryllium and boron often accommodate only 4 or 6 electrons, respectively, in
Lewis structures.
• Elements with available empty d orbitals in the valence shell (elements in the third period and below) can accommodate more than eight electrons (often 10 or
12).
When determining if elements have satisfied octets, assign lone pairs of electrons to the element they are placed around, and divide bonding electrons equally between the two atoms that are connected by the bond.
In addition to the octet rule, it is helpful to remember that only the elements C, N, O, P, and S typically form double and triple bonds with other elements. Lewis structures are drawn by following the five steps shown in the examples below.

Number of Bonds

Lewis Dot
Structure

Bond Order

Electron-Pair
Geometry

Resonance
Structures

Molecular
Shape

Chemical
Reactivity

Electron
Distribution

Molecular
Polarity

Physical Properties
Vapor Pressure
Solubility
Boiling and Melting Points

Chapter Goals Revisited
• Write Lewis symbols and
Lewis structures.
Use octet rule to write
Lewis symbols.

Chapter 8

Covalent Bonding and Molecular Structure

8-6

Example: PF3
Step 1: Count Valence Electrons
Count the total number of valence electrons in the molecule or ion. Anions have extra electrons, so add 1 electron for each negative charge. Cations have a deficiency of electrons, so subtract 1 electron for each positive charge.
Step 2: Arrange Atoms
The central atom is usually the one with the lowest affinity for electrons (the one farthest from fluorine on the periodic table).
Exception: H is never a central atom
Step 3: Add Single Bonds
Add one bond (using a line) between each terminal atom and the central atom. Each bond represents 2 electrons.
Step 4: Add Remaining Electrons
Assign any remaining electrons to the terminal atoms, in pairs, until the octet rule is satisfied for each terminal atom (except hydrogen). If additional electrons remain, add them to the central atom.
Step 5: Octet Rule Check
Use the octet rule to determine if multiple bonds are necessary between atoms. If there is an electron deficiency for an element, change a non-bonding electron pair (lone pair) on an adjacent atom into a bonding pair. Continue only until the octet rule is satisfied for all elements (other than the known exceptions).

Example: CO2

Phosphorus has 5 valence electrons and each fluorine has 7 valence electrons.
5 + (3 × 7) = 26 electrons
(or 13 electron pairs)

Carbon has 4 valence electrons and each oxygen has 6 valence electrons.
4 + (2 × 6) = 16 electrons
(or 8 electron pairs)

F

P

F
O

F

P
F

26 electrons – 6 bonding electrons
= 20 electrons remaining
F

P

F

O

O

C

O

F

Each F needs 3 pairs to satisfy the octet rule, and the one remaining pair is assigned to P

Each O needs 3 pairs to satisfy the octet rule. No other electrons remain to satisfy the octet rule for C.
O

F

P

C

O

O

F

C

O

F

All elements have complete octets.

The central atom is sharing only 4 electrons. Use one lone pair on each oxygen to make a second bond to carbon, satisfying the octet rule.

6 + (3 × 6) + 2 = 26 valence electrons (or 13 pairs)

2–

O

O
O

C

16 electrons – 4 bonding electrons
= 12 electrons remaining

Step 1: Sulfur has 6 valence electrons, each oxygen has 6 valence electrons, and the 2– charge on the ion adds two more electrons

S

O

O

F

The steps above can also be applied to the Lewis structures of ions. When drawing the
Lewis structure for an ionic species, the structure is placed within square brackets and the ion charge is shown outside the brackets. For example, the steps required to draw the
Lewis structure for SO32– are shown below.

Steps 2 and 3: O

C

F

O
Steps 4 and 5:

OWL Example Problems
8.5 Drawing Lewis Structures: Tutor

S

O

Chapter 8

Covalent Bonding and Molecular Structure

8-7

Exceptions to the Octet Rule
Beryllium and boron are often found with fewer than 8 electrons in electron deficient compounds, compounds in which an element has an incomplete octet. For example, BF3 is an electron deficient compound.

Each fluorine has a complete octet (6 nonbonding electrons + 2 bonding electrons), but boron has only 6 electrons (6 bonding electrons and no lone pairs). Changing a fluorine lone pair to a bonding pair would alleviate the electron deficiency, but neither element commonly forms multiple bonds. The electron deficiency affects the reactivity of BF3.
Consider the reaction between BF3 and NH3,

In this reaction, the lone pair on N is used to form a bond to B. Following the reaction, both N and B have full octets and the compound is not electron deficient.
A second exception to the octet rule that also involves elements with incomplete octets occurs in odd-electron molecules, also called free radicals. Nitrogen monoxide (nitric oxide, NO) is an odd-electron molecule with 11 valence electrons.
N

O

Free radicals are highly reactive species because they have an unpaired electron that can react with other molecules. Pure NO reacts readily with other molecules such as the halogens, O2, and other free radicals.
Compounds with a central atom that is an element in the third period or below in the periodic table occasionally have more than 8 electrons associated with the central atom.
The central atom in these compounds have an expanded valence. Consider SF4.

Each fluorine has a satisfied octet (6 nonbonding electrons + 2 bonding electrons), but the central sulfur atom has 10 electrons around it (8 bonding electrons and 2 nonbonding electrons). Chapter Goals Revisited
• Write Lewis symbols and
Lewis structures.
Recognize exceptions to octet rule when writing Lewis symbols.

Chapter 8

Covalent Bonding and Molecular Structure

8-8

EXAMPLE PROBLEM: Drawing Lewis Structures
Draw the Lewis structure for
(a) ClO–
(b) IBr3
SOLUTION:
(a) Step 1: 7 + 6 + 1 = 14 valence electrons (or 7 pairs)
O
Steps 2 and 3: Cl
Cl
O
Steps 4 and 5:
The Lewis structure includes the ion charge outside square brackets.
(b) Step 1: 7 + (3 × 7) = 28 valence electrons (or 14 pairs)
Br
I
Br

Br
Steps 2 and 3:
Iodine has a lower affinity for electrons that bromine, so it is the central atom.

Br

I

Br

Br
Steps 4 and 5:
Iodine is in the fifth period and therefore can have an expanded valence.

OWL Example Problems
8.6 Exceptions to the Octet Rule: Tutor
8.7 Drawing Lewis Structures: Exercise
8.8 Drawing Lewis Structures
8.9 Free Radicals
Resonance Structures
Some molecules have more than one valid Lewis structure. Two or more valid Lewis structures for a species that differ only in the arrangement of electrons, not the arrangement of atoms, are called resonance structures. A resonance hybrid is the actual electron arrangement for the molecule or ion, and it is intermediate between the resonance structures but not represented by any of the individual resonance structures.
There is a great deal of experimental evidence to support resonance, including bond distances and angles that cannot be explained by the existence of a single Lewis structure for a molecule or ion.
Consider the case of ozone, O3. Completing the first four steps of drawing its Lewis structure results in the following structure for ozone.

To complete the Lewis structure, the central oxygen needs one more pair of electrons.
Both terminal oxygens have lone pairs that can be changed to shared bonding pairs, completing the octet for the central oxygen. There are therefore two possible Lewis structures for O3, and they are drawn separated by a double-headed arrow.

The two resonance structures for ozone suggest that that the oxygen-oxygen bonds in this molecule are not single bonds or double bonds, but something in between. The resonance hybrid, which is not represented by either resonance structure, has oxygenoxygen bonds that have equivalent properties such as length and energy.

Chapter Goals Revisited
• Write Lewis symbols and
Lewis structures.
Understand
relationship bewteen resonance structures and resonance hybrid.

Chapter 8

Covalent Bonding and Molecular Structure

8-9

The ozone resonance structures are equivalent to one another because they contain the same number and type of chemical bonds. Not all resonance structures are equivalent, however. For example, carbon dioxide has three resonance structures and they are not all equivalent. The first and third resonance structures are equivalent because each contains one carbonoxygen triple bond and one carbon-oxygen single bond. The middle resonance structure is unique because it has two carbon-oxygen double bonds. Later in this chapter we will look more closely at the electron distribution in molecules and will see how to use that information to determine the most likely resonance structure for a molecule or ion

Flashforward
8.5 Electron Distribution

EXAMPLE PROBLEM: Resonance Structures
Draw all resonance structures for the carbonate ion, CO32–.
SOLUTION:
The incomplete Lewis structure (octet rule not satisfied for carbon) is
2–

O
O

C

O

The electron deficiency for carbon is corrected by changing a lone pair on an adjacent oxygen into a bonding pair. There are three oxygen atoms with lone pairs that can correct the carbon electron deficiency, so there are three equivalent resonance structures for the carbonate ion.
2–

O
O

C

2–

O

O

O

C

O

2–

O
O

C

O

OWL Example Problems
8.10 Resonance Structures: Tutor
8.11 Resonance Structures

8.4 Bond Properties
OWL Opening Exploration
8.12

The properties of the chemical bonds in a molecule affect its chemical and physical properties. For example, nitrogen, N2, is very unreactive because of the strong triple bond between nitrogen atoms. Hydrazine, N2H4, has a weaker nitrogen-nitrogen single bond that contributes to its high reactivity and its use as a rocket fuel. One factor that has a great influence on bond properties is bond order, the number of bonding electron pairs between two bonded atoms. The bond order of a carbon-carbon single bond is 1, for example, while a carbon-carbon double bond has a bond order of 2 and a carbon-carbon triple bond has a bond order of 3.
Bond Length
As we saw earlier, chemical bonds form when attractive forces between atoms are stronger than repulsive forces. The distance between the atomic nuclei when energy is minimized is the bond length (Figure 8.4).

Chapter 8

Covalent Bonding and Molecular Structure

8-10

Chapter Goals Revisited
• Predict bond properties.
Relate bond length to atom size and bond order. Figure 8.4 H—H bond formation

Accurate bond distances are determined from careful measurements using techniques such as X-ray crystallography. Bond lengths between two different elements vary slightly between compounds. For that reason, average bond lengths are reported in tables such as the one shown below (Table 8.2).
Table 8.2. Average Bond Lengths
The bond lengths in
Table 8.2 have units of picometers
–12
(1 pm = 10 m).
Other common units for bond lengths are nanometers –9
(1 nm = 10 nm) and Ångstroms
–10
(1 Å = 10 m).

The data in Table 8.2 demonstrate two general trends in bond lengths. First, bond lengths increase with increasing atom size. Consider the trend in H–X bond lengths
(where X is a halogen). Halogen atomic radii increase moving down the periodic table
(F < Cl < Br < I), and therefore the H–X bond lengths increase as the halogen radius increases. <

<

<

The second trend demonstrated in Table 8.x is the relationship between bond length and bond order. As bond order increases, there is an increase in electron density between two

Flashback
7.4 Properties of Atoms

Chapter 8

Covalent Bonding and Molecular Structure

nuclei. This results in stronger attractive forces between electrons and nuclei, decreasing the distance between the nuclei. A carbon-carbon single bond has a bond order of 1 and is longer than a carbon-carbon double bond with a bond order of 2. In general, for a series of bonds that differ only in bond order, an increase in bond order results in a decrease in bond length.
C—O 143 pm > C=O 122 pm > C≡O 113 pm
Bond Energy

8-11

For a series of bonds between like atoms, as bond order increases, bond length decreases.

As shown previously in Figure 8.x, a chemical bond forms when the attractive and repulsive forces between atoms results in an energy minimum. Bond energy is the energy required to break a chemical bond in a gas-phase molecule. For example, the bond energy of an H—H bond is 436 kJ/mol (at 298 K). This means that 436 kJ of energy is required to break one mole of H—H bonds, forming two moles of H atoms.
H—H(g) → H(g) + H(g)

ΔHo = +436 kJ/mol

Breaking bonds is an endothermic process, therefore bond energies are always positive values. Conversely, bond formation is an exothermic process that always releases energy. Just as bond lengths vary between compounds, so do bond energies. Some average bond energies are shown in Table 8.3.
Table 8.3 Average Bond Energies

Chapter Goals Revisited
• Predict bond properties.
Relate bond energy to bond order.

A clear trend that can be observed from the data in Table 8.x is the relationship between bond energy and bond order. As shown below, bond energy increases with increasing bond order.
C—O 351 kJ/mol < C=O 745 kJ/mol < C≡O 1075 kJ/mol
As bond order increases, the stronger attractive forces between bonding electrons and nuclei mean that more and more energy is required to separate the bonded nuclei.
Resonance Structures, Bond Order, Bond Length, Bond Energy
As shown earlier, compounds with resonance structures often have chemical bonds that are not easily described as single, double, or triple bonds. In order to describe the bond

For a series of bonds between like atoms, as bond order increases, bond energy increases.

Chapter 8

Covalent Bonding and Molecular Structure

8-12

order, bond energy and bond length of these bonds, the number of resonance structures and bonding pairs must be taken into account. Consider the nitrite ion, NO2–, which has two equivalent resonance structures.
O

N



O

O

N



O

Because neither resonance structure represents the actual electron arrangement, the nitrogen-oxygen bonds in this ion are not single bonds (bond order = 1) or double bonds
(bond order = 2). Instead, NO2– has two equivalent NO bonds where the three pairs of bonding electrons are distributed between the two equivalent NO bond locations. The best way to describe the bonding in the resonance hybrid is with a fractional bond order: total number of NO bonding pairs
3
=
Each NO bond order =
= 1.5 number of NO bond locations
2
The NO bonds in the nitrite ion have a bond order that is intermediate between a single and a double bond. The bond length and bond energy of the NO bonds in the nitrite ion are also intermediate between average NO single and double bond lengths and bond energies. Bond
Bond length
Bond energy
N—O
136 pm
201 kJ/mol
NO in NO2– 125 pm
174 kJ/mol
N=O
115 pm
143 kJ/mol

Chapter Goals Revisited
• Predict bond properties.
Predict bond properties of the resonance hybrid.

EXAMPLE PROBLEM: Bond Order, Length, Strength
(a) What is the bond order of the boron-oxygen bonds in the borate ion, BO33–?
(b) Which molecule has the shortest CC bond, C2H4, C2H2, or C2H6?
(c) Which ion has stronger nitrogen-oxygen bonds, NO3– or NO2–?
SOLUTION:
(a) The Lewis structure for the borate ion shows three boron-oxygen single bonds. Each B–O bond has a bond order of 1.
3–

O
O

B

O

(b) Bond length decreases with increasing bond order. C2H4 has the highest carbon-carbon bond order and the shortest carbon-carbon bond.
H

H

C

C

H

H

H

H

H

H

C

C

H

H

H

C

C

H

CC bond order
1
2
3
(c) Bond energy increases with increasing bond order. NO2– has the highest nitrogen-oxygen bond order and the strongest nitrogen-oxygen bond.


O
O

N

O

O

NO bond order in NO3– =



O
N

O

O

4 NO bonding pairs
= 1.3
3 NO bond locations



O

N

O

NO bond order in NO2– =



O

N

O

3 NO bonding pairs
= 1.5
2 NO bond locations



O
N

O

Chapter 8

Covalent Bonding and Molecular Structure

8-13

OWL Example Problems
8.13 Bond Order, Length, Strength

Bond Energy and Enthalpy of Reaction
Bond energy values can be use to calculate the enthalpy change for gas-phase reactions.
First assume all chemical bonds are broken in reactant molecules, creating isolated atoms. Next, rearrange the atoms to form products and new chemical bonds. The first step involves breaking bonds and is an endothermic process, and the second step involves forming bonds and is an exothermic process. The bond energies are summed as shown in equation 8.2.
ΔHº = Σ(energies of bonds broken) – Σ(energies of bonds formed)

Chapter Goals Revisited
• Predict bond properties.
Use bond energy to calculate reaction enthalpy. (8.2)

EXAMPLE PROBLEM: Bond Energy and Enthalpy of Reaction
Calculate the enthalpy change for the following gas-phase reaction.
C2H6(g) + Cl2(g) → C2H5Cl(g) + HCl(g)
SOLUTION:
Draw Lewis structures for reactants and products and use them to list all reactant bonds broken and product bonds formed, along with bond energies from Table 8.x.
H
H

Reactants:
Bonds broken
6 mol C—H
1 mol C—C
1 mol Cl—Cl

H

C

C

H

H

H
H

H

and

Cl

Cl

6 × 414 kJ/mol
347 kJ/mol
243 kJ/mol

Products:
Bonds formed
5 mol C—H
1 mol C—C
1 mol C—Cl
1 mol H—Cl

H

C

C

H

H

Cl

and

H

Cl

5 × 414 kJ/mol
347 kJ/mol
330 kJ/mol
431 kJ/mol

Use equation 8.x to calculate ΔHº for the reaction.
ΔHº = [6 mol(414 kJ/mol) + 1 mol(243 kJ/mol) + 1 mol(347 kJ/mol)]
– [5 mol(414 kJ/mol) + 1 mol(347 kJ/mol) + 1 mol(330 kJ/mol) + 1 mol(431 kJ/mol)]
= –104 kJ
Notice that five of the C–H bonds and the C–C bond in C2H6 are unchanged in the reaction. The enthalpy of reaction can also be calculated using only the energies of the bonds that are broken (1 mol C–H and 1 mol Cl–Cl) and the bonds that are formed (1 mol C–Cl and 1 mol H–Cl).
ΔHº = [1 mol(414 kJ/mol) + 1 mol(243 kJ/mol)] – [1 mol(330 kJ/mol) + 1 mol(431 kJ/mol)]
= –104 kJ
OWL Example Problems
8.14 Bond Energy and Enthalpy of Reaction: Simulation
815 Bond Energy and Enthalpy of Reaction

8.5 Electron Distribution in Molecules
OWL Opening Exploration
8.X

Lewis structures give chemists one method for visualizing the valence electron arrangement in molecules and ions. However, these structures do not fully represent the way electrons are distributed. Calculated formal charge and electronegativity are two

Chapter 8

Covalent Bonding and Molecular Structure

8-14

tools chemists use, in addition to a Lewis structure, to more accurately describe the electron distribution in a molecule.
Formal Charge
The formal charge of an atom in a molecule or ion is the charge it would have if all bonding electrons were shared equally. Formal charge is a “bookkeeping” method of showing electron distribution in a Lewis structure and while it does not give a completely accurate picture of charge distribution, it is helpful in identifying regions with a large positive or negative charge buildup. Consider the Lewis structure for BCl3, and in particular the electrons represented by one of the B-Cl bonds.

Chapter Goals Revisited
• Understand charge distribution in molecules.
Calculate formal charge. The formal charge on the chlorine atom in the B—Cl bond is a measure of the relationship between the chlorine valence electrons (its group number) and the number of bonding and nonbonding electrons assigned to chlorine in the Lewis structure. In calculating formal charge, we assume that all bonding electrons are shared equally between the atoms in the bond and therefore that each atom gets half of them (Equation
8.3).
Formal charge = (number of valence electrons)
– [(number of nonbonding electrons) + ½(number of bonding electrons)]

Cl formal charge = (7) – [6 + ½(2)] = 0
B formal charge = (3) – [0 + ½(6)] = 0

(8.3)

Imagine each bond is split in half and the bonding electrons are shared equally between bonded atoms.

In BCl3, all atoms have a formal charge of zero and we predict that there is no buildup of positive or negative charge in the molecule. Formal charge, however, assumes all bonding electrons are shared equally. As you will see in the following section, other methods that take into account the unequal sharing of bonding electrons can give a more accurate picture of electron distribution in a molecule.
The sum of the formal charges for all atoms in a molecule or ion is equal to the charge on the molecule or ion, as shown in the following example.
EXAMPLE PROBLEM: Formal Charge
Use the formal charge for each atom in the cyanide ion, CN–, to predict whether H+ is more likely to attach to carbon or nitrogen when forming hydrocyanic acid.
SOLUTION:
First draw the Lewis structure of CN–.

C

N

Use equation 8.x to calculate the formal charge for carbon and nitrogen.
C formal charge = (4) – [2 + ½(6)] = –1
N formal charge = (5) – [2 + ½(6)] = 0
Notice that the sum of the formal charges is equal to the overall charge on the ion (–1).
When H+ attaches to CN–, it is more likely to attach to the atom with the negative formal charge, carbon, forming H–C≡N.
OWL Example Problems
8.18 Formal Charge: Tutor

Chapter 8

Covalent Bonding and Molecular Structure

8-15

Bond Polarity
In a covalent bond, electrons are attracted to two nuclei, but sometimes one nucleus attracts the electrons more strongly than the other. When one nucleus attracts the electrons more strongly, the bonding electrons are located closer to one nucleus than the other. This creates an uneven distribution of bond electron density and a polar bond (or polar covalent bond). When electrons experience the same attractive force to both nuclei, the bond is nonpolar.
The term “polar” refers to the existence of a dipole, a separation of partial positive
(symbolized δ+) and partial negative (symbolized δ–) charge within a bond or a molecule. The partial charges are due to an uneven distribution of electron density, not a transfer of electrons between atoms to form ions. A dipole is indicated using an arrow that points at the negative end of the dipole and has a cross to indicate the positive end of the dipole.
!+
!–

X

Y

Consider a fluorine-fluorine bond and a carbon-fluorine bond.
!+

F

F

!–

C

F

In the fluorine-fluorine bond, the bonding electrons are attracted equally to both fluorine nuclei. The electron density is therefore evenly distributed between the fluorine atoms and the bond is nonpolar. The carbon-fluorine bond, in contrast, is polar because the bonding electrons are attracted more strongly to fluorine than carbon. This results in electron density that is closer to fluorine than to carbon. Recall that fluorine has a higher effective nuclear charge than carbon and its valence atomic orbitals are lower in energy than those of carbon. A partial negative charge occurs at fluorine due to this uneven electron density distribution, and a corresponding partial positive charge happens at the electron deficient carbon.
The ability of an atom in a molecule to attract electrons to itself is its electronegativity, χ
(Figure 8.5). Electronegativity, a concept first proposed by Linus Pauling in 1932, is a relative scale where the most electronegative element (fluorine) is assigned a value of
4.0.

Figure 8.5 Electronegativity Values

Chapter Goals Revisited
• Understand charge distribution in molecules.
Use electronegativity to predict bond polarity. Flashback
7.4 Properties of Atoms

Chapter 8

Covalent Bonding and Molecular Structure

8-16

Notice in Figure 8.5 that
• the noble gases (Group 8A) are not assigned electronegativity values (most of these elements do not form covalent bonds);
• electronegativity values increase moving left to right and decrease moving down the periodic table;
• hydrogen does not follow the periodic trend and has an electronegativity similar to that of carbon;
• metals generally have low electronegativity values and nonmetals generally have high electronegativity values; and
• fluorine, oxygen, nitrogen and chlorine are the most electronegative elements.
The polarity of a chemical bond is related to the difference in electronegativity (Δχ) of the two elements that make up the bond. For covalent bonds between nonmetals, a Δχ greater than zero indicates that the bond is polar and a zero Δχ indicates a nonpolar bond.
For example, the C—F bond described earlier has Δχ = (4.0 – 2.5) = 1.5. It is a polar bond. When Δχ is very large, the interaction between elements has more ionic character than covalent character.
For example, sodium and fluorine have very different electronegativities [F(4.0) – Na(1.0) = 3.0]. When sodium and fluorine react, electrons are transferred resulting in the formation of Na+ and F–.
There is not a defined electronegativity difference that separates ionic from polar bonds.
Instead, there is a continuum moving between these two extremes. All polar bonds have some “ionic character” and some “covalent character.”
EXAMPLE PROBLEM: Electronegativity and Bond Polarity
(a) Which of the following bonds are nonpolar?
C-Cl
H-Cl
H-H
F-F
P-H
S-O
B-F
(b) Which of the following bonds is most polar?
C-Cl
H-Cl
H-I
H-F
P-H
S-O
B-F
SOLUTION:
(a) H-H, F-F and P-H bonds are nonpolar. Any bond between two like atoms is nonpolar because Δχ = 0. The P-H bond is nonpolar by the coincidence that both P and H have the same electronegativity value of 2.1. C-Cl, H-Cl, S-O, and B-F bonds are polar because when two different elements are bonded, the bond is almost always polar.
(b) The B-F bond has the greatest Δχ (4.0 – 2.0 = 2.0) so it is the most polar bond.
OWL Example Problems
8.17 Electronegativity and Bond Polarity

Resonance Structures, Formal Charge, and Electronegativity
When nonequivalent resonance structures exist, the most likely resonance structure is the one with formal charges closest to zero. A general rule of chemical stability is that the localization of positive or negative charges within a molecule is destabilizing. Consider the two possible inequivalent resonance structures for BCl3 shown below along with calculated formal charges.

(0)

(+1)

Cl

Cl

Cl

B

Cl

Cl

B

Cl

(0)

(0)

(0)

(0)

(–1)

(0)

Chapter Goals Revisited
• Understand charge distribution in molecules.
Use charge distribution to evaluate resonance structures.

Chapter 8

Covalent Bonding and Molecular Structure

8-17

The resonance structure on the right is less likely than the one on the left because it has localized formal charges of +1 and –1. In addition, chlorine (χ = 3.0) is more electronegative than boron (χ = 2.0), so a resonance structure that has a positive formal charge on the more electronegative element (and a negative formal charge on the less electronegative element) is probably not stable.
The three nonequivalent resonance structures of CO2 are shown below along with calculated formal charges.

The central structure is the most likely resonance structure for CO2 because it has formal charges of zero on each atom, while both of the other resonance structures have formal charges of +1 and -1 on the O atoms.
The cyanate ion, OCN–, has three nonequivalent resonance structures.
O

C

N

(+1)

(0)



(–2)

O

C

N

(0)

(0)



(–1)

O

C

N

(–1)

(0)



(0)

None of the three resonance structures have formal charges of zero on all atoms. The first resonance structure is not very likely due to the large formal charges (+1 and –2).
The other two resonance structures, however, each have small formal charges. In this case, electronegativity can help determine the best resonance structure.
The
electronegativity of oxygen (3.5) is greater than that of nitrogen (3.0). Therefore, oxygen is more likely to carry a negative formal charge in a Lewis structure, and according to formal charges, the resonance structure on the right is most likely.
EXAMPLE PROBLEM: Resonance Structures and Formal Charge
Three inequivalent resonance structures for the chlorate ion are shown below. Assign formal charges to all atoms in the resonance structures and identify the more likely resonance structure.


O
O

Cl

O

O

A

O

O

O

Cl

O

(0)


O
O

Cl

C

(0)


O
Cl

Cl



O

B

SOLUTION:
Formal charges:
(–1)

O



O

O



O
O

Cl

O

(–1) (+2) (–1)
(–1) (+1) (–1)
(0) (–1) (0)
Structure B is the most likely resonance structure. The formal charges in B are close to zero and the highly electronegative oxygen atom carries a –1 formal charge. Structure A is unlikely because of the large positive formal charge on chlorine.
Structure C is unlikely because the least electronegative element in the ion (chlorine) has a negative formal charge while the most electronegative element in the ion (oxygen) has a formal charge of zero.

Chapter 8

Covalent Bonding and Molecular Structure

OWL Example Problems
8.19 Resonance Structures and Formal Charge

Partial Charge
Calculated formal charges and relative electronegativities are useful tools for assigning charge distribution in a Lewis structure because they help identify likely resonance structures and regions with a buildup of positive or negative charge in molecules and ions. They do not, however, give an accurate picture of the actual electron distribution in molecules. Modern computer modeling programs calculate partial charges on atoms that give a more accurate picture of electron distribution. For example, the calculated partial charges on the atoms in carbon dioxide and the cyanate ion are shown below. Red spheres indicate positive partial charge and yellow spheres indicate negative partial charge. The size of the sphere is proportional to the magnitude of the partial charge. As was true with formal charge, the sum of partial charges is equal to the charge on the molecule or ion.
[O=C=N]–

O=C=O

(–0.26)

(+0.52)

(–0.26)

(–0.54)

(+0.16)

(–0.62)

In CO2, each oxygen carries a negative partial charge and the carbon has a positive partial charge. This reflects the fact that oxygen is more electronegative than carbon but does not reflect the calculated formal charges of zero in the most likely resonance structure for
CO2. However, the fact that both oxygen atoms have negative calculated partial charges suggests that the two resonance structures that have a positive formal charge on oxygen are very unlikely. The calculated partial charges also show that the CO bonds are polar, results that are supported by electronegativity values but not by formal charge calculations. In the cyanate structure, both oxygen and nitrogen carry approximately the same negative partial charge and carbon has a slightly positive formal charge. None of the three OCN– resonance structures show a negative formal charge on both nitrogen and oxygen, but two of the resonance structures have a –1 formal charge on either nitrogen or oxygen.
The calculated partial charges for OCN– suggest that both of the resonance structures with –1 formal charges on nitrogen and oxygen are important and both contribute to the resonance hybrid for OCN–.

8.6 Valence-Shell Electron-Pair Repulsion Theory and
Molecular Shape
OWL Opening Exploration
8.20 Exploring Geometry 1
8.21 Exploring Geometry 2
Lewis electron dot structures show the atom connectivity and number of bonds and lone pairs in a molecule or ion but do not provide information about the three-dimensional shapes of molecules. For example, numerous experiments have shown that water, H2O, has a bent (non-linear) shape. The Lewis structure, however, can be drawn to show a linear arrangement of atoms.

8-18

Chapter 8

Covalent Bonding and Molecular Structure

H

O

8-19

H

The shape of a molecule or ion is one of its most important characteristics because it controls many other properties such as melting and boiling points and chemical reactivity. A variety of experimental techniques such as X-ray crystallography and spectroscopy are used to determine precise atom distances, bond angles, and the shape of a molecule. However, predicting the shape of a molecule or ion from its Lewis structure is one of the most useful tools a chemist has.
The valence shell electron pair repulsion (VSEPR) theory allows chemists to easily predict the shapes of molecules and ions made up of nonmetals. According to VSEPR theory, • positions around a central atom are occupied by either nonbonding electrons
(single or pairs of electrons are counted as one nonbonding position) or bonding electrons (single or multiple bonding pairs are counted as one bonding position);
• the number of structural positions around a central atom is equal to the total number of nonbonding and bonding positions occupied around the atom;
• the electrons in structural positions repel each other and are arranged so as to avoid one another as best as possible;
• the electron-pair geometry is the arrangement of the structural positions around the central atom; and
• the shape (also called molecular geometry) is the arrangement of atoms around the central atom (ignoring the nonbonding positions on the central atom).
Electron-Pair Geometry
The electron-pair geometry is defined by the electrons (bonding and nonbonding) around the central atom. In large molecules with multiple central atoms, we will describe the electron-pair geometry around each central atom. When two, three, four, five, or six pairs of electrons are arranged around a central atom, the electrons are arranged in one of the following ideal electron-pair geometries (Figure 8.6).
2 Structural Positions 3 Structural Positions 4 Structural Positions 5 Structural Positions 6 Structural Positions

Figure 8.6 Ideal Electron-Pair Geometries

Figure 8.6 also shows the bond angles that are characteristic for each ideal electron-pair geometry. Bond angle is the angle formed by the nuclei of two atoms with a central atom at the vertex. The bond angles shown in Figure 8.6 are ideal angles and most molecules have bond angles that are very close to these values.

Chapter Goals Revisited
• Use VSEPR theory
Identify electron-pair geometry. Chapter 8

Covalent Bonding and Molecular Structure

8-20

To determine the electron-pair geometry of a molecule with a single central atom,
1.

Draw the Lewis structure.

2.

Sum the number of structural positions (the total number of nonbonding and bonding positions) around the central atom.

3.

Use this total number of structural positions to identify the electron-pair geometry using the ideal geometries shown in Figure 8.6.
OWL Example Problems
8.22 Ideal Electron Pair Shapes: Tutor

Shape (Molecular Geometry)
The shape of a molecule (its molecular geometry) is defined by the atom positions in a molecule. The shape is the same as the electron-pair geometry when all structural positions are occupied by bonding electrons.
For example, boron trichloride (BCl3) has a trigonal planar electron-geometry and a trigonal planar shape. There are three structural positions around the central boron atom and all are occupied by bonding electrons.

Cl
Cl

B

Cl

Carbon dioxide, CO2, has a linear electron-geometry and a linear shape. There are two structural positions around the central carbon atom and both are occupied by bonding electrons. O

C

O

here are many molecules for which the shape that is not the electron-pair geometry. For example, ozone (O3) has a trigonal planar electron-pair geometry and a bent shape.
There are three structural positions around the central oxygen atom, but two are occupied by bonding electrons and one is occupied by nonbonding electrons. Because shape is defined by the positions of the atoms in a molecule not the electrons, ozone has a bent shape and not a trigonal planar shape.

O

Notice that a
Lewis
structure is not always drawn to show the shape of the molecule or the correct bond angles.

O

O

Figure 8.7 shows the shapes for molecules with 3–6 structural positions and varying numbers of nonbonding electron pairs on the central atom.

Chapter Goals Revisited
• Use VSEPR theory
Identify molecular geometry. Chapter 8

Covalent Bonding and Molecular Structure

8-21

Number of Nonbonding Positions
0
Trigonal planar

3

1

2

3

Bent

120º
Tetrahedral

5

Trigonal bipyramidal

6

Number of Structural Positions

4

Octahedral

Figure 8.7 Shapes of Molecules With 3–6 Structural Positions

EXAMPLE PROBLEM: Determining Molecular Shapes
Determine the shape of
(a) ICl5
(b) SO32–
SOLUTION:
Draw the Lewis structure of each species and use Figure 8.x to determine the shape.
(a) Square pyramid. ICl5 has 6 structural positions around the central atom. Five are occupied by bonding electrons and one is occupied by nonbonding electrons.
Cl
Cl

Cl
I

Cl

Cl

(b) Trigonal pyramid. SO32– has 4 structural positions around the central atom. Three are occupied by bonding electrons and one is occupied by nonbonding electrons.
2–

O
O

S

O

Chapter 8

Covalent Bonding and Molecular Structure

OWL Example Problems
8.23 Determining Molecular Shapes: Tutor
8.24 Determining Molecular Shapes: Exercise
8.23 Determining Molecular Shapes
When determining the shape of a molecule with lone pairs in structural positions, it is important to note the different steric requirements of lone pairs and bonding pairs. Lone pairs are diffuse because they are attracted to a single nucleus, while bonding pairs are localized between and attracted to two nuclei. Lone pairs therefore occupy more space around a central atom than bonding pairs. Consider SF4, which has four structural positions occupied by bonding pairs and one occupied by a lone pair. The five structural positions are arranged in a trigonal bipyramid. The lone pair occupies an equatorial position because this minimizes repulsive forces between the larger, diffuse lone pair and the bonding pairs (Figure 8.8(a)). If the lone pair occupies an axial position (Figure
8.8(b)), the close contact (90º angles) between the lone pair and bonding pairs increases repulsive forces making this configuration less favorable.

(a)

(b)
Figure 8.8 SF4 shape with the lone pair in (a) equatorial and (b) axial position

Thus lone pairs occupy equatorial positions in a trigonal bipyramidal electron-pair geometries Similarly, because lone pairs occupy positions opposite each other in an octahedron because this also minimizes repulsive forces.
Molecular shapes predicted using the VSEPR ideal electron pair shapes are usually very close to the shapes measured by experiment, especially when all terminal atoms are identical. However, when lone pairs are present on the central atom, bond angles in particular differ from predicted values. Consider the bond angles in NH3 and H2O
(Figure 8.9). Both molecules have the same electron-pair geometry (tetrahedral) but they differ in the number of lone pairs on the central atom. As the number of lone pairs on the central atom increases, the H-X-H bond angle decreases.

Figure 8.9 H–X–H bond angles CH4, NH3, and H2O

8-22

Chapter 8

Covalent Bonding and Molecular Structure

8-23

8.7 Molecular Polarity
OWL Opening Exploration
8.27

As we saw earlier, covalent bonds are polar when there is an uneven attraction for electrons between the bonded atoms. Polar bonds in a molecule can result in a polar molecule, which affects the physical properties of a compound. For example, polar molecules are often very soluble in water while nonpolar molecules are not.
For a molecule to be polar, it must contain polar bonds and those bonds must be arranged so that there is an uneven charge distribution. The following steps are used to predict whether a molecule is polar.
1.
2.
3.
4.

Draw the Lewis structure.
Determine the shape.
If the molecule has polar bonds, indicate them on the molecule.
Use the shape and bond dipoles to determine if there is an uneven distribution of bond electron density in the molecule.

For example, boron trifluoride, BF3, is a nonpolar molecule that has polar bonds.
F
B
F

F

The shape of BF3 is trigonal planar and the B—F bonds are polar (fluorine is the negative end of the bond dipole). As shown above, the three bond dipoles are arranged symmetrically in the molecule and do not result in an uneven distribution of electron density. BF3 is therefore nonpolar.
Trifluoromethane is a polar molecule.
H
C
F

F
F

The shape of CHF3 is tetrahedral and the C—F and C—H bonds are polar. Carbon is the negative end of the C—H bond dipole, and fluorine is the negative end of each C—F bond dipole. When we draw in the individual bond dipoles, we see that there is a net movement of bond electron density away from hydrogen and toward the fluorine atoms.
This uneven electron density distribution means that CHF3 is polar, and the molecule has a net dipole with the positive end near hydrogen and the negative end near the fluorine atoms. !+

!–

Chapter Goals Revisited
• Identify polar and nonpolar species.
Predict molecular polarity using molecular shape and electron distribution.

Chapter 8

Covalent Bonding and Molecular Structure

8-24

This method is also useful for larger, more complex molecules. For example, consider acetone, CH3COCH3.
O
H

C

H

C
H

C
H

H

H

The carbon-carbon bonds in acetone are nonpolar, the carbon-hydrogen bonds are slightly polar (carbon is the negative end of the bond dipole) and the carbon-oxygen bond is polar (oxygen is the negative end of the bond dipole). There is a net shift of electron density towards oxygen, so the molecule is polar. The net molecular dipole is shown below. EXAMPLE PROBLEM: Determining Polarity of Molecules
Determine if each of the following is polar.
(a) SO3
(b) BrF3
SOLUTION:
Draw the Lewis structure for each molecule, determine the shape, and label any polar bonds. Use shape and bond polarity to determine if the molecule is polar.
(a) Nonpolar. SO3 has a trigonal planar shape and all three SO bonds are polar. The three bond dipoles are arranged symmetrically in the molecule, so the molecule is not polar.
O
S
O

O

(b) Polar. BrF3 has a T-shaped shape and all three BrF bonds are polar. The three bond dipoles are not arranged symmetrically around the molecule, so the molecule is polar. The bromine is the positive end of the net dipole and one of the fluorine atoms is the negative end of the bond dipole.
F

F
F

Br

F

F

Br

F

OWL Example Problems
8.28 Determining Polarity of Molecules: Exercise
8.28 Determining Polarity of Molecules

Chapter 8

Covalent Bonding and Molecular Structure

8-25

Chapter Review
OWL Summary Assignments
8.29 Chapter Review
8.30 Challenge Problems

Key Equations
Force !

(qA )(qB ) r2 ΔHº = Σ(energies of bonds broken) – Σ(energies of bonds formed)

(8.1)
(8.2)

Formal charge = (number of valence electrons)
– [(number of nonbonding electrons) + ½(number of bonding electrons)]

Key Terms
8.1

Interactions Between Particles: Coulomb’s Law

Coulomb’s Law ionic bonding
Covalent bonding
Metallic bonding intermolecular forces

8.3
Lewis Structures
Lewis structure
Lewis symbol lone (electron) pairs single bond double bond triple bond octet rule electron deficient free radicals expanded valence resonance structures resonance hybrid
8.4
Bond Properties bond order bond length bond energy
8.5 Electron Distribution in Molecules formal charge polar bond polar covalent bond nonpolar dipole electronegativity, χ partial charges valence shell electron pair repulsion (VSEPR) structural positions electron-pair geometry shape (molecular geometry) electron-pair geometry bond angle

(8.3)

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...Centre for Foundation Studies, UTAR 1 2 Chapter Scopes • Endothermic & Exothermic reactions • Enthalpy changes: ∆H of formation, combustion, hydration, neutralization, atomization. CHAPTER 5 Chemical Energetic / Thermochemistry • Lattice energy, electron affinity • Heat of fusion and vaporization • Hess’ Law • Born-Haber cycles • Calorimetry © 2006 Brooks/Cole - Thomson © 2006 Brooks/Cole - Thomson 3 Energy & Chemistry 4 Thermochemistry • Thermochemistry is the study of heat (energy) change/transfer in a chemical reaction. • ENERGY is the capacity to do work or transfer heat. • HEAT is the transfer of thermal energy between two objects because of their difference in temperature. Heat energy is associated with molecular motions. Other forms of energy light electrical kinetic and potential Heat transfers until thermal equilibrium is established. ∆T measures energy transferred. © 2006 Brooks/Cole - Thomson © 2006 Brooks/Cole - Thomson System and Surroundings 5 System and Surroundings 6 Vacuum jacket • SYSTEM – The object under study • SURROUNDINGS – Everything outside the system © 2006 Brooks/Cole - Thomson FHSC1114 Physical Chemistry open Exchange: mass & energy closed energy isolated nothing © 2006 Brooks/Cole - Thomson 1 Centre for Foundation Studies, UTAR Directionality of Heat Transfer 7 Directionality of Heat Transfer • Heat always......

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Investigation of Action of Saliva and 3 M Hydrochloric Acid in Two Carbohydrate Solutions

...enzymes in human cells. Above this temperature the enzyme structure begins to break down (denature) since at higher temperatures intra- and intermolecular bonds are broken as the enzyme molecules gain even more kinetic energy.Also Enzymes have an active site. This is part of the molecule that has just the right shape and functional groups to bind to one of the reacting molecules. The reacting molecule that binds to the enzyme is called the substrate.At high temperature,the active site is said to be denatured.Therefore,the active site cannot bind with any subtrate anymore.Amylase breaking down the starch suspension into maltose and maltose into glucose when HCL was added into solution B .This is because solution B was hydrolyzed and the H+ ions present break down the bond in between molecules of the solution B The HCl will denature (unfold) the amylase (as noted at the bottom of the link you provided), disrupting...

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Investigation of Action of Saliva and 3 M Hydrochloric Acid in Two Carbohydrate Solutions

...Title: To investigate the trajectory of a small ball as it rolls off a surface which is inclined to the horizontal. Objectives: To investigate the trajectory of a two dimensional motion. Apparatus and materials: Ramp, wooden block, pendulum bob, plumb line, steel ball, wooden board, carbon paper, meter ruler, plasticine. Setup: 1. A ramp was set up at the edge of a bench. 2. A plum-line is suspended from the edge of the bench. 3. A wooden board is mounted horizontally using two clamps so that the board is situated about the bottom of the ramp. 4. A sheet of blank paper is placed on top of the board. 5. A piece of carbon paper is placed on the top of the blank paper. The ink-side of the carbon paper is facing down. 6. When a ball was released from the top of the ramp, the ball travelled through a trajectory. Theory: Let: g= 9.80ms-2 u= speed of the ball as it leaves the ramp k= constant y= vertical distance (between the bottom of the ramp and the top of the board) x= horizontal distance (between the plum-line and mark on the paper) The equation which relates x and y is yx=g1+k2x2u2+k Procedure : 1. The ball was positioned at the top of the ramp. The ball was released so that it rolls down the...

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Lop of Lop

...the two solutions with Iodine test and ... Discussion: ... Why You Add Hydrochloric Acid in Hydrolysis of Starch. Ingestion of saliva during carbohydrate feeding by ... - SciELO www.scielo.br/scielo.php?pid=S0074-02762006000100016&script... by RR Cavalcante - ‎2006 - ‎Cited by 5 - ‎Related articles Saliva ingestion by phlebotomine during the carbohydrate ingestion phase is ... The presence of saliva in each type of solution or substrate offered, as well as ... 0.2 ml of apyrase assay buffer pH 8 (50 mM Tris/HCl buffer containing 1 mM CaCl2, ... Saliva ingestion occurred under each condition investigated, as indicated by ... Experiment: investigation of action of saliva and hydrochloric ... https://answers.yahoo.com/question/index?qid... Jun 25, 2012 - 1) Name of enzyme involved 2)specific action(s) of enzymes involved. Two hydrolytic enzymes and an epistemological–historical ... www.scienceinschool.org/2007/issue4/enzymes Science in School Sep 3, 2007 - 5 M sodium hydroxide (NaOH); 5 M hydrochloric acid (HCl); Saliva. ... To demonstrate the test methods, test all four carbohydrate solutions ... Discussion ... more concentrated than is necessary for the activity and that using a ... The authors have recommended the Fehling’s test for this investigation. What is the objectives of investigation of action of saliva and ... www.answers.com › Wiki Answers › Categories › Science › Biology ... of investigation of action of saliva and diluted hydrochloric acid in two ...

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...Practical 3 Investigation of Action of Saliva and Hydrochloric Acid in Two Carbohydrate Solution | Objective: 1. To show the action of saliva in two carbohydrate solutions. 2. To show the action of hydrochloric acid in two carbohydrate solutions. Apparatus & Equipment’s: Boiling tubes Metal test tube racks Beaker Graduated plastic dropper Water bath,~37°C Water bath,~95°C Stop watch Test tube holder Materials: Carbohydrate solution A Carbohydrate solution B Benedict’s solution 3M Hydrochloric acid 3M Sodium hydroxide Procedures: 1. Prepared two boiling tubes with containing 1 ml solution A and 1 ml solution B respectively. 1 ml Benedict’s solution was added to each tube and heated both tubes together in the (~95°C) water bath for two minutes. Then, recorded the results in table 1. 2. Added a few drops of fresh solution A and B separately spaced on a white tile. On each solution, added 1-2 drops of iodine solution and mixed with pen cover. Recorded your observations in the table 1. 3. Pipetted 2 ml solution B into each of four boiling tubes. The tubes were labelled 1, 2, 3 and 4 respectively near mouth of tube. Labelled your group name. 4. Placed tubes 1 and 2 in a water bath of ~37°C. 5. Salivated into a small beaker until it reached about 5 ml. 6. At the same time, step (6) and (7) was to be done approximately. Measured out 4 ml of the saliva prepared in step (4) and pipetted 2 ml each into tubes 1 and 4. The contents of the tubes shook well to......

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...TITLE OF LAB REPORT | Practical 3: Investigation of Action of Saliva and 3M Hydrochloric Acid in Two Carbohydrate Solutions | LECTURER’S NAME | Ms. Bong Siew Mee | Title: Investigation of Action of Saliva and 3M Hydrochloric Acid in Two Carbohydrate Solutions Objective: To investigate the action of saliva and 3M hydrochloric acid in two carbohydrate solutions. Results: Table 1: Observations made when two carbohydrate solutions provided in laboratory were tested with Benedict and Iodine solution. | Observations | Conclusions | Solution A | Benedict’s test: An initial blue translucent mixture turned to brick-red opaque solution and moderate amount of precipitate settled after heated at a high temperature for two minute. | Presence of reducing sugar | | Iodine test: The translucent colouration of the mixture retained its yellowish-brown colour. | Absence of starch | Solution B | Benedict’s test: The translucent colouration of the mixture remained its blue colour. | Absence of reducing sugar | | Iodine test: The initial yellowish-brown translucent mixture turned to bluish-black opaque solution when solution was mixed. | Presence of starch | Table 2: Colour reactions of Benedict’s test for saliva and 3M hydrochloric acid in two carbohydrate solutions provided in laboratory. ......

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...Topic: Investigation of Action of Saliva and 3 M Hydrochloric Acid In Two Carbohydrate Solution Objective: To investigate the action of saliva and 3 M Hydrochloric Acide in two Carbohydrate Solution Apparatus & Equipment: Boiling tubes Beaker Graduated plastic dropper Water bath, (~ 37oC) Water bath, (~ 95oC) Materials: Carbohydrate solution A Carbohydrate solution B Benedict’s solution 3 M Hydrochloric acid 3 M Sodium hydroxide Iodine solution Procedures: Part 1 1. Two boiling tubes were prepared containing 1 ml solution A and 1 ml solution B respectively. 1 ml Benedict’s solution was added to each tube. Both tubes were heated together in the (~95oC) water bath for two minutes. The results were recorded in table 1. 2. Few drops of solution A and B were added separately on a white tile. On each solution, 1-2 drops of iodine solution were added. The observations were recorded in the table 1. Part 2 3. The boiling tubes were labelled 1, 2, 3 and 4. 2 ml of solution B were pipette into each of four boiling tubes. 4. Tubes 1 and 2were placed in a water bath of ~37oC to heat up the solution. 5. The saliva was salivated into a small beaker until it reaches approximate 5 ml. 6. This step was done approximately at the same time. 2 ml of saliva prepared were pipette into tubes 1 and 4. The contents of the tubes were shaken well to ensure thorough mixing. 2 ml of HCl were pipette into tubes 2 and 3. 7. Tubes 1, 2, 3 & 4 were incubated at......

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...Title Identification of Unknown Carbohydrate Solution and Investigation of Action of Saliva and Hydrochloric Acid in Carbohydrate Solution at Two Different Temperatures Objective To investigate the action of saliva and 3 hydrochloric acid in two carbohydrate solution Apparatus Boiling tubes, Metal test tube racks, Water bath, 37-40 Degree Celsius, Water bath, 90-95 Degree Celsius, Beaker, Dropper, Wooden Holder Materials Carbohydrate solution A , Carbohydrate solution B , Benedict’s solution , 3 M Hydrochloric acid, 3 M Sodium Hydroxide Procedures 1. Two boiling tubes are filled with 1 ml solution A and 1 ml solution B respectively. 1 ml of Benedict’s solution is added into each test tube. Both tubes are heated together in the hotter (90-95°C) water bath in two minutes. The results are recorded in Table 1. 2. A few drops of fresh solution A and B are spaced separately on a white tile. I2/KI solution (iodine) is added 1-2 drops on each solution. The solution is mixed with a glass rod on the tile. The observations are recorded in the Table 1. 3. 2ml solution B is added into each of four test tubes and the tubes 1, 2, 3, and 4 is labelled respectively with labelling paper near mouth of tube. 4. Tubes 1 and 2 is placed in a water bath of 37°C. 5. A measuring cylinder is salivated till it reached 5ml. 6. 2ml of saliva is added into tubes 1 and 4. The contents of the tubes are shaked well to ensure thorough mixing. 7. 4ml of HCL is......

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