chapter 8. basic concepts of chemical bonding - nina rodriguez - … · 2020. 12. 30. · chapter...

Chapter 8. Basic Concepts of Chemical Bonding

8.1 Chemical Bonds, Lewis Symbols, and the Octet Rule

Read p. 298-301

Learning Objective Essential Knowledge Unit 2.1 SAP-3.A Explain the relationship between the type of bonding and the properties of the elements participating in the bond.

SAP-3.A.4 The difference in electronegativity is not the only factor in determining if a bond should be designated as ionic or covalent. Generally, bonds between a metal and nonmetal are ionic, and bonds between two nonmetals are covalent. Examination of the properties of a compound is the best way to characterize the type of bonding.

The likelihood that the 2 elements will form a chemical bond is determined by the interactions between the VE and the nuclei of elements. When atoms or ions are strongly attracted to one another, we say that there is a chemical bond between them.

• In chemical bonds, electrons are shared or transferred(lost/gain) between atoms. Types of chemical bonds include: • IONIC bonds (electrostatic forces that hold ions together, e.g., NaCl, TiO2); • COVALENT bonds (result from sharing electrons between atoms, e.g., Cl2, CO2); • METALLIC bonds (refers to metal nuclei floating in a sea of electrons, e.g., Na) Ch 12!

Ch 8: Chemical Bonding 2

Lewis Symbols

The electrons involved in bonding are called VALENCE electrons. • Valence electrons are found in the incomplete, outermost shell of an atom. As a pictorial understanding of where the electrons are in an atom, we represent the electrons as dots around the symbol for the element.

• The number of valence electrons available for bonding are indicated by unpaired dots. • These symbols are called LEWIS SYMBOLS, or Lewis electron-dot symbols.

The Octet Rule

• OCTET RULE: Atoms tend to gain, lose, or share electrons until they are surrounded by EIGHT valence electrons.

o Exceptions: Hydrogen & Helium atoms has a full valence shell of 2e-. • An octet consists of full s and p subshells. • We know that s2p6 is a noble gas configuration. • We assume that an atom is stable (favorable) when surrounded by eight electrons (four

electron pairs). LDS of ATOMS: To draw the LDS (Lewis Dot Structure) for atoms, simply imagine a square around the element symbol and then place the valence electrons around the symbol. Remember Hund’s Rule: put one in each spot, then double up! LDS of IONS:

1. Determine how many valence electrons the ion will have! • Most metals will lose all their valence (“s” and “p”) electrons, or you will be told

what the charge will be. • Non-Metals will gain enough electrons to fill their “p” sublevel.

2. Place brackets around your Lewis Dot Structure. 3. Add the charge as a superscript on the upper right side outside of the brackets.

3

1. Write the Lewis symbol for atoms of each of the following elements:

(#13) a. Al b. Br c. Ar d. Sr

(#14) a. K b. As c. Sn +2 d. N-3

8.2 Ionic Bonding

Read p.301-306

Consider the reaction between sodium and chlorine:

Na(s) + ½Cl2(g) → NaCl(s) ΔH°f = –410.9 kJ/mol

• The reaction is violently exothermic. • We infer that the NaCl (s)is more stable than its constituent elements. • Sodium has lost an electron to become Na+ and chlorine has gained the electron to

become Cl–. • Note that Na+ has an Ne electron configuration and Cl– has an Ar configuration. • That is, both Na+ and Cl– have an octet of electrons.

Remember Ch 7 atomic properties that were discussed that gives us indication of how readily electron transfer occurs:

• Electron transfer to form oppositely charged ions occurs when one atom readily gives up a electron (metals since they have low ionization energy) and another atom readily gains an electron (nonmetals since they have high electron affinity).

• NaCl is a typical ionic compound – a metal Na (low IE) transfers a valence electron to nonmetal Cl (high EA).

Examples


IONIC COMPOUNDS LEWIS DOT STRUCTURES 1. Draw the Lewis dot structure for the positive ion with charge and brackets.

2. Draw the Lewis dot structure for the negative ion with charge and brackets.

3. Continue to draw positive and negative ions until the charges cancel themselves out to

ZERO = ELECTRICALLY NEUTRAL!

4. It is proper notation to alternate positive and negative ions.

1. Draw the Lewis Dot Structure for the following ionic compounds. (See p.301)

IONIC

COMPOUND

LEWS DOT Link to prior knowledge:

WRITE THE

FORMULA UNIT!

sodium

chloride

The arrow indicates the transfer of an electron from the Na atom to the Cl atom. Each ion has an octet of electrons, the octet being the 2s 2 2p6 electrons that lie below the single 3s

valence electron of the Na atom. We need to put a bracket around the chloride ion to emphasize that all eight electrons are located on it.

calcium

fluoride

aluminum

oxide

Examples

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2. Atoms of Mg combine with atoms of F to form a compound. Atoms of which other elements would you except to combine with atoms of F in the same ratio?

3. Element X has an electron configuration of 1s2

2s2

2p5

. What is the most likely compound

that will be formed between lithium and element X?

Structure determines function of Ionic bonds. Ionic bonds have the following characteristics:

PROPERTY EXPLANATION High melting (Tm) and boiling (Tb) points – most are SOLIDS at room temperature

Large difference in electronegativities

give elements in ionic compounds a

strong electrostatic force of attraction. A

large amount of energy is required to

break the strong coulombic forces

between oppositely charged ions. Brittle and Cleave when struck

Caused by ion-ion repulsion

Many (but not all) are SOLUBLE in water

When soluble, ions are attracted to the partial charges on water (Remember opposites attract)

Non-conductive as SOLIDS

Movement of electrons require MOBILE IONS - ionic solids are rigid

Conductive in LIQUID and AQUEOUS (dissolved in water) states

Ions become mobile and allow electron movement


IONIC Structure: NaCl forms a very regular structure in which each Na+ ion is surrounded by six Cl– ions. (See page 302)

• Similarly each Cl– ion is surrounded by six Na+ ions. • There is a regular arrangement of Na+ and Cl–. • Note that the ions are packed as closely as possible.

Energetics of Ionic Bond Formation

Recall from Ch 7.4 that the loss of electrons from an atom is always an endothermic process. Removing an electron from Na(g) to form Na + (g) , for instance, requires 496 kJ/mol. Recall from Section 7.5 that when a nonmetal gains an electron, the process is generally exothermic, as seen from the negative electron affinities of the elements.

• The heat of formation of NaCl(s) is exothermic:

Na(s) + ½Cl2(g) → NaCl(s) ΔH°f = –410.9 kJ/mol The principal reason ionic compounds are stable is the attraction between ions of opposite charge.

• This attraction draws the ions together, releasing energy and causing the ions to form a solid array, or lattice, such as that shown in Figure 8.3.

• A measure of how much stabilization results from arranging oppositely charged ions in an ionic solid is given by the LATTICE ENERGY, which is the energy required to completely separate ONE MOLE of a SOLID IONIC COMPOUND into its GASEOUS IONS.

• To envision this process for NaCl, imagine that the structure in Figure 8.3 expands from within, so that the distances between the ions increase until the ions are very far apart. This process requires 788 kJ/mol, which is the value of the lattice energy:

Separation of the NaCl into sodium and chloride ions is highly endothermic:

NaCl(s) → Na+(g) + Cl–(g) ΔH°lattice= +788 kJ/mol

The reverse process—to form NaCl(s)—is therefore highly exothermic (ΔH°lattice= -788 kJ/mol).

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Lattice energy depends on the CHARGE on the ions and the SIZE of the ions.

• Page 303 TABLE 8.2 lists the lattice energies for a number of ionic compounds. The large positive values indicate that the ions are strongly attracted to one another in ionic solids.

• The strong attractions also cause most ionic materials to be hard and brittle with high melting points—for example, NaCl melts at 801oC.

Let’s look at the table 8.2 and Figure 8.4 p. 304 to see some trends for lattice energy Same magnitude of charge Same magnitude of charge Same Cation +1, different anion -1 Different Cation +1, same anion -1 Different magnitude of charges Same magnitude of charge Different cation +2, same anion -1 Different Cation +2, same anion -2

Learning Objective Essential Knowledge Unit 1.8 SAP-2.B Explain the relationship between trends in the reactivity of elements and periodicity.

SAP-2.B.3 Typical charges of atoms in ionic compounds are governed by their location on the periodic table and the number of valence electrons.


Coulomb’s Law and Ionic Bonding L2 – Doc Dena Bonding Video

As two atoms approach one another, there may be an attraction between the electrons on one

atom and the positive nucleus on another atom. The inner, or ___________________ electrons

are quite content with their attraction to their own nucleus, but the outer, or

___________________ electrons can sometimes be kind of fickle – especially if the other

nucleus is exceedingly attractive!

Picture:1

1 http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch02/ljpot.jpg

Notice that energy when there is an attraction

between atoms to form a bond. Energy is

released when bonds form, energy is required

to break bond

After the optimal bond length –

the most stable “ r” – the energy

rapidly increases due to nucleus-

nucleus repulsion

Notice that energy when there is an attraction

between atoms to form a bond. Energy is

released when bonds form, energy is required

to break bond

After the optimal bond length –

the most stable “ r” – the energy

rapidly increases due to nucleus-

nucleus repulsion

d

d

d

9

The magnitude of the lattice energy of an ionic solid depends on the charges of the ions, their sizes, and their arrangement in the solid.

The specific relationship is given by Coulomb’s equation: (MATH NOT TESTED ON AP EXAM😊)

• where F is the potential energy (E) of the two interacting charged particles

• q1 and q2 are the charges on the particles • d is the distance between their centers • k is a constant (8.99 x 109 J M/C2).

Coulomb’s law indicates that the attractive interaction between two oppositely charged ions increases as the magnitudes of their charges increase and as the distance between their centers decreases. Force will decrease as the particles get further away!


The lattice energy increases as the charges on the ions (Q1 and Q2) increase and as their radii decrease. The magnitude of lattice energies depends predominantly on the ionic charges because ionic radii vary over only a limited range.

• Because lattice energy decreases as distance between ions increases, lattice energies follow trends of ionic radius learned from Ch 7.

• In particular, because ionic radius increases as we go down a group of the periodic table, we find that, for a given type of ionic compound, lattice energy decreases as we go down a group. (See FIGURE 8.4 on page 304 illustrates this trend for the alkali chlorides).

• The greater the charge on the ion(s) and smaller the ionic radius, the stronger the attractive force.

4. Without consulting Table 8.2, arrange the ionic compounds NaF, CsI, and CaO in order of increasing lattice energy. (p. 303)

Learning Objective Essential Knowledge Unit 2.2 SAP-3.B Represent the relationship between potential energy and distance between atoms, based on factors that influence the interaction strength.

SAP-3.B.3 Coulomb’s law can be used to understand the strength of interactions between cations and anions. a. Because the interaction strength is proportional to the charge on each ion, larger charges lead to stronger interactions. b. Because the interaction strength increases as the distance between the centers of the ions (nuclei) decreases, smaller ions lead to stronger interactions.

Examples

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5. AP Chemistry: 2.1-2.4 Chemical Bonds, Intramolecular Force, and Structure of Solids

WATCH THIS VIDEO BY MICHAEL FARABAUGH TO ASSIST IN THIS QUESTIONS (~33min)

Data for the lattice energy of NaF is given in the table below. Make predictions about the lattice energy of MgO and KCl. Do you predict that the lattice energy of each compound is less than 930 kJ/mol or greater than 930 kJ/mol? Justify your answer in terms of periodic properties and Coulomb’s law.

Reaction Lattice Energy (kJ/mol)

NaF(s) ⟶ Na+(g) + F－

(g) 930

MgO(s) ⟶ Mg2+(g) + O2－(g)

KCl(s) ⟶ K+(g) + Cl－

(g)

Examples

https://www.youtube.com/watch?v=BcRVEB_6PV0&list=PLoGgviqq4845Sy3UfnNh_PljzAptMR7MQ&index=3


Electron Configuration of Ions of the Representative Elements

The energetics of ionic bond formation helps explain why many ions tend to have noble-gas electron configurations.

• For example, sodium readily loses one electron to form , which has the same electron configuration as Ne:

Na: [Ne]3s1 Na+: [Ne]

• Even though lattice energy increases with increasing ionic charge, we never find ionic compounds that contain ions.

• The second electron removed would have to come from an inner shell of the sodium atom, and removing electrons from an inner shell (CORE e-) requires a very LARGE amount of energy.

• Similarly, adding electrons to nonmetals is either exothermic or only slightly endothermic as long as the electrons are added to the valence shell. Thus, a Cl atom easily adds an electron to form , which has the same electron configuration as Ar:

Cl: [Ne]3s23p5 Cl – : [Ne]3s23p6 = [Ar]

• To form a ion, the second electron would have to be added to the next higher shell of the Cl atom, an addition that is energetically very unfavorable. Therefore, we never observe Cl -2 ions in ionic compounds.

Transition-Metal Ions

Because ionization energies increase rapidly for each successive electron removed, the lattice energies of ionic compounds are generally large enough to compensate for the loss of up to only three electrons from atoms.

• We often encounter cations with charges of 1+, 2+ or 3+ in ionic compounds. • However, transition metals cannot attain a noble gas conformation (>3 electrons beyond a

noble gas core). • Transition metals tend to lose the valence shell electrons first (s&p) and then as many d

electrons as are required to reach the desired charge on the ion. • Thus, electrons are removed from 4s before the 3d, etc. • For instance, in forming from Fe +2, which has the electron configuration [Ar]3d6 4s 2 ,

the two 4s electrons are lost, leading to an [Ar]3d6 configuration. o Removal of an additional electron gives Fe +3, whose electron configuration is

[Ar]3d5 .

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8.3 Covalent Bonding

Read p. 306-309

The vast majority of chemical substances do not have the characteristics of ionic materials. • Most of the substances with which we come into daily contact—such as water—tend to

be gases, liquids, or solids with low melting points. • Many, such as gasoline, vaporize readily. • Many are in their solid forms—for example, plastic bags and paraffin.

For the very large class of substances that do not behave like ionic substances, we need a different model for the bonding between atoms.

• G. N. Lewis reasoned that atoms might acquire a noble-gas electron configuration by sharing electrons with other atoms.

• A chemical bond formed by SHARING a pair of electrons is a COVALENT bond. The hydrogen molecule, H2, provides the simplest example of a covalent bond.

• When two hydrogen atoms are close to each other, the two positively charged nuclei repel each other, the two negatively charged electrons repel each other, and the nuclei and electrons attract each other, as shown in FIGURE 8.6(a) p.307.

• Because the molecule is stable, we know that the attractive forces must overcome the repulsive ones.

• Thus, the atoms in H2 are held together principally because the two positive nuclei are attracted to the concentration of negative charge between them. See Figure 8.7 (b) p.307

• In essence, the shared pair of electrons in any covalent bond acts as a kind of “glue” to bind atoms together.


Lewis Structures

Formation of covalent bonds can be represented using Lewis symbols. • The structures are called Lewis structures. • We usually show each electron pair SHARED between atoms as a line (bond) – these

are called BONDING electrons. • Unshared electron pairs as dots (LONE PAIRS – these are NONBONDING

electrons). • Each pair of shared electrons constitutes one chemical bond – covalent bond!

Example: H + H → H : H → H – H The shared pair of electrons connect the two H nuclei.

1. Using Lewis symbols and Lewis structures, diagram the formation of SiCl4. (#33)

Draw the Lewis dot symbols for silicon and chlorine. Then draw molecule

a) How many valence electrons surround Si?

b) How many valence electrons surround each Cl?

c) How may bonding pairs of electrons are in the SiCl4 molecule?

2. Draw the Lewis dot symbols for nitrogen and fluorine. Predict the formula of the stable

binary compound (a compound composed of two elements) formed when nitrogen reacts with

fluorine and draw its Lewis structure for the molecule.

Examples

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3. Compare the Lewis symbol for neon with the Lewis structure for methane, CH4. In what

important way are the electron arrangements about neon and carbon alike? In what important

way are they different? (p.308)

Learning Objective Essential Knowledge Unit 2.2 SAP-3.B Represent the relationship between potential energy and distance between atoms, based on factors that influence the interaction strength.

SAP-3.B.2 In a covalent bond, the bond length is influenced by both the size of the atom’s core and the bond order (i.e., single, double, triple). Bonds with a higher order are shorter and have larger bond energies.

Multiple Bonds

It is possible for more than one pair of electrons to be shared between two atoms (e.g., multiple bonding). BOND ORDERS – the number of bonds shared between 2 atoms.

• One shared pair of electrons is a SINGLE bond (e.g., H2). • Two shared pairs of electrons form a DOUBLE bond (e.g., O2). • Three shared pairs of electrons form a TRIPLE bond (e.g., N2).

BOND LENGTH is the distance between the nuclei of the atoms in a bond. (measured in Angstrom - Å)

• Generally, bond distances decrease as we move from single through double to triple bonds.

• The move SHARED pairs of e-, the SHORTER the bond distance between the atoms.

4. The C – O bond length in carbon monoxide, CO, is 1.13 Å, whereas the bond length in

CO2 is 1.24 Å. Without drawing a Lewis structure, do you think that CO contains a

single, double, or triple bond?

Examples

Examples


8.4 Bond Polarity and Electronegativity

Read p. 309-315

Learning Objective Essential Knowledge

Unit 2.1 SAP-3.A Explain the relationship between the type of bonding and the properties of the elements participating in the bond.

SAP-3.A.1 Electronegativity values for the representative elements increase going from left to right across a period and decrease going down a group. These trends can be understood qualitatively through the electronic structure of the atoms, the shell model, and Coulomb’s law. SAP-3.A.2 Valence electrons shared between atoms of similar electronegativity constitute a nonpolar covalent bond. For example, bonds between carbon and hydrogen are effectively nonpolar even though carbon is slightly more electronegative than hydrogen. SAP-3.A.3 Valence electrons shared between atoms of unequal electronegativity constitute a polar covalent bond.

a. The atom with a higher electronegativity will develop a

partial negative charge relative to the other atom in the

bond.

b. In single bonds, greater differences in electronegativity

lead to greater bond dipoles.

c. All polar bonds have some ionic character, and the

difference between ionic and covalent bonding is not

distinct but rather a continuum.

• When two identical atoms bond, as in Cl2 or H2, the electron pairs must be shared

equally.

• When two atoms from opposites sides of the periodic table bond, such as NaCl, there is

relatively little sharing of electrons, which means that NaCl is best described as

composed of Na+ and Cl- ions.

BOND POLARITY is a measure of how equally or unequally the electrons in any covalent

bond are shared.

• A NONPOLAR COVALENT bond is one in which the electrons are shared equally, as

in Cl2 and N2.

• In a POLAR COVALENT bond, one of the atoms exerts a greater attraction for the

bonding electrons than the other.

• If the difference in relative ability to attract electrons is LARGE enough, an IONIC bond

is formed.

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Electronegativity

We use a quantity called electronegativity to estimate whether a given bond is nonpolar

covalent, polar covalent, or ionic.

Electronegativity is defined as the ability of an atom in a molecule to attract electrons to itself.

• The greater an atom’s electronegativity, the greater its ability to attract electrons to

itself.

The electronegativity of an atom in a molecule is related to the atom’s ionization energy and

electron affinity, which are properties of isolated atoms.

• An atom with a very negative electron affinity and a high ionization energy both

attracts electrons from other atoms and resists having its electrons attracted away;

it is highly electronegative.

• BEWARE: Electronegativity is NOT the same as electron affinity


Electronegativity values can be based on a variety of properties, not just ionization

energy and electron affinity.

• The American chemist Linus Pauling developed the first and most widely used

electronegativity scale, which is based on thermochemical data.

• The scale is based on the ABILITY to attract electrons, NOT the energy needed to gain

an electron (electron affinity).

• The Pauli scale assigns fluorine a value of 4 (HIGHEST)

• Electronegativity increases is you move from left to right across the

periodic table - that is, from the most metallic to the most nonmetallic elements.

o The effective nuclear charge is increasing faster than the electron shielding, so the

attraction that the atoms have for the valence electrons increases.

o As the number of protons INCREASES, attraction of outer electrons by the

nucleus INCREASES and the attraction for electrons INCREASES

• Electronegativity decreases as you move down a group on the periodic

table - With some exceptions (especially in the transition metals).

o There is an increasing number of energy levels and thus electrons are farther away

from the nucleus and are held less tightly.

o This is what we expect because we know that ionization energies decrease with

increasing atomic number in a group and electron affinities do not change very

much.

o Lower values for “n” indicate the outer electrons involved in the bond are

closer to the nucleus and thus experience a stronger attraction by the protons in

the nucleus.

What group/family was an exception to the trend in electronegativity value? Why are the values

so extremely different?

The noble gases, group 18. All of the noble gases have full outer energy shell.

Attracted electron(s) would enter a higher energy level which is too far away to be

strongly attracted to the nuclear charge.

Why would metals have fairly low electronegativity? Hint: Think about what metals want to do

during ion formation to become more energetically favorable.

They have a lower electronegativity because their nuclear charge is smaller and thus there

is less attraction for a pair of electrons in a bond.

19

What would non-metals (excluding Noble Gases) have fairly high electronegativity? Hint: Think

about what non-metals want to do during ion formation to become more stable.

They have a higher electronegativity because their nuclear charge is larger and thus there

is more attraction for a pair of electrons in a bond.

You do not need to memorize electronegativity values.

• Instead, you should know the periodic trends so that you can predict which of two

elements is more electronegative. • The Pauling electronegativity scale ranges from the most metallic to the most nonmetallic

elements - 0.7 (Cs) to 4.0 (F).


Electronegativity and Bond Polarity

We can use the difference in electronegativity between two atoms to gauge the polarity

of the bond the atoms form.

Consider these three fluorine-containing compounds:

In F2 the electrons are shared EQUALLY between the fluorine atoms and, thus, the covalent

bond is nonpolar.

• A NONPOLAR COVALENT bond results when the electronegativities of the

bonded atoms are EQUAL.

In HF the fluorine atom has a greater electronegativity than the hydrogen atom,

with the result that the electrons are shared UNEQUALLY—the bond is polar.

• In general, a POLAR COVALENT bond results when the atoms differ in

electronegativity.

• In HF the more electronegative fluorine atom attracts electron density away from the less

electronegative hydrogen atom, leaving a partial positive charge on the hydrogen atom

and a partial negative charge on the fluorine atom.

• We can represent this charge distribution as

• The δ+ and δ- (read “DELTA plus” and “DELTA minus”) symbolize the PARTIAL

POSITIVE and PARTIAL NEGATIVE, respectively.

In LiF the electronegativity difference is very LARGE, meaning that the electron density

is shifted far toward F.

• The resultant bond is therefore most accurately described as IONIC.

The shift of electron density toward the more electronegative atom in a bond can be seen in the

electron density distributions.

• For the three species in our example, the calculated electron density distributions are

shown in FIGURE 8.8.

• You can see that in F2 the distribution is symmetrical, in HF the electron density is

clearly shifted toward fluorine, and in LiF the shift is even greater.

• Therefore, that the greater the difference in electronegativity between two atoms, the

more polar their bond.

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When you determine whether bond is polar or nonpolar, you need to take the DIFFERENCE

between the 2 atoms.

General Rule:

1. In each case, which bond is more polar: (a) B – Cl or Cl – Cl (b) P – F or P – Cl?

Indicate in each case which atom has the partial negative charge. (p. 311)

2. Which of the following bonds is most polar: S – Cl, S – Br, Se – Cl, or Se – Br ? (p.311)

Examples

Examples


3. For each of these polar covalent bonds, which atoms has a partial positive charge (δ+)

and which atom has a partial negative charge (δ－)?

C – N F – Br Si – O

4. Arrange these three bonds in order from least polar to most polar. N – F P – F O – F

Dipole Moments

The difference in electronegativity between H and F leads to a polar covalent bond in

the HF molecule.

• As a consequence, there is a concentration of negative charge on the more electronegative

F atom, leaving the less electronegative H atom at the positive end of the molecule.

• A molecule such as HF, in which the centers of positive and negative charge do not

coincide, is a polar molecule.

• Thus, we describe both bonds and entire molecules as being polar and nonpolar.

• BEWARE: You can have POLAR covalent bonds, but the entire molecule could end up

being NONPOLAR….coming in Ch 9!

We can indicate the polarity of the HF molecule in two ways:

• Whenever two electrical charges of equal magnitude but opposite sign are separated by a

distance, a DIPOLE is established. (DIPOLE MOMENT)

• In the notation on the right, the arrow denotes the shift in electron density toward the

fluorine atom.

• The crossed end of the arrow can be thought of as a plus sign designating the positive end

of the molecule.

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Polarity helps determine many properties we observe at the macroscopic level in

the laboratory and in everyday life.

• Polar molecules align themselves with respect to one another, with the negative end of

one molecule and the positive end of another attracting each other. (Remember…opposite

charges attract)

• Polar molecules are likewise attracted to ions. The negative end of a polar molecule is

attracted to a positive ion, and the positive end is attracted to a negative ion.

o This is called Ion-Dipole….as you will see in Chapters 11!

See FIGURE 8.10: For these molecules, the change in the electronegativity difference has a

greater effect on the dipole moment than does the change in bond length.

Bond Types and Nomenclature

• Previously, we used two different approaches to naming binary compounds. • One is for ionic compounds and another is for molecular compounds (prefixes). • In both systems the less electronegative element is given first.

o The other element follows with the ending -ide.

• Both approaches are sometimes used with the same substance! • Metals with higher oxidation numbers tend to be molecular rather than ionic.

o Example: TiO2 ▪ The names titanium(IV) oxide and titanium dioxide are used but titanium

dioxide is more commonly used.


BOND POLARITY (14.5 min) L1 - Doc Dena Bonding Video

When you determine that a bond is “polar”, you indicate its direction of pull (also known as electron distribution) by drawing an arrow OVER the bond pointing to the atom which is the most electronegative. This shows that the electron is being shared MUCH closer to the more electronegative element, causing that side of the molecule to become “partially” NEGATIVE. Because there is a shift in charge within the molecule it causes the molecule to become “partially” POSITIVE on the other side. Since these are not true charges, we cannot use a + or – sign. Instead, we use the lower-case Greek letter DELTA + to indicate “partially positive” or — to indicate “partially negative”.

STEPS FOR INDICATING DIRECTION OF PULL (POLARITY)

OF A BOND…

(1) Label each element with its electronegativity (EN) value.

(2) Place an arrow above the bond pointing to the more EN element.

(3) Place a —

next to the more EN element, and a + next to the less

EN element. This shows the electron distribution in the bond!!!

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EXAMPLE: Show the direction of pull (electron distribution) for the bonds within F2 and HCl.

EXAMPLE: How many polar bonds and how many non-polar bonds are there in C2Cl6?

EXAMPLE: Use the periodic table to predict the rank for the following bonds from least polar

(#1) to most polar(#6).

Ca-Se Li-F S-P O-F Sn-I I-O

EXAMPLE: Use electronegativity values to rank the following bonds from least polar (#1) to

most polar(#6).

Ca-Se Li-F S-P O-F Sn-I I-O

8.5 Drawing Lewis Structures


Read p. 315-319 Here are some simple guidelines for drawing Lewis structures: “NAS” – Mrs Rodriguez helpful calculation to determine TOTAL number of bonds in molecule and hints on how to draw structure. N (Needed): # of e- needed to form octets for all elements.

The # needed is 8.

There are a few exceptions to the Octet Rule, WHICH YOU MUST MEMORIZE:

H only want 2 electrons (one bond)

Be only wants 4 electrons (two bonds)

B is satisfied with 6 electrons (three bonds)

A (Available): #of valence electrons available

S (Shared): Subtract the two numbers. S= N-A

Bond formed with two e-, so divide by two. This tells you how

many TOTAL bonds to draw between the elements.

So now your ready to Draw the molecule.

• Identify the central atom. When a central atom has other atoms bound to it, the central atom is usually written first.

• Write all other atoms around the central atom.

• According to NAS, how many TOTAL bonds in the molecule? o 1st : Place one bond (two electrons) between each pair of atoms. o 2nd: If you have MORE BONDS to add according to NAS, then you will have

multiple bonds (doubles or triples). This is where we need to look at valence e- to determine where.

o 3rd: After all bonds have been placed, complete the octets for all atoms connected to the central atom by adding LONE PAIR of e-.

▪ BEWARE of Exceptions: H, He, Be, B o H’s are ALWAYS terminal (outside) atoms. Hydrogen can only have two

electrons.

We aren’t very interested in which element brought which electron to the table. Think of the process like a potluck: each element brings its valence electrons.

Is the MAGIC NUMBER!

(Well, except for H, Be, B, and

sometimes S & P…..)

27

1. Let’s practice drawing Lewis structures (single bonds)

Br2

HI

NH3

PCl3

H2O

BeCl2

BI3

CH4

CH2Cl2

C 2H6

Examples

When you have

only carbons and

hydrogens, the

carbons will align

themselves in a

central chain and

the hydrogens

will go around

that chain!


2. Let’s practice drawing Lewis structures (multiple bonds) O2

N2

C2H4

C2H2

3. Let’s practice drawing Lewis structures (ions)

NH4 +1

OH －1

CN -1

NO+1

Examples

Examples

Don’t forget to

address the

charge shown!

29

Formal Charge

When we draw a Lewis structure, we are describing how the electrons are distributed in a

molecule or polyatomic ion.

• In some instances we can draw more than one Lewis structure and have all of them

obey the octet rule.

• All these structures can be thought of as contributing to the actual arrangement of the

electrons in the molecule, but not all of them will contribute to the same extent.

How do we decide which of several Lewis structures is the most important?

• The FORMAL CHARGE of any atom in a molecule is the charge the atom would have

IF all the atoms in the molecule had the same electronegativity (that is, if each

bonding electron pair in the molecule were shared equally between its two atoms).

• Formal charge is NOT a real charge; it’s a bookkeeping system.

It’s useful when comparing two Lewis structures in which each atom obeys the octet rule.

To calculate the formal charge on any atom in a Lewis structure, we assign electrons

to the atom as follows:

1. All unshared (nonbonding) electrons are assigned to the atom on which they are

found.

2. For any bond—single, double, or triple—half of the bonding electrons are assigned

to each atom in the bond. (It’s like if you cut the bond in half…half the e- go to one of the atoms

and the other half go to the other atom)

The formal charge of each atom is calculated by the following:

The number of VALENCE e- – the number of ASSIGNED e- for the atom.

Let’s practice on the cyanide ion, CN -1 Valence e-: 4 5 Assigned e-: 5 5 Formal Charge: -1 0

• Notice that the sum of the formal charges equals the overall charge on the ion, -1 .

• The formal charges on a neutral molecule must add to zero, whereas those on an ion add

to give the charge on the ion.


If we can draw several Lewis structures for a molecule, the concept of formal charge can

help us decide which is the most important, which we shall call the dominant Lewis structure.

• One Lewis structure for CO2 , for instance, has two double bonds.

• However,we can also satisfy the octet rule by drawing a Lewis structure having one

single bond and one triple bond.

• Calculating formal charges in these structures, we have

Formal Charge Guidelines

❏ If you have a molecule in which each atom obeys the octet rule, the most preferred structure

is the one in which each atom has a formal charge of ZERO.

❏ If you have a polyatomic ion, then the sum of all the formal charges in the Lewis structure

will be equal to the overall charge of the polyatomic ion.

❏ If it’s not possible for each of the atoms to have a formal charge of zero, the most favorable

structure is the one in which the NEGATIVE formal charge (if there is one) is on the atom

with the HIGHEST ELECTRONEGATIVE value.

• So, the FIRST Lewis structure of CO2 is the dominant one because the atoms carry no

formal charges and so satisfy the first guideline.

• The other Lewis structure shown (and the similar one that has a triple bond to the left O

and a single bond to the right O) do contribute to the actual structure but to a much

smaller extent.

Although the concept of formal charge helps us to arrange alternative Lewis structures

in order of importance, it is important that you remember that formal charges do not represent real charges on atoms.

31

4. (p.318)

5. The cyanate ion, NCO -1, has three possible Lewis structures. (a) Draw these three

structures and assign formal charges in each. (b)Which Lewis structure is

dominant? (p. 319)

Examples


AP Tip: The AP Exam may give you a skeleton structure an ask you to fill in the missing electrons (bonding and nonbonding)

Question 2d:

Question 1c:

Question 2d:

Examples

33

8.6 Resonance Structures

Read p.320-322

We sometimes encounter molecules and ions in which the experimentally determined

arrangement of atoms is not adequately described by a single dominant Lewis structure.

• Consider ozone, O3, which is a bent molecule

• Because each oxygen atom contributes 6 valence electrons, the ozone molecule has 18

valence electrons.

• This means the Lewis structure must have one O – O single bond and one O = O double

bond to attain an octet about each atom.

• However, this single structure cannot by itself be dominant because it requires that one

bond be different from the other, contrary to the observed structure—we would expect

the double bond to be shorter than the single bond.

• In drawing the Lewis structure, however, we could just as easily have put the bond on the

left:

• There is no reason for one of these Lewis structures to be dominant because they are

equally valid representations of the molecule.

• The placement of the atoms in these two alternative but completely equivalent Lewis

structures is the same, but the placement of the electrons is different.

• Lewis structures of this sort are called RESONANCE STRUCTURES.

o Resonance structures are attempts to represent a real structure that is a MIX

between several extreme possibilities.

o Resonance structures are Lewis structures that differ only with respect to

placement of the electrons.

o The “true” arrangement is a blend or hybrid of the resonance structures.

To describe the structure of ozone properly, we write both resonance structures and use a

DOUBLE-HEADED ARROW (NOT EQUILIBRIUM ARROWS!) to indicate that the real

molecule is described by an average of the two:


• Molecules such as ozone cannot be described as oscillating between the two individual

Lewis structures shown previously—there are two equivalent dominant Lewis

structures that contribute equally to the actual structure of the molecule.

• The ozone molecule always has two equivalent O – O bonds whose lengths are

intermediate between the lengths of an oxygen–oxygen single bond and an oxygen–

oxygen double bond.

o AP tip: Experimentally, ozone has two identical bonds whereas the Lewis

structure requires one single (longer) and one double bond (shorter). Based

on these 2 resonance structures, it’s an average of the 2 structures.

o We describe the bond order of the O-O bond in O3 as 1.5 bonds

on each side of central Oxygen.

1. Draw all possible Lewis structures for the following molecules/ions: CO3

2－

SO3

Examples

35

SO3

-2

Which is predicted to have the shorter sulfur–oxygen bonds, SO3 or SO3

-2? Explain (p. 321)

HCO2

-1

Examples


Another way of looking at it is to say that the rules for drawing Lewis structures do not allow us

to have a single dominant structure.

Questions to ask yourself when determining if a structure shows resonance:

(1) Do I have any double or triple bonds?

(2) Can the double or triple bonds be moved to other locations?

2. Which of these examples has resonance structures?

Resonance in Benzene

Resonance is an important concept in describing the bonding in organic molecules, particularly

aromatic organic molecules, a category that includes the hydrocarbon benzene, C6H6.

• The six C atoms are bonded in a hexagonal ring, and one H atom is bonded to each C

atom.

• We can write two equivalent dominant Lewis structures for benzene, each of which

satisfies the octet rule.

• These two structures are in resonance:

• Note that the double bonds are in different places in the two structures. Each of these

resonance structures shows three carbon–carbon single bonds and three carbon–carbon

double bonds.

Examples

37

• However, experimental data show that all six

bonds C – C are of equal length, 1.40 Å,

intermediate between the typical bond lengths

for a C – C single bond (1.54 Å) and a

C = C double bond (1.34 Å).

• Each of the C—C bonds in benzene can be

thought of as a blend of a single bond and a

double bond (FIGURE 8.14).

Benzene is commonly represented by omitting the

hydrogen atoms and showing only

the carbon–carbon framework with the vertices

unlabeled.

In this convention, the resonance in the molecule is

represented either by two structures separated by a

double-headed arrow or by a shorthand notation in

which we draw a hexagon with a circle inside: The shorthand notation reminds us that benzene is a blend of two resonance structures—

it emphasizes that the double bonds cannot be assigned to specific edges of the hexagon.

• Chemists use both representations of benzene interchangeably.

• The bonding arrangement in benzene confers special stability to the molecule.

• As a result, millions of organic compounds contain the six-membered ring characteristic

of benzene.

• Many of these compounds are important in biochemistry, in pharmaceuticals, and in the

production of modern materials.


LEWIS DOT STRUCTURES RESONANCE (6.5 min) L2 - Doc Dena Bonding Video

Resonance – To resonate means to match ___________________ (or vibrations). Resonance

occurs when more than one equivalent LDS structure can be drawn for a molecule. NOTE: The

molecule does not transition from one resonance structure to the next. The actual structure is

best described a ___________________ or average of the resonant structures. The shared

electrons are said to be ___________________.

𝐵𝑜𝑛𝑑 𝑂𝑟𝑑𝑒𝑟 = # 𝑏𝑜𝑛𝑑 (𝑒− 𝑝𝑎𝑖𝑟𝑠)

𝑝𝑒𝑟𝑖𝑝ℎ𝑒𝑟𝑎𝑙 𝑎𝑡𝑜𝑚𝑠 𝑠ℎ𝑎𝑟𝑖𝑛𝑔 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠

EXAMPLE: Draw the Lewis Dot structure(s) for SO2

EXAMPLE: Draw the Lewis Dot structure(s) for benzene

Picture2

NOTE: As bond order ___________________________, bond length

________________________________ & bond energy _______________________________!

(see cartoon for answers!)

2 http://www.newspaper.unsw.edu.au/archive/05_04_27/images/panel/images/VSU.jpg

39

8.7 Exceptions to the Octet Rule

Read p. 322-325

The octet rule is so simple and useful in introducing the basic concepts of bonding that you

might assume it is always obeyed. HOWEVER, there are some exceptions (and NAS will not

always work ).

These exceptions to the octet rule are of three main types:

1. Molecules and polyatomic ions containing an odd number of electrons

2. Molecules and polyatomic ions in which an atom has fewer than an octet of valence

electrons

3. Molecules and polyatomic ions in which an atom has more than an octet of valence electrons

Odd Number of Electrons

• In the vast majority of molecules and polyatomic ions, the total number of valence

electrons is even, and complete pairing of electrons occurs.

• However, in a few molecules and polyatomic ions, such as ClO2, NO, NO2, and O2 -1, the

number of valence electrons is odd.

• Complete pairing of these electrons is impossible, and an octet around each atom cannot

be achieved.

• For example, NO contains 5 + 6 = 11 valence electrons. (NAS still works )

• The two most important Lewis structures for this molecule are

1. Draw all possible Lewis structures for the following molecules/ions: NO2

Examples


Less than an Octet of Valence Electrons

• A second type of exception occurs when there are fewer than eight valence electrons

around an atom in a molecule or polyatomic ion.

• This situation is also relatively rare (with the exception of hydrogen and helium as we

have already discussed ONLY NEED 2 e-).

• Most often encountered in compounds of boron and beryllium. (NAS still works )

o Boron = needs 6e- (3 bonds)

o Beryllium = needs 4e- (2 bonds)

2. Draw the Lewis structures for the following: BH3 AlF3

More than an Octet of Valence Electrons

• This is the largest class of exceptions – NAS doesn’t work . o You have to draw the structures “old school”.

• Atoms from the third period and beyond can accommodate more than an octet. o Elements from the third period and beyond have unfilled d orbitals that can be

used to accommodate the additional electrons.

Examples

41

• Size also plays a role. o The larger the central atom, the larger the number of atoms that can

surround it. o The size of the surrounding atoms is also important. o Expanded octets occur often when the atoms bound to the central atom are

the smallest and most electronegative (e.g., F, Cl, O).

AP Tip: If you can draw a Lewis structure in which each atom obeys the octet rule, then do that!

The AP Exam will never ask students to decide, on their own, when drawing a Lewis structure, whether they should obey the octet rule OR expand the octet in order to get better formal

charges.

3. Draw the Lewis structures for the following: PF5 SF6

SF4 SF2

Examples


BrF5 ClF3

XeF2 AsF6

-1

ICl4 1－

Examples

43

8.8 Strengths of Covalent Bonds

Read p. 325-331

The stability of a molecule is related to the strengths of its covalent bonds.

• The strength of a covalent bond between two atoms is determined by the ENERGY

REQUIRED TO BREAK THE BOND.

• It is easiest to relate bond strength to the enthalpy change in reactions in which bonds are

broken. (See Ch 5.4)

• The BOND ENTHAPLY is the enthalpy change, ΔH , for the BREAKING of a

particular BOND in ONE MOLE of a GASEOUS substance.

TABLE 8.4 lists average bond enthalpies for a number of atom pairs.

• The bond enthalpy is always a positive quantity ; energy is always required to BREAK

chemical bonds.

o Units are listed as kJ/mol

• Energy is always RELEASED when a bond forms.

• The greater the bond enthalpy, the stronger the bond!

(Don’t memorize them…just know what they mean and how to use them)

Pay attention if any Lewis

structures have double or triple

bonds.

Stronger Bonds

= greater bond

enthalpy!


Bond Enthalpies and the Enthalpies of Reactions

We can use average bond enthalpies to estimate the ENTHAPLIES OF REACTIONS (ΔH rxn)

of reactions in which bonds are broken and new bonds are formed.

• This procedure allows us to estimate quickly whether a given reaction will be

endothermic (ΔH > 0) or exothermic (ΔH < 0) even if we do not know for all the

species involved. (Remember endo and exo from Ch 5 )

• Our strategy for estimating reaction enthalpies is a straightforward application of Hess’s

law. (Ch 5.6)

• We use the fact that breaking bonds is always endothermic and forming bonds is

always exothermic.

• We therefore imagine that the reaction occurs in two steps:

1. We supply enough energy to break those bonds in the reactants that are not

present in the products. The enthalpy of the system is increased by the SUM

of the bond enthalpies of the bonds that are broken.

2. We form the bonds in the products that were not present in the reactants.

This step releases energy and therefore lowers the enthalpy of the system by

the SUM of the bond enthalpies of the bonds that are formed.

The enthalpy of the reaction, ΔH rxn, is estimated as the sum of the bond enthalpies of

the bonds broken minus the sum of the bond enthalpies of the bonds formed:

ΔH rxn = Σ (bond enthalpies of bonds broken) – Σ (bond enthalpies of bonds formed)

Σ = means add them up!

Rodriguez Rules for calculating ΔH rxn :

1. Make sure all the Lewis structures are drawn in the WHOLE reaction.

• Just because you don’t see double or triples bonds doesn’t mean they aren’t there!

2. Make a T-chart: List all the bonds broken and formed that are different on BOTH sides.

• Only list differences! If you see the same bond broken and formed, then don’t list

it. Only list bonds that were broken that don’t appear to be formed, and vice

versa.

3. Use Table 8.4 to find the bond enthalpy values.

• Remember…The BOND enthalpy values are ALWAYS POSITIVE!

4. If you have more than one bond broken/formed, then multiply that number by the bond

enthalpy value.

5. Add up both sides and use the equation above to find the difference.

6. ENTHAPLIES OF REACTIONS (ΔH rxn) can be POSITIVE or NEGATIVE!

45

Let’s try an example reaction with methane, CH4, and chlorine to produce methyl chloride,

CH3Cl, and hydrogen chloride, HCl:

H – CH3 (g) + Cl – Cl (g) → Cl – CH3 (g) + H – Cl (g) ΔH rxn = ?

• Draw all Lewis Structures.

• Make the T-chart. Find the difference between both sides. Don’t list EVERY BOND broken or formed.

• Remember, if you have more than one of the same bond broken/formed you must multiply it with the bond enthalpies. Notice the moles listed with the number of each type. The moles will cancel when you multiply.

Bonds Broken Bond Formed

1 mol C – H 413 kJ/mol 1mol C – Cl 328 kJ/mol

1 mol Cl – Cl 242 kJ/mol 1 mol H – Cl 431 kJ/mol

Total = 655 kJ 759 kJ

ΔH rxn = Σ bond enthalpy (broken) – Σ bond enthalpy (formed)

ΔH rxn = 655 kJ – 759 kJ = -104 kJ

(Beware: If AP ask for the answer’s units to be in kJ/mol, then don’t cancel out the moles)

The reaction is exothermic because the bonds in the products (especially the H – Cl bond) are

stronger than the bonds in the reactants (especially the Cl – Cl bond).


1. See p.328

2. See p. 329

Examples

47

Bond Enthalpy and Bond Length

• The distance between the nuclei of the atoms involved in a bond is called the BOND LENGTH.

• Multiple bonds are SHORTER than single bonds. • We can show that multiple bonds are STRONGER than single bonds. • As the number of bonds between atoms INCREASES, the atoms are held CLOSER

and STRONGER.

This trend is illustrated in FIGURE 8.17 for N—N single, double, and triple bonds.

Shorter bond Triple bond = Stronger bond Sharing 6e- Higher bond enthalpy

Longer bond Single bond = Weaker bond Sharing 2e- Lower bond enthalpy

chapter 8. basic concepts of chemical bonding - nina rodriguez - … · 2020. 12. 30. · chapter...

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