Chapter #10 Chemical Bonding. CHAPTER 12 Forces Between Particles Noble Gas Configurations Ionic...

Chapter #10Chemical Bonding

CHAPTER 12Forces Between Particles

Noble Gas Configurations

Ionic Bonding

Covalent Bonding

VSEPR Theory and Molecular Geometry

Electronegativity

Polar Bonds and Molecules

It has been recognized for a long time that the noble gases have great chemical stability. With few exceptions they are unreactive or inert.The noble gases have 8 valence electrons with the exception of He which has 2.

He 1s2

Ne 1s22s22p6

Ar 1s22s22p63s23p6

Kr 1s22s22p63s23p64s23d104p6

Xe 1s22s22p63s23p64s23d104p65s24d105p6

Atomic Stability

The electronic configuration of the noble gases is described as being energetically stable.

We can draw a Lewis diagram to illustrate the number of valence electrons an atom has.

In a Lewis diagram valence electrons are represented by dots placed above, below and to the left and right of the atoms symbol.

Ee.g. element with 4 valence electrons

Lewis Diagrams

There are two simple rules to keep in mind when drawing Lewis diagrams:

• Place one dot in each of the four locations before doubling up.• There can be only a maximum of 2 dots in any one location.

E E E E

Lewis Diagrams

What is the Lewis diagram for H?

H

1. First write the electron configuration: 1s1

2. Identify the number of valence electrons. 1 valence electron.

For a representative element it is easy to identify the number of valence electrons as this is equal to the group number.

Lewis Diagrams

What is the Lewis diagram for S?

1. First write the electron configuration: [Ne]3s23p4

2. Identify the number of valence electrons. 6 valence electrons

SAlternatively you can recognize that S is in group VIA so has six valence electrons

Lewis Diagrams

What is the Lewis diagram for S?

1. First write the electron configuration:

[Ne]3s23p4

2. Identify the number of valence electrons.

6 valence electrons

SAlternatively you can recognize that S is in

group VIA so has six valence electrons

Lewis Diagrams

LEWIS STRUCTURES OF THE ELEMENTS

1 2 13 14 15 16 17 18

H

Li

Na

Be

Mg

B

Al

C

Si P

N O F

ClS

He

Ne

Ar

LEWIS STRUCTURES OF IONS(AFTER REMOVAL OR ADDITION OF

ELECTRONS)

1 2 13 14 15 16 17 18

2+ 4- 3- 2- 1-

H

Li

Na

Be

Mg

B

Al

C

Si P

N O F

ClS

He

Ne

Ar

1+

3+

The octet rule states that:

“Atoms interact in order to obtain a stable octet of eight valence electrons”

The octet rule works extremely well at describing the interactions of the representative elements.

Lewis Diagrams

One way in which atoms can interact to satisfy the octet rule is by transferring electrons between each other.

Transferring of electrons results in the atoms acquiring net positive and negative charges.

When an atom loses or gains electrons a simple ion is formed.Cations have more protons than electrons and are positive.Anions have more electrons than protons and are negative.

Lewis Diagrams

Na[Ne]3s1

Na+ + 1e-

Cl+ 1e-

[Ne]3s23p5

Cl

Consider a Na atom what happens if it loses one electron?

Consider a Cl atom would you expect it to lose or gain electrons?

[Ne]

[Ne]3s23p6

11 P and 11 e- 11 P and 10 e-

I.E.

E.A.

17 P and 17 e- 17 P and 18 e-

Ion Formation

-

Metals tend to lose electrons forming positively charged ions called cations.

• A representative metal will lose its group number of electrons to obtain a stable octet.

Na → Na+ + 1e- ( Isoelectronic with Ne)

Mg → Mg2+ + 2e- (isoelectronic with Ne)

What would the charge be of the ion formed by a Li atom?

And which Noble gas is it isoelectronic with?

+1 The ion formed would be Li+

Isoelectronic with He

Ion Formation

Non-metals tend to gain electrons forming negatively charged ions called anions.

• A representative non-metal will gain (8 - group number) electrons to obtain a stable octet.

O + 2e- → O2- (isoelectronic with Ne)S + 2e- → S2- (isoelectronic with Ar)

What would the charge be of the ion formed by a I atom?Which Noble gas is it isoelectronic with?

-1 The ion formed would be I-

Isoelectronic with Xe

Ion Formation

Lewis Structure of NaCl

Na+Cl-Na+Cl-Na+Cl-

Cl- Na+Cl-Na+Cl-Na+

Forces between oppositely charged ions are calledIonic bonds. Each ion is surrounded by an octet of Electrons, thus making the ions stable.

Ionic compounds do not exist as discrete molecules. Instead they exist as crystals where ions of opposite charges occupy positions known as lattice sites.

Ions combine in the ratio that results in zero charge to form ionic compounds. Which ions are the smaller ones?

Crystal Lattice of NaCl


Ionic compounds do not exist as discrete molecules. Instead they exist as crystals where ions of opposite charges occupy positions known as lattice sites.

Ions combine in the ratio that results in zero charge to form ionic compounds. Which ions are the smaller ones? Sodium



In our early lectures we defined a molecule as “as a compound made of nonmetals.”

Molecules exist as particles containing the number of atoms specified by their formula.

e.g. a water molecule is a particle containing 2 hydrogen atoms and one oxygen atom and has the formula H2O.

Molecular Compounds

Non-metals may also complete their octets by sharing electrons.

This may occur between non-metal atoms of the same type:

e.g. H2, O2, N2, Cl2, F2, I2, etc

Or between different types of non-metal atoms:

e.g. CO2, H2O, CH4, etc

Molecular Compounds

Consider two hydrogen atoms separated by a large distance. Each has 1 electron in a 1s atomic orbital.

+ - +-

Now lets bring the two atoms together so there orbitals overlap.

Why does the electron stay around the nucleus?

Molecular Compounds

+

-

+-

The atomic orbitals overlap to form a new molecular orbital. This is a stable configuration as each H atom can have a full 1s subshell (like He) where the electrons spend most of their time shared between the atoms. In this arrangement each nucleus feels an inwards attraction to the two electrons. This is called covalent bonding.

Molecular Compounds

This new arrangement of protons and electrons is more stable than separate hydrogen atoms since the attraction of a proton to two electrons is a stronger attraction compared to one proton to one electron of a hydrogen atom.

+

-

+-

Molecular Compounds

We can draw Lewis diagrams showing the arrangement of valence electrons in covalent compounds. In these diagrams we represent each pair of electrons between atoms as a line.

So for the H2 molecule discussed previously the Lewis diagram would be:

H – HAll other electrons are represented by dots as described previously.

Molecular Compounds

Lewis Structures

Draw Lewis Structures of the following molecular compounds

Ha. H2O O

H

HHO

Note each element has a Noble gas structure by electron sharing

b. NH3 NH H

H

NH

HH

Covalent bonding e’s

Nonbondingelectons

Simplified Lewis Structures

OH H

Straight lines are used to indicate a shared pair, or a covalent bond.

Nonbonding electrons

Covalent bond

Lewis Structure Construction

Step 1 Connect each element with a single lineStep 2 Use the “P” formula to determine extra bondsStep 3 Insert the extra bonds, to make double or triple bonds.Step 4 Give each atom an octet of electrons, except hydrogenStep 5 Determine the formal charge of each element

Examples: Give Lewis Structures for the following

P = 8(n-q) +2q - 2(n-1) - vN = number of atoms in moleculeQ = number of hydrogen atomsV = total number of valence electrons

CO2

H2CO3 SO3 NO2+

Lewis Structure of Carbon Dioxide

CO OFirst, connect atoms with lines


CO OFirst, connect atoms with lines

Second, use “p” formula to determine the number of extra bonds.

P = 8(n-q) + 2q – 2(n-1) - v

P = 8(3-0) + 2(0) – 2(3-1) - 16P = 24 + 0 – 4 - 16

P = 44 extra bonding electrons2 extra bonds2 extra lines


CO O

Third, add extra lines (Three possible locations)

O C OO C O


CO O

Third, add extra lines (three possible locations)

Fourth, give each atom an octet of electrons

O C OO C O


CO O



O C OO C O

O C O OOO OC C


CO O



O C OO C O

O C O OOO OC C

Fifth, give each an atom a formal charge


CO O

Third, add extra lines


O C OO C O

O C O OOO OC C


If the element owns less than its valence, then it is positive

If the element has more than its valence, then it is negative


CO O



O C OO C O

O C O OOO OC C




OOOOOO C C C


CO O



O C OO C O

O C O OOO OC C




OOOOOO C C C-


CO O



O C OO C O

O C O OOO OC C




OOOOOO C C C+-


CO O



O C OO C O

O C O OOO OC C




OOOOOO C C C+- -


CO O



O C OO C O

O C O OOO OC C




OOOOOO C C C+ +- -

There are actually three possible Lewis structures for SO3.

SO O

O

S OO

O

SO O

O

Each of these three structures is equivalent. We say they are in “resonance” or that they are “resonance structures”. Resonance demonstrates how the loosely held pi electrons are free to move about. Or another way to look at it is that the pi electrons are spread over the entire molecule, making it more stable

-

-

-

-

- -2+ 2+ 2+

So far we have been considering how electrons are distributed between atoms in molecules and polyatomic ions.

An important question is: How can we predict the shape of molecules and polyatomic ions?

Molecular Geometry

The simplest polyatomic ion or molecule is made of two atoms:

What is the shape of this type of molecule or ion?

The only one way to join two atoms is with a line. All diatomic molecules and ions have a linear geometry. Molecular shape is the geometry defined by the atoms making up the molecule.

Molecular Shape

Valence shell electron pair repulsion theory (VSEPR theory) allows us to predict the 3 dimensional shape of molecules and polyatomic ions with >2 atoms.

VSEPR theory states that electrons in lone pairs and bonds move as far away from one another as possible to minimize repulsive interactions.

Predicting Molecular Shape

Repulsion Angles

For a central atom with three electron regions there are two possibilities.


For a central atom with four electron regions there are three possibilities.


To determine the correct shape of a molecule we must first begin with the correct Lewis structure.

We then need to determine how many regions of electron density are around the central atom. This is the number of bonds and electron pairs.

O S O O HH N H

H

H



We then need to determine how many regions of electron density are around the central atom. This is

the number of bonds and electron pairs.

O S O O HH N H

H

H




O S O O HH N H

H

H


To determine the correct shape of a molecule we must first begin with the correct Lewis structure.We then need to determine how many regions of electron density are around the central atom. This is the number of bonds and electron pairs.

O S O O HH N H

H

H

3 regions


To determine the correct shape of a molecule we must first begin with the correct Lewis structure.We then need to determine how many regions of electron density are around the central atom. This is

the number of bonds and electron pairs.

O S O O HH N H

H

H

3 regions




O S O O HH N H

H

H

3 regions 4 regions




O S O O HH N H

H

H

3 regions 4 regions 4 regions


The arrangement of outer atoms is determined by how many nonbonding pairs there are around the central atom.

If there are 2 or less electron regions then the arrangement will always be linear.

The situation becomes more complex if there are more than 2 electron regions around the central atom.


Use VSEPR theory to determine as much as possible about the structure of N2, H2O, SO3, CH4, NH4

+ and NH3



+ and NH3

First we will start with the Lewis structure of nitrogen.



+ and NH3



N N

P = 8(n-q) + 2q -2(n-1) – vP = 8(2-0) + 2(0) -2(3-1) -10P = 4 , 2 extra lines, right?


+ and NH3


N N

P = 8(n-q) + 2q -2(n-1) – vP = 8(2-0) + 2(0) -2(3-1) -10P = 4 , 2 extra lines, right?


+ and NH3

First we will start with the Lewis structure of nitrogen.N N P = 8(n-q) + 2q -2(n-1) – v

P = 8(2-0) + 2(0) -2(3-1) -10P = 4 , 2 extra lines, right?N N


+ and NH3


N N P = 8(n-q) + 2q -2(n-1) – vP = 8(2-0) + 2(0) -2(3-1) -10P = 4 , 2 extra lines, right?

N N

Adding extra bonds

N N

Adding nonbonding pairs

Use VSEPR theory to determine as much as possible

about the structure of N2, H2O, SO3, CH4, NH4+ and NH


N N P = 8(n-q) + 2q -2(n-1) – vP = 8(2-0) + 2(0) -2(3-1) -10P = 4 , 2 extra lines, right?

N N Adding extra bonds

N N Adding nonbonding pairs

Consider either nitrogen to be the central atom. Notice there are two volumes of space for electrons

Use VSEPR theory to determine as much as possible

about the structure of N2, H2O, SO3, CH4, NH4+ and NH3


N NP = 8(n-q) + 2q -2(n-1) – vP = 8(2-0) + 2(0) -2(3-1) -10P = 4 , 2 extra lines, right?

N N Adding extra bonds

N N Adding nonbonding pairs

Consider either nitrogen to be the central atom. Notice there are two volumes of space for electrons

Two volumes of electrons repel to 180°, thus linear shape


+ and NH3

Now consider the Lewis Structure of water.

HH O


+ and NH3Now consider the Lewis Structure of water.

H In water there are two bonding and two nonbonding electron pairs

O H

Molecular Shape of Water


+ and NH3

H

In water there are two bonding and two nonbonding electron pairs.

HH


HO



+ and NH3

HIn water there are two bonding and nonbonding electron pairs

H


Bent or v-shape, bond angles less than

109.5°

O H

H


ElectronegativityElectronegativity is an atoms affinity for electrons. Affinity is a Greek word for loving. Since electrons are attracted to the nucleus, then it makes sense that small atoms with more protons should attract electrons stronger. Therefore, we should predict electronegativity should increase from left to right and bottom to top on the Periodic Chart.

Electronegative Chart

Molecule Polarity

• A molecule will be polar if– it has polar bonds, and– its centers of partial positive and partial

negative charges lie at different places within the molecule

• Carbon dioxide, CO2, has two polar bonds but, because of its geometry, is a nonpolar molecule

Water

• Water, H2O, has two polar bonds and, because of its geometry, is a polar molecule

OH H

-

+center of partial positivecharge is midway betweenthe two hydrogen atoms

Water(a polar molecule)

Ammonia• Ammonia, NH3, has three polar bonds and, because

of its geometry, is a polar molecule

N

HH

H

Ammonia(a polar molecule)

+

-center of partialpositive charge ismidway betweenthe three hydrogen atoms

Dichloromethane

• Both dichloromethane, CH2Cl2, and formaldehyde, CH2O, have polar bonds and are polar molecules

Dichloromethane

C

H H

Cl Cl+--

C

H HFormaldehyde

O+-

CHAPTER #10

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Chapter #10 Chemical Bonding. CHAPTER 12 Forces Between Particles Noble Gas Configurations Ionic...

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