Chemistry. The Lewis Theory 1.Valence electrons play a fundamental role in chemical bonding 2.Chemical bonding that results from the transfer of 1 or

LEWIS DOT STRUCTUREChemistry

The Lewis Theory

1. Valence electrons play a fundamental role in chemical bonding

2. Chemical bonding that results from the transfer of 1 or moreelectrons from 1 atom to another leading to the formation ofpositive and negative ions is called ionic

3. Chemical bonding that involves the sharing of electrons iscalled covalent (non-polar and polar)

4. Electrons are transferred/ shared to the extent that each atomacquires an especially stable electron configuration. Usuallythe configuration of a noble gas (8 outer shell electrons)

When atoms form compounds, it is their valence electrons that actually interact.

A Lewis dot symbol consists of the element’s symbol surround by dots.

Each dot represents a valence electron.

Boron 1s22s22p1

3 valence electrons

B•

• •

Lewis dot symbol for boron

B•

•• B•

•• B•

•

•

other reasonable Lewis dot symbols for boron

Lewis Dot Symbols5.2

Atoms combine in order to achieve a more stable electron configuration.

Maximum stability results when a chemical species is isoelectronic with a noble gas.

Na: 1s22s22p63s1 Na+: 1s22s22p6

10 electrons total, isoelectronic with Ne

Cl: 1s22s22p63s23p5 Cl‒: 1s22s22p63s23p6

18 electrons total, isoelectronic with Ar

Lewis Dot Symbols

Lewis Dot Symbols

Lewis dot symbols of the main group elements.

Lewis Dot Symbols

Dots are not paired until absolutely necessary.

Na •

O••

• •••

For main group metals such as Na, the number of dots is the number of electrons that are lost.

For nonmetals in the second period, the number of unpaired dots is the number of bonds the atom can form.

B•

• •

1s22s22p1

C•

• ••

1s22s22p2

N••

• ••

1s22s22p3

5 valence electrons; first pair formed in the Lewis dot symbol

Lewis Dot Symbols

Ions may also be represented by Lewis dot symbols.

Na•

Na 1s22s22p63s1 Na+ 1s22s22p6

Na+

O••

• •••

O 1s22s22p4 O2‒ 1s22s22p6

O••

••••••

2‒

Core electrons not represented in the Lewis dot symbol

Valence electron lost in the formation of the Na+ ion.

Remember the charge

Worked Example 5.1

Strategy Starting with the Lewis dot symbols for each element, add dots (for anions) or remove dots (for cations) as needed to achieve the correct charge on each ion. Don’t forget to include the appropriate charge on the Lewis dot symbol.

Write Lewis dot symbols for (a) fluoride ion (F-), (b) potassium ion (K+), and (c) sulfide ion (S2-).

Solution (a)

(b) K+

(c)

F

S2-

Think About It For ions that are isoelectronic with noble gases, cations should have no dots remaining around the element symbol, whereas anions should have eight dots around the element symbol. Note, too, that for anions, we put square brackets around the Lewis dot symbol and place the negative charge outside the brackets. Because the symbol for a common cation such as the potassium ion has no remaining dots, square brackets are not necessary.

Drawing Lewis Structures

Follow these steps when drawing Lewis structure for molecules and polyatomic ions.

1) Draw the skeletal structure of the compound. The least electronegative atom is usually the central atom. Draw a single covalent bond between the central atom and each of the surrounding atoms.

2) Count the total number of valence electrons present; add electrons for negative charges and subtract electrons for positive charges.

3) For each bond in the skeletal structure, subtract two electrons from the total valence electrons.

4) Use the remaining electrons to complete octets of the terminal atoms by placing pairs of electrons on each atom. Complete the octets of the most electronegative atom first.

5) Place any remaining electrons in pairs on the central atom.

6) If the central atom has fewer than eight electrons, move one or more pairs from the terminal atoms to form multiple bonds between the central atom and terminal atoms.

6.3

Write the Lewis structure of nitrogen trifluoride (NF3).

Step 1 – N is less electronegative than F, put N in center

F N F

F

Step 2 – Count valence electrons N - 5 (2s22p3) and F - 7 (2s22p5)

5 + (3 x 7) = 26 valence electrons

Step 3 – Draw single bonds between N and F atoms and complete octets on N and F atoms.

Step 4 - Check, are # of e- in structure equal to number of valence e- ?

3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons

Lewis Structures

A Lewis structure is a representation of covalent bonding.

Shared electron pairs are shown either as dashes or as pairs of dots.

Lone pairs are shown as pairs of dots on individual atoms.

H O H••••

H O H••••

•• ••O with 8 e−

H with 2 e− H with 2 e−

Shared electrons shown as dashes (bonds)

Multiple Bonds

In a single bond, atoms are held together by one electron pair.

In a double bond, atoms share two pairs of electrons.

H O H••••

C with 8 e−

O with 8 e−O with 8 e−

one shared pair of electrons results in a single bond

••O C O

•••• ••

••

••

••••

2 shared pairs of electrons result in double bonds

O C O

••

••

••••= =

H O H••••

•• ••

Multiple Bonds

A triple bond occurs when atoms are held together by three electron pairs.

each N has 8 e−

3 shared pairs of electrons result in a triple bond

••N N

•••••••• N N

•••• ≡

Multiple BondsBond length is defined as the distance between the nuclei of two covalently bonded atoms.

Multiple bonds are shorter than single bonds.

For a given pair of atoms: triple bonds are shorter than double bonds double bonds are shorter than single bonds

< <N≡N

110 pm

N=N

124 pm

N−N

147 pm

Multiple Bonds

The shorter multiple bonds are also stronger than single bonds.

We quantify bond strength by measuring the quantity of energy required to break it.

H2(g) → H(g) + H(g) Bond energy = 436.4 kJ/mol

Drawing Lewis Structures

Draw the Lewis structure for carbon disulfide (CS2).

Worked Example 6.4

Setup Step 1: C and S have identical electronegativities. We will draw the skeletal

structure with the unique atom, C, at the center.

Step 2: The total number of valence electrons is 16: 6 from each S atom and 4 from the C atom [2(6) + 4 = 16].

Step 3: Subtract 4 electrons to account for the bonds in the skeletal structure, leaving us 12 electrons to distribute.

Step 4: Distribute the 12 remaining electrons as 3 lone pairs on each S atom.

S C S

S C S

Draw the Lewis structure for carbon disulfide (CS2).

Worked Example 6.4 (cont.)

Setup Step 5: There are no electrons remaining after step 4, so step 5 does not apply.

Step 6: To complete carbon’s octet, use one lone pair from each S atom to make a double bond to the C atom.

S C S

Think About It Counting the total number of valence electrons should be relatively simple to do, but it is often done hastily and is therefore a potential source of error in this type of problem. Remember that the number of valence electrons for each element is equal to the group number of that element.

Solution

Formal Charges

Sometimes, while drawing Lewis structures, you don't have observational data to use.Formal charges can be used to find out what to choose.

Formal charge = (# of Valence shell e- of an atom) - (# Bond pair e-) - (# of unshared e-)

For example, lets take the first incorrect drawing of sulfuric acid. Formal charge on S = 6 - 4 - 0 = 2 Formal charge on H = 1 - 1 - 0 = 0 Formal charge on O = 6 - 2 - 4 = 0 (on the ones bonded to H) Formal charge on O = 6 - 1 - 6 = -1 (on the isolated ones)

We can disregard the ones with 0 formal charge. The ones that do have formal charge are the sulfur and the isolated oxygens. Since the 2 oxygens are -1 and sulfur is 2, another bond goes from each oxygen to the sulfur to cancel the formal charge. (2 + 2(-1) = 0)

When several Lewis structures are possible, those with the smallest formal charges are the most stable and are preferred.

Two possible skeletal structures of formaldehyde (CH2O)

H C O HH

C OH

An atom’s formal charge is the difference between the number of valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis structure.

formal charge on an atom in a Lewis structure

=1

2

total number of bonding electrons( )

total number of valence electrons in the free atom

-total number of nonbonding electrons

-

The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion.

H C O HC – 4 e-

O – 6 e-

2H – 2x1 e-

12 e-

2 single bonds (2x2) = 41 double bond = 4

2 lone pairs (2x2) = 4Total = 12

formal charge on C = 4 -2 - ½ x 6 = -1

formal charge on O = 6 -2 - ½ x 6 = +1


=1

2




-

-1 +1

C – 4 e-

O – 6 e-

2H – 2x1 e-

12 e-



HC O

H

formal charge on C = 4 -0 - ½ x 8 = 0

formal charge on O = 6 -4 - ½ x 4 = 0


=1

2




-

0 0

Formal Charge and Lewis Structures

1. For neutral molecules, a Lewis structure in which there are no formal charges is preferable to one in which formal charges are present.

2. Lewis structures with large formal charges are less plausible than those with small formal charges.

3. Among Lewis structures having similar distributions of formal charges, the most plausible structure is the one in which negative formal charges are placed on the more electronegative atoms.

Which is the most likely Lewis structure for CH2O?

H C O H

-1 +1 HC O

H

0 0

Lewis Structures and Formal Charge

Formal charge can be used to determine the most plausible Lewis Structure when more than one possibility exists for a compound.

To determine associated electrons:

1) All the atom’s nonbonding electrons are associated with the atom.

2) Half the atom’s bonding electrons are associated with the atom.

Formal charge = valence electrons – associated electrons

6.4


Determine the formal charges on each oxygen atom in the ozone molecule (O3).

O O O

••••

••

••= − ••••

e− associated with atom

Valence e−

Difference (formal charge)

6 6 6

6 5 7

0 +1 −1

4 unshared + 4 shared = 6 e−

2


2


2

The widespread use of fertilizers has resulted in the contamination of some groundwater with nitrates, which are potentially harmful. Nitrate toxicity is due

primarily to its conversion in the body to nitrite (NO2-), which interferes with the

ability of hemoglobin to transport oxygen. Determine the formal charges on each

atom in the nitrate ion (NO3-).

Worked Example 6.5

Strategy Follow the six steps to draw the Lewis structure of (NO3-). For each

atom, subtract the associated electrons from the valence electrons.

Setup

The N atom has five valence electrons and four associated electrons (one from each single bond and two from the double bond). Each singly bonded O atom has six valence electrons and seven associated electrons (six in three lone pairs and one from the single bond). The doubly bonded O atom has six valence electrons and six associated electrons (four in two lone pairs and two from the double bond.)

O N O

O


Solution

The formal charges are as follows: +1 (N atom), -1 (singly bonded O atoms), and 0 (doubly bonded O atom).

O N O

O

Think About It The sum of formal charges (+1) + (-1) + (-1) + (0) = -1 is equal to the overall charge on the nitrate ion.


When there is more than one possible structure, the best arrangement is determined by the following guidelines:

1) A Lewis structure in which all formal charges are zero is preferred.

2) Small formal charges are preferred to large formal charges.

3) Formal charges should be consistent with electronegativities.

O C O

••

••••••= = O C O

•• ••

••••≡ −

0 0 0 +1 0 −1Formal charge

better structure (based on formal charge)


Based on formal charge, identify the best and the worst structures for the isocyanate ion below:

Solution:Step 1 Assign formal charges on each atom using the formula

4 5 6

−2 +1 0

Ve−

C N O-

C N O-

C N O-

Formal charge = valence electrons – associated electrons

C N O-

C N O-

C N O-

Ae−

FC

4 5 6 4 5 6

6 4 6 5 4 7 7 4 5

−1 +1 −1 −3 +1 +1


Based on formal charge, identify the best and the worst structures for the isocyanate ion below:

Solution:Step 2 Determine the best and worst structure

−2 +1 0

C N O-

C N O-

C N O-

FC −1 +1 −1 −3 +1 +1

Worst structure:

Large formal charges

Formal charges inconsistent with

electronegativities

Best structure:

Small formal charges

Formal charges more consistent with

electronegativities

Formaldehyde (CH2O), which can be used to preserve biological specimens, is

commonly sold as a 37 percent aqueous solution. Use formal charges to determine which skeletal arrangements of atoms shown here is the best choice for

the Lewis structure of CH2O.

Worked Example 6.6

Strategy The complete Lewis structures for the skeletons shown are:

H C O H H C H

O

H C O H H C H

O



In the structure on the left, the formal charges are as follows:

Both H atoms: 1 valence e- – 1 associated e- (from single bond) = 0

C atom: 4 valence e- – 5 associated e- (two in the lone pair, one from the single bond, and two from the double bond) = -1

O atom: 6 valence e- – 5 associated e- (two from the lone pair, one from the single bond, and two from the double bond) = +1

H C O H H C H

O

H C O H

0 0+1-1Formal charges



In the structure on the right, the formal charges are as follows:

Both H atoms: 1 valence e- – 1 associated e- (from single bond) = 0

C atom: 4 valence e- – 4 associated e- (one from each single bond, and two from the double bond) = 0

O atom: 6 valence e- – 6 associated e- (four from the two lone pairs and two from the double bond) = 0

H C O H H C H

O

Formal charges all zero

H C H

O


Solution Of the two possible arrangements, the structure on the left has an O atom with a positive formal charge, which is inconsistent with oxygen’s high electronegativity. Therefore, the structure on the right, in which both H atoms are attached directly to the C atoms and all atoms have a formal charge of zero, is the better choice for the Lewis structure of CH2O.

Think About It For a molecule, formal charges of zero are preferred. When there are nonzero formal charges, they should be consistent with the electronegativities of the atoms in the molecules. A positive formal charge on oxygen, for example, is inconsistent with oxygen’s high electronegativity.

Write the Lewis structure of the carbonate ion (CO32-).

Step 1 – C is less electronegative than O, put C in center

O C O

O

Step 2 – Count valence electrons C - 4 (2s22p2) and O - 6 (2s22p4) -2 charge – 2e-

4 + (3 x 6) + 2 = 24 valence electrons

Step 3 – Draw single bonds between C and O atoms and complete octet on C and O atoms.

Step 4 - Check, are # of e- in structure equal to number of valence e- ?

3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons

Step 5 - Too many electrons, form double bond and re-check # of e-



The Double bond could be between C and any of the oxygens………this is known as resonance

Resonance StructuresSometimes, a single Lewis structure does not adequately represent the true structure of a molecule.

Carbon is the central atom, the three oxygens are bound to it and electrons are added to fulfill the octets of the outer atoms.

All the available electrons have been used but carbon is electron deficient - it only has six electrons around it. So………………………

Consider the carbonate ion, CO32-

carbon (C) has four valence electrons x 1 carbon = 4 e- oxygen (O) has six valence electrons x 3 oxygens = 18 e- The ion has an overall negative two charge so we add 2 e- to give a total of 24 e- to be placed in the Lewis structure.

we share a non-bonding electron pair on an oxygen with the carbon to create a double bond and thereby fulfill carbon's octet.

becomes

or

becomes

or

becomes

Three equivalent Lewis structures (formal charges are included) can be drawn for the carbonate ion. The true structure of the carbonate ion is an average of the three resonance structures connected by a double headed arrow and enclosed in brackets.

This can also be represented by a resonance hybrid where a dashed line is used to indicate the delocalized electrons.

Resonance

A resonance structure is one of two or more Lewis structures for a single molecule that cannot by represented accurately by only one Lewis structure.

Resonance structures are a human invention.

Resonance structures differ only in the positions of their electrons.

O O O

••

••

••

••= − ••••

O O O••••

•••• − = ••••

6.5

High oil and gasoline prices have renewed interest in alternative methods of producing energy, including the “clean” burning of coal. Part of what makes “dirty” coal dirty is its high sulfur content. Burning dirty coal produces sulfur

dioxide (SO2), among other pollutants. Sulfur dioxide is oxidized in the

atmosphere to form sulfur trioxide (SO3), which subsequently combines with

water to produce sulfuric acid – a major component of acid rain. Draw all possible resonance structures of sulfur trioxide.

Worked Example 6.7

Strategy Following the steps for drawing Lewis structures, we determine that a correct Lewis structure for SO3 contains two sulfur-oxygen single bonds and one sulfur-oxygen double bond.

But the double bond can be put in any one of three positions in the molecule.

O S O

O


Solution

O S O

O

O S O

O

OSO

O

Think About It Always make sure that resonance structures differ only in the position of the electrons, not in the positions of the atoms.

Expanded Octets

3) The central atom has more than eight electrons.

Atoms in and beyond the third period can have more than eight valence electrons.

In addition to the 3s and 3p orbitals, elements in the third period also have 3d orbitals that can be used in bonding.

SF

F F

F

F

F

Sulfur has 6 bonds corresponding to 12 electrons

Worked Example 6.9

Strategy The skeletal structure is

There are a total of 24 valence electrons (3 from the B and 7 from each of the three I atoms). We subtract 6 electrons to account for the three bonds in the skeleton, leaving 18 electrons to distribute in lone pairs on each I atom.

Solution

Draw the Lewis structure of boron triiodide (BI3).

B

I

II

B

I

II


Think About It Boron is one of the elements that does not always follow the octet rule. Like BF3, however, BI3 can be drawn with a double bond in order to satisfy the octet of boron. This gives rise to a total of four resonance structures.

B

I

II B

I

II B

I

I IB

I

II

Worked Example 6.10

Strategy The skeletal structure already has more than an octet around the As atom.

There are 40 total valence electrons [5 from As (Group 5A) and 7 from each of the five F atoms (Group 7A)]. We subtract 10 electrons to account for the five bonds in the skeleton, leaving 30 to be distributed. Next, we place three lone pairs on each F atom, thereby completing all their octets and using up all the electrons.Solution

Draw the Lewis structure of arsenic pentafluoride (AsF5).

As

F

F

F

F

F

As

F

F

F

F

F

Think About It Always make sure that the number of electrons represented in your final Lewis structure matches the total number of valence electrons you are supposed to have.

Worked Example 6.11

Strategy Begin by drawing the skeletal structure and counting the total number of valence electrons. Use the steps outlined in Section 6.4 to draw the first structure and reposition one or more lone pairs to adjust the formal charges for the second structure.

Note that each hydrogen in an oxoacid is attached to an oxygen atom, and not directly to the central atom. The total number of valence electrons is 26 (6 from S, 6 from each O, and 1 from each H).

Draw two resonance structures for sulfurous acid (H2SO3): one that obeys the

octet rule for the central atom, and one that minimizes the formal charges. Determine the formal charges on each atom in both structures.

H O S O H

O


Solution Following the steps we get the first structure:

From the top O atom, which has three lone pairs, we reposition one lone pair to create a double bond between O and S to get the second structure.

Incorporating the double bond results in every atom having a formal charge of zero.

H O S O H

O

+1

-1

H O S O H

O

Think About It In some species, such as the sulfate ion, it is possible to incorporate too many double bonds. Structures with three and four double bonds to sulfur would give formal charges on S and O that are inconsistent with the electronegativities of these elements. In general, if you are trying to minimize formal charges by expanding the central atom’s octet, only add enough double bonds to make the formal charge on the central atom zero.

Worked Example 6.12

Strategy Follow the steps for drawing Lewis structures. The skeletal structure is

There are 36 total valence electrons (8 from Xe and 7 from each of the four F atoms). We subtract 8 electrons to account for the bonds in the skeleton, leaving 28 to distribute. We first complete the octets of all four F atoms. When this is done, 4 electrons remain, so we place two lone pairs on the Xe atom.Solution

Draw the Lewis structure of xenon tetrafluoride (XeF4).

F Xe F

F

F

F Xe F

F

F

Think About It Atoms beyond the second period can accommodate more than an octet of electrons, whether those electrons are used in bonds or reside on the atoms as lone pairs.

Exceptions to the Octet Rule

Exceptions to the octet rule fall into three categories:

1) The central atom has fewer than eight electrons due to a shortage of electrons.

Elements in group 3A also tend to form compounds surrounded by fewer than eight electrons.

Boron, for example, reacts with halogens to form compounds of the general formula BX3 having six electrons around the boron atom.

H Be H only 4 total valence electrons in the system

6.6

2) The central atom has fewer than eight electrons due to an odd number of electrons.

Molecules with an odd number of electrons are sometimes referred to as free radicals.

O N O

••

••

••••= − ••

•

17 valence electrons in the system

Odd Numbers of Electrons

Draw the Lewis structure of chlorine dioxide (ClO2).

Worked Example 6.8

Strategy The skeletal structure is

This puts the unique atom, Cl, in the center and puts the more electronegative O atoms in terminal positions.

There are a total of 19 valence electrons (7 from the Cl and 6 from each of the two O atoms). We subtract 4 electrons to account for the two bonds in the skeleton, leaving us with 15 electrons to distribute as follows: three lone pairs on each O atom, one lone pair on the Cl atom, and the last remaining electron also on the Cl atom.

Cl OO

Cl OO

Solution

Think About It ClO2 is used primarily to bleach wood pulp in the manufacture of paper, but it is also used to bleach flour, disinfect drinking water, and deodorize certain industrial facilities. Recently, it has been used to eradicate the toxic mold in homes in New Orleans that were damaged by the devastating floodwaters of Hurricane Katrina in 2005.

Electronegativity and Polarity

There are two extremes in the spectrum of bonding: covalent bonds occur between atoms that share electrons ionic bonds occur between a metal and a nonmetal and involve

ions

Bonds that fall between these extremes are polar.

In polar covalent bonds, electrons are shared but not shared equally.

M:X

Pure covalent bondNeutral atoms held together by equally shared electrons

Mδ+Xδ−

Polar covalent bondPartially charged

atoms held together by unequally shared

electrons

M+X−

Ionic bondOppositely charged

ions held together by electrostatic attraction

6.2


Electron density maps show the distributions of charge. Electrons spend a lot of time in red and very little time in blue.

Electrons are shared equally

nonpolar covalent

Electrons are not shared equally and are more likely to be associated

with F

polar covalent

Electrons are not shared but rather transferred

from Na to F

ionic

Electronegativity

Electronegativity is the ability of an atom in a compound to draw electrons to itself.


Electronegativity varies with atomic number.

Dipole Moment, Partial Charges, and Percent Ionic Character

An arrow is used to indicate the direction of electron shift in polar covalent molecules.

The consequent charge separation can be represented as: Deltas (δ) denote a partial positive or

negative charge.

Regions where electrons spend little time

Regions where electrons spend a lot of time

H−F••••

••

H−F••••

••

δ+ δ−

Dipole Moment, Partial Charges, and Percent Ionic Character

A quantitative measure of the polarity of a bond is its dipole moment (μ).

Q is the charge.

r is the distance between the charges.

μ is always positive and expressed in debye units (D).

μ = Q x r

1 D = 3.336×10−30 C∙m

Documents

Chemistry. The Lewis Theory 1.Valence electrons play a fundamental role in chemical bonding 2.Chemical bonding that results from the transfer of 1 or