EXAMPLE 10.1

Copyright ©2009 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458

All rights reserved.

Introductory Chemistry, Third EditionBy Nivaldo J. Tro

Since phosphorus is in Group 5A in the periodic table, it has 5 valence electrons. Represent these as five dots surrounding the symbol for phosphorus.

EXAMPLE 10.1 Writing Lewis Structures for Elements

Write a Lewis structure for Mg.

SKILLBUILDER 10.1 Writing Lewis Structures for Elements

FOR MORE PRACTICE Example 10.12; Problems 27, 28.

Write a Lewis structure for phosphorus.

Solution:




Draw the Lewis structures of magnesium and oxygen by drawing two dots around the symbol for magnesium and six dots around the symbol for oxygen.

EXAMPLE 10.2 Writing Ionic Lewis Structures

Write a Lewis structure for the compound NaBr.

SKILLBUILDER 10.2 Writing Ionic Lewis Structures

FOR MORE PRACTICE Example 10.13; Problems 39, 40.

Write a Lewis structure for the compound MgO.

In MgO, magnesium loses its 2 valence electrons, forming a 2+ charge, and oxygen gains 2 electrons, forming a 2– charge and acquiring an octet.

Solution:




Draw the Lewis structures of calcium and chlorine by drawing two dots around the symbol for calcium and seven dots around the symbol for chlorine.

EXAMPLE 10.3 Using Lewis Theory to Predict the Chemical Formula of an Ionic Compound

Use Lewis theory to predict the formula for the compound that forms between magnesium and nitrogen.

SKILLBUILDER 10.3 Using Lewis Theory to Predict the Chemical Formula of an Ionic Compound

FOR MORE PRACTICE Example 10.14; Problems 41, 42, 43, 44.

Use Lewis theory to predict the formula for the compound that forms between calcium and chlorine.

Calcium must lose its 2 valence electrons (to effectively get an octet in its previous principal shell), while chlorine needs to gain only 1 electron to get an octet. Consequently, the compound that forms between Ca and Cl must have two chlorine atoms to every one calcium atom.

Solution:

The formula is therefore CaCl2.




EXAMPLE 10.5Writing Lewis Structures for Covalent Compounds

2. Calculate the total number of electrons for the Lewis structure by summing the valence electrons of each atom in the molecule.

EXAMPLE 10.4

Write a Lewis structure for CO2.

Write a Lewis structure for CCl4.

Solution:Following the symmetry guideline, we write:

O C O

1. Write the correct skeletal structure for the molecule.

Solution:Following the symmetry guideline, we write:

Total number of electrons forLewis structure =

Total number of electrons forLewis structure =




EXAMPLE 10.5Writing Lewis Structures for Covalent CompoundsContinued

3. Distribute the electrons among the atoms, giving octets (or duets for hydrogen) to as many atoms as possible. Begin with the bonding electrons, and then proceed to lone pairs on terminal atoms, and finally to lone pairs on the central atom.

EXAMPLE 10.4

4. If any atoms lack an octet, form double or triple bonds as necessary to give them octets.

Since all of the atoms have octets, the Lewis structure is complete.

Bonding electrons first.

(8 of 32 electrons used)Lone pairs on terminal atoms next.

(32 of 32 electrons used)

Bonding electrons first.

(4 of 16 electrons used)Lone pairs on terminal atoms next.

(16 of 16 electrons used)Move lone pairs from the oxygen atoms to bonding regions to form double bonds.




EXAMPLE 10.5Writing Lewis Structures for Covalent CompoundsContinued

EXAMPLE 10.4

SKILLBUILDER 10.4 SKILLBUILDER 10.5

Write a Lewis structure for CO.

Write a Lewis structure for H2CO.

FOR MORE PRACTICE

Example 10.15; Problems 49, 50, 51, 52, 53, 54.




Begin by writing the skeletal structure. Since hydrogen atoms must be terminal, and following the guideline of symmetry, put the nitrogen atom in the middle surrounded by four hydrogen atoms.

EXAMPLE 10.6 Writing Lewis Structures for Polyatomic Ions

Write the Lewis structure for the NH4+ ion.

Calculate the total number of electrons for the Lewis structure by summing the number of valence electrons for each atom and subtracting 1 for the positive charge.Next, place 2 electrons between each pair ofatoms.

Solution:

Since the nitrogen atom has an octet and since all of the hydrogen atoms have duets, the placement of electrons is complete. Write the entire Lewis structure in brackets and write the charge of the ion in the upper right corner.




EXAMPLE 10.6 Writing Lewis Structures for Polyatomic Ions

Write a Lewis structure for the ClO– ion.

SKILLBUILDER 10.6 Writing Lewis Structures for Polyatomic Ions

FOR MORE PRACTICE Problems 57bcd, 58abc, 59, 60.

Continued




Begin by writing the skeletal structure. Using the guideline of symmetry, make the three oxygen atoms terminal.

EXAMPLE 10.7 Writing Resonance Structures

Solution: O

O N O

Write a Lewis structure for the NO3– ions. Include resonance structures.

Sum the valence electrons (adding 1 electron to account for the –1 charge) to determine the total number of electrons in the Lewis structure.

Place 2 electrons between each pair ofatoms.Distribute the remaining electrons, first toterminal atoms.

Since there are no electrons remaining to complete the octet of the central atom, form a double bond by moving a lone pair from one of the oxygen atoms into the bonding region with nitrogen. Enclose the structure in brackets and write the charge at the upper right.




Notice that you could have formed the double bond with either of the other two oxygen atoms.


Write a Lewis structure for the NO2- ion. Include resonance structures.

SKILLBUILDER 10.7 Writing Resonance Structures


Continued

Since the three Lewis structures are equally correct, write the three structures as resonance structures.




EXAMPLE 10.9

The central atom (P) has four electron groups.

Predicting Geometry Using VSEPR

2. Determine the total number of electron groups around the central atom.Lone pairs, single bonds, double bonds, and triple bonds each count as one group.

EXAMPLE 10.8

The central atom (N) has threeelectron groups (the double bond counts as one group).

Predict the electron and molecular geometry of PCl3.

Predict the electron and molecular geometry of the [NO3]- ion.

1. Draw a Lewis structure for themolecule.

3. Determine the number of bonding groups and the number of lone pairs around the central atom.These should sum to the result from Step 2. Bonding groups include single bonds, double bonds, and triple bonds.

Solution:PCl3 has 26 electrons.

Solution:[NO3]– has 24 electrons.

Three of the four electron groups around P are bonding groups, and one is a lone pair.

All three of the electron groupsaround N are bonding groups.




EXAMPLE 10.9Predicting Geometry Using VSEPRContinued

EXAMPLE 10.8

SKILLBUILDER 10.8

Predict the molecular geometry of ClNO (N is the central atom).

SKILLBUILDER 10.9

Predict the molecular geometry of The SO3

2– ion.

FOR MORE PRACTICE

Example 10.17; Problems 67, 68, 71, 72, 75, 76.

4. Use Table 10.1 to determine theelectron geometry and moleculargeometry

The electron geometry istetrahedral (four electron groups), and the molecular geometry—the shape of the molecule—is trigonal pyramidal (four electron groups, three bonding groups, and one lone pair).

The electron geometry is trigonal planar (three electron groups), and the molecular geometry—the shape of the molecule—is trigonal planar (three electron groups, three bonding groups, and no lone pairs).




EXAMPLE 10.10 Classifying Bonds as Pure Covalent, Polar Covalent, or Ionic

Determine whether the bond formed between each of the following pairs of atoms is pure covalent, polar covalent, or ionic.(a) Sr and F(b) N and Cl(c) N and O

Solution:(a) From Figure 10.2, we find the electronegativity of Sr (1.0) and of F (4.0). The electronegativity difference (ΔEN) is:

ΔEN = 4.0 – 1.0 = 3.0

(b) From Figure 10.2, we find the electronegativity of N (3.0) and of Cl (3.0). The electronegativity difference (ΔEN) is:

ΔEN = 3.0 – 3.0 = 0

Using Table 10.2, we classify this bond as pure covalent.

(c) From Figure 10.2, we find the electronegativity of N (3.0) and of O (3.5). The electronegativity difference (ΔEN) is:

ΔEN = 3.0 – 3.0 = 0

Using Table 10.2, we classify this bond as polar covalent.




EXAMPLE 10.10 Classifying Bonds as Pure Covalent, Polar Covalent, or Ionic

Continued

Determine whether the bond formed between each of the following pairs of atoms is pure covalent, polar covalent, or ionic.(a) I and I(b) Cs and Br(c) P and O

SKILLBUILDER 10.10 Classifying Bonds as Pure Covalent, Polar Covalent, or Ionic

FOR MORE PRACTICE Problems 83, 84.




Begin by drawing the Lewis structure of NH3. Since N and H have different electronegativities, the bonds are polar

EXAMPLE 10.11 Determining Whether a Molecule Is Polar


Determine whether NH3 is polar.

The geometry of NH3 is trigonal pyramidal (four electron groups, three bonding groups, one lone pair). Draw a three dimensional picture of NH3 and imagine each bond as a rope that is being pulled. The pulls of the ropes do not cancel and the molecule is polar.

Determine whether CH4 is polar.

SKILLBUILDER 10.11 Determining Whether a Molecule Is Polar

Solution:




EXAMPLE 10.12 Lewis Structures for Elements

What is the Lewis structure of sulfur?

Solution:Since S is in Group 6A, it has 6 valence electrons. We draw these as dots surrounding its symbol, S.




EXAMPLE 10.13 Writing Lewis Structures for Ionic Compounds

Write a Lewis structure for lithium bromide.

Solution:




EXAMPLE 10.14 Using Lewis Theory to Predict the Chemical Formula of an Ionic Compound

Use Lewis theory to predict the formula for the compound that forms between potassium and sulfur.Solution:The Lewis structures of K and S are:

Potassium must lose 1 electron and sulfur must gain 2. Consequently, we need two potassium atoms to every sulfur atom. The Lewis structure is:

The correct formula is K2S.




EXAMPLE 10.15 Writing Lewis Structures for Covalent Compounds

Write a Lewis structure for CS2.

Solution:S C S





Write resonance structures for SeO2.

Solution:We can write a Lewis structure for SeO2 by following the steps for writing covalent Lewis structures. We find that we can write two equally correct structures, so we draw them both as resonance structures.




EXAMPLE 10.17 Predicting the Shapes of Molecules

Predict the geometry of SeO2.

Solution:The Lewis structure for SeO2 (as we saw in Example 10.16) is composed of the following two resonance structures.

Either of the resonance structures will give the same geometry.Total number of electron groups = 3Number of bonding groups = 2Number of lone pairs = 1

Electron geometry = Trigonal planarMolecular geometry = Bent




EXAMPLE 10.18 Determining Whether a Molecule Is Polar

Determine whether is SeO2 polar.

Solution:Se and O are nonmetals with different electronegativities (2.4 for Se and 3.5 for O). Therefore, the Se–O bonds are polar.

As we saw in Example 10.17, the geometry of SeO2 is bent.

The polar bonds do not cancel but rather sum to give a net dipole moment. Therefore the molecule is polar.

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EXAMPLE 10.1