Transcript Slide 1

EXAMPLE 10.1
Writing Lewis Structures for Elements
Write a Lewis structure for phosphorus.
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.
SKILLBUILDER 10.1
Solution:
Writing Lewis Structures for Elements
Write a Lewis structure for Mg.
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Example 10.12; Problems 27, 28.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
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EXAMPLE 10.2
Writing Ionic Lewis Structures
Write a Lewis structure for the compound MgO.
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.
Solution:
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.
SKILLBUILDER 10.2
Writing Ionic Lewis Structures
Write a Lewis structure for the compound NaBr.
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Example 10.13; Problems 39, 40.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
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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 calcium and chlorine.
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.
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.
SKILLBUILDER 10.3
Solution:
The formula is therefore CaCl2.
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.
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Example 10.14; Problems 41, 42, 43, 44.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Writing Lewis Structures
for Covalent Compounds
1. Write the correct skeletal structure
for the molecule.
EXAMPLE 10.4
EXAMPLE 10.5
Write a Lewis structure for CO2.
Write a Lewis structure for
CCl4.
Solution:
Following the symmetry
guideline, we write:
Solution:
Following the symmetry
guideline, we write:
OCO
2. Calculate the total number of
electrons for the Lewis structure by
summing the valence electrons of
each atom in the molecule.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Total number of electrons for
Lewis structure =
Total number of electrons for
Lewis structure =
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Writing Lewis Structures
for Covalent Compounds
EXAMPLE 10.4
EXAMPLE 10.5
Continued
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.
Bonding electrons first.
Bonding electrons first.
(4 of 16 electrons used)
Lone pairs on terminal atoms
next.
(8 of 32 electrons used)
Lone pairs on terminal atoms
next.
(16 of 16 electrons used)
4. If any atoms lack an octet, form
double or triple bonds as necessary to
give them octets.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Move lone pairs from the
oxygen atoms to bonding
regions to form double bonds.
(32 of 32 electrons used)
Since all of the atoms have
octets, the Lewis structure is
complete.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Writing Lewis Structures
for Covalent Compounds
EXAMPLE 10.4
EXAMPLE 10.5
SKILLBUILDER 10.4
SKILLBUILDER 10.5
Continued
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.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 10.6
Writing Lewis Structures for Polyatomic Ions
Write the Lewis structure for the NH4+ ion.
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.
Solution:
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 of
atoms.
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.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 10.6
Writing Lewis Structures for Polyatomic Ions
Continued
SKILLBUILDER 10.6
Writing Lewis Structures for Polyatomic Ions
Write a Lewis structure for the ClO– ion.
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Problems 57bcd, 58abc, 59, 60.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 10.7
Writing Resonance Structures
Write a Lewis structure for the NO3– ions. Include resonance structures.
Begin by writing the skeletal structure. Using the
guideline of symmetry, make the three oxygen atoms
terminal.
Solution:
O
ONO
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 of
atoms.
Distribute the remaining electrons, first to
terminal 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.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 10.7
Writing Resonance Structures
Continued
Notice that you could have formed the double bond
with either of the other two oxygen atoms.
Since the three Lewis structures are equally correct,
write the three structures as resonance structures.
SKILLBUILDER 10.7
Writing Resonance Structures
Write a Lewis structure for the NO2- ion. Include resonance structures.
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Example 10.16; Problems 57, 58, 59, 60.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Predicting Geometry Using
VSEPR
1. Draw a Lewis structure for the
molecule.
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.
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.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
EXAMPLE 10.8
EXAMPLE 10.9
Predict the electron and
molecular geometry of PCl3.
Predict the electron and
molecular geometry of the
[NO3]- ion.
Solution:
PCl3 has 26 electrons.
Solution:
[NO3]– has 24 electrons.
The central atom (P) has four
electron groups.
The central atom (N) has three
electron groups (the double
bond counts as one group).
Three of the four electron
groups around P are bonding
groups, and one is a lone pair.
All three of the electron groups
around N are bonding groups.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Predicting Geometry Using
VSEPR
EXAMPLE 10.8
EXAMPLE 10.9
Continued
4. Use Table 10.1 to determine the
electron geometry and molecular
geometry
The electron geometry is
tetrahedral (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).
SKILLBUILDER 10.8
Predict the molecular geometry
of ClNO (N is the central
atom).
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).
SKILLBUILDER 10.9
Predict the molecular geometry
of The SO32– ion.
FOR MORE PRACTICE
Example 10.17; Problems 67,
68, 71, 72, 75, 76.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
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.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 10.10
Classifying Bonds as Pure Covalent, Polar Covalent, or Ionic
Continued
SKILLBUILDER 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) I and I
(b) Cs and Br
(c) P and O
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Problems 83, 84.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 10.11
Determining Whether a Molecule Is Polar
Determine whether NH3 is polar.
Begin by drawing the Lewis structure of NH3. Since
N and H have different electronegativities, the bonds
are polar
Solution:
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.
SKILLBUILDER 10.11
Determining Whether a Molecule Is Polar
Determine whether CH4 is polar.
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Example 10.18; Problems 91, 92, 93, 94.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
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.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 10.13
Writing Lewis Structures for Ionic Compounds
Write a Lewis structure for lithium bromide.
Solution:
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
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.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 10.15
Writing Lewis Structures for Covalent Compounds
Write a Lewis structure for CS2.
Solution:
SCS
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 10.16
Writing Resonance Structures
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.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
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 = 3
Number of bonding groups = 2
Number of lone pairs = 1
Electron geometry = Trigonal planar
Molecular geometry = Bent
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
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.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.