NAME____________________________________ PER____________ DATE DUE____________
ACTIVE LEARNING I N C HEMISTRY E DUCATION
"ALICE"
CHAPTER 15
CHEMICAL
BONDING
(PART 2)
15-1
©1997, A.J. Girondi
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© 1997 A.J. Girondi, Ph.D.
505 Latshmere Drive
Harrisburg, PA 17109
[email protected]
Website: www.geocities.com/Athens/Oracle/2041
15-2
©1997, A.J. Girondi
sp 3 Hybridization of Atoms
SECTION 15.1
In this chapter we will continue the discussion from Chapter 14 concerning Lewis dot structures.
The electron dot structures for NH3 and H2O were discussed at the end of the last chapter. The formulas
of these compounds are in agreement with the numbers of bonding electrons in nitrogen and oxygen.
Nitrogen has three bonding electrons, while oxygen has two. The dot structures are presented again
(below) for your review.
H
and
N
H
H N H
H
O
and
The atoms
HO
H
The molecules
The next example presents us with a problem. It involves the compound CH4, which is methane
or natural gas. In this molecule there are four identical C–H covalent bonds. Each involves a shared pair of
electrons between a H atom and the C atom. Since all four bonds in this molecule are exactly alike, the
bonding electrons must be located in four identical orbitals. However, the dot notation of the carbon atom
indicates that it has only two bonding electrons. How can carbon bond to four hydrogen atoms?
bonding
nonbonding
C + 4 H -----> CH4 ?????
Solving this problem was a challenge for scientists. They knew that carbon almost always forms four
bonds, yet the orbital notation of the carbon atom showed that it had only two bonding electrons. In an
effort to explain this discrepancy, scientists devised the concept of hybrid orbitals. Figure 15.1 below
illustrates an ordinary carbon with the valence electron configuration: 2s22p 2.
py
py
pz
+
2 electrons in
the 2s orbital
+
pz
px
2 of the 2p orbitals contain 1
electron, while the third 2p
orbital is empty
=
=
px
The unhybridized carbon atom has 2
electrons in the 2s orbital and 1 unpaired
electron in 2 of its 2p orbitals
Figure 15.1 The Unhybridized Carbon Atom
The word hybrid is usually used in connection with things such as hybrid plants and animals like a
mule. In biology, the term hybrid refers to the offspring of two animals or plants of different species. The
hybrid animal or plant exhibits some characteristics of both "parents," yet it is different from both.
15-3
©1997, A.J. Girondi
In this same sense, we can mix the 2s and 2p orbitals in the carbon atom in order to end up with
hybrid orbitals. The hybrids are unique and different from the s and p orbitals from which they came. They
have a different shape, a different distance from the nucleus, etc. When two or more hybrid orbitals form,
however, they are identical to each other in every way.
The orbital notation below illustrates what happens when hybridization occurs in carbon.
Unhybridized Carbon
2s
(X)
2p
(/)(/)( )
Electron Moves
2s
(/)
Hybridized Carbon
2p
(/)(/)(/)
sp3 hybrids
(/)(/)(/)(/)
Note that unhybridized carbon has two "paired" electrons in its 2s
orbital and two of its three 2p orbitals contain a single "unpaired"
bonding electron.
Its other 2p orbital is empty.
When
hybridization occurs, the one 2s and three 2p orbitals change
their shapes and become four identical "hybrid" orbitals. These
hybrid orbitals are located at a distance from the nucleus
somewhere between where the 2s and 2p orbitals used to be.
You already know that when there is more than one
orbital in a sublevel, the electrons won't pair up in the orbitals until
each orbital has one electron. Well, since these new hybrids are
all in the same new sublevel, one of the two 2s electrons leaves
its orbital to enter one of the new empty hybrid orbitals.
C
Figure 15.2
The Hybridized Carbon Atom
Since these four hybrid orbitals were "born" out of one s and three p orbitals, they are
known as sp3 hybrid orbitals. The hybridized carbon atom now has four bonding
electrons, which explains the existence of molecules like CH4.
CH4
First
C
then finally,
and
C
C
+ 4H
15-4
2 H2
4H
H
H C H
H
©1997, A.J. Girondi
The electron density plot of the CH4 molecule is shown at right.
Notice that the hybrid orbitals are directed toward the corners
(vertices) of a geometric figure called a tetrahedron which
means four-sided. The angles between the bonds are 109.5o.
Name an element in row 3 of the periodic table that you would
expect to form a tetrahedral shape when bonded to four other
atoms: {1}_______________________.
H
C
H
C
C
unhybridized
carbon atom
SECTION 15.2
H
H
hybridized
carbon atom
Figure 15.3
The Methane Molecule, CH 4
sp and sp 2 Hybridization of Atoms
Below is a diagram of the orbitals of the beryllium atom. Beryllium is a family 2A element on the
periodic table, indicating that it has two valence electrons. In the unhybridized Be atom, these electrons
are found paired in the 2s orbital. Based on this information, how many bonds can the unhybridized Be
atom form using unpaired electron(s)?{2}________.
py
py
pz
+
2 electrons in
the 2s orbital
+
pz
px
=
=
All three of the 2p
orbitals are empty
px
The unhybridized beryllium atom has 2
electrons in the 2s orbital and no
electrons in any of its three 2p orbitals
Figure 15.4 The Unhybridized Beryllium Atom
This baffled scientists because compounds such as BeCl 2 and BeF2 were known to exist, and
these compounds should be forming two identical bonds. The hybridization theory can be expanded in
an effort to explain this.
In order to form two identical bonds, the Be atom needs two unpaired (lone) electrons each
present in a separate but identical orbital. So, while carbon needed four hybrid orbitals, Be needs only
two. Thus, the theory of hybridization dictates that the 2s orbital and only one of the three 2p orbitals in
15-5
©1997, A.J. Girondi
the Be atom, hybridize and become identical, assuming a new shape and distance from the nucleus. The
two 2s electrons become unpaired with each one entering one of the new hybrid orbitals. The two
remaining 2p orbitals remain unchanged. Since these two new hybrid orbitals were "born" out of one s
and one p orbital, they are called sp hybrid orbitals.
Unhybridized Be Atom
2s
(X)
Electron Moves
2p
( )( )( )
2s
(/)
2p
(/)( )( )
Hybridized Be Atom
sp hybrids
(/)(/)
2p
( )( )
How many unpaired bonding electrons does this give the hybridized Be atom? {3}_________ Beryllium
and other elements in family 2A of the periodic table can form two bonds that are 180o apart. The
molecules formed with family 2A elements at their center are thus linear in shape. Name an element from
row 3 of the periodic table that you would expect to form a linear molecule when bonded to two other
atoms: {4}____________________.
py
pz
a hybrid orbital
a hybrid orbital
Figure 15.5 The Hybridized Beryllium Atom
The Linear BeCl2 molecule:
Cl
Be
Cl
Family 3A elements can form hybrid orbitals, too. It is known that molecules such as BH 3 exist in
which the B atom (from Family 3A) forms three identical bonds with hydrogen. Yet, if you consider the
notation of the unhybridized B atom (1 s 2 2 s 2 2 p 1 ), it reveals one pair of non-bonding electrons (2s 2) in its
valence shell and only one unpaired bonding electron (2p1). (See Figure 15.6.) So, how could the boron
atom form three bonds with hydrogen? After all, each hydrogen atom with one valence electron is
"looking" for a single unpaired (lone) electron from another atom that it can share and, thereby attain the
stable helium configuration: 1s 2.
Again, the theory of hybridization can be expanded to provide an explanation. Since the boron
atom already has one bonding electron, it needs only two more to explain the three bonds. The theory
holds that the 2s orbital and two of the three 2p orbitals undergo hybridization and become identical. The
third 2p orbital is left unchanged.
Unhybridized B Atom
2s
(X)
2p
(/)( )( )
Electron Moves
Hybridized B Atom
2s
(/)
sp2 hybrids
(/)(/)(/)
2p
(/)(/)( )
15-6
2p
( )
©1997, A.J. Girondi
py
py
pz
+
2 electrons in
the 2s orbital
+
pz
px
1 of the 2p orbitals contains 1
electron, while the other two
2p orbitals are empty
=
=
px
The unhybridized boron atom has 2
electrons in the 2s orbital and 1 unpaired
electron in one of its 2p orbitals
Figure 15.6 The Unhybridized Boron Atom
How many unpaired electrons are available for forming bonds in the un hybridized B atom? {5}_________.
How many unpaired electrons are available to form bonds in the hybridized B atom?{6}__________.
The
new hybrid orbitals were "born" out of one s and two p orbitals, so they are called sp2 hybrid orbitals.
Boron and the other elements in family 3A of the periodic table all form three bonds which are
120 oC apart. The molecules formed with family 3A elements as their central atom are referred to as being
planar or trigonal planar in shape. Trigonal means three-sided, and planar means flat.
py
H
hybrid orbital
hybrid orbitals
H
B
H
Figure 15.7 The Hybridized Boron Atom and the BH 3 Molecule
Name an element from row 3 of the periodic table that you would expect to form a trigonal planar molecule
when bonded to three other atoms.{7}_______________________
15-7
©1997, A.J. Girondi
SECTION 15.3
Summary Of sp, sp 2 , and sp 3 Hybridization
Study Table 15.1. It summarizes much of the information about hybridization theory that you
have learned up to this point.
Table 15.1
Bonding Electrons in Elements of Families 1A – 8A
Family
1A
2A
3A
4A
5A
6A
7A
8A
Dot Notation
Li
Be
B
C
N
O
F
Ne
Li
Be
B
C
N
O
F
--
1
2
3
4
5
6
7
8
1
2
3
4
3
2
1
0
(Unhybridized)
Dot Notation
(Hybridized)
No. of
Valence
Electrons
No. of
Bonds
Our discussion of hybridization has not included elements in families 5A to 8A. Actually, they can
be considered to undergo hybridization, but that is a subject that will be left for a more advanced course.
The theory of hybridization does not consider family 1A elements. You will note in Table 15.1 that in
families 5A - 8A, the number of bonds which atoms will usually form is equal to their number of unpaired
valence electrons.
SECTION 15.4
The Concept of Electronegativity
As you can see, electron dot formulas are quite useful for illustrating the polar covalent bonds in
molecules. There is another way of describing just how polar a bond is. To do this, we can use a concept
called electronegativity. The idea of electronegativity was first developed by Nobel prize winner Linus
Pauling, perhaps the foremost American chemist of the 20th century. The electronegativity of an atom
reflects its tendency to "attract" electrons to itself.
A numerical value for electronegativity is assigned to atoms of every element. The value assigned
to atoms of each element was determined by scientists who observed the behavior of the atoms in various
chemical reactions. Don't be concerned about the details of how the numbers were determined. You will
use these numbers to estimate the degree of ionic and covalent character in certain bonds.
15-8
©1997, A.J. Girondi
Problem 1. Use the information in Table15.1 to complete and balance the equations below, written in
electron dot form. The products should also be written in electron dot form. To save you some work, two
steps have already been done. They are: (a) any diatomic or polyatomic elements involved have already
been broken into individual atoms which are ready to bond to other atoms; and, (b) the dot notation shows
atoms in the hybridized form, where necessary, so that they are ready to bond to other atoms. The
first equation is done for you as an example.
a.
__2__ H
+
__1__ O
---------->
b.
_____
H
+
_____
C
---------->
c.
_____
Br
+
_____
Be
---------->
d.
_____
F
+
_____
Al
---------->
e.
_____
H
+
_____
Cl
---------->
f.
_____
H
+
_____
S
---------->
g.
_____
F
+
_____
P
---------->
H O
H
Since electronegativity reflects an atom's "attracting power" for electrons, the higher the
electronegativity value, the stronger the attraction. Consider a reaction between two elements. Their
relative attraction for electrons determines how they react. We can use the electronegativity scale to
determine this attraction. See the scale in Table 15.2 (or Table R-8 in your ALICE reference notebook).
Refer to those electronegativity values when answering the questions that follow.
1. In general, which group (metals or nonmetals) has the lowest electronegativity values? {8}____________
2. Arrange these elements in order of increasing electronegativity: bismuth (#83), chlorine (#17),
tellurium (#52), gallium (#31), thallium (#81): {9}____________________________________________
3. Which element on the periodic table has the highest electronegativity?
{10}_____________________
4. Which two elements has (have) the lowest electronegativity?{ 11}_____________________________
5. Which family of elements has the highest electronegativities?
{12}____________________________
6. Which family of elements has the lowest electronegativity values?
{13}_________________________
When the difference in electronegativity between atoms is high, electrons are considered to be
transferred from one atom to another. In questions 3 and 4 above, you located the elements with the
highest and lowest electronegativities. Now let's see what happens when two of these atoms, cesium and
fluorine, combine:
Cs
electronegativity values:
0.7
+
F
----->
Cs1+F1-
4.0
15-9
©1997, A.J. Girondi
Because fluorine's electronegativity is so great, it "steals" an electron from cesium. The result is that
cesium becomes a positive ion and fluorine becomes a negative ion. The attraction of opposite charges
holds the two ions together. This force forms what we have called a 100% ionic bond. Ionic bonds involve
the actual transfer of electrons between atoms.
When two atoms with identical electronegativities combine to form a bond, we describe such a
bond as being 100% covalent. Our previous examples of H 2, F 2, N2, and so forth are cases where this
occurs. These molecules have atoms in them with equal electron–attracting power, and we expect the
atoms to share valence electron pairs equally in the bonds that are formed.
We can use the scale of electronegativity values to estimate the degree (percent) of ionic and
covalent character in polar covalent bonds which are formed between atoms not having identical
electronegativities. Below Table 15.2 is a scale showing electronegativity differences. Below each
electronegativity difference is a listing of percent ionic character.
The higher the percent ionic character, the more ionic the bond. The
lower the percent ionic character, the more covalent the bond.
Study the procedure below for determining the percent ionic and percent covalent character for NaCl.
Electronegativity of Na = 0.9
Electronegativity of Cl = 3.0
Difference = 3.0 – 0.9 = 2.1
Referring to Table 15.2 (or Table R-8 ), a difference of 2.1 corresponds to a 67% ionic bond. The percent
ionic and covalent characters must add up to 100%, so the percent covalent character is 100% – 67% =
33% covalent. The bond between Na and Cl is 67% ionic and 33% covalent.
Problem 2. Using this same procedure, calculate the percent ionic and percent covalent characters of
the compounds listed below. In the compounds with multiple atoms, AlCl3, for example, it is only
nececssary to find the electronegativity difference between one Al atom and one Cl atom. Remember,
you are determining the character of each bond separately.
Percentages of Ionic and Covalent Characters in Selected Compounds
1 = Value of most electronegative element
2 = Value of least electronegative element
3 = Electronegativity Difference
Compound
MgBr2
4 = Percent Ionic character
5 = Percent covalent character
1
2
3
4
5
Br = 2.8
Mg = 1.2
1.6
47
53
NaF
CO
NH3
FrCl
PBr 3
15-10
©1997, A.J. Girondi
15-11
©1997, A.J. Girondi
Sodium chloride (NaCl) is an ionically bonded substance. You have seen sodium chloride many
times - it is table salt. Each grain of salt that you pour out of the shaker is certainly more than one ion of
sodium and one ion of chlorine. From our studies of atomic size, you found that it takes a great number of
atoms or ions for us to be able to see them. The salt crystals are actually huge numbers of Na1+ and Cl1ions bonded together to form an ionic bond. The diagram below helps to show that ionic solids form
crystals that are held together by attraction of opposite charges. There are no NaCl molecules as such.
The ionic crystal is just an array of ions. The arrangement of ions is called a crystal lattice. The nature of a
lattice varies from one ionic solid to another, depending on the ratio and size of positive and negative
ions in the compound.
Ionic and polar covalent compounds have properties that are
characteristic of their bond types. In Activity 15.5 you will be examining the
properties of an ionic compound and of a compound with polar covalent
bonds. You may assume that the properties of the ionic and covalent
substances which you will examine are typical of ionic and covalent
compounds.
Figure 15.8
NaCl Crystal Lattice
ACTIVITY 15.5
Comparing Properties Of Ionic & Covalent Compounds
We will use rock salt, NaCl, as a typical ionic compound for all of the tests. For the first test,
naphthalene will be used as a typical covalently–bonded substance. For the second test, paraffin will
serve as the covalent substance. Table sugar (sucrose), C12H22O11, will be used as the covalent
substance in tests three and four. Get small samples of each of these from the materials shelf. Subject
each substance to the test outlined below and record your observations in Table 15.3. Be sure to wear
safety glasses!
Use NaCl (ionic) and naphthalene (covalent) for test 1:
Test 1. Smell each compound. If you detect an odor, you may assume the substance is volatile (which
means it evaporates easily at room temperature).
Use NaCl (ionic) and paraffin (covalent) for test 2:
Test 2. Test the hardness of each compound by grinding a small piece of each with a metal file. Keep
each substance dry as you do this. Try to distinguish between the two. There is a difference.
Use table sugar (sucrose) for the covalently–bonded substance in tests 3 and 4:
Test 3. Place a very small sample of NaCl and sugar (sucrose) in separate piles on a piece of metal such as
a can lid (from the materials shelf). DO NOT use naphthalene! It is highly flammable! Place the metal on an
iron ring on a ring stand and heat the materials by applying heating with your burner under the metal. Note
which material melts first. Stop heating as soon as one material begins to melt; otherwise, you will cause
burning to occur.
Test 4. Dissolve a little NaCl (about enough to cover a quarter) in a 100 mL beaker about 1/2 full of tap
water. Stir well and test the solution for electrical conductivity using the apparatus designed for that
purpose. (Your teacher will guide you.) Dip the electrodes of the apparatus into a beaker of water to rinse
them and repeat this procedure using a little sugar instead of NaCl. Indicate whether the substance, when
dissolved, conducts a current.
Results and Conclusions:
15-12
©1997, A.J. Girondi
From your results, state the properties that are characteristic of ionic substances:
_________________
______________________________________________________________________________
State the general properties of a covalent substance: ______________________________________
______________________________________________________________________________
Using your observations from this activity and the table of electronegativity values, determine which of
these compounds, BaCl2 or CO2, would conduct more electric current when dissolved in water. Explain.
{14}____________________________________________________________________________
______________________________________________________________________________
Drawing on your observations in this activity and Table 15.2, which substance would you expect to have
the lowest melting point: AlBr3, CS2, or CaCl2? Explain your reasoning.
{15}____________________________________________________________________________
______________________________________________________________________________
Table 15.3
Properties of Ionic and Covalent Compounds
Property
Ionic Substance
Covalent Substance
Volatility
(high or low)
Hardness
Melting Point
(high or low)
Electrical
Conductivity
SECTION 15.6
More About Molecular Geometries
You already know something about the shapes of certain molecules. Knowing the shapes can
help you to determine many of the chemical properties of molecules. Lewis electron dot structures are
very useful in helping us to predict geometries of molecules. Ordinary chemical formulas don't help much
in this respect. Let's now expand your knowledge of molecular shapes.
In a previous chapter, we described electron probability density plots as a means of indicating the
15-13
©1997, A.J. Girondi
three-dimensional spaces in which electrons spend most of their time. The shapes and arrangement of
these orbitals determine the overall shapes of molecules. We can use the compound called ammonia,
NH3, as an example to illustrate how the probability plots can be used to explain the geometry (shape) of a
molecule. Observe the electron dot structures in the equation below. (Assume N2 and H2 have been
broken apart into N and H.)
N
+
H
H N H
3H
According to the equation above, how many valence electrons does H have? {16}_________
How many valence electrons does N have?{17}____________
How many bonding electrons does N have?{18}____________
How many nonbonding electron pairs does N have?{19}__________
N atom
Valence Orbitals:
2s
(X)
2p
(/)(/)(/)
Lewis Dot Symbols:
N
Lewis Dot Structure:
H N H
H
H atom
1s
(/)
H
Molecular Geometry:
(pyramidal)
The shape of the NH3 molecule can be illustrated by using lines to represent bonds. The shape of a
molecule depends on the bond angle between the bonds which join the H atoms to the central N atom.
The shapes of molecules can often be determined by looking at the valence electrons of the central atom.
Remember, because electrons are all negatively charged, they repel each other. To achieve a condition
of maximum stability, electrons locate themselves as far away from each other as possible. This serves to
minimize the electrostatic repulsion between the pairs of electrons. The position of the electrons
depends of the number of electrons present. The bond angle for the ammonia, NH3, molecule has been
determined to be 107o, forming what is referred to as a pyramidal–shaped molecule. The elements in
family 5A of the periodic table (such as nitrogen) all tend to form pyramidal-shaped molecules when they
are bonded to three other atoms. The unshared (nonbonding) electron pair on nitrogen tends to repel
the three shared pairs, causing the pyramidal arrangement.
Bond angles are determined by studying molecules with x-rays. Name another element (from row
3) on the periodic table that you would expect to be the center of pyramidal–shaped
molecules:{20}_______________________.
Now we will follow a similar procedure as we study the shape of molecules which have elements
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©1997, A.J. Girondi
from family 6A at their centers. Oxygen will be used as an example of a typical family 6A element. Water's
formula is H2O. The formula equation and the dot structure equation for the formation of water are shown
below.
O
2 H2 + O2 -----> 2 H2O
OR
2 H
+
O
----->
H O
H
H
H
104.5 o bond angle
The two unshared (nonbonding) pairs of electrons and the two shared pairs in the molecule all tend to
repel each other. The result is a bent–shaped molecule with a bond angle of 104.5o. Bent molecules are
typically formed when family 6A elements become bonded to two other atoms. Name an element from row
3 on the periodic table that you would expect to form the center of a bent molecule: {21}______________.
If you take a more advanced course in chemistry, you will learn that it is possible for unshared
electron pairs of electrons to become involved in bonding, and thereby, become shared pairs. Consider
the formation of the ammonium, NH41+, ion. A hydrogen ion (H1+) forms when a hydrogen atom loses its
electron. The hydrogen atom's electron configuration is 1s1. The hydrogen ion (H 1+) configuration is 1s0.
It has no electrons, but it "wants" the helium configuration (1s2) to gain stability. So, the hydrogen ion is
looking for two electrons to share. When this H 1+ ion meets a molecule of NH 3, it bonds to the unshared
pair on the N atom as follows:
1+
H
H
H1+
+
N
H
H
H
N
H
H
Since the hydrogen ion carries a positive charge, the entire particle assumes a positive charge when the
ammonium ion is formed. (Don't confuse ammonia, NH 3, with ammonium, NH 41+.) We can show this by
enclosing the whole structure inside brackets with the plus sign outside. This also happens when the H1+
ion reacts with a water molecule. The product (below) is an ion (the hydronium ion) that is very important in
the study of acids.
1+
H1+
+
There are many other examples. Whenever an atom or ion that needs a pair of electrons in order to
become stable meets another particle which has one or more unshared pairs of electrons, this kind of
reaction is possible. Table 15.4 summarizes the shapes of the molecules you have studied in this chapter.
Notice in the molecular drawings that a dash is often used to represent a pair of shared electrons (a bond).
Also notice that while all unshared pairs are shown in the dot structure, in the drawings of the molecules
only the unshared pairs around the central atoms are shown. Any other unshared pairs are "assumed" to
be there - as needed.
Problem 3 requires that you complete the blanks in Table 15.5. The drawings of the molecules
use dashes for bonds, and unshared pairs are only shown if they exist around the central atom.
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©1997, A.J. Girondi
Table 15.4
Geometries of Molecules Containing Family 1A – 8A Elements
Family
1A
Shape
Drawing
of
Molecule
2A
3A
4A
linear
linear
planar
tetrahedral
H Cl
I – Be – I
Cl
Cl
B
C
Cl
Cl
H Cl
Dot
Structure
Formula
I Be I
Cl
B
Cl
HCl
BeI 2
Cl
Cl
5A
pyramidal
H
linear
O
H
Cl Cl
H
H
Cl
Cl Cl
H N H
H
Cl C Cl
BCl3
H
7A
bent
N
Cl
Cl
6A
Cl
CCl4
NH3
H2O
Cl2
Problem 3. Complete Table 15.5.
Table 15.5
Dot Structures and "Stick" Drawings of Molecules
Compound
CHCl3
Dot Structure
Stick Drawing
Cl
Cl
Cl C H
C
Cl
Cl
Cl
Geometry
tetrahedral
H
LiI
PCl3
GaF3
CH2Cl2
H2S
15-16
©1997, A.J. Girondi
ACTIVITY 15.7
Determining the Formula of a Hydrate
In this activity you will be determining the formula of a hydrated compound. Remember that you
have seen the formulas of hydrates before. An example is CuSO4•5H 2O. This formula states that for
every 1 mole of CuSO 4 in the crystals of the compound, 5 moles of water are bonded to it. (The 1 in front
of the CuSO4 is assumed to be there.) The raised period means "plus". The water is bonded to the metal
in hydrates by a special kind of covalent bond which is called a coordinate covalent bond. We will save the
discussion of these bonds for a more advanced course.
When you heat a hydrate, you break the covalent bonds which bond the water to the compound,
and the water leaves in the form of a vapor. The product that is left is the anhydrous (without water) form of
the compound. The formula of the anhydrous form is obtained simply by leaving the water out of the
formula of the hydrate.
You are going to heat a sample of the hydrated form of barium chloride. We will represent the
formula temporarily as BaCl2 •?H2O. Your task is to determine the coefficient that should go in front of the
H2O in the formula. Remember, there is already an assumed "1" in front of the BaCl 2. You will be looking
for the mole ratio between the BaCl2 and the H2O in the compound.
Procedure:
1. Obtain a clean porcelain crucible and cover. Measure the mass of the empty crucible and cover to the
nearest 0.01 g. Add approximately 3 g of BaCl2 •?H2O crystals to the crucible, replace the cover, and
measure the mass to the nearest 0.01g. Calculate the mass of BaCl2 •?H2O crystals in the crucible, and
record this data in Table 15.7.
2. Begin heating slowly. Increase the heat until you have heated the crucible strongly for about 10
minutes. Remove the crucible from the triangle, let it cool, and measure the mass of the crucible, cover,
and contents. Record data in Table 15.7.
3. Reheat with a hot flame for a few minutes, allow to cool, and remeasure the mass. If the mass is more
than 0.01 g different from what it was after the first heating, reheat and measure again.
4. Dispose of the material in the crucible in the container provided by your teacher. Clean and dry the
crucible and cover.
5. After you have completed Table 15.7, show your results to your instructor who will provide you with the
actual answer. Comment on how your result compares to the actual formula for the hydrate: __________
______________________________________________________________________________
15-17
©1997, A.J. Girondi
Table 15.7
Determination of the Formula of a Hydrate
a. Mass of crucible and cover:
__________ g
b. Mass of crucible, cover, & contents before heating:
__________ g
c. Mass of BaCl2 •? H2O:
__________ g
d. Mass of crucible, cover, & contents after heating:
__________ g
e. Mass of H2O driven off (b–d):
__________ g
f. Mass of BaCl2 left in crucible (d–a):
__________ g
g. Moles of H2O driven off:
__________ moles
h. Moles of BaCl2 left in crucible:
__________ moles
Whole number mole ratio of moles of BaCl2 to moles of H2O: 1 to ____
(To get this whole number ratio, divide the moles of water and the moles of
BaCl2 by the smallest of the two. Round the results to whole numbers)
Experimentally determined formula for the hydrated compound: BaCl2 • ____ H 2O
SECTION 15.8
Learning Outcomes
This is the end of Chapter 15. Review the learning outcomes below. Check each one when you
think you have mastered it. Arrange to take any quizzes or exams on Chapter 15 and move on to Chapter
16.
_____1. Explain what is meant by the electronegativity of an element.
_____2. Describe the general properties of ionic and covalent compounds.
_____3. Determine the percent ionic and covalent characters of a bond.
_____4. Use hybridization theory to explain why atoms in families 2A, 3A, and 4A form the number of
bonds that they do.
_____5. Given the formula of a compound which has an atom from family 2A, 3A, or 4A in the center,
identify the type of hybridization that it exhibits.
_____6. Predict the geometry (shape) of simple molecules given the molecular formula.
_____7. Be able to calculate the formula of a hydrate, given the needed experimental data.
15-18
©1997, A.J. Girondi
SECTION 15.9
Answers to Questions and Problems
Questions:
{1} silicon; {2} none; {3} two; {4} magnesium; {5} one; {6} three; {7} aluminum; {8} metals;
{9} gallium, thallium, bismuth, tellurium, chlorine; {10} fluorine; {11} cesium and francium;
{12} family VII - the halogens; {13} family IA - the alkali metals; {14} BaCl 2 - it is more ionic, and ionic
compounds form solutions which conduct electricity; {15} CS 2 - because it is covalently bonded.
Covalent compounds tend to have lower melting points than ionic ones; {16} one; {17} five; {18} three;
{19} one; {20} phosphorus; {21} sulfur
Problems:
1.
a.
__2__ H
+
__1__ O
---------->
H O
H
b.
__4__ H
+
__1__ C
---------->
H
H C H
H
c.
__2__ Br
+
__1__
---------->
Br Be Br
d.
__3__ F
+
__1__ Al
---------->
F
F Al
F
e.
__1__ H
+
__1__ Cl
---------->
H Cl
f.
__2__ H
+
__1__ S
---------->
H S
H
g.
__3__ F
+
__1__
---------->
2.
MgBr2;
NaF;
CO;
NH3;
FrCl;
PBr 3;
Mg = 1.2; 1.6; 47; 53
F = 4.0; Na =0.9; 3.1; 91; 9
O = 3.5; C = 2.5; 1.0; 22; 78
N = 3.0; H = 2.1; 0.9; 19; 81
Cl = 3.0; Fr = 0.7; 2.3; 74; 26
Br = 2.8; P = 2.1; 0.7; 12; 88
3.
LiI
PCl3
GaF3
CH2Cl2
H2S
(same as HCl as shown in Table 15.4, page 15-16)
(same as NH3 as shown in Table 15.4, page 15-16)
(same as BCl3 as shown in Table 15.4, page 15-16)
(same as CCl4 as shown in Table 15.4, page 15-16)
(same as H2O as shown in Table 15.4, page 15-16)
Be
P
15-19
F
F P
F
©1997, A.J. Girondi
SECTION 15.10
Student Notes
15-20
©1997, A.J. Girondi