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Module 25 — Bonding *****
Module 25 – Bonding ....................................................................................................683
Lesson 25H: Covalent Bonds .................................................................................................. 683
Molecular Shapes and Bond Angles ............................................................... 688
Molecular Polarity ............................................................................................. 699
Solubility ............................................................................................................. 706
Double and Triple Bonds.................................................................................. 710
Ion Dot Diagrams .............................................................................................. 715
Orbital Models for Bonding ............................................................................. 717
For additional modules, visit www.ChemReview.Net © 2010 ChemReview.Net v.1p Page i Module 25 — Bonding Module 25 — Bonding
Prerequisites: For this topic, you will need a set of molecular models. These can be
purchased at college bookstores or online. In some courses, models are provided in your
“lab drawer.” As an alternative, patterns for cardboard models are provided in Lesson 25B,
but commercial models are recommended.
Pretests: If you believe that you know the material in a lesson, try two problems at the end
of the lesson. If you can do those, you can skip the lesson.
Bonds are forces that hold atoms together. The nature of the chemical bond is a question at
the heart of chemistry, but the answer is not completely understood. An explanation of
bonding must take into account protons and electron pairs, wave equations and orbitals,
electrical attraction and repulsion, neutral molecules, and polyatomic ions. A theory that
successfully unites all those factors does not yet exist.
However, a variety of models predict bond behavior. We will begin with two simple
models, Lewis diagrams and VSEPR, that allow us to predict the composition and shape of
a significant percentage of the molecules within and around us.
***** Lesson 25A: Lewis (Electron Dot) Diagrams
Ionic Versus Covalent Bonds
I In ionic bonding, charged particles (ions) are held together by electrical attraction. Ions
that are monatomic usually have the valence electron configuration of the nearest noble
Covalent bonding is often described as electron sharing. As in ionic bonding, each atom is
often found surrounded by valence electrons in a noble gas configuration, but in covalent
bonding, a pair of electrons between two atoms can play the role of valence electrons for
both atoms. These shared electrons are the covalent bond that holds the two atoms
In reality, bonds are not “either ionic or covalent.” All ionic bonds have some covalent
character that is evident under certain conditions. Covalent bonds often have some ionic
character. Whether a molecule should be considered as primarily ionic or covalent is best
determined by its behavior.
Atoms that are bonded covalently in a molecule or a polyatomic ion do not easily separate
when melted or dissolved. The forces holding atoms together inside a covalent molecule
are strong compared to the forces between the molecules. Compared to the ions in ionic
compounds, covalently bonded molecules are more easily pulled apart from each other
when heated, so covalent molecules typically melt and boil at temperatures much below
that of ionic compounds. © 2011 ChemReview.Net v.1z Page 684 Module 25 — Bonding Lewis (Dot) Diagrams
Lewis diagrams (also called Lewis formulas or structures, or electron dot diagrams) can
be drawn to represent covalent bonds in covalent molecules and polyatomic ions.
Example: The Lewis diagram of H2 is H:H Lewis diagrams are useful for predicting the bonding, shape, and solubility of substances.
To draw the Lewis (dot) diagrams for molecules, we begin by drawing the dot diagrams for
atoms. Rules for Drawing Lewis Diagrams for Neutral Atoms
1. Write the symbol for the atom.
2. Determine the number of valence electrons for the atom.
Recall that the number of valence electrons for an atom is equal to the number for the
main group (the tall columns) of the periodic table in which the atom is located.
• First column neutral atoms have one valence electron. • Neutral atoms in the carbon family have four valence electrons. • Noble gases have 8 valence electrons (except helium which has 2). 3. Assume that each atom symbol has four sides. On each side can go at most two
electrons. Using dots to represent the valence electrons, draw the valence electrons
around the atom symbol. Put one electron on each of the four sides of the symbol before
you start to pair electrons. The four sides are equivalent: you may place the paired and
unpaired electrons on any side. • Examples: Boron, with 3 valence e─, is • B • • Nitrogen (5 valence e─) is • N •
•• The dot diagram for a neutral boron atom has three unpaired electrons. Nitrogen has
three unpaired electrons and one pair of electrons. The Octet Rule
To draw dot diagrams for molecules and ions, apply the octet rule: in achieve maximum
stability, an atom needs to be surrounded by eight valence electrons. (Hydrogen, however,
needs only two.)
The octet rule is related to the orbital energy level diagram for atoms: filled clusters in the
diagram have high stability, and (except for row 1 atoms), filled clusters have 8 valence (s
and p) electrons.
Combinations of atoms that result in all atoms being surrounded by eight valence electrons
(two for H atoms) tend to be stable combinations: those that are likely to be found in nature
and formed in chemical reactions. A species that does not have a satisfied octet may exist,
but it is likely to be unstable: it will tend to be a very reactive species. © 2011 ChemReview.Net v.1z Page 685 Module 25 — Bonding Using Dot Diagrams To Predict Bonding
To predict the bonding in stable molecules, depending on the type of problem, there are
two methods for drawing dot diagrams.
This method predicts the molecular formula that combine one or two kinds of atoms, if the
formula is not supplied but the compound has all single bonds (one pair of electrons per
Let’s learn Method 1 with an example.
Q. Draw the Lewis diagram for a stable molecule that contains only chlorine atoms
with single bonds. Complete the following steps, then check your answer below.
Method 1 Steps. If you do not know the molecular formula, but you know which atoms
are combining and that the molecule has single bonds:
1. Draw the dot diagram for each neutral atom.
2. Combine the diagrams of the atoms so that the unpaired electrons pair and are shared
between two atoms. Combine the atom diagrams until each symbol is surrounded by
eight valence electrons (except H which needs two).
* ** * *
A neutral chlorine atom has seven valence electrons. Place one on each of
the four sides of the symbol, then start to pair electrons. This results in 3
pairs and one unpaired dots representing the valence electrons around
chlorine. The chlorine has seven valence electrons and it needs eight to be
stable. •• : Cl • •• To make a stable molecule, slide two chlorines together so that their unpaired electrons
pair. Each chlorine is now surrounded by eight valence electrons. The octet rule is
satisfied. The two shared electrons are a bond between the two chlorines. Each chlorine
atom has 3 lone pairs: pairs around the atom that are not part of a bond.
•• : Cl
•• • + Two atom dot diagrams ↑ • •• Cl :
•• •• •• •• •• : Cl : Cl : = Cl ─ Cl molecule dot diagram ↑ = Cl2 structural ↑ and molecular ↑ formulas In a particle with more than one atom, the lone pairs may also be termed unshared pairs or
Method 1 is over-simplified, but it does predict the bonding in cases when the molecular
formula (and therefore the total number of valence electrons) is not known.
. © 2011 ChemReview.Net v.1z Page 686 Module 25 — Bonding Practice A: Use a periodic table. If needed, check your answers after each part. The
atoms in covalent compounds will most often be the non-metals found toward the top right
of the periodic table, plus hydrogen. 1. How many valence electrons are in these neutral atoms?
a. Silicon b. Phosphorous c. Bromine d. Sulfur 2. Draw the Lewis diagram for each atom in #1.
a. Si b. P c. Br d. S 3. Using Method 1, draw a Lewis diagram for the predicted stable molecules formed by
a. Fluorine atoms b. Hydrogen and chlorine atoms 4. For each of the molecules in Problem 3, list the number of covalent bonds and
the total number of lone pairs of electrons.
Bonds: 3a. _____ 3b. _____ Lone Pairs: 3a. _____ 3b. _____ Method 2:
If the formula for a molecule or polyatomic ion is supplied, a model that better predicts the
nature of bonds in both simple and more complex molecules is to combine the valence
electrons without regard to which atom contributes the electrons.
In simple molecules with more than two atoms, the central atom is usually the atom with
the most unpaired electrons in its atom dot diagram. This means that the central atom in a
formula is usually the atom closest to Group 4A (the carbon family) in the periodic table.
Carbon family atoms have 4 unpaired electrons.
Let’s try a simple example using Method 2. Use the following steps.
Q. Draw the Lewis diagram for a water molecule (H2O).
Method 2. When you know the molecular formula:
1. Count the total number of valence electrons in the neutral atoms of the molecule.
2. Arrange the valence electrons around the atoms to satisfy the octet/duet rule: For
maximum stability, each symbol needs to be surrounded by eight valence electrons (H
needs two). Do not worry about which atom contributes which electrons.
* ** * *
1. Each neutral hydrogen has one valence electron (and wants 2). The one neutral oxygen
has six valence electrons (and wants 8). The total for the molecule is 8 valence electrons. © 2011 ChemReview.Net v.1z Page 687 Module 25 — Bonding 2. The Lewis diagram for water can be written in two equivalent ways.
•• •• H:O: or •• •• = H─O─H
•• H ^ Two equivalent
dot diagrams ^ or •• H─O :
H = H─O
H ^ Two equivalent ^
bond-dot diagrams ^ A structural
formula The first two structures above are the “electron dot” form of the Lewis diagram,
showing all of the valence electrons. Because all four sides of the oxygen are
equivalent, the 90° and 180° drawings of the two Lewis diagrams are equivalent: both
represent the same molecule.
Between adjacent atoms, shared electrons are termed bonding pairs. In addition, the
oxygen has two lone pairs of electrons.
The third and fourth structures show an alternate way of writing a Lewis structure,
with the bonding pairs (two shared electrons) written as a line to represent a bond, but
the lone pairs (the unshared or non-bonding pairs) represented by dots.
The last formula is a structural formula. A structural formulas provides some
information about the location of the atoms in a molecule, but often does not include
the location of the lone pairs.
All of the diagrams show that in water, there are two bonds, and the oxygen atom is in
the middle. The Lewis diagrams also show the lone pairs that will be needed to explain
the shapes of molecules.
The atom which had the most unpaired electrons in its atom dot diagram will be
surrounded by the highest number of bonds, and in most cases is considered the central
atom in the molecule.
The Lewis diagram predicts that in water, two H and one O are a stable, favored
combination because by sharing electrons, all of the atoms can be surrounded by the
number of valence electrons wanted: 2 for H, 8 for other atoms.
Many of the frequently encountered covalent molecules in first-year chemistry, as well as
organic chemistry, consist of hydrogen plus the second-row non-metals. In general, neutral
atoms in the second row of the periodic table have the following characteristics when they
bond covalently. These patterns apply, with many additions and exceptions, for atoms
below the second row. Learn this table so that given the terms in the first column you can
fill in the blanks.
For neutral single-bonded atoms: Second Row Symbol Li Be B C N O F Ne Main Group Number 1 2 3 4 5 6 7 8 Valence Electrons - - 3 4 5 6 7 8 or 0 3 4 3 2 1 0 0 0 1 2 3 4 Bonds Ionic Bonds Lone Pairs
© 2011 ChemReview.Net v.1z - Page 688 Module 25 — Bonding In predicting typical formulas for covalent molecules, it is helpful to remember that
“carbon bonds 4 times, nitrogen 3 times, oxygen twice, and hydrogen and halogens once.” P ractice B: Use a periodic table. 1. Using Method 2, draw a Lewis diagram and then a structural formula for these.
a. CH4 b. PCl3 2. For each of the molecules in Problem 1, list the number of covalent bonds and
the total number of lone pairs of electrons.
Bonds: 1a. _____ 1b. _____ Lone Pairs: 1a. _____ 1b. _____ 3. Predict how many bonds will typically be found around these neutral atoms.
a. Selenium (34) b. Iodine (53) c. Silicon (14) d. Nitrogen (7) ANSWERS
1. Valence electrons: a. Silicon 4
2. Dot diagrams: a. • Si •
• b. Phosphorus 5
b. • P •
• c. c. Bromine 7 •• : Br : d. • d. Sulfur 6
• :S :
• It does not matter which of the four sides have the paired or unpaired electrons.
3a. •• •• :F : F : = F─F 3b. •• • • 4. Bonds: 3a. 1 3b. 1 •• H : Cl : =
•• Lone Pairs: 3a. 6 H ─ Cl
3b. 3 Practice B H
H : C : H = H─C─H
H 1a. 2. Bonds: 1a. 4 1b. 3 3. a. Selenium 2 (main group 6) •• 1b. •• •• : Cl : P : Cl :
•• •• ••
: Cl : = •• Lone Pairs:
b. Iodine 1 1a. 0 Cl─P─Cl
Cl 1b. 10 c. Silicon 4 d. Nitrogen 3 ***** © 2011 ChemReview.Net v.1z Page 689 Module 25 — Bonding Lesson 25B: Molecular Shapes and Bond Angles
A key factor that determines the behavior of molecules is their shape: how the atoms and
electrons are arranged in three-dimensional space.
The shapes and bond angles of most molecules can be predicted with reasonable accuracy
using Lewis diagrams. This technique is called valence shell electron-pair repulsion
theory (VSEPR). The term simply means that all electron pairs, whether they are lone pairs
or bonds, will repel each other, and they will separate by the maximum possible angle
around the nucleus of an atom.
Below, we will learn to predict how atoms and electrons are arranged based on the
columns of the periodic table. A chart at the end will summarize these rules. To Predict the Shape of a Covalent Molecule
1. Draw the Lewis diagram for the molecule.
2. The general shape of a molecule is named based on the number of directions in which the
electron pairs are found around the central atom. For the general shape, it will not
matter whether electrons are in bonds or lone pairs, or single, double, or triple bonds.
All of the electron pairs must be considered to determine the shape and bond angles,
but the shape is named based only on the positions of the atoms. The lone pairs help to
determine the shape, but they are ignored in naming the shape, of a covalent molecule.
a. One pair: If a bonded atom is surrounded by only one electron pair, it has one
bond to a second atom. The shape around this atom is said to be linear. Since it
takes three points to determine an angle, and two atoms are two points, an atom
with only one bond has no bond angles.
Example: H : H = H─H = H2 Each H has one bond. The molecule has a linear shape with no bond angles.
b. Two pairs: If an atom in a molecule is surrounded by two electron pairs, both will
be bonds. The two electron pairs will separate as much as possible by assuming a
linear shape around the central atom. This shape results in three atoms in a line.
With three points to determine an angle, the bond angle around the central atom is
•• •• Example: •• •• : Cl : Be : Cl : = Cl ─ Be ─ Cl = BeCl2 The Be in BeCl2 is surrounded by two electron pairs, and both are bonds. The
arrangement of the bonds around the central atom, and the shape of the
molecule, is linear with 180° bond angles.
Note that BeCl2 is electron deficient: it violates the octet rule. BeCl2 does form,
but as an electron-deficient molecule it has some unusual properties. © 2011 ChemReview.Net v.1z Page 690 Module 25 — Bonding c. Three pairs: If a central atom is surrounded by electron pairs in three directions,
the shape that allows the electron pairs to get as far apart as possible is termed
trigonal planar. The three bonds are in a plane (flat) with 120° bond angles.
•• •• :F: B:F: Example: •• •• •• :F:
120° = BF3 The shape of the BF3 molecule is trigonal planar. All bond angles are 120°.
Like BeCl2, a BF3 Lewis diagram can be drawn using single bonds, but it
violates the octet rule. BF3 does form, but as you might predict with its electron
deficient structure, it has some unusual properties.
d. Four pairs: Due to the octet rule, most stable atoms are surrounded by four electron
pairs. The three-dimensional tetrahedral shape allows those four pairs to get as far
apart as possible. In a tetrahedron, all of the angles are 109.47°.
You will need a tetrahedral molecular model for the sections below. If you have not
purchased models, build the cardboard model on the next page, then return here. © 2011 ChemReview.Net v.1z Page 691 Module 25 — Bonding Building A Cutout Tetrahedral Model
If you do not have access to a commercial molecular model kit, a tetrahedral
model can be constructed from the patterns below.
1. Obtain a sheet of foamboard or thick or corrugated cardboard at least onehalf the size of this sheet of paper.
2. Either copy this page, or cover this page with thin paper and trace onto the
paper, the three shapes below. Cut out the three paper patterns.
3. Using the patterns and blunt scissors, carefully cut the foamboard or
cardboard to make 2 circles, 4 rectangles, and 3 ovals.
4. Cut slots in the 9 pieces at the thick lines. The slots should be to the depth
shown by the thick lines. Cut the slots to a width that matches the thickness
of the cardboard, so that the pieces slide together in the slots tightly, but
with minimal binding.
5. On the four bonds, round off the corners of the atom ends just a bit.
6. Push together the two circles using the deep slot in each. Arrange them so
that they are at right angles, simulating a spherical shape.
7. Add two bonds to each circle to give four bonds total. Push the slots on the
bonds into the shallow slots on each circle. Try to get the bonds to be
perpendicular to the circle to which they are attached.
With four bonds, the model represents central atoms in the carbon family. The
four electron pairs around the central atom are in a tetrahedral shape. Since all of
the electron pairs are bonds to atoms, the atoms around the central atom are in a
tetrahedral shape. Since the position of the atoms decides the shape of the
molecule, the molecule is tetrahedral, and the angle between any two bonds is
Models for other families will be made by substituting lone pairs for bonds. © 2011 ChemReview.Net v.1z •• Lone Pair Atom Bond Page 692 Module 25 — Bonding a. For a single-bonded central atom in the carbon family (main group 4), all four
electron pairs around a central atom are bonds, and the arrangement of the bonds
is said to be tetrahedral, with ~109° angles between all of the bonds. H
C H •• Example: H:C:H =
H 109° = H C H4 = tetrahedral, 109° angles A three-dimensional tetrahedron is difficult to represent on two-dimensional
paper. In the diagram above, the - - - - line represents a bond going behind the
plane of the paper, and the ∆ represents a bond coming out of the paper. A 3-D
model will assist in working with this important shape.
Build a CH4 molecule from your molecular model pieces. Place the
assembled model on a flat surface, then flip it so that it rests on three
different points. Flip it again. Note the high symmetry of a threedimensional tetrahedron: the shape of the molecule should be the same no
matter which three atoms the model sits upon.
b. A single-bonded central atom in the nitrogen family (main group 5) most often is
surrounded by three bonds and one lone pair.
There are four electron pairs around the central atom, and the pairs assume a
tetrahedral shape to get as far apart as possible. Because this electronic
geometry is tetrahedral, the angles between all of the electron pairs, bonds, and
the atoms are tetrahedral (~109°).
However, the lone pairs, though they count in determining the shape around a
central atom, are not considered when naming the shape. The shape is named
based on the position of only bonds and atoms.
For this case of one lone pair and three bonds around a central atom, the four
atoms are in the shape of a low pyramid. The central atom is above the plane of
the three atoms to which it bonds. Since the pyramid rests on 3 points, the
shape of the atoms is called a trigonal pyramid, and the molecular geometry is
termed trigonal pyramidal.
N •• Example: H:N:H =
•• H H ~109° H = NH3 H Build this NH3 molecule from your kit. Starting from CH4 , replace one
bond with a lone pair. The four electron pairs are still in a tetrahedral shape.
Then take off the lone pair to look at just the shape and angles of the bonds
and atoms. With the central nitrogen atom on top, check that the atoms form
a low pyramid with tetrahedral (~109°) angles.
© 2011 ChemReview.Net v.1z Page 693 Module 25 — Bonding The shape of NH3 is trigonal pyramidal with ~109° bond angles.
c. A single-bonded neutral atom in the oxygen family (main group 6) is most often
surrounded by two bonds and two lone pairs. These four electron pairs repel to
assume a tetrahedral shape with ~109° angles around the central atom.
As always, the lone pairs count in deciding the shape, but do not count when
naming the shape of the bonds around the central atom or the molecule. The
two bonds and the three atoms are said to be in a bent shape, with ~109° angles.
•• Example: H :O:
•• •• = H |
• ~109° H = H2O H Build this water molecule. Place two bonds and two lone pairs around the
central atom. This puts the four electron pairs into a tetrahedral shape.
Switch the position of one bond and one lone pair. Does this create a new
No. Due to the symmetry of a tetrahedron, all four electron pairs around
the central atom are in equivalent positions. The same molecule results no
matter where the two bonds and two lone pairs are attached.
In Lewis diagrams, we treat four sides around an atom symbol as
equivalent because four electron pairs repel into a tetrahedral shape, and
the four sides of a tetrahedron are equivalent.
Now remove the two lone pairs. The geometric shape of the bonds and
atoms, and of the H2O molecule, is bent. Its one bond angle is ~109°.
d. If four electron pairs surround a central atom, but only one is a bond, the
electronic geometry is tetrahedral. However, since there are only two atoms,
and the atoms determine the name of the shape, the shape of the molecule is
linear. Since there is only one bond, there is no bond angle.
Cl •• Example: H : Cl : =
•• H •• ~109°
• = H ─ Cl = HCl The shape around the chlorine, and of the HCl molecule, is linear with no
bond angles. © 2011 ChemReview.Net v.1z Page 694 Module 25 — Bonding 3. FINE TUNING: We can increase the accuracy of VSEPR bond-angle predictions with
the following rule:
Lone pairs repel slightly more than bonds. The lone pairs need more room.
If one or two lone pairs are present in a tetrahedral shape, the angles around the lone
pairs will be slightly larger than 109°. This will push the angles between bonds to be
slightly less than 109°.
Lone pairs tend to occupy slightly more space than bonds, because bonding pairs are
more localized along the axis between the atomic nuclei. This means that the lone pairs
repel other pairs slightly more than bonds. The angle between the lone pairs is
therefore slightly larger than the angle between bonds.
For example, the general model predicts that in a water molecule, the shape is bent
with bond angles of ~109°. However, because water has two lone pairs, they repel
each other and the bonds slightly more than the bonds repel each other. This “lone
pair scrunch” forces the bonds into an angle slightly smaller than 109°.
In water, the shape is bent as predicted, but the actual measured bond angle is
104.5°, slightly less than the tetrahedral angle that the general rules predict. The
angle in water is a typical value for central atoms surrounded by two lone pairs and
Angle more than 109° between lone pairs >
< Angle less than 109° between atoms
In what cases for single-bonded compounds will bond angles be less than 109°? Only
cases with one lone pair and three bonds, or two lone pairs and two bonds: those with
central atoms in the nitrogen or oxygen family, which are those in the 5th or 6th tall
column of the periodic table.
For single-bonded carbon family neutral atoms, the bond angles are 109° rather than less
than 109°. When there are no lone pairs around the central atom, there is no lone pair
effect on the angles.
We will call this the “lone pair scrunch” rule:
In neutral molecules containing only single bonds, around central atoms in the
nitrogen family (column 5) or oxygen family (column 6), the bond angles are
usually 103° to 109°: slightly less than 109°. © 2011 ChemReview.Net v.1z Page 695 Module 25 — Bonding Summary
For the second row of the periodic table, and with frequent exceptions for rows below the
second row, central atoms in neutral compounds will generally have the characteristics
listed in the table below. Learn this table so that given the terms in the first column you can
fill in the blanks, based on the rules for the behavior of electron pairs.
For neutral, single-bonded atoms: Second
Symbol Li Be B C N O F Ne Main
Number 1 2 3 4 5 6 7 8 Valence
Electrons 1 2 3 4 5 6 7 8 3 4 3 2 1 0 0 0 1 2 3 4 Bent Linear No
Bonds <109° None Bonds
Pairs Ionic Bonds
- - Shape Linear Linear Bond
Angles None Practice: 180° Trigonal Tetrahedral Trigonal
120° 109° <109° Use a periodic table, plus models if needed. Check answers after each part. 1. For a molecule in which the central atom is surrounded by two bonds and two lone
a. What is the shape of the electron pairs?
b. What is the shape of the atoms?
c. What is the bond angle?
2. From memory, list the predicted shapes of the bonds around central atoms for the
atoms in the second row of the periodic table, in order.
3. From memory, under each shape in Problem 2, write the predicted bond angles. © 2011 ChemReview.Net v.1z Page 696 Module 25 — Bonding 4. Complete this table based on VSEPR predictions for neutral, single-bonded atoms.
NF3 Molecule SiH4 AlCl3 SI2 Lewis
5. Which molecule in Problem 4 would likely be the least stable and most reactive? Why? ANSWERS
1. The four electron pairs repel to into a tetrahedral shape. The three atoms are bent, with a bond angle of
slightly less than 109°.
2. Linear, Linear,
3. None Trigonal Planar, Tetrahedral, Pyramidal, 180° 120° ~109° <109° Bent, Linear, <109° No Bonds None 4.
Diagram •• •• •• :F:N :F:
•• •• •• :F:
•• SiH4 H •• H : Si : H
•• : Cl : Al : Cl :
•• •• H : Cl : •• • •• •• •• •• •• •• :I : S : I :
(also could be
drawn at 90°) e─ Pairs Tetrahedral Tetrahedral Trigonal Planar Tetrahedral Molecule
and Bonds Trigonal
Pyramidal Tetrahedral Trigonal Planar Bent <109° 109° 120° <109° Bond Angles 5. AlCl3 would be predicted to be the least stable and most reactive because Al has an unsatisfied octet.
***** © 2011 ChemReview.Net v.1z Page 697 Module 25 — Bonding Lesson 25C: Electronegativity
The Electronegativity Scale
What decides if a particular bond will be ionic or covalent? Our previous “rule of thumb”
has been that if a bond is between a metal and a non-metal atom, it will likely be ionic, but
if it is between two non-metals, it will be covalent. A more precise view is that bonds have
a mixture of ionic and covalent character. This latter model for bonding will allow more
accurate predictions of the properties and behavior of bonds and substances.
Atoms have differing attractions for electrons. Electronegativity (a model developed by the
American chemist Linus Pauling) predicts how strongly each atom attracts the electrons in
The electronegativity scale assigns each atom a value between 0.7 and 4.0. Fluorine (EN =
4.0) is the strongest electron attractor of all the atoms. Cesium and francium (EN = 0.7) are
the weakest electron attractors.
The following table lists the electronegativity values of the atoms. (These numbers are
termed the Pauling values. Other models may use slightly different EN values.)
The electronegativity values (EN) for the second row atoms should be memorized. This is
easy, since the 2nd row numbers start at 1.0 and increase by 0.5 for each atom to the right.
The frequently used values for hydrogen (2.1) and chlorine (3.0) should also be memorized.
It will speed your work if the electronegativity values (EN) for the second row atoms are
memorized. This is easy, since the 2nd row numbers start at 1.0 and increase by 0.5 for
each atom to the right. The values for hydrogen (2.1) and chlorine (3.0) are also
encountered frequently and should be committed to memory. Electronegativity Values
Row 1 2.1 Row 2 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Row 3 0.9 1.2 1.5 1.8 2.1 2.5 3.0 Row 4 0.8 1.0 1.3-1.9 1.6 1.8 2.0 2.4 2.8 Row 5 0.8 1.0 1.2-2.2 1.7 1.8 1.9 2.1 2.5 Row 6 0.7 0.9 1.0-2.4 1.8 1.9 1.9 2.0 2.2 Row 7 0.7 0.9 In the table, note that
• hydrogen’s value of 2.1 is in the middle range of values. • Only four atoms have EN values of 3.0 and above: N, O, F, and Cl. • Values generally (but not always) increase toward the top right corner of the
periodic table: to the right across a row and up a column. © 2011 ChemReview.Net v.1z Page 698 Module 25 — Bonding To predict bond behavior, the electronegativity model divides bonds into three types:
Ionic, polar covalent, and non-polar covalent. Ionic Bonds
In general, if the difference between the electronegativities of two bonded atoms
• is greater than 1.7, the bond will generally have ionic character; • is 1.7 or less, the bond is likely to have covalent character. (Different texts may use different cutoff values, but 1.7 is a typical choice.)
An ionic bond can be thought of as a bond in which the difference in electron attraction is
so strong that the more electronegative atom removes valence electrons from the other
atom to form two charged particles. Polar Versus Non-Polar Covalent Bonds
Covalent bonds are divided into two types: polar and non-polar.
For a covalent bond between two atoms that have the same or very similar
electronegativity values, the electrons on average will be found at an equal distance
between the two nuclei.
Examples: F─F C≡C EN: 4.0 4.0 N ─ Cl 2.5 2.5 3.0 3.0 A covalent bond in which the electrons are equally shared is said to be non-polar.
Whether bonds are single, double, or triple bonds (as in C ≡ C above) does not affect the
As the difference in the electronegativity of two bonded atoms increases, the bond becomes
more polar. The electrons are still shared, but they tend to be found closer to the more
This uneven electron sharing creates a dipole: an uneven distribution of electric charge. The
more electronegative atom takes on a partial negative charge, while the weaker electron
attractor takes on a partial positive charge.
In chemistry, dipoles are generally represented using two types of notation. In math and
science, a δ (a lower-case Greek delta) is often used as a symbol meaning partial. In this
notation, a polar bond is labeled with two deltas: The atom that is the stronger electron
attractor, with the higher electronegativity value, has its partial negative charge labeled δ –
(pronounced delta minus), and the weaker electron attractor is labeled δ+ (delta plus).
Examples: δ+ C — O δ— δ+ N — F δ— δ— O — H δ+ An alternate way to represent the dipole in a bond is to use an arrow in place of the bond.
The arrow points toward the end of the bond with the stronger electron attractor; toward
the side where on average the electrons are more likely to be found.
Examples: © 2011 ChemReview.Net C v.1z O N F O H Page 699 Module 25 — Bonding As the difference in electronegativity in bonds rises, from zero to as high as 3.3, the
character of the bond changes gradually from non-polar to polar to ionic. However, in
general, if the difference in electronegativity between two atoms is
• from 0 to 0.4, the bond is considered to be covalent and non-polar; • from 0.5 to 1.7, the bond is considered to be covalent but polar; • above 1.7, the bond is considered to be ionic. The choice of the breakpoints at 0.4 and 1.7 is arbitrary. Some textbooks use values such as
0.3 or 0.7 and 2.0. In any case, the electronegativity differences are best understood as
gradually shifting as they increase: from non-polar, to polar, to ionic character. Practice: Memorize the electronegativity values for H and Cl. Note the pattern for the
values for the second row atoms. Then use a periodic table that does not include
electronegativity values on the problems below. Check your answers after each part.
1. For each of the bonds below,
a. write the electronegativity value above each atom from memory.
b. Below the bond, label each atom as δ+ or δ—.
c. On the next line down, calculate the electronegativity difference.
d. On the next line down, label the bond as non-polar, polar, or ionic.
e. On the next line down, re-write the bond using an arrow in place of the bond to
show the direction of the dipole.
i. C—H ii. N — Cl iii. C — F iv. O — B ANSWERS
1b. 2.5 2.1
N — Cl
no δ 2.5 4.0
δ— 3.5 2.0
δ+ 1c. 0.4 0 1.5 1.5 non-polar polar polar 1d.
1e. non-polar or
C H N — Cl (no dipole) C F O B ***** © 2011 ChemReview.Net v.1z Page 700 Module 25 — Bonding Lesson 25D: Predicting Polarity
Polar versus Non-Polar Substances
• Why do salad oil and water-based vinegar, after being shaken, separate into two
layers, yet most alcohols and water dissolve into each other without forming two
layers? • Why do soaps and detergents dissolve in water, but also dissolve oils from food and
skin that normally do not dissolve in water? • Why do salts and sugars dissolve in water, but most rocks do not? • Can we predict formulas for pharmaceuticals that will relieve pain and cure
disease? The answers to these practical and important questions are often found in the shapes and
polarities of bonds and of substances.
In the previous lesson, electronegativity was used to classify bonds as ionic, polar and nonpolar. Molecules can also be classified as having ionic, polar, and non-polar character.
In classifying the polarity of combinations of atoms, the rules are:
1. A compound with just one ionic bond will generally (but not always) have ionic
behavior, even if it also has many non-polar bonds.
2. A molecule with all covalent bonds will be polar if
• it has polar bonds and • the dipoles do not cancel due to molecular symmetry. 3. A molecule will be non-polar if
• it has all non-polar bonds, or • it has polar bonds, but the dipoles cancel due to symmetry. Flow Chart: Predicting the Polarity of Molecules
Knowing the chemical formula for a substance, we can often make general predictions about
whether its whether molecules will have ionic, polar, or non-polar behavior. The rules
below comprise a simplified model, but they provide reasonably good predictions for the
polarity of most substances.
Compounds that are combinations of metals and non-metals usually (but not always)
display ionic behavior. Formulas that you recognize as ionic solids will also be ionic.
In other cases, to predict whether a substance will have ionic, non-polar, or polar behavior,
apply the rules in the following flow chart in order.
1. Write the formula for the substance. If the substance includes a metal and a non-metal,
the substance is ionic. In addition, if any one bond is ionic, the substance is ionic.
2. If rule 1 does not apply, list each type of bond in the substance, then
a. add electronegativity (EN) values to each atom.
b. Based on the EN differences, label the bonds as ionic, non-polar, or polar. © 2011 ChemReview.Net v.1z Page 701 Module 25 — Bonding c. If all of the bonds are non-polar, the substance is non-polar.
3. If rules 1and 2 do not apply to the particle, one or more of the bonds must be polar.
a. Draw the Lewis diagram and sketch the shape of the molecule.
b. On the sketch, replace the bonds with arrows representing the dipoles. Use
geometry and symmetry to see if the dipoles cancel. If needed, make a 3-D model.
c. If the dipoles cancel, the molecule is non-polar. If the dipoles do not cancel, there is a
net dipole, and the molecule is polar. Dipole Cancellation
You may have had practice adding vectors in math or physics classes. Dipoles are one of the
types of quantities that add in two or three dimensions using vector addition. Even if you
have not practiced vector addition, dipole addition can often by simplified by this rule:
Equal but opposite dipoles cancel.
Let’s learn the method by example. Based on the flow chart rules above, apply the steps
above to the following cases, and then check your answers below.
Q. Label these substances as ionic, polar, or non-polar. Use a periodic table without EN
values (all of these atoms have values you should know). Because these particles
are two-dimensional, you should not need models to evaluate symmetry.
2. LiCl 1. Cl2 3. O=C=O (linear) 4. HCl *****
1. In Cl2, the shape must be Cl—Cl . Both have the same electronegativity value, so the
difference in EN values between the two atoms is zero. When the EN difference is 0 to
0.4, the bond is non-polar, and the molecule is covalent non-polar.
When the electronegativity difference between two atoms is zero, there is equal sharing
of the electrons in the bond between the two atoms. On average, the two electrons in
the bond will be found half-way between each atom, so there is no bond dipole.
2. Li has a 1.0 EN and Cl has a 3.0 EN. The difference is 2.0, which is above 1.7, so the
bond is likely to have ionic behavior. If one bond (or more) in a compound is ionic, the
compound is ionic. The more electronegative atom will tend to take the two electrons
in the bond. The result is an Li+ ion and a Cl─ ion.
3. CO2 is a linear molecule with two double bonds. In calculating an EN difference, it
does not matter whether the bond is single, double, or triple. The carbon EN is 2.5,
oxygen’s EN is 3.5, the EN difference is 1.0, so both bonds are polar.
When bonds are polar, the symmetry test must be applied to see if the dipoles cancel.
Add the dipoles to the molecular shape: O C O . When dipoles are equal but in
opposite directions, they cancel due to symmetry. That is true in this case. O C O has
polar bonds but is a non-polar molecule because the dipoles cancel. © 2011 ChemReview.Net v.1z Page 702 Module 25 — Bonding *****
4. H has a 2.1 EN and Cl a 3.0 EN. The difference is 0.9, which is in the range of 0.5 to 1.7,
so the bond is polar. The shape for this molecule must be H—Cl . Because Cl is more
electronegative than H, the dipole points toward Cl: H Cl . Since this bond is polar
and there are no other bonds to cancel its dipole, the molecule has a dipole and is polar. Practice A: Use a periodic table that does not include electronegativity values (you
should know these from their table position). If needed, check answers after each part. Based on VSEPR and electronegativity, predict whether these compounds will be ionic,
covalent polar, or covalent non-polar.
1. BeH2 H
C=O (flat shape)
H 2. LiF 4. BCl3 Polarity In 3-D Molecules
In the section above, we considered two-dimensional molecules. For compounds that are
three-dimensional, it helps to make a model to judge the dipoles and symmetry. For threedimensional molecules with tetrahedral pairs, the following are general rules.
1. If a central atom in the carbon family has single bonds to four atoms that are the same
kind of atom, even if the bonds are polar, the dipoles will cancel due to symmetry. The
molecule will be non-polar.
Examples: CH4 and SiF4 are non-polar molecules.
2. For a single-bonded central atom that obeys the octet rule in the nitrogen or oxygen
family, the molecular shape is trigonal pyramidal or bent. If any of the three bonds are
polar, and if all of the dipoles point to, or all point away from, the central atom, the
molecule is polar because the dipoles cannot cancel.
Examples: NH3, OF2 are polar molecules.
3. In more complex cases, a model should be made and the dipoles analyzed.
Some examples will help with these rules. Do the parts below one at a time, checking your
answers after each part.
Q. Using the flow chart and symmetry rules, label these compounds as ionic, polar, or
non-polar. Use a periodic table without electronegativity values. Make molecular
models if needed.
1. H2O 2. CCl4 3. NH3 4. CHF3 *****
To evaluate molecular polarity, use the flow chart. First evaluate bond polarity. If the
bonds are polar, evaluate symmetry to see if the dipoles cancel. © 2011 ChemReview.Net v.1z Page 703 Module 25 — Bonding 1. H2O: H has an EN of 2.1, O has an EN of 3.5. The difference of 1.4 makes the bond
polar. When the bond is polar, evaluate the symmetry. H •• Water = : O: H =
•• <109° H H = H
added = H
O H2O has tetrahedral 3-D electron pairs but 2-D bonds and atoms. In this model, assume
that bonds, not lone pairs, contain the dipoles. The bonds in water have a bent shape
with <109° angles. Both of the dipoles point toward oxygen, so they do not cancel.
The bonds are polar and the water molecule is polar. The dipole in water can be
represented by an arrow (above) or using δ+ notation below.
δ+ The net dipole points from the H side toward O.
O δ— The H side of water is δ+ and the O side is δ—. H
Even with its polar bonds, if water were H-O-H linear in shape, it would not have a net
dipole. However, the bonds in water are bent rather than linear. This means that
Water is polar.
The polarity of water is an important factor in many reactions in chemistry and biology.
2. CCl4: C is EN 2.5 and Cl is EN 3.0, so the C—Cl bond weakly polar, and the dipoles
point toward Cl.
CCl4 is tetrahedral with 4 equal bond dipoles. Turning the model so that two bonds are
up and two down, the top and the bottom two dipoles cancel side to side. The
resultants are two dipoles, one pointing up and the other down. These two resultant
dipoles also cancel, because they are equal but in opposite directions.
•• : Cl :
CCl4 = •• •• •• •• •• •• : Cl : C : Cl : =
: Cl :
109° Cl = C Cl
Cl Cl When a central atom is surrounded by four tetrahedral bonds to the same atom, the
molecule is always non-polar due to symmetry.
***** © 2011 ChemReview.Net v.1z Page 704 Module 25 — Bonding 3. NH3: First evaluate bond polarity. The EN of N is 3.0, and of H is 2.1. The difference
of 0.9 means that the bonds are polar, with the dipoles pointing toward N.
Polar bonds can mean polar or non-polar molecules, depending on whether the dipoles
cancel. To check for dipole cancellation, draw the Lewis diagram: NH3 has tetrahedral
electron pairs with one lone pair and 3 bonds. Make the tetrahedral model, then take
off the lone pair, to focus on the bonds that determine the polarity. If the model is
placed so that the central N is up, all of the dipoles point upward from the H’s toward
N. The dipoles are equal but not opposite: they do not cancel. NH3 is polar.
N •• NH3 = H:N:H =
•• H H = H H <109° H N H H In pyramidal molecules, three dipoles in the same direction, pointing either to or
from a central atom, always result in polar molecules.
4. CHF3: The bonds between C (2.5) and F (4.0) are strongly polar toward F, with an EN
difference of 1.5. The C—H bond is only slightly polar. The molecule may be non-polar
if the C—F dipoles cancel. To check for dipole cancellation, draw the Lewis diagram.
CHF3 has tetrahedral electron pairs with 4 bonds. Then assemble the tetrahedral
model. If the model is held so that the H atom is up, all of the dipoles point down. The
dipoles are not equal and opposite: they do not cancel. By VSEPR and electronegativity
rules, CHF3 is predicted to be a polar molecule. H
•• CHF3 = •• •• •• •• •• :F:C:F:
•• = F H
F H 109° F = C F F F Exceptions
The above rules classify substances as non-polar, polar, and ionic. In reality, polarity is not
• The polarity of substances can be measured numerically, and those measurements
show a continuum of values from totally non-polar to highly ionic. • There are factors that affect polarity in addition to the electronegativity and atom
geometry considered in our model above. • Our rules classify as non-polar molecules that are slightly polar. That said, our simplified model is a starting point for evaluating polarity. It does provide
predictions of properties based on polarity, such as whether or not the substance will be
soluble in water, that hold true for a wide variety of substances. © 2011 ChemReview.Net v.1z Page 705 Module 25 — Bonding Practice B: On these, use a periodic table and a table of electronegativity values. Be
prepared to build tetrahedral models. If needed, check your answers after each part.
Based on VSEPR and electronegativity, predict whether these compounds will be ionic,
polar, or non-polar.
2. PCl3 1. SF2 3. SiH4 4. SiH3Cl ANSWERS
All of the molecules in Practice A are two dimensional: their shapes can be drawn on paper.
1. BeH2 First assign EN values to categorize the bonds. 2.1 H – 1.5 Be = 0.6 > 0.4 = polar bonds.
If bonds are polar, draw the Lewis diagram and shape to see if the dipoles cancel.
The central atom Be has a linear shape for its bonds and 180° bond angles.
Add the dipoles by vector addition. Because they are equal and in opposite directions: they cancel. The
VSEPR prediction is that the molecule is non-polar.
EN: 2.1 1.5 2.1
BeH2 = H : Be : H = H─Be─H =H Be H = non-polar 2. LiF First assign EN values to categorize the bonds. 4.0 F – 1.0 Li = 3.0 > 1.7 = ionic bonds.
If one bond is ionic, the compound is ionic.
3. H2C=O Assign EN values to categorize the bonds.
2.1 H – 2.5 C = 0.4 = slightly polar bond toward C. 3.5 O – 2.5 C = 1.0 = a polar bond toward O.
Electronegativity differences apply in the same way to single and double bonds.
If one or more bonds are polar, draw the Lewis diagram, sketch the shape, add the dipoles, and see if the
dipoles cancel. Since this molecule is flat, it can be analyzed on paper.
4. BCl3 H
C==O = H
C H O H C= O= a polar molecule. Assign EN values to categorize the bonds. 3.0 Cl – 2.0 B = 1.0 > 0.4 = polar bonds. If bonds are polar, draw the Lewis diagram and shape to see if the dipoles cancel.
BCl3 = •• •• : Cl : B : Cl : =
•• •• •• : Cl :
•• Cl B
120° = Cl B Cl Cl The central atom B is predicted by VSEPR to have a trigonal planar shape for its bonds and 120° bond
angles. Adding the dipoles by vector addition, they are equal and in opposite directions: they cancel. The
VSEPR prediction is that the molecule is non-polar. © 2011 ChemReview.Net v.1z Page 706 Module 25 — Bonding Practice B
1. SF2 : Assign EN values to categorize the bonds. 4.0 F – 2.5 S = 1.5 > 0.4 = polar bonds.
If bonds are polar, either draw the Lewis diagram and shape, add the vectors and see if they cancel.
Since sulfur is in the oxygen family, it generally bonds twice. Fluorine, a halogen, generally bonds once.
Sulfur, with more bonds, is therefore the central atom. Central atoms in the oxygen family form neutral
covalent molecules that are bent, with slightly less than 109° bond angles. Bent molecules are two
dimensional; their polarity can be evaluated on paper.
Since the two dipoles are 109° apart and not 180°, they are equal but not opposite. Adding the two dipoles
by vector addition gives a net, resultant dipole. The molecule is predicted to be polar.
S—F = F
S Shape F == dipoles S— F = a polar molecule. net resultant dipole Or use the rule that bent molecules with two or trigonal pyramidal molecules with three of the same polar
bonds are always polar molecules.
2. PCl3 : First label the bonds. 3.0 Cl – 2.1 P = 0.9 = polar bonds.
If any of the bonds are polar, draw the Lewis diagram and shape, then add the dipole vectors to see if they
Since phosphorus is in the nitrogen family, it typically bonds 3 times. Chlorine, a halogen, usually bonds
once. Phosphorus, bonding more, is therefore the central atom. Single bonded central atoms in the
nitrogen family form neutral covalent molecules that are trigonal pyramids with slightly less than 109° bond
PCl3 = •• •• •• : Cl : P : Cl : =
•• •• •• : Cl : Cl
P Cl = Cl P Cl <109°
Lone pairs are not bonds and do not have dipoles. The three bond dipoles are equal but not in opposite
directions, so they do not cancel. The molecule is polar.
Bent and trigonal pyramidal molecules with the same polar bonds are always polar molecules.
3. SiH4: For a central atom in the carbon family, whenever four of the same atoms are attached, any dipoles
will cancel due to tetrahedral symmetry. SiH4 is a non-polar molecule.
4. SiH3Cl Assign EN values to categorize the bonds.
2.1 H – 1.8 Si = 0.3 = non-polar Si—H bonds. 3.0 Cl – 1.8 Si = 1.2 = a polar Si—Cl bond with the
dipole toward Cl. If any bonds are polar, draw the Lewis diagram and shape, add the vectors and see if
Since silicon is in the carbon family, it bonds four times. Chlorine and hydrogen bond once. Silicon,
bonding more, is the central atom. For neutral covalent compounds with their central atoms in the carbon
family, molecules are tetrahedral with 109° bond angles. Make a model to investigate the symmetry. © 2011 ChemReview.Net v.1z Page 707 Module 25 — Bonding H H
Si •• H : Si : H = SiH3Cl = H •• : Cl : H 109° = Cl Si H H •• Cl H SiH3Cl has one polar bond un-cancelled by others. The molecule is predicted to be polar.
***** Lesson 25E: Solubility
How much of a substance will dissolve in a given liquid is complex: it depends on the
size, geometry, electronic properties, temperature, and relative amounts of the particles
of the substance and the liquid. However, some useful general rules can predict
solubility for a large number of substances and solvents.
If a liquid composed of non-polar molecules (such as a salad oil) is shaken with a liquid
composed of polar molecules (such as vinegar, which is primarily water), when the shaking
stops, the two liquids will slowly separate into two layers.
When two liquids composed of polar molecules, such as water and alcohols, are mixed, the
liquids dissolve in each other. One solution without layers is the result.
Why the difference?
For solubility, the general rule is: like dissolves like.
Polar liquids tend to dissolve polar or ionic particles.
Non-polar liquids tend to dissolve non-polar molecules.
Polar and non-polar substances tend not to dissolve in each other.
The most common polar solvent is water. H
O H •• Water = : O: H =
•• <109° H = H H
added = H
O An example of a non-polar liquid is carbon tetrachloride, in which the four equal and
opposite dipoles cancel to give a non-polar molecule.
•• : Cl :
CCl4 = •• •• •• •• •• •• : Cl : C : Cl : =
: Cl :
•• © 2011 ChemReview.Net v.1z Cl Cl
Cl 109° Cl Cl
Cl Cl Page 708 Module 25 — Bonding If H2O and CCl4 are shaken together, after the shaking stops, the two liquids separate
into two layers, just as with oil and vinegar dressing. In oil and vinegar, the salad oil
will rise above the denser water. If water is mixed with CCl4, the denser CCl4 will be
the bottom layer, with the water on top.
When mixed liquids separate into layers, they are said to be immiscible (pronounced
em-MISS-ible). When liquids dissolve in each other, as in the case of water and ethanol,
they are termed miscible (MISS-ible). Choosing a Solvent
A solvent can be any liquid that dissolves other materials, but different liquids dissolve
different substances. By analyzing the polarity of substances and solvents, we can
generally predict whether a solvent will dissolve a substance.
Because water is polar, it tends to dissolve ionic solids, polar sugars, and polar alcohols.
Carbon tetrachloride at one time was used as a “dry cleaning fluid.” It dissolves nonpolar oils from clothing without the use of the water that could damage some fabrics.
(Modern dry cleaning liquids are also non-polar but are less hazardous than CCl4.)
Compounds that are classified as oils do not dissolve well in water. For this reason,
water is a poor solvent for the oils produced by skin that can soil clothing. Soaps and
detergents are generally long-chain molecules that have a polar group on one end of a
long non-polar chain. The polar group on one end allows soaps and detergents to
dissolve in water, while the non-polar segment of the molecules can attract the skin oils
coating fabrics. This “both polar and non-polar” structure allows soaps to dissolve in
water and dissolve oils at the same time.
To choose a solvent to dissolve a substance, we begin by analyzing whether its particles
are ionic, polar, or non-polar, then apply the solubility rule: “like dissolves like.”
Let’s try a few examples. Below, complete the PH3 column first, check your answers on
the next page, and then complete the remaining columns. © 2011 ChemReview.Net v.1z Page 709 Module 25 — Bonding H2 PH3 Molecule HBr Lewis
Diagram Shape of
Oil or Water?
PH3 : PH3 is predicted to have tetrahedral ••
•• H:P:H = •• H H |
P H H electron pairs, a trigonal pyramidal
shape, and bond angles of < 109°. <109° For the bond polarity: EN of P = 2.1, EN of H = 2.1 . The P—H bond is non-polar.
Because all of the bonds are non-polar, the molecule is predicted to be non-polar. Nonpolar molecules dissolve in non-polar solvents, such as oils, gasoline, or CCl4. They
tend not to dissolve in water.
H2 : = H:H •• Two atoms in a molecule always have a linear shape with no bond angles.
For bond polarity: The EN difference is zero, so the bond is non-polar. If the only bond
is non-polar, then the particle must be non-polar. © 2011 ChemReview.Net v.1z Page 710 Module 25 — Bonding Non-polar compounds tend to dissolve in polar solvents like oils more than in water. You
would predict that hydrogen gas is not very soluble in water.
HBr = •• H : Br :
•• = H—Br (linear) HBr has tetrahedral electron pairs around bromine, a linear shape, and no bond angles.
For the bond polarity: The EN of H = 2.1, the EN of Br = 2.8.
The difference is 0.7, in the range of 0.5 to 1.7, so the bond is polar.
Adding the dipole to the linear shape gives H Br . The molecule is polar. Polar compounds tend to dissolve in polar solvents like water, but not in non-polar oils.
For solubility, like dissolves like.
***** The Reliability of VSEPR and Solubility Predictions
“Like dissolved like” is a simplified solubility rule, and there are many exceptions.
Factors other than bond polarity and geometry, including molecular size, affect
solubility. Most molecules are soluble to at least a slight extent in all solvents. Bond and
molecular polarities are better described as a continuum than as a simple case of “polar
versus nonpolar.” The VSEPR and electronegativity models are useful in predicting
shape and solubility, but there are many exceptions.
For example, in PH3 above, the VSEPR bond angles would be predicted to be about 107°
as in NH3, but actual bond angles are about 94°. For some substances, more
sophisticated models will be needed to explain experimental results. That said, the rules
for VSEPR, electronegativity, and “like dissolves like” will generally predict the shape,
polarity, and solubility of most substances. © 2011 ChemReview.Net v.1z Page 711 Module 25 — Bonding Practice. You may use a periodic table and a table of electronegativity values. Be
prepared to build models. If needed, check your answers after each part. Fill in the following chart for the substances shown. Based your predictions on the
general rules for VSEPR, electronegativity, and solubility.
S F2 SeI2 SF2 Molecule BCl3 SeI2 Lewis
Diagram Shape of
Oil or Water? ANSWERS
Lewis •• •• •• •• •• •• BCl3 •• •• •• •• •• •• •• •• : I : Se : I : : Cl : B : Cl : (also could be
drawn at 90°) Diagram : F : S : F: (also could be
drawn at 90°) : Cl : Tetrahedral Tetrahedral Trigonal Planar Shape of
Electron Pairs © 2011 ChemReview.Net v.1z •• •• •• •• Page 712 Module 25 — Bonding Shape of
Sketch) Bent Bent
Polarity I F—S Bond
Bond Polarity Trigonal Cl Planar B I — Se Cl Cl <109° <109° 120° 4.0 – 2.5 = 1.5 = polar 2.6 – 2.4 = 0.2 = nonpolar 3.0 – 2.0 = 1.0 = polar Polar Non-Polar Non-Polar (bent with polar bonds) (non-polar bonds) (the 3 dipoles cancel) Water Oil Oil Dissolves in
Oil or Water?
***** Lesson 25F: Double and Triple Bonds
Atoms in main groups 4 and above of the periodic table can satisfy the octet rule by
forming double bonds that have two pairs of electrons between the atoms. In the Lewis
diagram, the valence electrons for each double-bonded atom are placed on three sides of
the atom symbol instead of the four sides used for single bonds. Drawing Double Bonded Lewis Diagrams
If you know the molecular formula for a substance and you know that the molecule
contains a double bond, draw the Lewis diagram using these steps.
1. Count the total number valence electrons in the particle.
2. Determine which atoms in the formula have the double bond between them.
Because hydrogen bonds only once, it will not double bond. (The octet rule can be
used for double-bonded halogens, but in most cases halogens will bond only once).
3. Write the two double-bonded atoms next to each other so that there are two pairs of
valence electrons (four electrons total) between them.
Two double-bonded atoms = :: 4. Distribute the remaining valence electrons around each atom symbol. If the
remaining bonds are single bonds, each bond will have two valence electrons.
Double-bonded atoms may have lone pairs of electrons.
Satisfy the octet rule for the remaining atoms (but duet rule for H). © 2011 ChemReview.Net v.1z Page 713 Module 25 — Bonding An example of a compound with one double bond is ethene (also called ethylene), C2H4.
Use the rules above to draw the Lewis diagram for ethene. Try steps one and two, and
then check your answer below.
1. C2H4 has 4 valence electrons from each carbon and one from each hydrogen, for a
total of 12 valence electrons.
2. Since hydrogen is in column one, it can bond only once and it can form only single
covalent bonds. That leaves the two carbons to form the double bond.
Try step 3.
3. Combine the two double-bonded atoms to form the double bond. It will have two
pairs: four valence electrons. Note how the triangles faces meet in arranging the
valence electrons around the double-bonded atoms.
Two double-bonded carbons C::C = Try step 4.
4. Distribute the remaining 8 valence electrons and four atoms. Satisfy the octet for
each carbon and the duet for each hydrogen. H
Double-bonded C2H4 = • H H C::C • • • H Double Bond Shape
The shape around a double-bonded atom can be predicted using the fundamental
VSEPR rule: The directions with the electron pairs around each atom get as far apart as
In determining shape, the pairs in a double bond have about the same repulsion as a
single lone pair, only slightly more repulsion than a single bond. In determining shape,
what matters primarily is the number of directions in which electron pairs are found
around the central atom.
The shape that lets the electron pairs get as far apart as possible around each carbon is
trigonal planar. The shape of the bonds around a double-bonded atom is flat, with
~120° bond angles. H Double-bonded C2H4 = C == C
H © 2011 ChemReview.Net v.1z H H The H-C-H angles are < 120°.
All atoms are in the
plane of the paper. Page 714 Module 25 — Bonding Practice A: Use a periodic table. If needed, check your answers after each part.
Problems 1 and 3, and 2 if you need more practice. Do Draw Lewis diagrams, sketch the shape, and add the bond angles for each of these
2. N2H2 with one double bond. 1. H2CO with one double bond. 3. Try CO2 with two double bonds and C in the middle. Follow the octet rule. Triple Bonds
Triple bonds most often occur for atoms in main groups 4 and 5 (the carbon and
nitrogen families). In triple bonds, the valence electrons are put on two sides of the atom
symbol, with three pairs of valence electrons between the two triple-bonded atoms.
To draw Lewis diagrams for triple-bonded atoms, use these steps.
1. Total the valence electrons for the neutral atoms in the molecule.
2. Determine the two atoms that triple bond.
3. Write symbols for the two triple-bonded atoms. Place the valence electrons on two
sides, with three pairs of valence electrons between them.
4. Distribute the remaining atoms and valence electrons to satisfy the octet/duet rule.
An example of a triple-bonded compound is ethyne (also called acetylene), C2H2 , a gas
that is used in torches that cut steel.
Try steps one and two above for C2H2, and then check your answer below.
1. C2H2 has 4 valence electrons from each carbon and one from each hydrogen.
4+4+1+1= a total of 10 valence electrons.
2. H can only form single bonds, so the triple bond must be between the two C’s.
3. Place three electron pairs between the two triple-bonded C atoms.
Two triple-bonded carbons = C:::C Try steps 4 and 5.
4. Match the unpaired electrons on the single-bonded atoms to the remaining unpaired
electrons on the double bonded atoms. Satisfy the octets (and duet for H).
Triple-bonded C2H2 = H:C:: :C:H Check that each carbon is surrounded by 8, and each hydrogen by 2, valence
electrons. © 2011 ChemReview.Net v.1z Page 715 Module 25 — Bonding Triple Bond Shape
The shape that lets electron pairs in two directions around a central atom get as far apart
as possible is linear with 180° bond angles. = C2H2 Practice B: H — C ≡≡ C — H = linear shape with all 180° bond angles. Try to do these without a periodic table. If needed, check your answers after each part.
Draw Lewis diagrams, sketch the shape, and add the bond angles for each of these
1. HCN with a triple bond. 2. N2 with a triple bond. Predicting Single, Double, or Triple Bonds From the Formula
For many relatively simple molecules, if you know the formula, you can use the
octet/duet rule to predict whether the molecule will have single, double, or triple bonds.
The steps are: count the valence electrons, and then draw Lewis diagrams that satisfy
the octet/duet rule.
Q. Draw a Lewis diagram and then a structural formula for O2, and then check your
O2 has 12 total valence electrons. Lewis diagrams that satisfy the octet rule are O::O O == O = Lewis Dot
Diagram = Lewis Dot-Bond
Diagram O == O
Formula = O2
Formula ***** Practice C: Try to do these without a periodic table. If needed, check your answers after each part.
Draw a Lewis diagram and then a structural formula for
1. C2Cl2 © 2011 ChemReview.Net 2. O3 v.1z 3. H2S Page 716 Module 25 — Bonding ANSWERS
Practice A H 1. C::O = H2CO H
C == O = H All bond angles are 120°.
All atoms are in the
plane of the paper. H The double bond must be between C and O, since H only bonds once.
The electron pairs in a double bond are placed on two triangle shapes ◄► around the atoms. The
shape lets electron pairs on 3 sides of an atom get as far apart as possible is trigonal planar. H 2. N::N = N2H2 H
N == N = H H Note that octets and duets are satisfied. The shape can also be drawn as 3. C O2 O::C::O = = O == C == O All bond angles are 120°.
All atoms are in the
plane of the paper.
N == N
The bond angle is 180°.
The molecule is linear.
All octets are satisfied. Practice B
1. Triple-bonded HCN = H:C:::N: = H — C ≡≡ N The molecule is linear with a 180° bond angle between the three atoms
2. Triple-bonded N2 = :N:::N: = N ≡≡N Both N’s have satisfied octets. The molecule is linear with no bond angle (it takes three atoms to have
an angle in the molecule).
1. C2Cl2 has 4+4+7+7 = 22 total valence electrons. The chlorines generally bond only once because
they are halogens. A Lewis diagram that satisfies the octet rule for all four atoms is
•• •• •• •• : Cl : C : : : C : Cl : © 2011 ChemReview.Net v.1z = Cl — C ≡≡ C — Cl Page 717 Module 25 — Bonding 2. O3 has 6+6+6 = 18 total valence electrons. The chlorines must bond only once because they are
halogens. One Lewis diagram that satisfies the octet rule for all four atoms is O3 O::O = O == O = O O An equally valid diagram (called a resonance structure) can be drawn with the single and double
bonds switched in position. When resonance structures are possible for a particle, the properties
of the particle will be a blend of the two resonance structures, and the particle will have special
3. H2S has 1+1+6 = 8 total valence electrons. A Lewis diagram that satisfies the octet rule for S and duet
rule for H is
•• H2S = H:S:H
•• •• or H :S:
•• = / H S ─H H These two Lewis diagrams are not two different structures; they represent the same molecule. If four
sides are shown around a central atom, the four sides are equivalent. H2S will have tetrahedral
electron pairs and a bent shape, with bond angles of slightly less than 109°.
***** Lesson 25G: Dot Diagrams For Ions
Ion Dot Diagrams
Dot diagrams for ions are drawn by adding and subtracting electrons from the Lewis
diagrams for neutral particles.
1. To make Lewis diagrams for monatomic ions,
a. draw the Lewis diagram for the neutral atom.
b. Take away electrons to form positive ions. Add electrons to form negative ions.
Most often, a positive ion shows no valence electrons around the symbol, and a
negative ion has 8 valence electrons (hydride ion, H─, has 2 valence electrons).
c. Put brackets [ ] around the Lewis diagram, then add the charge as a superscript.
Examples: Neutral Na atom =
Neutral Al atom =
Neutral S atom = © 2011 ChemReview.Net v.1z Na
:S: ; Na+ ion = [ ; Na ] + ; Al3+ ion = ; S2─ ion = [ Al ]3+, [ : S : ]2 ─
Page 718 Module 25 — Bonding 2. To make Lewis diagrams for polyatomic ions,
a. Count the valence electrons as you would for a neutral particle, then add valence
electrons to form a negative ion, and take away valence electrons to form a
b. Construct the Lewis diagram as you would for a neutral particle, but use the
corrected number of valence electrons.
c. Put the Lewis diagram in brackets, and add the charge as a superscript.
Let’s try an example. Represent the following reaction by drawing Lewis diagrams for
NH + H+
3 4 *****
•• + H + [ H ]+ H •• H: N:H
•• H Note that in NH4+, the octet rule for N and the duet rule for H are satisfied.
• Using VSEPR, predict the shape and bond angles for NH3.
Using VSEPR, predict the shape and bond angles for NH +.
For NH3, with one lone pair, the shape is a trigonal pyramid with bond angles of
<109° (build the model if needed). The actual bond angles in NH3 are ~107°.
In NH4+, the N is surrounded by four electron pairs. When four pairs surround a
central atom, the electron pairs are tetrahedral. Since all four pairs are bonds, the
molecular shape is also tetrahedral. Since there are no lone pairs, the bond angles
are 109°, rather than <109°.
Let’s try a negative ion. Draw the Lewis diagram for the hydroxide ion, OH─.
A neutral OH particle would be unstable due to its unsatisfied octet:
An OH─ is stable , with a satisfied octet and duet: © 2011 ChemReview.Net v.1z •• H:O: ─ • H:O:
•• •• Page 719 Module 25 — Bonding Practice: Try these without a periodic table. Check your answers after each part. Draw Lewis diagrams, sketch the shape, and write the bond angles for these ions.
1. The hydronium ion, H3O+, formed by the reaction H2O + H+ H3O+ . 2. CN─, the cyanide ion, with a triple bond. ANSWERS
1. •• •• H : O : + [ H ]+ H:O:H •• •• H 2. CN─ + H = [:C: : :N:]─ = ─C ≡≡ N If needed, build a model.
With one lone pair and 3
bonds, the shape is a trigonal
pyramid; bond angles <109° . (linear, no bond angle) ***** Lesson 25H: Orbital Models For Bonding
The Hybridized-Orbital Model For Bonding
The octet rule is easy to use, and it accurately predicts the formulas and shapes for most
covalently bonded molecules. A bonding model that is a bit more complex, but explains
many facets of bonding that the octet rule does not, is based on the wave equation model of
The orbital configuration for neutral atoms (Lesson 24A), based on the wave equation
model for the atom, explains the electron configuration in single neutral atoms.
For example, according to the wave equation model, the configuration of the valence
electron orbitals for carbon as a single neutral atom is :
2s ↑↓ 2p ↑ ↑ This prediction, that carbon has two unpaired electrons, is consistent with
measurements of single carbon atoms. However, in Lewis diagrams for bonding, a
single-bonded carbon is assigned four equivalent unpaired electrons. • •C•
• Carbon: © 2011 ChemReview.Net v.1z 4 unpaired electrons Page 720 Module 25 — Bonding This configuration accurately predicts carbon’s bonding behavior in millions of carbon
compounds. How do we explain the apparent discrepancy between number of
unpaired electrons predicted the wave equation model and the octet rule?
One mathematical solution to the wave equation does predict separate s and p orbitals.
That solution accurately predicts much of the behavior of electrons in isolated single atoms.
However, as with quadratic equations, wave equations can have more than one correct
solution. An alternate but valid solution to the wave equation predicts that stable orbitals
can form in molecules if the single s and the three p orbitals are hybridized.
Hybridization results in four equivalent orbitals termed sp3 hybridized orbitals (pronounced
“s p three”). The wave equation predicts that these four hybridized orbitals will have the
same energy and will be equally spaced around the central atom, matching the behavior of
carbon when it forms single bonds. For a series of orbitals at the same energy, the electrons
go into the orbitals one at a time before they start to pair.
For a single-bonded carbon, the hybridized configuration is: 2sp3 ↑ ↑ ↑↑ Lower potential energy is favored in physical and chemical systems. Hybridized orbitals
are not the lowest energy arrangement for single carbon atoms, but bonded atoms have
lower energy than the atoms by themselves. Hybridized orbitals allow more unpaired
electrons and therefore more bonds. The bonds reduce the energy of the system. Because
the hybridized orbitals allow more bonds, they are usually favored when atoms bond. Hybrid Orbitals For Double Bonds
Other mathematical solutions to the wave equation explain double and triple bonds.
In molecules with one double-bond, the wave equation permits the hybridization of a
single s and the two of the three p orbitals to form three sp2 orbitals (pronounced “s p two”).
This leaves one p valence orbital that does not hybridize.
In the case of a molecule with one double-bond, the bond has two parts. A single bond
called a σ (sigma) bond is created by the overlap of an unpaired electron from an sp2 orbital
on each atom. A second bond between those atoms, called a π (pi) bond, is formed by the
pairing of an unpaired electron from the p orbital of each atom. A single π bond has
electron density both above and below the plane of the σ bonds.
Double-bonded C2H4: 120° H C pi
sigma p C H
H Hybrid Orbitals For Triple Bonds
In each triple-bonded atom, the valence s and one of the three valence p orbitals are
hybridized to form two sp hybrid orbitals. This leaves two p orbitals that are not hybridized.
Between two atoms that are triple-bonded are one σ bond, created by the overlap of
unpaired electrons in an sp orbital, and two π bonds at right angles to each other around the © 2011 ChemReview.Net v.1z Page 721 Module 25 — Bonding σ bond. The π bonds are formed by pairing the unpaired electrons in the two p orbitals of
The following table compares the terminology for multiple bonding for these two models
If two atoms are
attached by a In the dot (Lewis) diagram
model In the hybridized-orbital model Double bond Two pairs of shared
valence electrons between
two atoms. From each atom, overlap of one sp2
orbital to form one σ bond, and one p
orbital to form one π bond. Triple bond Three pairs of shared
valence electrons between
two atoms. From each atom, overlap of one sp
orbital to form one σ bond, and two p
orbitals to form two π bonds. The shapes and bond angles predicted by the hybridized orbital and Lewis diagram models
are the same. Exceptions to the Octet Rule and spx Hybridization
Both the octet rule and spx hybridization models successfully predict many of the
characteristics of most covalent molecules, but there are molecules that are quite stable
that these models do not explain.
Row 3 Exceptions
Frequently encountered exceptions to the octet rule occur for non-metal central atoms in
rows 3 and above. For example,
• When combining with chlorine, phosphorus forms both PCl3, with the lone pair
predicted by the octet rule, and PCl5, which violates the octet rule. In PCl5, all 5
of the phosphorus valence electrons are used for bonds.
In PCl5, the shape predicted by VSEPR that allows the 5 electron pairs to get as
far apart as possible is a trigonal bipyramid (which can be described as a Y in the
plane of the paper with a pin stuck down through the middle). PCl5: 120° Cl
90° PCl5 is often labeled as having dsp3 hybridization, suggesting that the empty 3d
orbitals in phosphorus may participate in hybridization with the 3s and 3p to
maximize the opportunity for bonding. © 2011 ChemReview.Net v.1z Page 722 Module 25 — Bonding • Combining with chlorine, sulfur forms both SCl2, with the two lone pairs
predicted by the octet rule, and SCl6, in which all 6 valence electrons are used for
In SCl6, the shape that allows the 6 electron pairs to get as far apart as possible is
octahedral, which can be described of as an X in the plane of the paper with a
pin stuck down through the middle. All bond angles are 90°. SCl6: 90° Cl
Cl 90° These single-bonded octahedral molecules are sometimes referred to as examples of
d2sp3 hybridization, suggesting the bonding of six unpaired electrons using the
empty 3d orbitals in sulfur.
Noble Gas Exceptions
The octet rule predicts that a noble gas, which has a filled valence cluster, should be stable
without bonding. This rule holds for the noble gases helium and neon, which form no
stable compounds. However, noble gases in rows 3 and above form a few compounds.
One is XeO3, a molecule which can be isolated (but tends to decompose explosively). XeO3
has several possible single and double-bonded Lewis-diagrams that do not violate the octet
rule. Practice: Use a periodic table. 1. Based on VSEPR, sketch the shape and name the shape would you predict for
a. Arsenic pentabromide b. Selenium hexafluoride 2. For molecules that contain at most one double or triple bond between two atoms,
label the following types of bonds in those molecules as single, double, or triple bonds.
a. One σ bond and one π bond b. One σ bond and two π bonds c. σ bonds but no π bonds © 2011 ChemReview.Net v.1z Page 723 Module 25 — Bonding ANSWERS
AsBr5: Br Trigonal
Bipyrimid Br Br
As 1b. Br SeF6:
Octahedral Br F
F 2. a. One σ bond and one π bond -- a double bond
b. One σ bond and two π bonds -- a triple bond
c. σ bonds but no π bonds -- a single bond
***** Summary: Bonding
1. The Octet Rule: Most atoms want to be surrounded by 8 valence electrons (H and
He want 2). The electrons can be shared, as in covalent bonds, or gained or lost from
neutral atoms, as in ionic bonds.
2. In covalent molecules, atoms in the carbon family tend to bond 4 times, nitrogen
family 3 times, oxygen family two times, and halogen family one time. A covalent
hydrogen forms one bond.
3. In drawing dot diagrams, if the bonds around an atom include
• All single bonds, place the valence electrons on 4 equivalent sides around the
atom; • One double bond, place the electrons on 3 sides; • One triple bond, place the electrons on 2 sides. 4. The Valence Shell Electron Pair Repulsion (VSEPR) model for predicting shapes:
Electron pairs tend to get as far apart as possible around an atom. Lone pairs and
double bonds repel other pairs slightly more than single bonds.
5. When there are 4 electron pairs around an atom, the pairs tend to be in a tetrahedral
shape, but the shape of the molecule is named based on where the atoms are.
6. Electronegative atoms have more attraction for electrons. Electronegativity tends to
increase as you go toward the top right corner of the periodic table. Across row 2,
the EN values increase by 0.5 per atom, from Li (1.0) to F (4.0).
7. A molecule will tend to be polar IF its bonds are polar, and if the dipoles do not
cancel due to symmetry when added by vector addition.
8. The Solubility Rule: like dissolves like. Polar solvents such as water tend to dissolve
polar and ionic compounds. Non-polar molecules tend to dissolve in non-polar
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