Bonding
Metals versus non-metals
AP Question 1984
Discuss some differences in physical and chemical properties of metals
and nonmetals. What characteristic of the electronic configurations of
atoms distinguishes metals from nonmetals? On the basis of these
characteristics, explain why there are many more metals than nonmetals.
Answer:
Physical properties:
Melting points
Electricity
Luster (shine)
Physical state
solids
metals
relatively high
conductivity
(due to mobile
electrons)
high
most solids***
Ductile
yes
non-metals
relatively low
good insulators
(due to a lack of
mobile electrons)
little or none
gases, liq. or
(mostly gases)
no
*** Most metals are solids due to the strong attraction of the positive
metal ions being so very attracted to the free mobile electrons
Chemical properties:
Oxides
Reacts with
metals
reducing
electropositive
basic or amphoteric
nonmetals
Electronegativity
generally low
Redox agents
non-metals
oxid. or reducing
electronegative
acidic
metals &
nonmetals
generally high
Electron configurations:
Metals:
Outermost electrons filling in s or d sublevels of their atoms. (A few heavy
elements have atoms with a couple of electrons in p sublevels.)
Nonmetals:
Valence electrons in the s and p sublevels of their atoms.
There are more metals than nonmetals because filling d orbitals in a given
energy level involves the atoms of ten elements and filling the f orbitals
involves the atoms of 14 elements. In the same energy levels, the maximum number of elements with atoms receiving p electrons is six.
Types of Bonding
Atoms form molecules because atoms want to have a full outer valence
shell. This means that they usually will have eight electrons in their outer
shells (s and p sublevels). If atoms gain or lose electrons, then they will
form ionic bonds. If atoms share electrons, then they have covalent bonds.
The electronegativity difference determines the bond character. A
difference of 00.3 is designated as a NON-POLAR covalent bond while
a difference of 0.311.67 is considered POLAR covalent and a
difference greater than 1.67 is IONIC since the electrons are spending
most of their time around only one of the nuclei.
Ionic bonds occur between metals and non-metals (high difference
between electronegative values). Metal loses electrons (cation) and nonmetals gain electrons (anion).
Covalent bonds occur when two atoms are sharing electrons. Each atom
counts the shared pair of electrons in its valence shell, thereby fulfilling
the octet rule. THE NUMBER OF COVALENT BONDS THAT AN
ATOM CAN FORM IS THE SAME AS THE NUMBER OF
ELECTRONS IN ITS VALENCE SHELL.
The first covalent bond formed between two atoms is a sigma (σ) bond
and the second and/or third is a pi (π) bond. Pi’s are formed from the side
to side overlap of unhybridized p orbitals.
SINGLE AND MULTIPLE BONDS:
•
•
•
•
single bond–one pair of electrons shared = sigma (σ) bond
•
sigma’s form from the overlap of an s and a p, or an s and
an s, OR a p and a p that are UNHYBRIDIZED
MULTIPLE BONDS ARE MOST OFTEN FORMED BY
C,N,O, P and S ATOMS [pronounced “SEE-NOPS”]
double bond–two pairs of electrons shared. One σ bond and
one π bond
triple bond–three pairs of electrons shared. One σ bond and
two π bonds
Coordinate covalent bond: When one atom provides both electrons
to form a sigma bond. Some atoms such as N and P tend to share a lone
pair with another atom that is short of electrons, leading to the formation
of a coordinate covalent bond.
All complex “coordinated” ions contain a coordinate covalent bond - this
is also where the “coordination” number concept comes from.
THINK: Lewis Acid-Base reactions – see the definition of Lewis acids and bases.
“Non-octets”
•
Fewer than eight –
H has, at most, only 2 electrons!
BeH2, only has 4 valence electrons around Be!
Boron compounds, only have 6 valence electrons!
Hydrogen = 2 electrons
Beryllium = 4 electrons
Boron = 6 electrons
•
Expanded Valence–3rd or higher period can be surrounded by more
than four valence pairs in certain compounds. The number of bonds
depends on the balance between the ability of the nucleus to attract
electrons and the repulsion between the pairs.
•
Odd-electron compounds–A few stable compounds contain an odd
number of valence electrons and thus cannot obey the octet rule:
NO, NO2 , and ClO2
Hybridization:
Hybridization occurs when an atom moves electrons around within the
SAME energy level to provide more electrons for bonding; this may
include a central atom expanding its octet to include the d sublevel.
We can think of this in terms of “sites” of electron density surrounding an
atom. A site can be a lone pair, or a bond of any variety.
Double and triple bonds count as ONE site—they are just more electron
dense than singles or lone pairs. THIS IS WHY I ALWAYS SAY,
“MULTIPLE BONDS COUNT AS A SINGLE BOND FOR
GEOMETRY”.
We can also think of this as pairs of electrons around the central atoms.
Pairs of electrons = “sites” and multiple bonds count as a single bond i.e.
only one site.
Hybridization—
use as many
letters as there
are sites starting
with s, then p’s
then d’s
sp
sp2
sp3
sp3d
# of “Sites” of Electron
Density
-or# of pairs of electrons
around central atom
Bond
Angle
2
3
4
5
sp3d2
6
180
120
109.5
3 @ 120
& 2 @ 90
all 90
Example
Carbon in CO2
Carbon in CO32Carbon in CH4
Phosphorous in
PF5
Sulfur in SF6
To determine the hybridization:
1. Draw the Lewis structure then count the pairs of electrons around
the central atom (or sites).
2. Use as many letters as there are pairs of electrons (remember
multiple bonds count as a single bond for this) or count the “sites”.
4 sites, 4 letters therefore sp3  tetrahedral.
3. Only expanded octet elements exhibit hybrids involving d orbitals.
This is WHY they had to be from at least the 3rd principal energy
level
If you see d2sp3 it just means the d’s that hybridized were from the
LOWER energy level than the s and/or p’s.
Lewis Dot Diagrams
Rules for Lewis Dot Diagrams
1)
2)
3)
4)
5)
6)
7)
8)
9)
Determine the central atom. The central atom is:
a) The least electronegative element.
b) The atom with the smallest number of valence electrons.
c) The “oddball” element.
d) The central atom can never be hydrogen.
Count the number of valence electrons in each atom.
Include the number of electrons lost or gained due to charge.
Total the number of valence electrons.
Draw the skeletal structure with the central atom in the middle.
Draw the first sigma bonds connecting all atoms to the central atom.
Subtract the total number of electrons involved in the sigma bonds
(two electrons for each bond) from the total number of electrons.
Finish drawing lines around the atoms up to the number of electrons
left. Work from the outside in.
Make sure that all atoms satisfy the octet rule. If they don’t, move
electrons around to form multiple bonds (double or triple) so that
everything does have a full octet. See-NOPS only.
10) If it is an ion, put [square brackets] around the diagram and put the
charge outside.
Almost always atoms will have an EVEN number of electrons, but
compounds of nitrogen may not – nitrogen may have an odd number of
electrons around it.
Resonance forms:
When a molecule has more than one way to draw its structure it will have
resonance structures. This occurs ONLY with multiple bonds. A double
arrow (⇔) separates the pictures in the resonance diagrams.
Valence Shell Electron Pair Repulsion
Theory
(VSEPR Theory)
VSEPR is a theory that allows students to determine the shape of the
molecule and the bond angles in the molecule. This is done by:
1. Drawing the Lewis Dot Diagram.
2. Determining the number of bonding pairs of electrons around the
central atom.
3. Determining the number of unbonded pairs of electrons around
the central atom.
Reference chart on next page →
V alence S hell E lectron P air R epulsion Theory
sp2
sp
Structural Bonded Lone
Pairs
Pairs (σ) Pairs .
Geometry
2
0
Linear
180
3
3
0
120
2
1
Triangular
planar
Bent
< 120
4
0
Tetrahedral
109.5
3
1
< 109.5
2
2
Triangular
pyramidal
Bent
5
0
4
sp3
sp3d1
5
6
sp3d2
Bond angle
2
4
Hybridization
Molecular
<< 109.5
1
Triangular
bipyramidal
See-saw
120 & 90
< 120 & 90
3
2
T-shape
90
2
3
Linear
180
6
0
Octahedral
90
5
1
90 & < 90
4
2
Square
pyramidal
Square planar
3
3
T-shape
< 90
2
4
Linear
180
90
Practice
Draw the Lewis Dot Diagrams for CCl4, NH3, PCl5, PCl3, SF6, SF4, and
SF2. Explain each of their shapes and their hybridization.
Typical AP question: Account for the fact that PCl3, NCl3 and PCl5
exist, but NCl5 does not.
Answer: Draw the arrow diagram for nitrogen and for phosphorus and
use hybridization.
Other Terms
Bond order = The number of bonding electron pairs shared by two atoms
in a molecule.
Easiest way to find this is use the formula:
Bond Order =
# of bonds in a " site"
# of resonance structures
Bond length = The distance between the nuclei of two bonded atoms.
Bond energy = The energy supplied to a gaseous molecule to separate
two of its atoms. This is also called the bond dissociation energy. The
process of breaking bonds in a molecule is endothermic, as energy must be
added. THIS IS ANOTHER WAY TO DETERMINE ΔHrxn :
ΔHrxn = Σ bonds broken – Σ bonds formed
Bond polarity – Determines if the electrons of a bond are being shared
equally or unevenly. If a bond does not share its electrons equally it is
said to be polar. One end of the bond must be positive and the other must
be negative. If the electrons in the bonds are shared evenly, then the bond
is said to be non-polar. Determines the Intermolecular Forces involved in
the molecule and explains the behavior of the molecule.
Determining the
Electronegativity Values For Elements
H = 2.1
Li = 1.0
Be = 1.5
B = 2.0
C = 2.5
N = 3.0
O = 3.5
F = 4.0
Na = 0.9
Mg = 1.2
Al = 1.5
Si = 1.8
P = 2.1
S = 2.5
Cl =3.0
K = 0.8
Ca = 1.0
Ga = 1.6
Ge = 1.8
As = 2.0
Se = 2.4
Br =2.8
Rb = 0.8
Sr = 1.0
Y = 1.2
Sn = 1.8
Sb = 1.9
Te = 2.1
I = 2.5
Cs = 0.7
Ba = 0.9
****** Pb = 1.9
Bi = 1.9Po = 2.0
At = 2.2
****** La  Lu = 1.0  1.2
Another way of looking at the type of bond that you have is to look at the
difference in the electronegativity values. There are two types of covalent
bonds: non-polar and polar. Although different textbooks classify them
differently, we will use the point breakdown below.
Non-polar
covalent
0.0
0.3
Polar covalent
weakly
IONIC
1.67
4.0
strongly
100% covalent
0.0 % ionic
0.0 % covalent
100% ionic
The differences in the electronegativity values determine the type of bond.
The shape of the molecule and the type of bond determine if the molecule
is polar or non-polar.
•
•
A molecule can be non-polar with symmetrical polar bonds.
This can only be in a molecule that has ZERO unbonded pairs of
electrons around the central atom.
Molecular polarity = if a molecule has dipoles on its surface, due to
its shape and the polarity of the bonds that make it up.
Dipole Moment = a measurement of the polarity of a bond or
molecule. Generally, the more polar the bond, the greater the dipole
moment.
Intermolecular Forces
Polar bonds = When electrons are not shared equally, one end of the bond
has a positive dipole and the other end has a negative dipole.
Must use electronegativity values to determine the type of bond.
Polar molecules = molecules are polar if one end of the molecule is
positive and the other is negative.
Molecules that are symmetrical are non-polar. The only shapes of
molecules that COULD be non-polar are linear (all three forms),
triangular planar (trigonal planar), tetrahedral, triangular bipyramidal
(trigonal bipyrimidal), square planar and octahedral. Just because a
molecule is one of these shapes does not mean that it will be non-polar. It
must be symmetrical. Take for instance, CCl4 and CCl3F – both are
tetrahedral, but only CCl4 is non-polar because it is symmetrical down
each one of the axes (the bond between the central atom (carbon) and the
attached atoms (either chlorine or fluorine).
CCl4
Symmetrical
Non-polar
CCl3F
Non-symmetrical
Polar
Intermolecular Attractions
The weak attraction between molecules (within a molecule it is called
“intramolecular”) due to arrangement of electrons. Dipole to dipole
interactions only has about 1% the strength of a covalent bond. Named
after Johannes van der Waals, these are called van der Waals forces.
Weakest to the strongest:
•
•
•
•
London Dispersion
•
Induced dipole to induced dipole
Induce dipole to dipole
Dipole to dipole
Hydrogen bonding
Dipole interactions
•
London Dispersion Forces = caused by the motion of the electrons. As
electrons move around the nucleus, they can create an asymmetrical
distribution that results in temporary dipoles. If the atom or molecule
is moving slowly enough (i.e. very low temperature) then these dipoles
can cause a neighboring atom/molecule to form a dipole. It is this
interaction that causes the noble gases to liquefy. Named for Fritz
London and not the city in England.
•
•
•
Strength increases with increased number of electrons (more
electrons means that there are more polarizable electrons and
the stronger the resulting dipoles are)
•
Strength increases with more mass.
explains behavior of noble gases
explains behavior of the halogens
•
Dipole interactions occur when polar molecules are involved
•
Hydrogen bonding = attractive forces in which a hydrogen atom that is
bonded to a highly electronegative atom has a very strong attraction to
an unbonded pair of electrons.
IMF allows gases to become a liquid.
IMF is involved with the Ideal Gas Law
The Ideal Gas Law says gas particles have no volume and they don’t
interact with each other (intermolecular forces). This is an
oversimplification of the following equation. You can notice the terms for
the intermolecular forces and the volume of the particles. When using the
Ideal Gas Law, the terms “a” and “b” are both equal to zero and then the
equation just becomes PV = nRT
[P + an2][V- bn] = nRT
v2
Correction for
Intermolecular
forces
Correction for
molecular
volume
PHASES OF MATTER AND THE KINETIC MOLECULAR THEORY
(a) The volume of the molecules themselves [when in the gas phase] is
negligible
(b) No attractive forces between molecules
(c) Molecules are in constant random motion
(d) KE is determined by temperature only
•
•
compressibility—due to (a) above—not appreciable in the liquid
or solid phase
intermolecular forces—KT assumes weak forces
•
expansion—occurs when KE overcomes attractive IMF’s;
gases that resist have strong IMF’s
•
•
pouring—due to IMF’s in solids and liquids
condensation—molecules lose their KE [especially upon
collision with each other and with cooler surfaces] and the
attractive forces “kick” in.
INTERMOLECULAR FORCES
Collectively called van der Waals forces. These are the attractive forces
between molecules, between ions, or between ions and molecules
•
ION-DIPOLE forces—weaker than ion/ion interactions; occurs when a
polar molecule is attracted to an ion
•
•
•
Distance—the larger the interacting species, the weaker this force
between them
Charge—follow Coulomb’s Law; the greater the charge the
stronger the force of attraction
Magnitude—the greater the magnitude of the dipole [larger dipole
moment], the stronger the attraction
•
•
•
hydrated—metals bond to several water molecules; porcelain
also holds water
heat or energy of hydration—when the water sets up an IMF
with the metal, energy is released in the process
•
energy of solvation—when “hydration” occurs with something
other than water
•
acids—H3O+ is a hydrated proton
DIPOLE-DIPOLE forces—Energy is released upon interaction;
Energy is required to separate interacting dipoles [high MP, high BP,
high ΔHvap]
•
•
Energy released when polar molecules interact with one another,
and this is one of the reasons you must cool a gas consisting of
polar molecules to convert it into a liquid, also reduces KE.
Hydrogen bonding—really, it’s not a bond, it’s an attractive force
that is NEVER to be confused with a bonded hydrogen!!
•
•
•
•
Two requirements for this IMF.
It’s a dipole-dipole interaction that involves an H that is
bonded to a highly electronegative atom
•
•
N, O, or F—very highly electronegative atoms
•
That THEN interacts with a pair of unbonded electrons on a
highly electronegative atom on an adjacent molecule.
Cl or S—weaker, yet still high in electronegative, however,
size of the atoms (since period 3 elements) can offset the
effects.
•
It’s a really big deal in Biology! That highly electronegative
atom requirement keeps lots of hydrocarbons from exhibiting
H-bonding
•
About 10% the strength of a covalent bond.
melting point vs. intermolecular forces—the stronger the IMF the
higher the MP
Interactions Involving Induced Dipoles—without this, Br2 wouldn’t be
a liquid nor would I2 be solid!
•
Dipole-Induced Dipole forces—a polar molecule induces a
temporary situation where a NON-polar molecule becomes slightly
polar since the electron cloud is mushy
•
Induced Dipole-Induced Dipole forces—since the electron cloud is
mushy and electrons move all the time, a molecule may build up a
concentration of electrons on one side of the surface leaving a
slight positive charge on the opposite side—this sets off a chain
reaction! Now very nonpolar molecules are behaving as if they
were slightly polar—happens in iodine all the time! The larger the
molecule, the more likely this will happen.
•
London or dispersion forces—another name just to confuse you!
Melting points vs. dispersion forces—again, the stronger the IMF, the
higher the MP since the attractive forces must be overcome by increasing
the KE [temperature]
Although the term “van der Waals forces” usually refers to ALL IMF,
other terms may be used interchangeably with it. Those terms are:
•
•
•
dispersion forces
London forces
dipole-induced dipole forces
AP Multiple Choice Questions:
1984
Questions 8 - 9
(A)
(B)
(C)
(D)
(E)
A network solid with covalent bonding
A molecular solid with zero dipole moment
A molecular solid with hydrogen bonding
An ionic solid
A metallic solid
8. Solid ethyl alcohol, C2H5OH
9. Silicon dioxide, SiO2
18.
Hydrogen Halide
HF
HCl
HBr
HI
Normal Boiling Point, °C
+19
- 85
- 67
- 35
The liquefied hydrogen halides have the normal boiling points given
above. The relatively high boiling point of HF can be correctly explained
by which of the following?
(A)
(B)
(C)
(D)
(E)
HF gas is more ideal.
HF is the strongest acid.
HF molecules have a smaller dipole moment.
HF is much less soluble in water.
HF molecules tend to form hydrogen bonds.
40.
The geometry of the SO3 molecule is best described as
(A)
(B)
(C)
(D)
(E)
41.
trigonal planar
triangular pyramidal
square pyramidal
bent
tetrahedral
Which of the following molecules has the shortest bond length?
(A) N2
51.
(D) Br2
(E) I2
(B) C2H4
(C) CN-1
(D) C6H6
(E)CH4
Which of the following has no dipole moment?
(A) HCN
80.
(C) Cl2
Pi bonding occurs in each of the following species EXCEPT
(A) CO2
60.
(B) O2
(B) NH3
(C) SO2
(D) NO2
(E) PF5
For which of the following molecules are resonance structures
necessary to describe the bonding satisfactorily?
(A) H2S
(B) SO2
(C) CO2
(D) OF2
(E) PF3
1989
a.
b.
c.
d.
e.
hydrogen bonding
hybridization
ionic bonding
resonance
van der Waals forces (London dispersion forces)
11. Is used to explain why iodine molecules are held together in the solid
state.
12. Is used to explain why the boiling point of HF is greater than the
boiling point of HBr.
13. Is used to explain the fact that the four bonds in methane are
equivalent.
14. Is used to explain the fact that the carbon-to-carbon bonds in benzene,
C6H6, are identical.
17. The Lewis dot structure of which of the following molecules shows
only one unshared pair of valence electron?
(A) Cl2
47.
(B) N2
CCl4,
CO2,
(C) NH3
PCl3,
(D) CCl4
PCl5,
(E) H2O2
SF6
Which of the following does not describe any of the molecules above?
(A)
Linear
(B)
Octahedral
(C)
Square planar
(D)
Tetrahedral
(E)
Triangular pyramidal
57. Which of the following compounds is ionic and contains both sigma
and pi covalent bonds?
(A)
(D)
Fe(OH)3
NO2
(B)
(E)
HClO
NaCN
(C)
H2 S
1994
Questions 8-10 refer to the following diatomic species.
(A)
(B)
(C)
(D)
(E)
Li2
B2
N2
O2
F2
8. Has the largest bond-dissociation energy
9. Has a bond order of 2
10. Contains 1 sigma and 2 pi bonds
15. In a molecule in which the central atom exhibits sp3d2 hybrid orbitals,
the electron pairs are directed toward the corners of
(A)
(B)
(C)
(D)
(E)
a tetrahedron
a square-based pyramid
a trigonal bipyramid
a square
an octahedron
57. Which of the following molecules have planar configurations?
I. BCl3
II. CHCl3
III. NCl3
(A)
(B)
(C)
(D)
(E)
I only
III only
I and II only
II and III only
I, II, and III
I2(g) + 3 Cl2(g) 
60.
2 ICl3(g)
According to the data in the table below, what is the value of ΔHO for
the reaction represented above?
Bond
I–I
Cl – Cl
I – Cl
Average Bond Energy (Kilojoules/mole)
149
239
208
(A) -860 kJ
(B) -382 kJ
(C) +180 kJ
(D) +450 kJ
(E) +1,248 kJ
62. The electron-dot structure (Lewis structure) for which of the following
molecules would have two unshared pairs of electrons on the central
atom?
(A)
(B)
(C)
(D)
(E)
H2 S
NH3
CH4
HCN
CO2
68. Which of the following molecules has a dipole moment of zero?
(A)
(B)
(C)
(D)
(E)
C6H6 (benzene)
NO
SO2
NH3
H2 S
1999
Questions 13-16 refer to the following descriptions of bonding in different
types of solids.
(A)
(E)
Lattice of positive and negative ions held together by electrostatic
forces
Closely packed lattice with delocalized electrons throughout
Strong single covalent bonds with weak intermolecular forces
Strong multiple covalent bonds (including π-bonds) with weak
intermolecular forces
Macromolecules held together with strong polar bonds.
13.
Cesium chloride, CsCl(s)
14.
Gold, Au(s)
15.
Carbon dioxide, CO2(s)
16.
Methane, CH4(s)
32.
Types of hybridization exhibited by the C atoms in propene,
CH3CHCH2, include which of the following?
(B)
(C)
(D)
I. sp
II. sp2
III. sp3
(A)
(B)
(C)
(D)
(E)
I only
III only
I and II only
II and III only
I, II, and III
40. Of the following molecules, which has the largest dipole moment?
(A) CO
(B) CO2
(C) O2
(D) HF
(E) F2
68. In which of the following processes are covalent bonds broken?
(A)
(B)
(C)
(D)
(E)
I2(s) I2(g)
CO2(s) CO2(g)
NaCl(s)  NaCl(l)
C(diamond)  C(g)
Fe(S)  Fe(l)
74. Which of the following gases deviates most from ideal behavior?
(A) SO2
(B) Ne
(C) CH4
(D) N2
(E) H2
AP Multiple Choice Answers
Question
Answer
% correct
1984
8
9
18
40
41
51
60
80
C
A
E
A
A
E
E
B
66
56
65
54
49
56
34
45
1989
11
12
13
14
17
47
57
E
A
B
D
C
C
E
57
72
44
55
76
50
38
1994
8
9
10
15
57
60
62
68
C
D
C
E
A
B
A
A
21
47
57
50
46
47
64
41
1999
13
14
15
16
32
40
68
74
A
B
D
C
D
D
D
A
60
64
65
61
38
54
54
48
AP Free Response Questions
1990 D – Lewis Dot Diagrams
Use simple structure and bonding models to account for each of the following.
(a) The bond length between the two carbon atoms is shorter in C2H4 than in C2H6.
(b) The H-N-H bond angle is 107.5o‚ in NH3.
(c) The bond lengths in SO3 are all identical and are shorter than a sulfur-oxygen single bond.
(d) The I3-1 ion is linear.
(e) Molecules of NF3 are polar, but those of BF3 are not.
1989 D – Lewis Dot Diagrams
CF4
XeF4
ClF3
(a) Draw a Lewis electron–dot structure for each of the molecules above and identify the shape
of each.
(b) Use the valence shell electron–pair repulsion (VSEPR) model to explain the geometry of
each of these molecules.
1982 D – Lewis Dot Diagrams
(a) Draw the Lewis electron–dot structures for CO32-, CO2, and CO, including resonance
structures where appropriate.
(b) Which of the three species has the shortest C-O bond length? Explain the reason for your
answer.
(c) Predict the molecular shapes for the three species. Explain how you arrived at your
predictions.
1979 D – Lewis Dot Diagrams
Draw Lewis structures for CO2, H2S, SO3 and SO32- and predict the shape of each species.
1990 – Lewis Dot Diagrams and Bond Energy
______________________________________________ Average Bond Dissociation
Energies at 298 K
______________________________________________ Bond
C-H
C-C
C-Cl
Cl-Cl
H-Cl
Energy, kJ mol-1
414
347
377
243
431
______________________________________________ The tables above contain information for determining thermodynamic properties of the reaction below.
C2H5Cl(g) + Cl2(g)   C2H4Cl2(g) + HCl(g)
(a) Calculate the ΔHrxn for the reaction above, using the table of average bond dissociation
energies.
(b) Calculate the ΔSrxn for the reaction at 298 K, using data from either table as needed.
(c) Calculate the value of Keq for the reaction at 298 K.
(d) What is the effect of an increase in temperature on the value of the equilibrium constant?
Explain your answer.
1992 D – Lewis Dot Diagrams and IMF
Explain each of the following in terms of atomic and molecular structures and/or
intermolecular forces.
(a) Solid K conducts an electric current, whereas solid KNO3 does not.
(b) SbCl3 has measurable dipole moment, whereas SbCl5 does not.
(c) The normal boiling point of CCl4 is 77oC, whereas that of CBr4 is 190oC.
(d) NaI(s) is very soluble in water, whereas I2(s) has a solubility of only 0.03 gram per 100
grams of water.
1996 – Lewis Dot Diagram and IMF
Explain each of the following observations in terms of the electronic structure and/or bonding of
the compounds involved.
(a)
(b)
(c)
(d)
At ordinary conditions, HF (Boiling point = 20oC) is a liquid, where as HCl (B. Pt. = 114oC) is a gas.
Molecules of AsF3 are polar, whereas molecules of AsF5 are non-polar.
The N-O bonds in NO2-1 ion are equal in length, whereas they are unequal in HNO2.
For sulfur, the fluorides SF2, SF4, and SF6 are known to exist, whereas for oxygen only
OF2 is know to exist.
AP Free Response Answer
1992 Answer:
(a) K conducts because of its metallic bonding - or - “sea” of mobile electrons (or free
electrons). KNO3 does not conduct because it is ionically bonded and has immobile ions (or
immobile electrons).
(b) SbCl3 has a measurable dipole moment because it is asymmetrical due to the lone pair of
electrons which causes a dipole - or - its dipoles do not cancel - or - it has a trigonal
pyramidal structure - or - a clear diagram illustrating any of the above.
(c) CBr4 boils at a higher temperature than CCl4 because it has stronger intermolecular forces
(or van der Waal or dispersion). These stronger forces occur because CBr4 has more
polarizable electrons than CCl4.
(d) NaI has greater aqueous solubility than I2 because NaI is ionic (or polar), whereas I2 is nonpolar (or covalent). Water, being polar, interacts with the ions of NaI but not with I2. (Like
dissolves like accepted if polarity of water is clearly indicated.)
1996
(a) hydrogen bonding is much larger in HF than in HCl (hydrogen bonding).
(b) AsF3 has triangular pyramidal shape and a lone pair of electrons, while AsF5 is triangular
bipyramidal and is symmetrical.
(c) The N-O bond in nitrite ion has resonance structure (bond order = 1½) while that of HNO2
is prevented to have resonance structures by the addition of the hydrogen.
(d) Hybridization of sulfur can go to sp3d2 while that of oxygen is only sp3.
.
1979 Answer:
or
or
bent with sp3 hybridization
Linear
or
Trigonal pyramidal
or
triangular planar
1982 Answer:
(a)
(b) CO has the shortest bond because there is a triple bond. OR because there is the greatest
number of electrons between C and O in CO.
(c) CO32- trigonal planar (planar and triangular). C bonding is sp2 hybrid - or - C has three
bonding pairs and no lone pair.
CO2 linear. C bonding is sp hybrid - or - C has two bonding pairs and no lone pairs - or CO2 is nonpolar and must be linear.
CO linear. Two atoms determine a straight line.
1989 D Answer:
(a)
Tetrahedral
square planar
T shape
(b) CF4 - 4 bonding pairs around C at corners of regular tetrahedron to minimize repulsion
(maximize bond angles).
XeF4 - 4 bonding pairs and 2 lone pairs give octahedral shape with lone pairs on opposite
sides of Xe atom (along the same axis)
ClF3 - 3 bonding pairs and 2 lone pairs give trigonal bipyramid with one pairs in equatorial
positions 120O apart.
1990 Answer:
(a) C2H4 has a multiple bond; C2H6 has a single bond. Multiple bonds are stronger and,
therefore, shorter than single bonds.
(b) NH3 has 3 bonding pairs of electrons and 1 lone pair. Bonding pairs are forced together
because repulsion between lone pair and bonding pairs is greater than between bonding
pairs.
(c) The bonding in SO3 can be described as a combination of 3 resonance forms of 1 double and
2 single bonds.
The actual structure is intermediate among the 3 resonance forms, having 3 bonds that are
equal and stronger (therefore, shorter) than an S-O single bond.
(d) The central I atom has 3 lone pairs and 2 bonding pairs around it.
To minimize repulsion, the 3 lone pairs on the central atom are arranged as a triangle in a plane
and are right angles to the I-I-I- axis (equatorial positions)
1990 Answer:
(a) ΔH = energy of bonds broken - energy of bonds formed
C2H5Cl + Cl2  C2H4Cl2 + HCl
ΔH = (2794 + 243) - (2757 + 431) kJ = -151 kJ
OR
CH + Cl-Cl  C-Cl + HCl (representing the changes)
ΔH = (414) + 243) - (377 + 431) = -151 kJ
(b) ΔG = ∑ ΔG°(products) − ∑ ΔG °(reactants)
= [-80.3 + (-95.3)] - [-60.5 + 0] = -115 kJ
ΔS° =
ΔH ° − ΔG °
T
=
−151 − ( −115 )kJ
298 K
= −0 .120 kJ K
(c) Keq = e-G/RT = e-(-115100/(8.3143)(298)) = 1.50 x1020
(d) Keq will decrease with an increase in T because the reverse (endothermic) reaction will be
favored with the addition of heat. OR
ΔG will be less negative with an increase in temperature
(from ΔG = ΔH - TΔS), which will cause Keq to decrease.