Midterm Test
Tuesday, October 10th (This Tuesday) at 7:30pm
Practice is test on website.
Section A, M.W.F. 8:00 – 8:50 am, Room D634.
90 minute test
Chapters: 1, 2, 7, 8
Review
Chapter 1
Units
SI System
Dimensional Analysis
Scientific notation
Prefixes and suffixes
Significant figures
Propagation through addition and multiplication
Precision/Acuracy
Chapter 2
Dalton’s atomic theory of matter
Elemental Forms
Chemical & physical properties
Subatomic particles - protons, neutrons, and electrons
Models of the atom – Rutherford's model
Atomic number and mass number - # of n’s, p’s, & e’s
Isotopes - calculating average atomic mass and percent
abundance
Elements - names and symbols
Avogadro’s number and the mole
Periodic table (groups and periods) – Common properties in
each group
Chapter 7
Properties of waves - wavelength, frequency, amplitude, speed
Electromagnetic spectrum - speed of light
Planck’s equation and Planck’s constant
Wave-particle duality for light , electrons, etc.)
Atomic line spectra: Balmer and Rydberg Series
Ground vs. excited states
Heisenberg uncertainty principle
Bohr and Schrödinger models of the atom
Quantum numbers (n, l, ml)
Shells (n), subshells (s,p,d,f)
Shapes and properties of atomic orbitals –#nodes, # lobes
Chapter 8
Diamagnetic vs. paramagnetic vs. ferromagnetic substances
Electron spin
Pauli exclusion principle
Orbital box diagrams
Electron configurations
Aufbau order and its exceptions.
Predicting ions using electron configurations
Core vs. valence electrons
Effective nuclear charge
Periodic trends (atomic radius, ionization energy, electron
affinity, ionic radius)
Chapter 9
Bonding and Molecular Structure:
Fundamental Concepts
Countless arrangements of atoms are possible!!
Bonding behaviour:
What does it depend on?
How do atoms form bonds?
What kind of bonds exist?
How do we predict bonding
behaviour?
Valence Electrons
Given an electron configuraton one can group the electrons into
the core and valence e’s
Ex) C
6 e’s
1s2 2s22p2
core valence
Atoms can form bonds by sharing or exchanging valence e’s.
The core e’s don’t participate, just like the noble gas is unreactive.
Then main objective for each atom is to achieve a noble gas
configuration for its valence electrons
Ex) Ne
1s2 2s22p6
Has a full valence shell with 8 e’s
C needs 4 more e’s to achieve a full shell.
How?
Electron Dot Structures (G. N. Lewis)
The first row elements will either want to achieve He or
Ne configurations.
Those who attain Ne configurations need 8 e’s in their
valence shell
The number of valence electrons can neatly be depicted
by arranging them around the atomic symbol
Ex) C has 4 valence e’s
•
•C •
•
Ex) N has 5 valence e’s
••
•N •
•
Electron Dot Structures
Electrons placed around the four sides of the atom symbol
individually or in pairs depending on the # of valence e’s.
Initially on e’s is placed on each side and are paired up when
all four sides are occupied, after which they are paired until
the valence shell is full.
Octet Rule
Notice that all noble gas configurations have an outer
shell with ns2.. . np6.
Ne = 2s22p6
Kr = 4s23d104p6
Ar = 3s23p6
Xe = 5s24d105p6
Rn = 6s25d104f146p6
Ignoring the d and f e’s the valence shell contains 8 e’s.
The d and f subshell contributions can be ignored if
they are full
Therefore the valence e’s in groups 13(3A) to 18(8A) can
be described in terms of ns and np e’s
In each case atoms want to have 8 electrons hence the
Octet rule.
Octet Rule
An element can achieve octet status in one of three ways:
1. It can gain valence electrons to make an anion:
Cl- [Ne]3s23p6=[Ar]
Ex) Cl [Ne]3s23p5
2. It can lose valence electrons to make an cation:
K+ [Ar]
Ex) K [Ar]4s1
3. It can share valence electrons with another atom to make
a covalent bond. This will generally involve two atoms of
similar electronegativity (ability to attract electrons)
••
••
••
•
••
••
Cl • + Cl
••
••
Cl • Cl
••
••
••
••
••
•
Ex) Cl2
Octet Rule
Nonmetals normally gain electrons to obtain a complete octet.
Ex) Cl to Cl-
O to O-2
S to S2-
P to P3-
Metals normally lose electrons to obtain a complete octet.
Ex) Na to Na+
Ca to Ca2+
Al to Al3+
Atomic charge = # protons – # electrons
Ex) Al3+ has 13 protons and 10 electrons q = 13 – 10 = +3
Elements in groups 1, 2 and 3, ionic charge is group #:
Ex) Al: 3(A) → q = +3 Ca: 2(A) → q = +2
Li: 1(A) → q = +1
Elements in groups 15, 16 and 17, ionic charge is group # -18:
Ex) As: 15 → q = -3
S: 16 → q = -2
F: 17 → q = -1
Ionic Compounds
Recall that opposite charges attract each other.
A cation and an anion, experience an electrostatic
force, given by:
e
= unit of charge
 q1eq2 e  k = 8.988 × 10 N m /C
F k 2 
9

r

2
Coulonb’s Constant
This force pulls them together to make an ionic bond.
q1
+
Na
r
q2
r
Cl
+
Na
Cl
2
The energy released to form an ionic bond:
 q1eq2 e 
E k

 r 
Exercise
Compute the force that a sodium cation and chloride
anion experience when 10.00 nm apart
F = ke2q1q2/r2
q1 = +1
q2 = -1
r = 10 nm = 1.000*10-8 m
F = (8.988 × 109 N m2/C2)(1.6022 *10-19 C)2(1)(-1)
(1.000*10-8 m)2
F = -2.307*10-12 N
Compute the energy of formation of the ionic bond in
NaCl, where the bond length is 279 pm
E = ke2q1q2/r
q1 = +1
q2 = -1
r = 279 pm = 2.79*10-10 m
E = (8.988 × 109 N m2/C2)(1.6022 *10-19 C)2(1)(-1)
(2.79*10-10 m)
E = -8.27*10-19 N m = J
How energy for one mole of NaCl bonds?
E(total) = E(bond) *(# of bonds)
= (-8.27*10-19 J/bonds)(6.022*1023 bonds/mol)
= 498,000 J/mol = 498 kJ/mol
LiF formation
Li .
Li
Li .
..
:F .
..
Li
+
Li loses all e’s in its valence shell
Large reduction in atomic radius:
Li – 152 pm
Li+ - 78 pm Vol = 1/7th
Ionization Energy consumed
As an atomic gas.
+
.. :F :
..
F gains electron its valence shell
Expansion in the radius because
effective Z* is decreased due to
extra electron.
F – 71 pm
F- - 133 pm Vol = 6x
Electron Affinity Energy released
As an atomic gas.
LiF formation
Energy is released to form the ionic bond
Ionic Materials
Lattice - 3-D pattern of ions
- Minimize repulsive forces
-Maximize attractive forces
- No net charge
F
Na
F
Na
Na
- Large lattice energy
released
- High melting point
- Brittle
F
Na
F
Na
F
Na
F
Na
F
Na
F
F
Na
Formation of
ionic materials
Whether element from
ionic compounds
depends on the
balance between:
1) Ionization energy
2) Electron affinity
3) Lattice energy
4) Phase transition
energies
5) Bond energies
Electron Affinity
Dissociation
energy
Ionization
energy
vaporization
Formation energy
Lattice
energy
Covalent Compounds
Covalent bond - valence electrons are shared between two or
more atoms. i.e. bonding electrons are “co-valent”.
The combination of electrons forms one entity.
In Ionic systems - the cation and anion are separate entities
The wavefunctions for each electron constructively interfere
and form a combined wavefunction, called a molecular
orbital
1s(1)
1s(2)
Molecule
1s
Atom 1
Atom 2
Nucleus 1
Nucleus 2
Electrons are shared to complete the valence shell of each atom
Covalent Compounds
Covalent bonds form if energy is released when atoms bond:
H. +
.H
H
H
E = -435 kJ/mol
The negative energy of reaction means that the product (H2) is
more stable than the reactants (2 × H).
2 electrons are shared completing the 1s orbital for each H
H
H
H.
+
.H
Edissociation = +435 kJ/mol
Bond dissociation energy Energy is required to break
the bond.
The lowest energy point is at
the average bond length.
Lewis structure diagrams
The exact solutions of molecular quantum mechanics are highly complex
•
••
Cl • Cl
••
Cl • + Cl
••
•
••
The Lewis dot diagrams of atoms can be combined to depict bonding in
molecules
••
••
••
••
••
••
These Lewis diagrams reflect the underlying quantum mechanics and
••
••
serve primarily as a bookkeeping device for the valence electrons in the
molecule
The guiding principle behinds Lewis diagrams is that each atom in molecules
achieve noble gas electron configuration by sharing electrons
This is known as the octet rule because the majority of noble gases have 8
valence e-, except for H which only requires two e’s to complete its valence.
••
Cl
••
Cl- Cl
••
••
••
•
••
••
••
Cl •
••
••
••
••
The electrons of a chemical bond are represented by a dash
There can also be non-bonding electrons, which are written as dots ●●
Steps for Drawing Lewis Electron Dot Structures
1. Determine the central atom. The rest are terminal atoms.
2. Determine the total number of valence electrons.
3. Use one pair of electrons to make a single bond between
each pair of bonded atoms.
4. Use any remaining electrons as lone pairs around each
terminal atom (except H) so that each terminal atom has a
complete octet, if possible.
5. Allow for any deficit or excess of e’s only for the central atom.
6. Check the central atom for too few or too many electrons.
7. If there are too few electrons, increase the bond order of
one or more bonds by sharing non-bonded electrons. (i.e.
make double or triple bonds, as necessary)
8. Calculate formal charge for all atoms, and indicate any
which are not zero.
•
•C
•
Central
4 v.e’s
F
F
••
••
••
F
••
F
••
F
••
••
F C
••
•
••
••
••
••
•F
••
••
••
• C•
•
••
F
••
F
•
•
••
•
••
•
••
Terminal
7 v.e’s
••
••
••
••
••
•
••
Ex) CF4
••
Tips
When finished review it to verify that correct number of atoms
and electrons were used and that the octet rule is obeyed
Remember that Lewis dot structures do not show a molecule’s
shape.
If there is more than one acceptable solution, the true electron
distribution is a hybrid of the possible distributions. This is called
resonance.
If it is impossible to avoid having one atom with too few or too
many electrons, make sure it is the central atom. Elements in
the 1st or 2nd period can never have more than eight
electrons under any circumstance.
Molecules with odd numbers of electrons form free radicals
and cannot fully obey the rules.
Tips
Larger molecules are treated as a sequence of central atom
problems.
The central atom can be generally be chosen using a few rules:
1. The central atom is never H or F.
2. If you have many atoms of one element and one atom of
another, the lone atom is the central one.
3. Given a choice, the central atom is not O.
4. Given a choice, C is central.
Ex) Choose the central atom for the following molecules:
(a) BBr3
(b) CH3Cl
(c) CH2O (d) POCl3
Draw Lewis dot structures for the following molecules
(a) NH3
a)
(b) C2H6
••
H •• N •• H
••
H
••
H - N -H
H
(c) C2H4
b)
(d) C2H2
H H
•• ••• •
H •• C • C • H
••
••
H H
H H
H C C H
H H
Draw Lewis dot structures for the following molecules
(a) NH3
H
•
• •
•
•• C • C • H
•• ••
(c) C2H4
d)
H
(d) C2H2
•
• •
•
•• C • C • H
•
•
H H
H
C
••
H H
•
•H
H C C H
H H
H •• C C •• H
•••
•••
•• C
••
••
••
c)
(b) C2H6
H C C H
We have miscounted the e’s
-
Ex) ClO
Cl • O
•
••
••
••
••
[
••
••
q
]
Extra e added to complete
Valence on O.
#p = 17
#e = 18
10 core
8 valence
#p = 8
#e = 9
2 core
8 valence
qCl = #p - #e qO = #p -#e
= 17 – 18
= 8 – 10
= -1
= -2
Total charge = -3 ????
The bonding e’s have been double
counted!!!
The bonding e’s are shared and
therefore count only as half an
electron for each atom
#e = core e’s + ½(bond e’s)
+ non-bond e’s
# e’s Cl = 10 + ½ (2) + 6 = 17
qCl = 17 – 17 = 0
# e’s O = 2+ ½ (2) + 6 = 9
qCl = 8 – 9 = -1
Therefore the –ve charge is on O
It needs it to complete its valence
Formal Charge
Formal charge (Qf) is the charge on an atom assuming
that every bond is completely covalent.
All bonding electrons are shared equally
Qf = #p’s – [ (# core e’s) + ½ (# bonded e’s) + (# non-bonded e’s)]
Qf = (# valence e’s) – [ ½ (# bonded e’s) + (# non-bonded e’s)]
••
••
••
••
As a general rule, we want to keep the formal charge on
each atom as close to 0 as possible (without giving any
atom more than a complete octet).
••
••
Ex) BF3
Qf(B) = 3 –[½ (6) + 0] = 0
•
•
F •B •F
•• •• ••
Qf(F) = 7 –[½ (2) + 6] = 0
F
Note: valence of B is incomplete!!
••
••
••
Let’s try another arrangement taking an extra set of bonding
electrons from one of the F’s
-1
Qf(B) = 3 –[ ½ (8) + 0] = -1
•• 0
0 ••
•F
•
•
Qf(F) = 7 –[ ½ (2) + 6] = 0
F •B ••
• • •• ••
or Q (F) = 7 –[ ½ (4) + 4] = +1
••
••
F
f
+1
Total q = -2 + 0 + 1 + 1 = 0
Large charge separation!!!
••
0
•• 0
••
F - B -F
••
••
F
••
••
••
0
Therefore the previous
arrangement is preferred
••
0
0
]
Consider the anion NO3-.
0
••
•
•
O•N •O
• ••
••
O
•
•
0
••
••
••
Incomplete valence on two O’s
-1
Lets try to fill them using the
remaining 2 e’s on N
•
••
••
• • -1
•
•
•
••
•• N •
O
O
•• ••
••
••
O
0
••
0 ••
Charge separation is the same
As above
Total charge is -1
All valences are complete
••
O
••
••
N
••
O
O
••
••
0
0
-
]
Resonance
O
••
••
N
O
••
••
]
O
••
••
O
-
••
••
••
N
O
••
]
••
••
O
-
This arrangement is also correct?
This one too!!!
When there are several permutations that are equivalently
correct, the actual situation is an average between them.
This phenomenon is known are resonance, which can be
depicted using bidirectional arrows
-1
-1
-1
O=N=O
O = N _O
O _N = O
O
O
O
Ex) O3
..
..
_
:
O
=
O
O
..
.. ..
+1
-1
.. _ .. ..
O
=
O
:O
..
..
-1
+1
-1
+2
C
C
X
..
..
O = O =O
..
..
X
_
-1
Ex) Benzene, C6H6
. .C
C.
?
. .
C . C
.. _ .. ..
:O
O O
.. ..
..:
C
C
C
C
C
C
C
C
C
C
C
C
Bond polarity and electronegativity
Li+
F-
F
F
In a symmetric molecule such as H2, the concept of covalence is unambiguous:
the electrons are evenly shared by the two atoms.
But in LiF, we have complete electron transfer: Li+ and :F-, i.e. ionic bonding
There is a continuum of behaviour between
these two extremes
d
Ionic bonds are completely polarized towards the
opposite charged ions.
As the electronegativity difference decreases, e’s are more likely to be shared,
but unequally: polar covalent bonds
In polar covalent bonds, there is a bond dipole, which is indicated by a vector
There are also partial positive and negative charges, indicated by δ+ and δ-
Pauling Electronegativity
We use Pauling electronegativity values to determine bond polarity.
Electronegativity is the ability of an atom in a molecule to attract electrons to
itself.
A difference in Pauling electronegativity between elements of more than 2
units is enough to cause ionic bonding
General trend in element electronegativity
Evaluating Lewis diagrams - Revisited
How do you chose between several possible Lewis diagrams?
1. Most important is achieving the Octet rule: any structure that obeys the
octet rule is better than any structure that does not
2. Any structure that minimizes the sum of the absolute values of the
Formal Charge is better
3. The diagram that associates negative Formal Charge with more
electronegative elements and positive Formal Charge with electronegative
elements is to be preferred over others
Let us now apply this rule to some more complex cases:
1) OCN- (cyanate)
.. _
O C N:
-1 : ..
2) CO2 (carbon dioxide) 3) H2CSO (sulfine)
..
..
O=C=O
..
..
..
H _C = S _ O:
.. ..
H
+1
-1
Let us now apply this rule to some more complex cases:
1) OCN- (cyanate)
.. _
O C N:
-1 : ..
2) CO2 (carbon dioxide) 3) H2CSO (sulfine)
..
.. -1
_ ..
:
O C N:
O
=
C
=
N
..
..
..
+1
..
..
O=C=O
..
..
..
_
_ :
H C=S O
.. ..
H
+1
-1
-1
-2
-1
-1
.. _
O C O:
:..
-1
_ ..
:O C O
:
..
-1
..
..
_
H _C S = O
.. ..
.. _ .. _ ..
: ..
:
O C
O
.. ..
H
+1
-1
Exercise
State whether the bond is ionic or covalent and draw the
dipole vectors and partial charges for:
2.1 2.8
H - Br
d+ d-
C
2.5 2.8 C
C - Br
d+ d-
3.5
2.5 C
2.1
4.0 C
O-C
d+
d-
H-F
d+ d-
4.0 2.8 C
F-I
d+
d-
3.5 0.9 I
O - Na
d+
d-
0.8 4.0
K-F
d+ d-
I
3.5 2.5
O-S
C
d-
d+
%-ionic character
Continuum in behaviour
between pure covalent and
ionic character
Molecules such as H2, F2
are at 0 on this scale
Common “ionic”
compounds such as NaCl
cover a wide range on this
scale but are all over 2
The formal charge separation gives rise to ionic forces of attraction
contributing to the bond energy.
Therefore all bonding must be considered as having ionic and covalent
contributions.
Partial Charge
The bond dipole can be quantified by calculating the partial charge (Q) on
each atom.
Partial charge, similar to formal charge, except that the electronegativities of
the atoms, govern the distribution of the bonding electrons between the
atoms.
Q (X)  evalence -


 enonbonding



X


e

 X  Y bonding 

Recall the formula for formal charge
Q (X)  evalence f

e
 nonbonding


1

 e
2 bonding 
Therefore, pure covalency assume electronegativities are the same
Ex) H-Cl
Q(Cl) = 7 – [6 + {3.0/(2.1+3.0)}*2] = -0.2
Q(H) = 1 – [0+ {2.1/(2.1+3.0)}*2] = 0.2
Bond order: length and energy
We have seen that chemical bonds between the same pairs of
elements may be single, double or triple bonds
Quadrupole bonds also exist between some of the d-elements
How do these bonds differ?
The higher the bond order, the shorter will be the bond will be.
The higher the bond order the larger the bond dissociation
energies
Predicting Bond Lengths
Bond lengths shows the same trends as atoms size
Size decreases from left to right across the Periodic Table, and increases
down any group
154
C—C
147
143
136
C—N C—O N—O
194
187
181
C—Si
C—P
C—S
234
227
Si—Si
221
214
Si—P Si—S P—S
Increasing the bond order always shortens the bonds; however the %shortening is not a very regular parameter, therefore not simply predicted.
Lewis diagrams and Molecular Shape
There is a direct link between Lewis diagrams
and molecular shape
The Valence Shell Electron Pair Repulsion
theory states a molecule’s shape is determined
by the electron pairs that surround central
atoms
These electron pairs (EP) include both bond
pairs (BP) and lone pairs (LP)
We define five shape families based on the need to
accommodate mutually repelling electron pairs around
a central atom
Four electron pairs
define the tetrahedral
shape family, as in
SiCl4
The balloons
illustrate these
“natural” shapes
The 5 Shape Families
The five shape families that electron pairs develop are illustrated above
Linear – Trigonal-planar – Tetrahedral – Trigonal-bipyramidal – Octahedral
In this case each molecule has all its electron pairs as bond pairs around
the central atom
Not all the shape-determining electrons pairs need be bond pairs
General Shape Family Scheme (1)
0 LP 2 BP
2 EP’s
X
1 LP 1 BP
..
X
0 LP 3 BP
3 EP’s
X
1 LP 2 BP 2 LP 1 BP
X
X
Lone pairs and shape: 4 Electron Pairs
When there are four electron pairs, the shape family is tetrahedral
Four possibilities exist: with 0, 1, 2 or 3 lone pairs
0 LP – tetrahedral shape
2 LP – bent shape
1 LP – trigonal-pyramidal shape
3 LP – linear shape
Linear
Hydrogen Fluoride, HF
1 bond pair
3 lone pairs
General Shape Family Scheme
4 BP 0 LP
X
3 BP 1 LP
X:
2 BP 2 LP
1 BP 3 LP
:
X
..
..
:
X
..
General Shape Family Scheme
5 BP 0 LP
4 BP 1 LP
3 BP 2 LP
2 BP 3 LP
X
X
X
X
..
..
..
1 BP 4 LP
..
X
Special features of 5 EP shapes
In the trigonal bipyramidal geometry, alone among the 5 shape families,
the sites are not identical
There are three equivalent sites which are identical to those in trigonalplanar, with bond angles of 120 . These are called the equatorial sites
Axial Sites - above and below
the plane in a mutually linear
relationship.
Note - the angles are different
Note - if there are LP’s, these will
always first occupy equatorial
sites
General Shape Family Scheme
6 BP 0 LP 5 BP 1 LP 4 BP 2 LP 3 BP 3 LP 2 BP 4 LP 1 BP 5 LP
X
..
..
..
X
X
X
..
X
..
..
X
..
Some special features of VSEPR
multiple bonds, are found to occupy almost the same volume of space
Hence, multiple bonds can be considered as if they are single BP of electrons
Ex) NO2
The N has 4 EP’s - 1 LP and 2 BP’s
bent shape
LP are considered to be larger than BP because constrained by only a single
positively charged nucleus as opposed to two for a BP
For 6 EP case LP prefer to be opposite since otherwise at 90o they would
experience strong repulsive forces
For 5 EP since neighboring LP’s at 120 experience much less repulsion than
at 90o they tend to cluster on the equatorial sites of the trigonal-bipyramidal
geometry
Geometry of large molecules
VSEPR method extends to larger molecules by treating them as a chains
central atoms
The only thing that remains is to twist the bits of the molecule about any
single bonds to minimize congestion between the bonded atoms
The geometry at the S, O and N are similar to those in water and ammonia,
i.e. bent and trigonal pyramidal
The geometry at the C1 and C2 atoms are tetrahedral, while at C3 it is
trigonal planar
The complete molecule simply twists the bits and pieces about any given
single bond
Shape and polarity
Recall that most covalent bonds are polar
Therefore molecules must be polar Not always!!!
Symmetry in a molecule can cause bond
dipole vectors to cancel each other making
the molecule non-polar even though its
individual bonds are very polar
Permanent Dipole Moments
Molecules that have permanent dipole moment are called polar
The unit of dipole moment is the Debye
Note that for molecules like H2 and CCl4, there is no permanent dipole
moment!
Such molecules are non-polar.
Measuring dipole moments
Molecules which have permanent dipole moments respond to an
applied electric field
This is an arrangement similar to that of a capacitor
The dipole moments in the molecule tend to align with the applied electric
field
By measuring the degree of alignment as a function of applied field strength,
the size of the dipole moment in Debye units can be determined
Concepts from Chapter 9
DRAWING LEWIS ELECTRON DOT DIAGRAMS
Octet rule
Resonance structures
Bond polarity (ionic, polar covalent and covalent bonds)
Ionic vs. covalent compounds
Electronegativity
Dipole vectors
Calculating formal charges and partial charges
Bond order
Bond lengths
VSEPR and predicting shapes of molecules
Polarity of molecules