Transcript IX: Chemical Bond Formation

Topic 11: Chemical Bond Formation
LECTURE SLIDES
• Valence electrons
• Ionic Bonding
• Covalent Bonding
• Lewis Structures
• Acid, Anion Relationships
• Resonance Structures
• Bond Length vs Bond Order
• Octet Violators
• Formal Charge
Kotz & Treichel, 9.1-9.6
CHAPTER 9: BONDING AND MOLECULAR
STRUCTURE
Now that we have examined the structure of the atom
and the arrangement of its electrons, we are ready to
turn to the molecules and compounds they form.
Our next studies will center on the bonds that hold
the atoms together in compounds, molecules and
polyatomic ions. We will also examine the three
dimensional shape of these species, and their polar
or non polar nature.
BOND FORMATION
The interactions between atoms which lead to bond
formation are all centered around the electrons
in incomplete subshells and in incomplete outer shells:
the valence electrons...
The atoms of the elements lose, gain or share these
electrons to achieve, where possible, the noble gas
configurations we have met.
For the “main group elements”, the s and p block
members, the electrons available for bonding, the
“valence electrons”are the outer shell s and p
electrons (except those of the noble gases!)
In forming compounds from these elements, only
these electrons will be used, in two ways:
They may be transferred to form ions so that
incomplete subshells are completed or removed;
They may be shared so that two atoms together
have complete subshells
For the transition metals, the valence electrons
include both the s electrons from their outermost
shell and also electrons from their inner,
incomplete d subshell.
The PT column number of the main group and transition
metals gives the sum of valence electrons for each
element in the family.
Note that the column number indicates the maximum
positive charge (or oxidation state) these metals
can achieve in a compound through loss of e’s in a
chemical reaction.
For all main group elements (the columns 1A-8A),
it is convenient for bonding purposes, to represent
the elements as “Lewis Dot Symbols”, which include
one dot for each of their valence electrons.
The tendency of these elements to achieve an outer
shell configuration of eight electrons (the “octet
rule”) is easily visualized through use of these
symbols.
The valence electrons for transition elements (columns
3B-8B, 1B, 2B) are not represented by dot symbols.
LEWIS DOT STRUCTURES FOR PERIOD 2
2A
1A
ns
Li
1
ns
2
Be
3A
4A
5A
6A
7A
8A
2 3
2 5
2 6
2 4
ns2np1 ns2np2 ns np ns np ns np ns np
B
C
N
O
F
All elements, same column: same dot structure
Ne
The elements come together to form compounds
so that each element can achieve a more satisfactory
outer shell electronic configuration.
Elements may lose or gain electrons resulting
in cation and anion formation and the attraction
between the two which we call the “ionic bond”
Elements may share one or more pairs of electrons.
The attraction of both nuclei for the same pair of
electrons results in the force we call the “covalent
bond”.
The “ionic bond”: attraction of opposite charges
when transfer of electrons cause formation of
positive and negative species: cations and anions.
The individual ions radiate charge in all directions
and cluster in geometric patterns which are described
as crystal lattices.
Note in the following slide that each ion has many
neighbors and the compound itself is not molecular
in nature: no discrete “formula units” exist.
Ionic compounds are all solids at room temperature
with elevated melting points. Their melting points
reflect the very high degree of attraction exhibited by
these fully charged particles, which depends on the
magnitude of their charge and the ionic size:
The larger each charge and the smaller each ion, the
greater the attraction.
Energy = n (+) X
d
n= magnitude of charge
n (-)
d=distance between ions
Ion formation and the resulting ionic bond
occurs when metals of sufficiently low
electronegativity (X) react with non-metals of
sufficiently high X values.
The most ionic of compounds are those formed
between the active s block metals (X < 1)with the
non-metals whose X values are 3.0 or larger.
All compounds we have met containing
“polyatomic” anions or ammonium are also of
course truly ionic type compounds.
Most active non-metals
INCREASES
>2
1.5-1.9
.8-1.4
Most active metals
ELECTRONEGATIVITY
ELECTRONEGATIVITY VALUES, MAIN GROUP
ELEMENTS
H
2.1
Li
1.0
Na
1.0
K
0.9
Rb
0.9
Cs
0.8
METALS
Non-metals
Be
1.5
Mg
1.2
Ca
1.0
Sr
1.0
Ba
1.0
B
2.0
Al
1.5
Ga
1.7
In
1.6
Tl
1.6
C
2.5
Si
1.8
Ge
1.9
Sn
1.8
Pb
1.7
N
3.0
P
2.1
As
2.1
Sb
1.9
Bi
1.8
O
3.5
S
2.5
Se
2.4
Te
2.1
METALLOIDS
F
4.0
Cl
3.0
Br
2.8
I
2.5
THE COVALENT BOND
The second mode of bond formation occurs when
elements share one or more pairs of electrons to
achieve where possible an outer shell octet.
The attraction of both nuclei for the same pair of
shared electrons is the basis of the covalent
bond.
Covalent bonds are directed between two atoms,
sharing together one or more pairs of electrons.
This type of bonding leads to formation of
discreet molecules, individual units made up of
two or more atoms covalently bonded together.
Any formula consisting solely of nonmetals and
metalloids can assumed to molecular and
covalent in nature.
Covalent bonds hold together the atoms within
a polyatomic ion. Occasionally, metals with higher
electronegativity values will form a compound more
covalent than ionic in nature.
Group Work, 11.1: Bond Type
Compound
NaCl
CaCl2
AlCl3
SiCl4
BeCl2
CCl4
 en ( X )
Bond type
Ionic compounds,  en >1.6 or 1.7
LEWIS DOT STRUCTURES: MOLECULES AND
COMPOUNDS
We are next going to use the Lewis Dot Symbols for
the “main group elements” to represent the
bonding and structure for various species, molecules,
compounds and polyatomic ions.
• all valence electrons for every atom will be included
• all shared pairs of e’s will be indicated by a “dash”
• all unshared pairs of e’s will be indicated by a dot
We will use the “octet rule” as our guiding principle.
Diatomic Elements:
H
H
H
H2 Cl2 N2 O2
H
H
H
H
H
H
H
Chlorine, Cl2 (same for Br2, F2, I2)
Cl
Cl
Cl
Cl Cl
Cl Cl
Cl Cl
Cl
Cl
Cl Cl
Check octets!
Cl
N2, Nitrogen:
N
N
N N
N
N
NO OCTET
STILL NO
OCTET
N
N
N
N
N
N
Covalent “triple bond”
N
N
O
O
O O
O O
OXYGEN ACTS
LIKE THIS*
O O
O
O
O
O
CORRECT
LEWIS STRUCTURE,
Covalent double bond
*Required whole new bonding theory to explain...
Lewis Structures: Compounds and Polyatomic Ions
GUIDELINES
First step:
Decide on arrangement of atoms. For most species, the
element written first in the molecule or ion is the central
atom and the remainder of the atoms are grouped around
it.
Hydrogen is a problem in “oxo acids” where it is written
first in the formula. Ignore H, start with the next atom in
formula and place the H or H’s on the O or O’s.
NH3
H
N
H
P
Cl
PCl3
H
Cl
H
CHBr 3
Br
C
Cl
O
Br
CH2O
H
C
H
Br
O
O
HNO3
H O
N
O
SO42-
O
S
O
O
Second Step
•Add up all available valence electrons.
• If species is cation, subtract positive charge from total.
• If species is anion, add negative charge to total.
•Divide total by two to determine available number of
electron pairs
Third Step
Place a pair of electrons between each pair of bonded
atoms to represent a single bond (use a “dash”!)
Step 2
Step 1
NH3
H
N
H
N: 1 x 5= 5 valence electrons
3H: 3 x 1= 3 valence e's
H
8 valence e's
8 e's / 2 = 4 electron pairs
H
N
H
Step 3
H
Fourth Step
Place leftover electron pairs around “terminal” atoms
to achieve their octet (except H). Do central atom last.
Fifth Step
Examine central atom to determine if a double or triple
bond is required to achieve the central atom’s octet.
Do so using unshared pairs, IF central atom is:
C, N, P, O, S
Step 4:
H, duet
8 e's / 2 = 4 electron pairs
H
N
H
H
H
N
H
H
N
H
H
H
N, octet
No Step 5 needed
Step 1
Cl
PCl3
Step 2
P
Cl
P: 1 x 5= 5 valence electrons
3Cl: 3 x 7=21 valence e's
Cl
26 valence e's
26e's / 2 = 13 electron pairs
Cl
P
Cl
Step 3
Cl
26 e's / 2 = 13 electron pairs -3, BONDS= 10 e pairs
Cl
P
Cl
Cl
Cl
P
Cl
Step 4
Cl
Cl
P
Cl
Cl
H
CHBr 3
Br
C
Br
C:
1 x 4 = 4 valence electrons
3Br: 3 x 7 = 21 valence e's
Br
1H: 1 x 1 = 1 valence e
26 valence e's
26e's / 2 = 13 electron pairs
H
Br
C
Br
Br
26 e's / 2 = 13 electron pairs -4, BONDS= 9 e pairs
H
H
Br
C
Br
Br
Br
C
Br
Br
GROUP WORK 11.2: Lewis Structures
Use 5 steps:
Arrange;
adds up e’s;
draw bonds;
assign unshared pairs
double bonds if needed
to draw Lewis structures for following species:
NBr3
CH2Cl2
Now let’s examine situations requiring the double bond:
C:
O
CH2O
C
H
H
1 x 4 = 4 valence electrons
1O: 1 x 6 = 6 valence e's
2H: 2 x 1 = 2 valence e's
12 valence e's
12e's / 2 = 6 electron pairs
O
H
C
H
12 e's / 2 = 6 electron pairs -3, BONDS=3 e pairs
O
H
C
O
H
H
C
No octet
O
O
H
H
C
H
H
C
H
N:
O
HNO3
H O
N
1 x 5= 5 valence electrons
3 O: 3 x 6 = 18 valence e's
O
1H: 1x 1 =
1 valence e's
24 valence e's
24e's / 2 = 12 electron pairs
O
H
O
N
O
24 e's / 2 =12 electron pairs -4, BONDS=8 e pairs
O
H
O
N
O
O
O
H
O
N
H
O
O
N
O
No octet
Either one
O
H
O
N
O
S:
O
SO42-
O
S
O
1 x 6= 6 valence electrons
4 O: 4 x 6 = 24 valence e's
2-: 2 x 1 = 2 valence e's
O
32 valence e's
32e's / 2 =16 electron pairs
O
O S O
O
32 e's / 2 =16 electron pairs -4, BONDS=12 e pairs
2-
O
O
S
O
O
O
O
S
O
O
Be sure to include charge
on finished product
GROUP WORK, 11.3: Lewis Structures # 2
Use 5 steps:
Arrange;
adds up e’s;
draw bonds;
assign unshared pairs;
double bonds if needed
to draw Lewis structures for following species:
H3PO4
NO21+
ClO41-
Let’s explore the relationship between various
“oxo” acids (H, Non metal element, O) and the
charge and formula of their anion relative.
Recall that acids, by definition, ionize in water to
lose one or more H’s as H+. The anion left
behind is named according to the name of its
“parent” acid.
In an acid/base reaction, as we met last unit, acids(H+)
react with bases (OH-) to form water, leaving behind
the anion of the acid and the cation of the base to form
a salt.
Recall that acids “ionize” in water, or react with a base to
form water, in either case leaving behind some “anion”:
Cl-, NO3-, SO42- etc...
H- “Anion”
H2O
H+ + Anion-
H- “Anion” + NaOH
Acid: HCl, HNO3 H2SO4 etc...
H2O + Na+ An-
CO32-
H2CO3
O
H
O
C
O
H
O
Carbonic Acid
H3PO4
2-
O
C
O
Carbonate
PO43-
O
3-
O
H
O
P
O
H
O
O
P
O
H
O
Phosphoric Acid
Phosphate
Group Work, 11.4: Acids and Anions
Give missing structures, next four slides!
Base structure on accompanying acid or anion...
Group Work 11.4a
NO 31-
HNO 3
O
H
O
N
Nitric Acid
O
Nitrate
NO 21-
HNO 2
O
O
Nitrous Acid
N
Nitrite
1-
Group Work 11.4 b
SO42-
H2SO4
O
O
S
2O
O
Sulfate
Sulfuric Acid
SO32-
H2SO3
O
H
O
S
O
Sulfurous Acid
H
Sulfite
GW 11.4c
ClO31-
HClO3
O
H
O
Cl
O
Chloric Acid
Chlorate
ClO41-
HClO4
O
H
O
Cl
O
O
Perchloric Acid
Perchlorate
Group Work, 11.4d
ClO1-
HClO
1O
Hypochlorous Acid
HClO2
Cl
Hypochlorite
ClO211O
Chlorous Acid
Cl
Chlorite
O
RESONANCE THEORY: WHERE TO PLACE THE
DOUBLE BOND...
EQUIVALENT
O3, OZONE: 3 O = 3x 6 = 18 e's
O
O
O
No OCTET
O
O
O
OR
O
O
O
EQUIVALENT
NO31-, NITRATE ION: 5 + 18 + 1 = 24 e's
1-
O
O
O
N
N
O
1-
O
1-
O
O
N
1-
O
O
O
N
O
O
CO32-, CARBONATE ION: 4 + 18 + 2 = 24 e's
2-
O
O
O
C
2-
O
O
C
O
2-
O
C
O
2-
O
O
C
O
O
In all three cases, O3, NO3 -, CO32-, when forming a
double bond from a “terminal oxygen” one has a
choice of moving e’s from several different O’s to
makeup the “central atom’s” octet.
Examination of experimental evidence (x ray) shed an
interesting light on this topic:
When two atoms are bonded together, the distance
between their nuclei, their “bond length,” depends on
whether the bonds between the two are single,
double, or triple.
Note that triple bonds are shorter than double and also
double shorter than single, as well as being characteristic
between any two given atoms.
TYPICAL BOND LENGTHS
C
C
154 pm
C
N
147 pm
C
O
143 pm
C
C
134 pm
C
N
127 pm
C
O
122 pm
C
C
121 pm
C
N
115 pm
C
O
113 pm
X ray evidence of bond lengths in ozone, nitrate and
carbonate ions should therefore prove interesting...
Predicted, “usual” bond lengths:
121 pm
O
O
O
132
pm
Instead of the predicted bond lengths observed in
other compounds, both bonds in x ray showed identical
lengths of 127.8 pm, close to an average of 1 1/2 bonds
to each O.
Linus Pauling proposed the “theory of resonance” to
describe this situation:
When two or more equivalent Lewis structures
can be drawn for a species, differing only in the
position of electron pairs, then none are correct:
The real structure is a hybrid of all structures
drawn.
The Lewis structures drawn are called “contributing” or
“resonance structures” needed to describe the makeup of
the hybrid, which resembles all but is none of the above.
A special double headed arrow is drawn between the
contributing structures to indicate their hypothetical
nature:
O
O
O
O
O
O
The hybrid structure, with two equivalent bonds to
the central atom, are said to have a bond order
of “1.5” or an average of 1 and 1/2 bonds between
each O:
THE HYBRID STRUCTURE OF OZONE
O
O
O
Bond Order describes the number of bonds between
two atoms in a molecule. Normally, the bond number
is 1 (a single bond) or 2 (a double bond) or 3 (a triple
bond.)
When hybrid structures and resonance situations
exist, one must average the number of bonds between
all atoms affected, and fractional values
arise.
In the case of the nitrate and the carbonate ions, the
number of bonds to the central atom is averaged out
over 3 atoms, and 4 bonds/3 atoms= 1.33 bond order.
In both cases, x ray data confirms this theory.
2-
O
O
C
2-
O
O
O
C
O
O
O
C
C
O
2-
O
CO32-
2-
O
O
The carbonate ion has three equivalent C-O bonds, of a
length typical of 1 and 1/3 bond, for a 1.33 bond order.
1-
O
O
N
O
NO 3
1-
O
O
N
O
N
N
O
1-
O
1-
O
O
1-
O
O
The nitrate ion also has three equivalent N-O bonds, of a
length typical of 1 and 1/3 bond, for a 1.33 bond order.
Group Work 11.5: Resonance Structure and
Bond Order
Draw two acceptable Lewis Structures for SO2 and
a resonance hybrid. What is the bond order for
the bonds in this compound?
OCTET VIOLATORS
Another aspect of drawing correct Lewis structures
involves the handling of compounds that do not have
an octet around the central atom. Three situations exist:
1. More than 4 e- pairs around central atom
2. Less than 4 e- pairs around central atom
3. Molecules with odd number of electrons
In all cases we will handle, the irregularity occurs at
the central atom; all “terminal atoms” will have normal
octet.
EXAMPLE:
PF5: 5 + (5x7) = 40 valence e's = 20 e pairs
F
F
F
F
F
P
F
F
P
F
F
F
Note: Only the central atom, P, is an “octet violator”
ClF3 : 4 x 7 = 28 e's / 2 = 14 e pairs
F
Cl
F
F
F
F
+
9 PAIRS
Cl
F
F
F
3 PAIRS
Cl
+
Note again: Only the central atom
exceeds the octet rule.
2 PAIRS
F
Case #2: Less than 4 e- pairs around central atom
This category specifically applies to the metalloid
Boron, but also to metals that form salts that are more
covalent in nature than ionic: Beryllium, Aluminum,
for example.
These elements use their valence e’s to form
compounds but do not form an octet in the process and
do not accept double bonds to compensate.
These “octet deficient” species will react with other
atoms however to form polyatomic ions or compounds
which relieve the deficiency.
EXAMPLE:
BF 3: 3+ (3 x 7) = 24 e's / 2 = 12 e pairs
F
F
B
F
F
F
B
No Octet,
octet rule violator
F
Group work 11.6: Identify “Octet Violators”
Compound
BH3
SF6
BrF5
AlI3
NH3
Octet Violator?
While B will not form a double bond to F to achieve
an octet (F’s “don’t do” double bonds), it will accept
electron pairs readily from other sources to do so:
F
H
H
N
H
+
B
F
F
H
F
H
N
H
B
F
F
When one atom donates two electrons for a pair of
atoms to share, the bond is called a “coordinate
covalent bond” and introduces “charge buildup”
in the species formed.
To keep track of this kind of charge within a molecule
or polyatomic ion, the concept of “FORMAL CHARGE”
is introduced.
Formal charges look within a molecule or polyatomic ion
and determine how the charges are distributed by
considering for each atom:
• the number of valence e’s it started with
• the number of bonds formed
• the number of unshared electrons leftover
FORMAL CHARGE:
For each atom in species:
formal charge = # valence e’s - (#bonds + #unshared e’s)
The “formal charge” system requires a Lewis Dot
Structure and assigns an individual “formal” charge
to each atom in the species.
Formal charge is an alternate “bookkeeping method”
for tracking electron distribution to the “oxidation
number” system we met previously.
Now let’s return to the compound formed between
ammonia and boron trifluoride, and determine
formal charges:
F
H
H
N
H
+
B
F
F
H
F
H
N
H
B
F
F
Formal charges
H
H
F
N
B
H
F
F
FORMAL CHARGE = VALENCE E'S - (# BONDS + # UNSHARED E'S)
Each
N 5 - (3 + 2) = 0
B 3 - (3 + 0) = 0
H 1 - (1 + 0) = 0
Each F 7 - (1 + 6) = 0
NH3
BF3
F
H
H
N
H
N 5 - (4 + 0) = 1
Each
H 1 - (1 + 0) = 0
B
F
F
B 3 - (4 + 0) = -1
Each F
7- (1 + 6) = 0
OXIDATION NUMBERS:
“Ox #’s” are assigned or calculated based on known fixed
positive and negative charges, and can be determined
by examination of the formula for the species.
Oxidation numbers are useful to identify how charges
change in a redox (oxidation-reduction) reaction.
To see how both work, let’s look at chloric acid,
HClO3, and see how its charge distribution would
be described using both the oxidation number
and the formal charge systems.
Known ox #’s
per atom
?
+1
-2
HClO3
+1
?
-6
Sum of all charges
in compound = 0
1 + ( ?)+ ( -6) = 0
( ?) = 6 - 1 = +5
Valence e's
Cl 7
3O 18
H
1
HClO3
26 e's =13 prs
O
H
O
Cl
O
O
H
O
Cl
O
Lewis Structure
formal charge = # valence e’s - (#bonds + #unshared e’s)
FORMAL CHARGE
O
H
O
Cl
O
Cl
7 - (3 + 2)
O 2[ 6 - (1 + 6) ]
O
6 - (2 + 4)
H
1 - (1 + 0)
OVERALL
= +2
= -2
= 0
= 0
=0
Like ox #’s, the sum of all formal charges in a compound
must equal 0.
Oxidation numbers
?
+1
-1
-2
0
O
HClO 3
+1
+5
-6
0
H
O
Cl
O
-1
Formal Charges
+2
Finally, both oxidation numbers and formal charges
must add up the same way:
For compounds, which are always electrically neutral,
the sum of all oxidation numbers or the sum of all
formal charges must equal zero.
For polyatomic ions, which always have a specific
charge, the sum of all oxidation numbers or the sum
of all formal charges must equal the charge on the ion.
MnO4
Valence e's
Mn 7
4O 24
-1
1
1-
32 e's =16 prs
1-
O
O
Mn
O
O
FORMAL CHARGE
Mn
7 - (4 + 0)
O 4[ 6 - (1 + 6) ]
= +3
= -4
OVERALL
= -1
Group Work 11.7: Oxidation Numbers vs Formal Charge
H2SO4
Sulfuric Acid
Sulfuric Acid
H2SO4
O
H
O
S
O
O
H
H
O
O
Correct Lewis Structure
S
O
H
O
Expanded octet version
1: Do Ox #’s from formula (same for both!)
2: Do formal charge from Lewis Structure for all atoms:
formal charge = # valence e’s -( # bonds+ # unshared e’s)