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Lecture 22 © slgCHM 151
RESONANCEOCTET VIOLATORS FORMAL CHARGES MOLECULAR SHAPES
TOPICS:
No OCTET EQUIVALENT
EQUIVALENT
OO
O
O3, OZONE: 3 O = 3x 6 = 18 e's
OO
OOO
OOR
RESONANCE THEORY: WHERE TO PLACE THE DOUBLE BOND...
NO31-, NITRATE ION: 5 + 18 + 1 = 24 e's
O
N OO
1-
O
N OO
1-
O
N OO
1-
O
N OO
1-
CO32-, CARBONATE ION: 4 + 18 + 2 = 24 e's
O
C OO
2-
O
C OO
2-
O
C OO
2-
O
C OO
2-
In all three cases, O3, NO3 -, CO3
2-, when forming adouble bond from a “terminal oxygen” one has a choice of moving e’s from several different O’s tomakeup 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.
TYPICAL BOND LENGTHS
Note that triple bonds are shorter than double and alsodouble shorter than single, as well as being characteristicbetween any two given atoms.
X ray evidence of bond lengths in ozone, nitrate andcarbonate ions should therefore prove interesting...
OO
O
132 pm
121 pm
Predicted, “usual” bond lengths:
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 bondsto each O.
Linus Pauling proposed the “theory of resonance” todescribe 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 thecontributing structures to indicate their hypotheticalnature:
OO
OO
OO
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: (Bond order: # bonds/# atoms bonded to central atom)
THE HYBRID STRUCTURE OF OZONE
OO
O
Bond Order describes the number of bonds between two atoms in a molecule. Normally, the bond numberis 1 (a single bond) or 2 (a double bond) or 3 (a triplebond.)
When hybrid structures and resonance situationsexist, one must average the number of bonds between all atoms affected, and fractional valuesarise.
In the case of the nitrate and the carbonate ions, thenumber of bonds to the central atom is averaged outover 3 atoms, and 4 bonds/3 atoms= 1.33 bond order.
In both cases, x ray data confirms this theory.
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.
O
C OO
2-
O
C OO
2-O
C OO
2- O
C OO
2-
CO 32-
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.
O
N OO
1-
O
N OO
1-O
N OO
1- O
N OO
1-
NO 31-
OCTET VIOLATORS
Another aspect of drawing correct Lewis structuresinvolves the handling of compounds that do not havean octet around the central atom. Three situations exist:
1. More than 4 e- pairs around central atom2. Less than 4 e- pairs around central atom3. 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:
PF 5: 5 + (5 x7 ) = 4 0 vale nce e 's= 2 0 e pair s
P
F
F
F
FF
P
F
F
F
FF
Note: Only the central atom, P, is an “octet violator”
ClF3 : 4 x 7 = 28 e's / 2 = 14 e pairs
ClF F
F
ClF F
F
3 PAIRS +9 PAIRS
ClF F
F
+ 2 PAIRS
Note again: Only the central atom exceeds the octet rule.
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 anddo 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.
B
F
F F
BF 3: 3 + (3 x 7 ) = 2 4 e's / 2 = 1 2 e p airs
B
F
F F
EXAMPLE:
No Octet, octet rule violator
B
F
F
FN
H
H
H
+ B
F
F
FN
H
H
H
While B will not form a double bond to F to achievean octet (F’s “don’t do” double bonds), it will acceptelectron pairs readily from other sources to do so:
When one atom donates two electrons for a pair ofatoms to share, the bond is called a “coordinatecovalent 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 ionand 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
For each atom in species:
formal charge = # valence e’s - (#bonds + #unshared e’s)
FORMAL CHARGE:
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.
B
F
F
FN
H
H
H
+ B
F
F
FN
H
H
H
Now let’s return to the compound formed betweenammonia and boron trifluoride, and determineformal charges:
Formal charges
B
F
F
FN
H
H
H
+
N 5 - (3 + 2) = 0
F 7 - (1 + 6) = 0
B 3 - (3 + 0) = 0
H 1 - (1 + 0) = 0Each Each
FORMAL CHARGE = VALENCE E'S - (# BONDS + # UNSHARED E'S)
OXIDATION NUMBERS:
“Ox #’s” are assigned or calculated based on known fixed positive and negative charges, and can be determinedby examination of the formula for the species. Oxidation numbers are useful to identify how chargeschange 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.
Sum of all chargesin compound = 0
HC lO3
+1 -2?
+1 ? -6
1 + ( ?)+ ( -6) = 0
( ?) = 6 - 1 = +5
Known ox #’s per atom
HC lO3
Valence e's
Cl 73O 18H 1
26 e's =13 prs
Cl
O
O
O
HCl
O
O
O
H
Lewis Structure
Cl
O
O
O
H
FORMAL CHARGE
Cl 7 - (3 + 2) = +2O 2[ 6 - (1 + 6) ] = -2O 6 - (2 + 4) = 0H 1 - (1 + 0) = 0
OVERALL =0
formal charge = # valence e’s - (#bonds + #unshared e’s)
Like ox #’s, the sum of all formal charges in a compoundmust equal 0.
HClO 3
+1 -2?
+1 +5 -6
Cl
O
O
O
H
-1
-1
+2
0
0
Oxidation numbers
Formal Charges
Finally, both oxidation numbers and formal chargesmust add up the same way:
For compounds, which are always electrically neutral,the sum of all oxidation numbers or the sum of allformal charges must equal zero.
For polyatomic ions, which always have a specificcharge, the sum of all oxidation numbers or the sum of all formal charges must equal the charge on the ion.
MnO 41-
Valence e's
Mn 74O 24-1 1
32 e's =16 prs
Mn
O
OO
O
1-
FORMAL CHARGE
Mn 7 - (4 + 0) = +3O 4[ 6 - (1 + 6) ] = -4
OVERALL = -1
GROUP WORK
For each of the species pictured on the next slide,please determine:
The oxidation number for each element in the formula;
The formal charge for each atom in the Lewis dot structure.
H2SO 4
S
O
O O
O
HH
Sulfuric Acid
S
O
O O
O
HH
H2SO 4 Sulfuric Acid
Correct Lewis Structure 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)
S
O
O O
O
HH
H2SO 4 Sulfuric Acid
Correct Lewis Structure
FORMAL CHARGE
S 6 - (4 + 0) = +2O 2 [ 6 - (1 + 6) ] = -2O 2 [ 6 - (2 + 4) ] = 0H 1 - (1 + 0) = 0
OVERALL =0
H2SO 4
+1 -2?
+2 +6 -8
S
O
O
O
H O
-1
-1
+2
0
0
H0
0
OxidationNumbers Formal Charge
H2SO 4 Sulfuric Acid
S
O
O O
O
HH
Expanded octet version
FORMAL CHARGE
S 6 - (6 + 0) = 0O 2 [ 6 - (2 + 4) ] = 0O 2 [ 6 - (2 + 4) ] = 0H 1 - (1 + 0) = 0
OVERALL =0
H2SO 4
+1 -2?
+2 +6 -8
S
O
O
O
H O
0
0
0
0
0
H0
0
OxidationNumbers
Formal Charge
Case #3: Molecules with odd number of electrons
This situation occurs primarily with oxides of nitrogen, NO and NO2: as ever, give the terminalatoms their octet, and let the central atom, N, handle the lone electron.
Species involving unpaired single electrons are generally not stable and react quickly to formoctet observing compounds with all e’s paired.
Single electron species are called “free radicals.”
NO 2
N, 5e's 2 O, 12 e's
17 valence e's
8prs + 1 single
NO O NO O
NO O NO O
NO O
NO OOR
NO OHYB RID:
NO N, 5e's O, 6 e's
11 valence e's
5 prs + 1 single
N O N O
Like O2, this “correct Lewis Structure” is inadequateto describe its reactivity, better done by alternate“molecular orbital theory”.
Two tasks remain to describe the molecules andpolyatomic ions for which we have drawn Lewis Structures:
• their three dimensional shapes
•the charge distribution within the species that arises from unequal sharing of bonded pairs which we will call “polarity”
Both of these properties require startingwith a correct Lewis structure, a skill we arenow ready to handle!
Molecular and Polyatomic Ion Shapes
Once a Lewis structure is drawn, the three -dimensional geometry of the species can easilybe determined by utilizing the “valence shell electronpair repulsion theory” called “VSEPR”:
“VSEPR” theory is based on the tendency of negativelycharged regions to repel each other and align as farapart as possible, resulting in predictable shapes forany covalently bonded species.
To utilize “VSEPR”, the number of regions of electron density around the central atom in the species iscounted.
Count as “one region”:
• Single Bonds
• Unshared Pairs
• Multiple bonds between same two atoms
Examples of “four regions”:
A AA
“three regions”:
A A
“two regions”:
A A
Basic Shapes predicted by VSEPR:
Two regions: Three Regions:
A
120o
trigonal planar
A
180o
linear
Bond Angles
Geometry
Four Regions: Five Regions:
Six Regions:
A
trigonal-bipyramidal
90o
120o
A
90o
octahedral
A
tetrahedral
109.5o
trigonal tetrahedral
trigonal-bipyramidal octahedral
