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Lewis Structures: Compounds and Polyatomic IonsGUIDELINES
Decide on arrangement of atoms. For most species, the element written first in the molecule or ion is the centralatom and the remainder of the atoms are grouped around it.
Hydrogen is a problem in “oxo acids” where it is writtenfirst in the formula. Ignore H, start with the next atom informula and place the H or H’s on the O or O’s.
First step:
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 bondedatoms to represent a single bond (use a “dash”!)
NH3
NH H
H
N: 1 x 5= 5 valence electrons
3H: 3 x 1= 3 valence e's
8 valence e's
8 e's / 2 = 4 electron pairs
NH H
H
Step 1Step 2
Step 3
Fourth Step
Place leftover electron pairs around “terminal” atomsto 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
P: 1 x 5= 5 valence electrons
3Cl: 3 x 7=21 valence e's
26 valence e's
26e's / 2 = 13 electron pairs
PCl Cl
Cl
PCl 3
PCl Cl
Cl
Step 1Step 2
Step 3
CHBr 3 C
H
Br
Br
BrC: 1 x 4 = 4 valence electrons
3Br: 3 x 7 = 21 valence e's
1H: 1 x 1 = 1 valence e
26 valence e's
26e's / 2 = 13 electron pairs
CBr Br
Br
H
Now let’s try some: GROUP WORK
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
NBr 3 N Br
Br
Br
N: 1 x5= 5 valence electrons
3 Br: 3 x 7 = 21 valence e's
26 valence e's
26 e's / 2 =13 electron pairs
N Br
Br
Br 13 electron pairs -3, BONDS=10 e pairs
N Br
Br
Br N Br
Br
Br
key
CH2Cl 2 C H
Cl
Cl
C: 1 x4= 4 valence electrons
2 Cl: 2 x 7 = 14 valence e's
2 H: 2 x 1 = 2 valence e's
20 valence e's
20 e's / 2 =10 electron pairs
C H
Cl
Cl 10 electron pairs -4, BONDS=6 e pairs
C H
Cl
Cl
H
H
H
C H
Cl
Cl
H
key
CH2O C
O
HH
C: 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
C
O
HH
Now let’s examine situations requiring the double bond:
HNO3 N
O
OO H
N: 1 x 5= 5 valence electrons
3 O: 3 x 6 = 18 valence e's
1H: 1x 1 = 1 valence e's
24 valence e's
24e's / 2 = 12 electron pairs
O N
O
OH
24 e's / 2 =12 electron pairs -4, BONDS=8 e pairs
O N
O
OH O N
O
OH O N
O
OH
O N
O
OH
No octetEither one
SO42- S
O
O
O
O
S: 1 x 6= 6 valence electrons
4 O: 4 x 6 = 24 valence e's
2-: 2 x 1 = 2 valence e's
32 valence e's
32e's / 2 =16 electron pairs
S
O
O
O
O
32 e's / 2 =16 electron pairs -4, BONDS=12 e pairs
S
O
O
O
O
2-
S
O
O
O
O
Be sure to include charge on finished product
GROUP WORK
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+ ClO4
1-
H3PO4 P O
O
O
P: 1 x 5 = 5 valence electrons
4 O: 4 x 6 = 24 valence e's
3 H: 3 x 1 = 3 valence e's
32 valence e's
32 e's / 2 =16 electron pairs
P O
O
O 16 electron pairs -7, BONDS=9 e pairs
O
O
H H
H
HH
H
P O
O
O
O
HH
H
P O
O
O
O
HH
H
Key:
NO21+ N OO
N: 1 x 5 = 5 valence electrons
2 O: 2 x 6 = 12 valence e's
1+: 1 x ( -1e) = -1 valence e
16 valence e's
16 e's / 2 =8 electron pairsN OO
8 electron pairs -2, BONDS=6 e pairs
N OO N OO N OO
N OO1+
key
Cl O41- Cl
O
O
O
O
Cl: 1 x7= 7 valence electrons
4 O: 4 x 6 = 24 valence e's
1-: 1 x 1 = 1 valence e
32 valence e's
32e's / 2 =16 electron pairs
Cl
O
O
O
O16 electron pairs -4, BONDS=12 e pairs
Cl
O
O
O
O
1-
Cl
O
O
O
O
key
Let’s explore the relationship between various“oxo” acids (H, Non metal element, O) and thecharge and formula of their anion relative.
Recall that acids, by definition, ionize in water tolose one or more H’s as H+. The anion leftbehind 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 behindthe 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”:
H- “Anion” + NaOH H2O + Na+ An-
Acid: HCl, HNO3 H2SO4
etc...
Cl-, NO3-, SO4
2- etc...
H- “Anion” H+ + Anion-
H2O
H2CO 3 CO 32-
O
CO OH H
O
CO O
2-
Carbonic Acid Carbonate
H3PO 4
P
O
O O
O
HH
H
Phosphoric Acid
P
O
O O
O
3-
PO 43-
Phosphate
H2SO 4
S
O
O O
O
HH S
O
O O
O
2-
SO 42-
H2SO 3
S
O
O O HH S
O
O O
2-
SO 32-
Sulfuric Acid
Sulfurous Acid
Sulfate
Sulfite
We turn next to the structure of the “oxo acids” of Cl, which are identical to those for Br and I.
HClO Hypochlorous AcidHClO2 Chlorous Acid HClO3 Chloric AcidHClO4 Perchloric Acid
Recall that Cl, Br, I and also F form acids with H, no O’s included:
HCl Hydrochloric AcidHF Hydrofluoric Acid
ClOH
ClO 1-
1-
HClO 2
ClO OH1-
ClO 21-
Hypochlorous Acid
Chlorous Acid Chlorite
Hypochlorite
HClO
ClO
ClO O
HClO 3
Cl
O
O OH Cl
O
O O
1-
ClO 31-
Chloric Acid Chlorate
HClO 4
Cl
O
O O
O
H Cl
O
O O
O
1-
ClO 41-
Perchloric Acid Perchlorate
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...
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:
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-