Lecture 21 © slg CHM 151 Lewis Structures: Molecules Ions Oxy Acids Resonance Structures TOPICS:

Lecture 21 © slg CHM 151

Lewis Structures: Molecules Ions Oxy Acids

Resonance Structures

TOPICS:

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:

NH3

CH2O

SO42-

HNO3

CHBr 3

PCl 3

NH H

H

PCl Cl

Cl

C

H

Br

Br

BrC

O

HH

N

O

OO H S

O

O

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 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

NH H

H

NH H

H

8 e's / 2 = 4 electron pairs

NH H

H

N, octet

H, duetStep 4:

No Step 5 needed

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

26 e's / 2 = 13 electron pairs -3, BONDS= 10 e pairs

PCl Cl

Cl

PCl Cl

Cl

PCl Cl

Cl

Step 4

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


CBr Br

Br

H

26 e's / 2 = 13 electron pairs -4, BONDS= 9 e pairs

CBr Br

Br

H

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


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


C

O

HH

Now let’s examine situations requiring the double bond:

12 e's / 2 = 6 electron pairs -3, BONDS=3 e pairs

C

O

HH C

O

HH C

O

HH C

O

HH

No octet

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


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


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


3 H: 3 x 1 = 3 valence e's

32 valence e's


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


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


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

HNO 3

O

N OOH

O

N OO

NO 31-

1-

Nitric AcidNitrate

HNO 2

O

NOH

O

NO

NO 21-

1-

Nitrous Acid Nitrite

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...

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:

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-

Lecture 21 © slg CHM 151 Lewis Structures: Molecules Ions Oxy Acids Resonance Structures TOPICS:

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