IIIIII Molecular Structure Part I: Lewis Diagrams CH. 6 – MOLECULAR STRUCTURE LPChem:Wz (after...

I II III

Molecular StructurePart I: Lewis DiagramsCH. 6 – MOLECULAR STRUCTURE

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:Wz (after Johannesson)

I. Lewis Diagrams(p. 170 – 175)

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Molecular Compounds:

Are made of nonmetalsNonmetals have high

electronegativity, so they do NOT release their electrons.

Two nonmetals share some of their valence electrons (in bonds) to achieve full octets.

The atoms are CO-valent-ly bonded!

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Shared Valence

I. Lewis Dot Structures

Are designed to show the placement of valence electrons in covalently bonded compounds.Covalent compounds

= Molecules

= Compounds with shared electrons

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Step 1: Add up all valence electrons in the compound.This is the total number of

electrons available to bond the molecule.

(Remember: inner electrons do not participate in bonding.)

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Step 1: Add up all valence electrons in the compound.NO2

If the formula carries a CHARGE, add or subtract electrons accordingly.

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(e- tally)

1 N × 5e- = 5e-

2 O × 6e- = + 12e-

17e-

NO2- = 17e-

+1e -

= 18e-


Step 1: Add up all valence electrons in the compound.If the formula carries a CHARGE,

add or subtract electrons accordingly:Negative charge: ADD electronsPositive charge: SUBTRACT

electrons

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Step 1: Add up all valence electrons in the compound.(This is the “electron tally.”)

CF4

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1 C × 4e- = 4e-

4 F × 7e- = +28e-

32e-1 2 3 4 5 6 7 8


Step 1: Add up all valence electrons in the compound.

NH3

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1 N × 5e- = 5e-

3 H × 1e- = + 3e-

8e-1 2 3 4 5 6 7 8



SO22-

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1 S × 6e- = 6e-

2 O × 6e- = 12e-

Charge = +2e-

20e-1 2 3 4 5 6 7 8



SO22-

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NOTE: Charged dot structures (like SO2

2- or PO43-) will

always be drawn inside square brackets!

1 2 3 4 5 6 7 8


Step 2: Draw a skeleton structure.A)The first element in the

formula goes in the center.*B)The second element goes

around the first element.Left, right, top, bottom(Don’t choose weird angles.)

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C

F

F

F

F


A) The first element in the formula goes in the center.*

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S HH

*BUT NOT HYDROGENH cannot be the center atom.If H is first in the formula, skip

it.Put the second element in the

center.Treat H as the second element.

H2S


Step 2: Draw a skeleton structure.C) Draw a bond (line)

connecting each secondary atom to the center.

(Do not connect secondary atoms to each other.)

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CF

F

F

F


Step 3: Calculate remaining electrons.A) Each bond (line) represents

two electrons that are shared between two atoms.Number of bonds x 2e- = # e- in bonds

B) Subtract the bonded e- from the total.

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Step 3: Calculate remaining electrons.CF4

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CF

F

F

F

= 32e-

4 bonds × 2e- = - 8e-

24e-

These electrons will appear as dots.


Step 4: Distribute remaining electrons.A) Octet rule: every atom needs

8 electrons in its valence.*A bond is two valence electrons

that count as valence for both elements involved (at the same time).

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CO-valence!

I. Lewis Dot StructuresA) Octet rule: every

atom needs 8 electrons in its valence.*

*BUT NOT HYDROGEN.H can only have two

electrons (because 1s is its only orbital)

When it has a bond, H is “full.”

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S HH

H2S

Both H’s are full.No dots for H!

I. Lewis Dot StructuresA) Octet rule: every atom needs

8 electrons in its valence.**But not hydrogen.

B) Add electron dots to atoms as needed:You must use up all the e-’s

available.You may NOT use more e-’s than

that!

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Step 4: Distribute remaining electrons.CH4

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CF

F

F

F

= 32e-

4 bonds × 2e- = 8e-

24e-



Step 4: Distribute remaining electrons.

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CF

F

F

F

24e- dots:C: 4 bonds x 2e- = 8 v.e-.

No dots needed on C

F: 1 bond x 2e- = 2 v.e-.

6 dots needed on each F



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24e- dots:C: 4 bonds x 2e- = 8 v.e-.

No dots needed on C

F: 1 bond x 2e- = 2 v.e-.

6 dots needed on each F

CF

F

F

F



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CF

F

F

F

24e- dots:We used exactly 24e-! Octet check:

Do all atoms* have 8 v.e-.?

YES! We win.


Step 4: Distribute remaining electrons.NH3

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N

H

HH

= 8e-

3 bonds × 2e- = -6e-

2e-


N

H

HH



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We used exactly 2e-! Octet check:

Do all atoms* have 8 v.e-.? N has 8 H has 2 We win.


Step 4: Distribute remaining electrons.C) If you run out of electrons:

Share more!2 e- short = 1 double bond4 e- short = 2 double bonds

or 1 triple bond

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Let’s try CO2

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(e- tally)

1 C × 4e- = 4e-

2 O × 6e- = + 12e-

16e-

16e- (e- tally)- 4e- (in 2 bonds)12e- (as dots)

C OO

But 12 dots won’t be enough.

We’re 2 e- pairs short.

That means we need 2 more bonds. (4 bonds total)


Let’s try CO2

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(e- tally)

1 C × 4e- = 4e-

2 O × 6e- = + 12e-

16e-

Here we go again:16e- (e- tally)- 8e- (in 4 bonds) 8e- (as dots)

C OO

We’re 2 e- pairs short.

That means we need 2 more bonds. (4 bonds total)


Let’s try CO2

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(e- tally)

1 C × 4e- = 4e-

2 O × 6e- = + 12e-

16e-

Here we go again:16e- (e- tally)- 8e- (in 4 bonds) 8e- (as dots)

C OO

Octet check! Remember,

each bond counts as 2e-

4 dots, 2 bonds 4 bonds 4 dots, 2 bonds

WIN


Step 4: Distribute remaining electrons.D) If you have left-over electrons:

Make an Expanded Octet This is a fancy name for “more than 8

electrons on the central atom.”

Central atom must be at an energy level ≥ 3. Expanded octets cannot exist in the 1st or 2nd E.L.

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4. Distributing Electrons

Make an Expanded Octet1) If there are simply left-over

electrons, put them on the central atom.

No multiple bonds allowed!The limit for expanded octets is 12

electrons total.ONLY the central atom gets more

than 8 electrons.

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Let’s try SeCl4

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(e- tally)

1 Se × 6e- = 6e-

4 Cl × 7e- = + 28e-

34e-

34e- (e- tally)- 8e- (in 4 bonds)26e- (as dots)

But 26 dots won’t fit!

We’ve got 1 extra e- pair.

That pair goes on Se. (Se gets 10 e- total)Each Cl gets an octet.

Se

Cl

Cl

Cl

Cl

4. Distributing Electrons

Make an Expanded OctetIf there are more than 4 atoms of

the secondary element, they still bond to the central atom.

Still no multiple bonds allowed!The limit is 6 secondary atoms bonded

to the central atom (Still 12 e- total)Still ONLY the central atom gets more

than 8 electrons.

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Let’s try PF5

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What’s your e- tally?

1 P × 5e- = 5e-

5 F × 7e- = + 35e-

40e-

How many e- in bonds and how many in dots?

40e- (e- tally)- 10e- (in 5 bonds) 30e- (as dots)

Even before doing a tally we can tell that P must expand its octet in order to bond 5 F atoms.

The 5 Fs are distributed in a pentagon.(If there were six, it would be a hexagon.)

Octet check: P gets 10 e- total Each F gets an

octet.

P

F

FF

F F

5. Finishing:

A) If your dot structure was for an ion:Place square brackets around

the ion and write the charge outside the brackets:

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ClO4-

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1 Cl × 7e- = 7e-

4 O × 6e- = 24e-

OO Cl O

O

Charge + 1e-

32e-

32e-

- 8e-

24e-

5. Finishing

B) Resonance Structures Some molecules with double bonds

can’t be correctly represented by a single Lewis diagram.

The actual structure is an average of all the possibilities.

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B) Resonance Structures

Truth: the electrons are evenly distributed, but there isn’t a way to draw that in a Lewis dot structure. (Half a bond? Half a dot?)

Show all possible structures separated by a double-headed arrow.

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D. Resonance Structures LP

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OO S O

OO S O

OO S O

SO3 has 1 double bond which could be

in any of these three places:

HOMEWORK:

Molecular Geometry WS1st two columns ONLY!(We’ll learn the others next time.)

Skyward!Do the worksheet first– it will give

you the answers to the Skyward problems.

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IIIIII Molecular Structure Part I: Lewis Diagrams CH. 6 – MOLECULAR STRUCTURE LPChem:Wz (after...

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Transcript of IIIIII Molecular Structure Part I: Lewis Diagrams CH. 6 – MOLECULAR STRUCTURE LPChem:Wz (after...