Ch. 13 - Molecular Structure
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Transcript Ch. 13 - Molecular Structure
Molecular
Structure
Part I: Lewis Diagrams
CH. 6 – MOLECULAR STRUCTURE
I
II
III
I. Lewis Diagrams
(p. 170 – 175)
Molecular Compounds:
Are
made of nonmetals
Nonmetals
have high
electronegativity, so they do NOT
release their electrons.
Two
nonmetals share some of their
valence electrons (in bonds) to
Shared Valence
achieve full octets.
The
atoms are CO-valent-ly bonded!
I. Lewis Dot Structures
Are designed to show the
placement of valence
electrons in covalently
bonded compounds.
Covalent
compounds
= Molecules
= Compounds with shared electrons
I. Lewis Dot Structures
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.)
I. Lewis Dot Structures
Step
1: Add up all valence
electrons in the compound.
NO2
(e- tally)
1 N × 5e- =
5e2 O × 6e- = + 12e17e-
If
the formula carries a CHARGE,
add or subtract electrons
accordingly. NO2- = 17e- +1e = 18e-
I. Lewis Dot Structures
Step
1: Add up all valence
electrons in the compound.
If
the formula carries a CHARGE,
add or subtract electrons
accordingly:
Negative
charge: ADD electrons
Positive charge: SUBTRACT
electrons
I. Lewis Dot Structures
Step
1: Add up all valence
electrons in the compound.
(This
1 2
is the “electron tally.”)
1 C × 4e- = 4eCF4 4 F × 7e- = +28e32e
3 4 5 6 7 8
I. Lewis Dot Structures
Step
1: Add up all valence
electrons in the compound.
NH3
1 2
3 4 5 6 7
1 N × 5e- = 5e3 H × 1e- = + 3e8e
8
I. Lewis Dot Structures
Step
1 2
1: Add up all valence
electrons in the compound.
- = 6e21
S
×
6e
SO2 2 O × 6e- = 12eCharge = +2e3 4 5 6 7 8
20e-
I. Lewis Dot Structures
Step
1: Add up all valence
electrons in the compound.
NOTE: Charged dot structures
2SO2
(like SO22- or PO43-) will
always be drawn inside
square brackets!
1 2
3 4 5 6 7 8
I. Lewis Dot Structures
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.)
F
F C F
F
I. Lewis Dot Structures
A)
The first element in the
formula goes in the center.*
*BUT NOT HYDROGEN
H
cannot be the center atom.
If H is first in the formula, skip it.
the second element in the
center.
Treat H as the second element.
H2S
Put
H S H
I. Lewis Dot Structures
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.)
F
F C F
F
I. Lewis Dot Structures
Step
3: Calculate
remaining electrons.
A)
Each bond (line) represents
two electrons that are shared
between two atoms.
Number
B)
of bonds x 2e- = # e- in bonds
Subtract the bonded e- from
the total.
I. Lewis Dot Structures
Step
3: Calculate
remaining electrons.
CF4
= 32e4 bonds × 2e- = - 8e24eThese electrons
will appear as dots.
F
F C F
F
I. Lewis Dot Structures
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).
CO-valence!
I. Lewis Dot Structures
A)
Octet rule: every
atom needs 8 electrons
in its valence.*
*BUT NOT HYDROGEN.
H2S
can only have two
electrons (because 1s is its
only orbital)
H S H
When it has a bond, H is
“full.”
Both H’s are full.
H
No dots for H!
I. Lewis Dot Structures
A)
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!
I. Lewis Dot Structures
Step
4: Distribute remaining
electrons.
CH4
= 32e4 bonds × 2e- = 8eF
24eF C F
These electrons
will appear as dots.
F
I. Lewis Dot Structures
Step
4: Distribute remaining
electrons.
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
F
F C F
F
I. Lewis Dot Structures
Step
4: Distribute remaining
electrons.
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
F
F C F
F
I. Lewis Dot Structures
Step
4: Distribute remaining
electrons.
24e- dots:
We used exactly 24e-!
Octet
check:
Do all atoms* have 8 v.e-.?
YES! We win.
F
F C F
F
I. Lewis Dot Structures
Step
4: Distribute remaining
electrons.
NH3
= 8e3 bonds × 2e- = -6eH
2eN
H
H
These electrons
will appear as dots.
I. Lewis Dot Structures
Step
4: Distribute remaining
electrons.
We used exactly 2e-!
Octet
check:
H
Do all atoms* have 8 v.e-.?
H N H
N has 8
H
has 2
We
win.
I. Lewis Dot Structures
Step
4: Distribute remaining
electrons.
C)
If you run out of electrons:
Share
more!
2
e- short = 1 double bond
4
e- short = 2 double bonds
or 1 triple bond
I. Lewis Dot Structures
Let’s
try CO2
(e- tally)
1 C × 4e- =
4e2 O × 6e- = + 12e16e-
16e- (e- tally)
- 4e- (in 2 bonds)
12e- (as dots)
But 12 dots won’t be enough.
We’re 2 epairs short.
That means
we need 2
more bonds.
(4 bonds total)
O C O
I. Lewis Dot Structures
Let’s
try CO2
(e- tally)
1 C × 4e- =
4e2 O × 6e- = + 12e16eHere we go again:
16e- (e- tally)
- 8e- (in 4 bonds)
8e- (as dots)
We’re 2 epairs short.
That means
we need 2
more bonds.
(4 bonds total)
O C O
I. Lewis Dot Structures
Let’s
try CO2
(e- tally)
1 C × 4e- =
4e2 O × 6e- = + 12e16eHere we go again:
16e- (e- tally)
- 8e- (in 4 bonds)
8e- (as dots)
Octet check!
Remember,
each bond
counts as 2e-
O C O
4 dots, 2 bonds
4 bonds
4 dots, 2 bonds
WIN
I. Lewis Dot Structures
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.
4. Distributing Electrons
Make
an Expanded Octet
1)
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.
I. Lewis Dot Structures
Let’s
try SeCl4
(e- tally)
1 Se × 6e- =
6e4 Cl × 7e- = + 28e34e-
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.
Cl
Cl Se Cl
Cl
4. Distributing Electrons
Make
an Expanded Octet
If
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.
I. Lewis Dot Structures
Let’s
try PF5
What’s your e- tally?
1 P × 5e- =
5e5 F × 7e- = + 35e40eHow 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.
F
F P F
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:
ClO4
1 Cl × 7e- = 7e4 O × 6e- = 24eCharge
+ 1e32e32e- 8e24e-
O
O Cl O
O
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.
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.
D. Resonance Structures
SO3 has
1 double bond which could be
in any of these three places:
O
O S O
O
O S O
O
O S O
HOMEWORK:
Molecular
1st
Geometry WS
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.