Covalent Electrostatic Interactions
Chapter 7.b:
Increase attractive energies by allowing
electrons to be attracted to TWO nuclei
BUT… two atoms can only overlap and share
electrons if they have partially empty orbitals
Lewis Diagrams
&
VSEPR Model
-e1
-e2
Attract
Attract
+ZB
+ZA
1s1
1s1
Chem 111
Dr. Gentry
For SiH4 Molecule
Silicon
Si (Z=14)
Si has room
for 4 more
electrons
H’s fill those
spots by
sharing e’s
Si:
1s2 2s2 2p6 3s2 3p2
4 x H:
1s1
• Valence Electrons
Electrons with highest principal quantum number (n)
(Electrons in upper most ns and np orbitals)
• Core Electrons
3p
Valence Electrons
3s
(reactive)
3s23p2
Electrons in shells having lower quantum-number n values
Valence electrons are farthest outside – most
likely to overlap with other atoms and form bonds
2p
1s22s22p6 3s23p2
Si:
Core Electrons
2s
(non-reactive)
1s22s22p6 = [Ne]
core
electrons
1s
Lewis Dot Structures provide simple, but useful,
way of representing chemical reactions
- Show where valence electrons are on each atom
valence
electrons
Atomic Lewis Electron-Dot Structures
• Valence electrons shown as dots around atom
- Electrons exist as pairs of electrons in different orbitals
- 4 pairs of dots ⇒ 4 valence orbitals
(s, px, py, pz … or hybrids)
electron
transfer
Li
• Ionic
+
- 4 pairs
Li+
F
2s1
Li: [He]
F: [He] 2s22p5
• Covalent
H
+
O
+
H
+
F-
⇒ 8 possible electrons
• Hund’s rule: put one electron in each of 4 pairs before putting
2nd electron in each pair
Li+:
[He]
F‒: [He] 2s22p6
H O H
Li
Be
2s1
2s2
B
C
2s22p1
2s22p2
N
2s22p3
O
F
Ne
2s22p4
2s22p5
2s22p6
shared
electron pair
1
Covalent Bonds
Bonds Formed by Overlapping Orbitals
Each Hydrogen atom starts with unfilled 1s orbital
Both would like to fill orbital ( 1s1 1s2 )
• One electron from each atom combine to complete
unfilled pairs (share half-filled orbitals)
But neither is strong enough to take electron from other
e-
• Requires orbitals to overlap with one another
e-
H
2 H atoms:
H
1s1
nonbonding
pair of e–’s
1s1
Covalent bond
N
Share electrons so both can fill orbitals at same time
Requires half-filled orbitals that overlap
H
H2 molecule:
1s2
+ 3x
H
H
N
H
ee-
H
H N H
H
1s2
Hybridization
How can the bonding in CH4 be explained?
C
H
y
H
Would appear that only 2 halffilled orbitals available to
overlap with H orbitals.
Standard Carbon
Configuration
bonding
pair of e–’s
Problem with 4 bonds in CH4 if forming
bonding overlaps with 2s and 2p orbitals
H
H
H
2px
2py
2pz
?
x
+
3d empty 2p orbital could
accept a 3d H electron, but 2s
orbital already filled so cannot
share with a 4th H.
4x H
z
2s
Problems:
- 3 atoms bond to 2p orbitals,
90º from each other
- 4th atom unable to overlap
with s orbital
- different energies for each
How do we bond 4 H’s to C?
Hybridization
sp3 Hybridization
s + 3xp
Mix 2s + (3 x 2p) orbitals together to form 4 new sp3 hybrids
4 x sp3
4 old atomic orbitals give 4 new hybrid orbitals (4 x sp3)
2p
2sp3
2s
1s
2px
2py
+
2pz
mix to
make new
orbitals
1s
2s
Standard Carbon
Configuration
for lone atoms
Hybrid Carbon
Configuration
for atoms in molecules
2
Constructing Lewis Structures for Molecules
Forming CH4 from sp3 Hybrids
1) Find total number of valence electrons in molecule
– assign any ionic charge to central atom
H
2) Identify central atom and draw on paper w/ electrons
⇒ one with fewest valence e–’s (but not H)
could be transition element or heavy halogen
=
C
3) Match other atoms with half-filled pairs on central atom
– add hydrogens and halogens last
– exceed octet rule (if needed) if central atom beyond 2nd row
H
H
H
4) If central atom still has unfilled pairs, move e–’s from
side atom and form multiple bonds
Tetrahedral Shape
5) Move remaining e–’s if unfilled pairs on side atoms
(angle between bonds = 109.5°)
– move from one atom to another if needed for octets
– never exceed 2 e–’s around H … or 8 if 2nd row atom
Lewis Dot Diagrams
NF3 :
val. elec. :
N
3x
F
1x
N
H2 O :
(3 x 7) + (1 x 5) = 26
F― N ―F
F
F
O
O central atom
(can’t be H)
10 lone pairs = 20
3 bonds = 6
Total = 26 val.elec’s
Lewis Dot Diagrams
1x O
2x H
val. elec. :
F― N ―F
match e– from F’s
to unfilled pairs
on N
N central atom
since fewest
val. elec’s
Lewis Dot Diagrams
(2 x 1) + (1 x 6) = 8
O ―H
O ―H
H
H
match H’s with
unfilled pairs on O
(H can only have 1
pair since only 1sx)
2 lone pairs = 4
2 bonds = 4
Total = 8 val. elec’s
Hybridization
- Polyvalent Ions Mix 2s + (3 x 2p) orbitals together to form 4 new sp3 hybrids
NH4+ :
1x
N+
4x H
4 old atomic orbitals give 4 new hybrid orbitals (4 x sp3)
[ionic charge assigned to central atom]
val. elec. :
N+
N+ central atom
(1 x 4) + (4 x 1) = 8
2p
H
H
H – N+ – H
H– N –H
H
H
match e– from H’s
to unfilled pairs
on N+
0 lone pairs = 0
4 bonds = 8
Total = 8 val.elec’s
+
2sp3
2s
1s
Standard Carbon
Configuration
for lone atoms
1s
Hybrid Carbon
Configuration
for atoms in molecules
3
Moving Beyond Octet Rule
Moving Beyond Octet Rule
( less than 8 e‒ ’s around central atom )
( more than 8 e‒ ’s around central atom )
• 3d Column Elements often finish with 6 instead of 8 electrons
All electrons are paired, but with an empty orbital
p
sp3
s
4 hybrid
sp3 orbitals
H
H C H
H
CH4 →
empty
p orbital
p
p
3 hybrid
sp2 orbitals
H B H
H
BH3 →
sp2
s
• Molecules with central atoms in third row or beyond can go
beyond 8 electrons if necessary :
1x
Br
3x
F
28
valence
electrons
F Br F
F
• Put extra electrons around
central bromine
3d
d
• Add d orbitals to hybrids to
get additional sp3d or sp3d2
orbitals
3p
sp3d2
3s
Lewis Dot Diagrams
- Double Bonds-
• Single
Bonds
2x O
CO2 : 1 x C
val. elec. :
(1 x 4) + (2 x 6) = 16
C
O - C- O
C = central
atom
Match e– from O’s
to unfilled pairs
on C
C still has
unfilled
pairs
O =C = O
• Double
Bonds
2(
C
2(
)+
C
6(
H)
) + 4(
H)
H
H C
H
H
C H
H
H
H
C C
H
H
H C
C H
Move 2nd e–’s
from O’s to form
double bonds
4 lone pairs = 8
4 bonds = 8
Total = 16 val.elec’s
Moving Electrons From One
Atom to Another
• Method of last resort
• Triple
Bonds
2(
C
)+
2(
H)
Valence-Shell Electron-Pair Repulsion
(VSEPR Model)
• Use Lewis structures to predict shape of molecules
If 2nd row element, cannot have more than 8
But wants full 8 if possible
Wants all orbitals to be filled with pairs
• For geometry around the central atom
- Side groups try to get as far apart as possible
- Side groups could be side atoms
OR could be nonbonding e– pairs
O3
O
O - O- O
O = central
atom
Match e– from
other O’s to
unfilled pairs on
central O
O‒O= O
Move an e‒
off of central O
on to O on left
2 side groups
3 side groups
4 side groups
4
Total #
Electron
Groups
(LP+BP)
Hybridization of
Central
Atom
2
sp
(180°)
A
2
linear
CO2
C2H2
3
sp2
(120°)
A
2
angular (bent)
NO2-
3
trigonal planar
BF3
sp3
(109.5°)
2
angular (bent)
H2O
4
A
3
pyramidal
NH3
4
tetrahedral
CH4
2
linear
XeF2
3
T-shaped
ICl3
4
seesaw
SF4
5
trigonal
bipyramidal
PF5
5
6
Hybridized
Orbital
Geometry
sp3d
(90°, 120°,
180°)
#
Bonding
Groups
(BP)
A
sp3d2
(90°, 180°)
A
Molecular
Geometry
Examples
4
square planar
XeF4
5
square pyramidal
BrF5
6
octahedral
SF6
Hybridization
Mix 2s + (3 x 2p) orbitals together to form 4 new sp3 hybrids
4 old atomic orbitals give 4 new hybrid orbitals (4 x sp3)
for C on a CH4 molecule
2p
2sp3
2s
1s
1s
Standard Carbon
Configuration
for lone atoms
Four Electron Groups, sp3 : Electron groups
point to corners of a tetrahedron
Hybrid Carbon
Configuration
for atoms in molecules
Examples of Tetrahedral Shapes
tetrahedral
p
CH4
sp3
s
4 bonding pairs
trigonal pyramid
NH3
3 bonding pairs
1 lone pair
bent
H2O
2 bonding pairs
2 lone pairs
Hybridization
p
p
sp1
sp1
sp2
s
Three Electron Groups, sp2: Electron groups lie in
same plane and form a flat triangle (trigonal planar)
p
A
2 groups
s
linear
s
p
# of Electron
Groups
CH2O
p
sp2
A
3 groups
(3 bonding pairs)
p
sp2
SO2
trigonal planar
(2 bonding pairs,
1 nonbonding pair)
bent
trigonal planar
p
A
sp3
s
4 groups
sp3
tetrahedral
5
Two Electron Groups, sp1: Electron groups
point in opposite directions
p
d-Orbital Hybridization
(only allowed for elements in 3d row and beyond)
p
sp1
d
d
s
sp3d1
p
sp3d1
s
linear
trigonal bipyramidal
d
sp3d2
d
p
sp3d2
s
octahedral
Five Electron Groups, sp3d1 : Electron groups
point to the corners of a trigonal bipyramid
d
p
Examples of sp3d1 Trigonal Bipyramidal Shapes
d
sp3d1
s
trigonal bipyramid
seesaw
SF4
PCl5
4 bonding prs
1 lone pair
5 bonding prs
T-shaped
Six Electron Groups, sp3d2 : Electron groups
point to the corners of a regular octahedron
d
p
I3–
3 bonding prs
2 lone pairs
2 bonding prs
3 lone pairs
Examples of sp3d2 Octahedral Shapes
octahedral
d
sp3d2
linear
ClF3
SF6
6 bonding pairs
s
square pyramidal
SbCl5
5 bonding pairs
1 lone pair
square planar
XeF4
4 bonding pairs
2 lone pairs
6
Double Bonds From sp2 Systems
Mixture of s and p orbitals
sp1
A
linear
H
C
H
O
pz
A
sp2
trigonal planar
p
C: 3 x sp2 orbitals
sp3
tetrahedral
sp3d1
Overlap 1
sp3d2
A
trigonal
bipyramidal
pi (π) bond
rather than 1 single & 1 double bond
• The average is called a resonance hybrid.
O
O
O
O
O
–
N
O
O
Solving the Lewis structure required moving
an electrons from N to one of the O’s
Change in the Formal Charge
O
Formal Charge on Each Atom
# of v.e–’s
on free atom
–
O
O
(does atom lose / gain / stay even when sharing?)
–
O
O
Definition: resonance structures have same physical
arrangement of atoms but with different
interchangeable arrangements of electrons
• The nitrate ion, NO3–, has three equivalent oxygen
atoms, and its electronic structure is a resonance
hybrid of three electron-dot structures. Draw them.
O
sp2
H
Molecule has two 1.5 bonds
O
Resonance Structures
O
sp2
• When multiple structures can be drawn, the actual structure
is an average of all possibilities.
• The correct answer is that both are correct,
but neither is correct by itself.
N
sp2
O
C
OR ?
O
O
O
sp2
Resonance Structures
• Where is the double bond formed in O3?
O
pz
sp2
PLUS overlap pz from each
Resonance Structures
O
from each atom
sigma (π) bond
octahedral
O
sp2
pz
H
C=O Double Bond
Inclusion of d orbitals
O
sp2
s
sp2
sp2
A
p
C
+ 1 standard 2p
A
Cve: 2s2 2p2
sp2
N
O
O
Formal Charge =
( # of Valence e- )
 # of bonding e- 
 −
− 


2


( # of nonbonding e- )
minus # of v.e–’s
on atom in molecule
7
Formal charge can be used to determine
the preferred Lewis structure.
Formal Charge on Each Atom
 # of bonding e- 
 −  # of nonbonding e- 
Formal Charge =  # of Valence e-  − 
 

 

2


# of v.e–’s
on free atom
 # of bonding e- 
 −  # of nonbonding e- 
Formal Charge =  # of Valence e-  − 
 


 
2


# of v.e–’s
on free atom
# of v.e–’s
on atom in molecule
–
hydrogen isocyanide
carbon monoxide
C
O
4 v.e.
6 v.e.
C
δ+
# of v.e–’s
on atom in molecule
–
C
O
hydrogen cyanide
H C
N H
N
preferred
C: 4 – (6/2) – 2 = – 1
H: 1 – (2/2) – 0 = 0
H: 1 – (2/2) – 0 = 0
O: 6 – (6/2) – 2 = + 1
C: 4 – (3/2) – 2 = –1
C: 4 – (8/2) – 0 = 0
N: 5 – (4/2) – 0 = +1
N: 5 – (6/2) – 2 = 0
δ–
δ+ = partial
positive charge
H–Cl
Polar
Covalent
M+
:Y–
Ionic Bond
(full transfer)
Electron cloud
δ+
X :Y
Polar Covalent
(partial transfer)
δ+
Formal Charge:
Artificially assigns electrons evenly to both sides of bond
H = 0, Cl = 0
Oxidation Number:
Artificially assigns all electrons to stronger atom in bond
H = +1, Cl = –1
Partial Charge
The true distribution of electrons on each atom
H = + 0.3,
Cl = – 0.3
δ–
Nonpolar Covalent
(even distribution)
δ–
H–Cl
Polar
Covalent
(HCl)
Y:Y
δ+ = partial
positive charge
Electron cloud
Electrons attracted more strongly to Cl than H
… But not enough for complete transfer
… Average e– position closer to Cl than to H
Polar Bonds vs. Polar Molecules
H2O
CCl4
Bent
Tetrahedral
Cl
O
H
Cl
H
C
Cl
µ = 1.85 D
δ-
δ+
H-O bonds
are polar
Molecule is
polar
δ+
Cl
µ=0D
Each C-Cl bond
is polar
But bonds cancel
so molecule is
nonpolar
8