Transcript H 2 - Milwaukie High

Molecular Geometry
and
Bonding Theories
The properties of a molecule depend on its shape and
and the nature of its bonds. In this unit, we will
discuss three models.
(1) a model for the geometry of molecules
-- valence-shell electron-pair
repulsion (VSEPR) theory
(2) a model about WHY molecules form bonds
and WHY they have the shape they do
-- valence-bond theory
(3) a model of chemical bonding that deals with
the electronic structure of molecules
-- molecular orbital (MO) theory
bond angles: the angles made by the lines joining
the nuclei of a molecule’s atoms
carbon dioxide
CO2
180o
methane
CH4
109.5o
formaldehyde
CH2O
120o
VSEPR
electron domain: a region in which at least two
electrons are found
-- they repel each other because… they are all (–)
bonding domain: 2-to-6 e– that are shared by two
atoms; they form a… covalent bond
nonbonding domain: 2 e– that are located on a single
atom; also called a… lone pair
For ammonia, there are
NH3
..
three bonding domains and
H– N –H
one nonbonding domain.
..
H
N
H
Domains arrange themselves so
H
H
as to minimize their repulsions.
..
The electron-domain geometry is one
of five basic arrangements of domains.
N
H
H
H
-- it depends only on the total # of e– domains,
NOT the kind of each domain
The molecular geometry describes the
orientation of the atoms in space.
-- it depends on how many of each
kind of e– domain
..
Total # of Electron-Domain
Domains
Geometry
2
3
4
5
6
linear
Possible Molecular Geometries
linear (CO2)
trigonal
planar
trigonal planar (BF3), bent (NO2)
tetrahedral
tetrahedral (CH4),
trigonal pyramidal (NH3), bent (H2O)
trigonal
bipyramidal
trig. bipyramidal (PCl5), linear (XeF2)
seesaw (SF4), T-shaped (ClF3)
octahedral
octahedral (SF6), sq. pyr. (BrF5),
square planar (XeF4)
“atoms – axial”
To find the electron-domain geometry (EDG) and/or
molecular geometry (MG), draw the Lewis structure.
Multiple bonds count as a single domain.
Predict the EDG and MG of each of the following.
SeCl2
20 e–
..
..
.. .. ..
Cl–Se–Cl
.. .. ..
..
..
18 e–
]
–
..
O3
[
..
..
26 e–
..
SnCl3
–
.. .. ..
Cl–
.. Sn–Cl
..
Cl
..
..
O–
..
EDG: tetrahedral
MG: trig. pyramidal
EDG: trig. planar
MG: bent
EDG: tetrahedral
MG: bent
IF5
EDG: trig. planar
MG: trig. planar
..
..
..
..
..
..
..
..
34 e–
2–
]
F
..
..
SF4
[
..
24
e–
(res.)
CO32–
..
O– C= O
..
O
..
..
.. F ..
..F– S –F
..
..F
..
EDG: octahedral
MG: sq. pyramidal
I
42 e–
EDG: trig. bipyr.
MG: seesaw
28 e–
..
..
..
ClF3
..
..
..
..F
.. .. ..
..F– Cl –F
..
..F
EDG: trig. bipyr.
MG: T-shaped
..
..
..
[
–
]
..
Cl
..
..
..
36 e–
..
ICl4–
Cl
.. ..
Cl–
.. I –Cl
..
EDG: octahedral
MG: sq. planar
For molecules with more than one central atom,
simply apply the VSEPR model to each part.
Predict the EDG and MG around the three interior
atoms of ethanoic (acetic) acid.
CH3COOH
PORTION
EDG
MG
H O
..
H–C–C–O–H
..
H
–CH3
tetra.
tetra.
C=O
trig. plan.
trig. plan.
..
tetra.
bent
–OH
..
Nonbonding domains are attracted to only one nucleus;
therefore, they are more spread out than are bonding
domains. The effect is that nonbonding domains
(i.e., “lone pairs”) compress bond angles. Domains for
multiple bonds have a similar effect.
e.g., the ideal bond angle for the tetrahedral
EDG is 109.5o
H
H
H2O
N
O
..
C
H
109.5o
H
H
H
H
COCl2
124.3o
H
O=C
124.3o
107.0o
104.5o
..
Cl
..
..111.4
Cl
..
..
NH3
o
..
CH4
H
EDG = trig. plan.
ideal = 120o
Polarity of Molecules
A molecule’s polarity is determined by its overall
dipole, which is the vector sum of the dipoles of
each of the molecule’s bonds.
Consider CO2 v. H2S...
bond dipoles
O=C=O
bond dipoles
S
H
H
overall dipole = zero
overall dipole =
(NONPOLAR)
(POLAR)
Classify as polar or nonpolar:
BCl3
..
..
..
Cl
B
..
..
26
e–
..
PCl3
.. .. ..
Cl–
.. P–Cl
..
Cl
..
..
24 e–
Boron can be extracted by
the electrolysis of molten
boron trichloride. Boron is
an essential nutrient for
plants, and is also a primary
component of control rods
in nuclear reactors.
polar
nonpolar
Valence-Bond Theory: merges Lewis structures
w/the idea of atomic orbitals
(2s, 3p, etc.)
Lewis theory says… covalent bonding occurs when
atoms share electrons
V-B theory says… covalent bonding occurs when
valence orbitals of adjacent
atoms overlap; then, two e–s
of opposite spin (one from
each atom) combine to form
a bond
V-B theory is like Velcro: “No overlap, no bond.”
Consider H2, Cl2, and HCl...
(unpaired e– is in 1s orb.)
H
Cl
[Ne]
H2
(unpaired e– is in 3p orb.)
Cl2
HCl
= orbital overlap region
(responsible for bond)
There is always an optimum distance between two
bonded nuclei. At this optimum distance,
attractive (+/–) and repulsive (+/+ and –/–) forces
balance.
Energy
0
H2 molecule
optimum dist.
(min. PE)
H–H distance
Hybridized Orbitals
V-B theory can’t explain some observations about
molecular compounds without the concept of
hybridized orbitals.
i.e., covalent (NM/NM) compounds
:F Be F:
: :
BF3
: :
BeF2
: :
In…
(LS)
…the central atom …but is ACTUALLY
SHOULD be… hybridized to be…
2s
2p
sp
2s
2p
sp2 2p
2 hyb. sp orbs.
2 unhyb. 2p orbs.
3 hyb. sp2 orbs.
1 unhyb. 2p orb.
2p
B F:
H2O
H O:
H
2s
2p
4 hyb. sp3 orbs.
sp3
:
CH4
H
H C H
H
…the result
then being…
: :
PF5
2s
P F:
XeF4
:F:
:F Xe F:
:F:
: :
:F Xe F:
:
: :
XeF2
3s
5s
2p
3p
5p
4 hyb. sp3 orbs.
sp3
3d
5d
sp3d
sp3d
3d
5 hyb. sp3d orbs.
4 unhyb. 3d orbs.
5d
5 hyb. sp3d orbs.
4 unhyb. 5d orbs.
:
: :
: :
:
5s
5p
5d
sp3d2
5d
6 hyb. sp3d2 orbs.
3 unhyb. 5d orbs.
KEY: EDG  hybridization of central atom
linear  sp
trig. planar  sp2
tetrahedral  sp3
trig. bipyr.  sp3d
octahedral  sp3d2
sp hybridization
sp hybridization occurs when one s and one of the three p
orbitals hybridize. It results in the creation of two sp hybrid
orbitals (orange). The other two p orbitals (purple) are
unhybridized; they retain the “figure-8” or “dumbbell” shape.
Multiple Bonds
s (sigma) bonds are bonds in which the e– density is
..
along the internuclear axis.
N
-- These are the single bonds we have
H
H
H
considered up to this point.
-- e.g., s-s, s-p, or p-p overlap,
The bonds
and also p-sp hybrid overlap
in ammonia
are s bonds.
sigma (s) = single
Multiple bonds result from the overlap of two p orbitals
(one from each atom) that are oriented perpendicularly
to the internuclear axis. These are p (pi) bonds.
pi, multiple, unhybridized
p bonds are generally weaker than s bonds because
p bonds have less overlap.
For ethene (C2H4)…
H
H
H
C=C
H
H
For each carbon atom, there
are 3 sp2 hybridized
orbitals; these form s
H
bonds (- - - -) with C or H.
Where unhybridized
orbitals overlap,
a p bond ( ) is formed.
H
C
C
H
Single bonds
are s bonds.
e.g., C2H6
Double bonds consist of
one s and one p.
e.g., C2H4
Triple bonds consist of
one s and two p.
e.g., C2H2
Experiments indicate that all
of C2H4’s atoms lie in the
same plane. This suggests
that p bonds introduce
rigidity (i.e., a reluctance
to rotate) into molecules.
-- p bonding does NOT occur with sp3 hybridization,
only sp and sp2
-- p bonding is more prevalent with small molecules
(e.g., C, N, O) (Big atoms don’t allow enough
p-orb. overlap for p bonds to form.)
Delocalized p Bonding
Localized p bonds are between… two atoms only.
-- These are common for
molecules with…
resonance structures.
..
..
Delocalized p bonds are
“smeared out” and
shared among… > two atoms.
[
..
O– N=O
..
O
..
..
-- e.g., C2H2, C2H4, N2
]
–
The nitrate ion (NO3–) has
delocalized electrons.
-- The electrons involved in
these bonds are delocalized electrons.
H
Consider benzene, C6H6.
-- Each carbon atom is
sp2 hybridized.
___
H
C
H
C
C
C
-- This leaves... a 2p
orbital on each C
C
C
that is oriented
H
H
to plane of m’cule.
-- 6 e– shared equally by 6 C atoms
H
Which of the following exhibit delocalized bonding?
=
– –
=
– –
H O H
H–C–C–C–H
H
H
propanone
..
SO2
18 e–
sulfur dioxide
..
O–S=O
..
“YEP.”
(res.)
..
..
..
[
.. O ..
O– S –O
..
..
O
..
]
..
..
..
sulfate ion
32
e–
..
SO4
2–
“NOPE.”
..
O
H3C–C–CH3
24 e–
2–
“NOPE.”
Molecular Orbitals
-- The VSEPR and valence-bond theories don’t
explain the excited states of molecules, which
come into play when molecules absorb and
emit light.
-- This is one thing that the molecular orbital (MO)
theory attempts to explain.
Molecules respond to the
many wavelengths of light.
The wavelengths that are
absorbed and then re-emitted
determine an object’s color,
while the wavelengths that are
NOT re-emitted raise the
temperature of the object.
molecular orbitals: wave functions that describe the
locations of electrons in molecules
-- these are analogous to atomic orbitals in atoms
(e.g., 2s, 2p, 3s, 3d, etc.), but MOs are possible
locations of electrons in molecules (not atoms)
-- MOs, like atomic orbitals,
can hold a maximum of
two e– with opposite spins
-- but MOs are for
entire molecules
MO theory is more powerful than valence-bond theory;
its main drawback is that it isn’t easy to visualize.
Hydrogen (H2)
The overlap of two atomic
orbitals produces two MOs.
(antibonding MO)
s*1s
1s
+
1s
H atomic orbitals
s1s
(bonding MO)
H2 molecular orbitals
-- The lower-energy bonding molecular orbital
concentrates e– density between nuclei.
-- For the higher-energy antibonding molecular orbital,
the e– density is concentrated outside the nuclei.
-- Both of these are s molecular orbitals.
Energy-level diagram (molecular orbital diagram)
s*1s
1s
1s
H atom s1s
H atom
H2 m’cule
(antibonding MO)
s*1s
1s
+
1s
H atomic orbitals
s1s
(bonding MO)
H2 molecular orbitals
Consider the energy-level diagram for the
hypothetical He2 molecule…
s*1s
1s
1s
He atom s1s
He atom
He2 m’cule
2 bonding e–, 2 antibonding e–
No energy benefit to bonding.
He2 molecule won’t form.
bond order = ½ (# of bonding e– – # of antibonding e–)
-- the higher the bond order,
the greater the bond stability
0 = no bond
1 = single bond
2 = double bond
3 = triple bond
007
7 = James Bond
-- MO theory allows for fractional bond orders as well.
-- a bond order of...
What is the bond order of H2+?
1 e–
total
1 bonding e–,
zero antibonding e–
BO = ½ (1–0) = ½
Second-Row Diatomic Molecules
3
4
Li Be
6.941
9.012
B
5
10.811
C
6
12.011
7
8
N
O
14.007
15.999
F
9
18.998
10
Ne
20.180
1. # of MOs = # of combined atomic orbitals
2. Atomic orbitals combine most effectively
with other atomic orbitals of similar energy.
3. As atomic orbital overlap increases, bonding
MO is lowered in energy, and the antibonding
MO is raised in energy.
4. Both the Pauli exclusion principle and
Hund’s rule apply to MOs.
Use MO theory to
predict whether Li2
and/or Be2 could
possibly form.
s*2s
2s
2s
s2s
s*1s
1s
1s
Li
BO = ½ (4–2) = 1
s1s
Li
Li2
“YEP.”
s*2s
2s
2s
s2s
Bonding and antibonding e– cancel each
other out in core energy levels, so any
bonding is due to e– in bonding orbitals of
outermost shell.
Be
Be
BO = ½ (4–4) = 0 Be2
“NOPE.”
Molecular Orbitals from 2p Atomic Orbitals
The 2pz orbitals overlap in head-to-head fashion,
so these bonds are... s bonds.
-- the corresponding MOs are: s2p and s*2p
y
x
z
The other 2p orbitals (i.e., 2px and 2py) overlap in
sideways fashion, so the bonds are... p bonds.
-- the corresponding MOs are:
p2p (two of these) and p*2p (two of these)
Rule 3 above suggests that, from low energy to
high, the 2p MOs SHOULD follow the order:
LOW
HIGH
s2p < p2p < p*2p < s*2p
ENERGY
ENERGY
General energy-level diagrams for MOs
of second-row homonuclear diatomic molecules...
don’t fit on this slide.
(And so, they’re on the next one…
…if that’s all right with you.)
(And if not, you can…
…quit and go make pancakes.)
For B2, C2, and N2...
s*2p
“Mr. B”
(or Mr. C)
(or Mr. N)
same for
both
p*2p
s2p
p2p
For O2, F2, and “Ne2”...
s*2p
p*2p
p2p
s2p
s*2s
s*2s
s2s
s2s
(1s MOs are down here)
Here, the interaction is weak.
Here, the interaction between
the 2s of one atom and the 2p The energy distribution is
as expected.
of the other is strong. The
orbital energy distribution is
altered.
paramagnetism: describes the
attraction of molecules with
unpaired e– to a magnetic field
diamagnetism: describes substances
with no unpaired e–
(“di-” = two; diamagnetic ~
= “dielectron”)
-- such substances are VERY weakly (almost
unnoticeably) repelled
by a magnetic field
Use the energy diagrams
above to tell if diatomic
species are paramagnetic
or diamagnetic.
paramagnetism of liquid oxygen
Paramagnetic or diamagnetic?
B2
(10)
P
C2
(12)
D
N2
(14)
D
s*2p
p*2p
s2p
p2p
s*2p
p*2p
p2p
s2p
s*2s
s*2s
s2s
s2s
O2
(16)
P
F2
(18)
D
s*1s
s*1s
O2+ (15)
P
s1s
s1s
O22– (18)
D
C22– (14)
D