Post on 20-Jan-2018
description
Molecular Geometryand
Bonding Theories
The properties of a molecule depend on its shape andand the nature of its bonds. In this unit, we willdiscuss 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
CO2
bond angles: the angles made by the lines joining the nuclei of a molecule’s atoms
CH4 CH2Ocarbon dioxide methane formaldehyde
180o 120o109.5o
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 arethree bonding domains andone nonbonding domain. N–HH–
H
..
NHH
H
..
4 e – domains
NH3
Domains arrange themselves soas to minimize their repulsions.
The electron-domain geometry is oneof five basic arrangements of domains.
-- it depends only on the total # of e– domains, NOT the kind of each domain
The molecular geometry describes theorientation of the atoms in space.
-- it depends on how many of each kind of e– domain
NHH
H
..
..
Total # of Domains
Electron-Domain Geometry Possible Molecular Geometries
2
3
4
5
6
linear linear (CO2)
trigonalplanar
trigonal planar (BF3), bent (NO2)
tetrahedral tetrahedral (CH4),trigonal pyramidal (NH3), bent (H2O)
trigonalbipyramidal
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/ormolecular geometry (MG), draw the Lewis structure.Multiple bonds count as a single domain.
Predict the EDG and MG of each of the following.
SnCl3–
26 e– [ ]Sn–ClCl–....
..
....
....
....Cl
–..EDG: tetrahedral
MG: trig. pyramidal
O3
18 e–
EDG: trig. planarMG: bent
..O–OO= .... ..
....
SeCl2
20 e–
EDG: tetrahedralMG: bent
Cl–Se–Cl....
..
...... ....
..OO–O
=.. ...... ..
CO32–
24 e– [ ]C=OO–....
..
.... ..
.... O
2–EDG: trig. planar
MG: trig. planar
SF4
34 e–
EDG: trig. bipyr.MG: seesaw
..
S–FF–....
..
....
....
....
F
F.. ..
..
IF5
42 e– I
.. .... F–F.. .
...–F.. ..
..F–......
F.. ....
.. EDG: octahedralMG: sq. pyramidal
ClF3
28 e–
Cl–FF–....
..
....
....
.... F
..EDG: trig. bipyr.
MG: T-shaped
..
(res
.)
ICl4–
36 e–
EDG: octahedralMG: sq. planar[ ]..
I–ClCl–....
..
....
....
....
Cl
Cl
.. ..
.. ..–
For molecules with more than one central atom,simply apply the VSEPR model to each part.
Predict the EDG and MG around the three interioratoms of ethanoic (acetic) acid.
H–C–C–O–H..H
..H O....
CH3COOH PORTION EDG MG
tetra. tetra.–CH3
trig. plan. trig. plan.
tetra. bent
C=O–– ..
..–OH
..
..
Nonbonding domains are attracted to only one nucleus;therefore, they are more spread out than are bondingdomains. The effect is that nonbonding domains(i.e., “lone pairs”) compress bond angles. Domains formultiple bonds have a similar effect.
e.g.,
CH4 NH3 H2O
NHH
H
..C
HH
H
H OH
..H
..
..ClO=C
Cl...... ..
..
....
the ideal bond angle for the tetrahedralEDG is 109.5o
111.4o
EDG = trig. plan.ideal = 120o
COCl2
124.3o
124.3o
109.5o107.0o 104.5o
Polarity of Molecules
A molecule’s polarity is determined by its overalldipole, which is the vector sum of the dipoles ofeach of the molecule’s bonds.
Consider CO2 v. H2S...
O=C=O....
.... S
H
..H
..bond dipoles
overall dipole =
bond dipoles
overall dipole =(NONPOLAR) (POLAR)
zero
Boron can be extracted bythe electrolysis of moltenboron trichloride. Boron isan essential nutrient forplants, and is also a primarycomponent of control rodsin nuclear reactors.
Classify as polar or nonpolar:
PCl3
BCl3
26 e–P–ClCl–
..
......
..
....
....Cl
..polar
..
B–ClCl–.... .... ..
..Cl .... nonpolar
24 e–
Valence-Bond Theory: merges Lewis structures w/the idea of atomic orbitals (2s, 3p, etc.)
V-B theory says… covalent bonding occurs whenvalence orbitals of adjacentatoms overlap; then, two e–sof opposite spin (one fromeach atom) combine to forma bond
Lewis theory says… covalent bonding occurs whenatoms share electrons
V-B theory is like Velcro: “No overlap, no bond.”
Consider H2, Cl2, and HCl...
H
Cl [Ne]
H2 Cl2
HCl
(unpaired e– is in 1s orb.)
(unpaired e– is in 3p orb.)
= orbital overlap region (responsible for bond)
There is always an optimum distance between twobonded nuclei. At this optimum distance, attractive (+/–) and repulsive (+/+ and –/–) forcesbalance.
H–H distance
Energy
0
optimum dist.(min. PE)
H2 molecule
Hybridized Orbitals
V-B theory can’t explain some observations aboutmolecular compounds without the concept of hybridized orbitals.
i.e., covalent (NM/NM) compounds
In…
BeF2
(LS) …the central atomSHOULD be…
…but is ACTUALLYhybridized to be…
…the resultthen being…
2 hyb. sp orbs.2 unhyb. 2p orbs.2s 2p
BF3 3 hyb. sp2 orbs.1 unhyb. 2p orb.2s 2p
2psp
sp2 2p
CH4 4 hyb. sp3 orbs.2s 2p sp3
H2O 4 hyb. sp3 orbs.2s 2p sp3
PF53s 3p sp3d3d 3d
5 hyb. sp3d orbs.4 unhyb. 3d orbs.
XeF2sp3d5s 5p 5d 5d
5 hyb. sp3d orbs.4 unhyb. 5d orbs.
XeF4sp3d25s 5p 5d 5d
6 hyb. sp3d2 orbs.3 unhyb. 5d orbs.
: BeF :
::F
::
B :
::F
:F
::
:F:
:
CH HH
H
OHH
:
:
P :
::F
:F
::
:F:
:
:: :F
:
::F
: XeF :
::F
::
: ::
: XeF :
::F
::
::F: :
F::
::
KEY: EDG hybridization of central atom
linear sptrig. planar sp2
tetrahedral sp3
trig. bipyr. sp3doctahedral sp3d2
sp hybridization
sp hybridization occurs when one s and one of the three porbitals hybridize. It results in the creation of two sp hybridorbitals (orange). The other two p orbitals (purple) areunhybridized; they retain the “figure-8” or “dumbbell” shape.
Multiple Bonds
s (sigma) bonds are bonds in which the e– density isalong the internuclear axis.
-- These are the single bonds we have considered up to this point. -- e.g., s-s, s-p, or p-p overlap, and also p-sp hybrid overlap
Multiple bonds result from the overlap of two p orbitals(one from each atom) that are oriented perpendicularlyto the internuclear axis. These are p (pi) bonds.
sigma (s) = single
pi, multiple, unhybridized
NHH
H
..
The bondsin ammoniaare s bonds.
p bonds are generally weaker than s bonds becausep bonds have less overlap.
C C
H
H
H
H
For ethene (C2H4)…
C=C–––
– H
H
H
H
For each carbon atom, thereare 3 sp2 hybridizedorbitals; these form s
bonds (- - - -) with C or H.
Where unhybridizedorbitals overlap,
a p bond ( ) is formed.
Single bondsare s bonds.
Double bonds consist ofone s and one p.
Triple bonds consist ofone s and two p.
e.g., C2H2
e.g., C2H4
e.g., C2H6
Experiments indicate that allof C2H4’s atoms lie in thesame plane. This suggeststhat p bonds introducerigidity (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.)
resonance structures.
Delocalized p Bonding
Localized p bonds are between…
-- e.g., C2H2, C2H4, N2
Delocalized p bonds are“smeared out” andshared among…
-- These are common for molecules with…
-- The electrons involved in these bonds are delocalized electrons.
The nitrate ion (NO3–) has
delocalized electrons.
[ ]N=OO–....
....
..
..
....O
–
> two atoms.
two atoms only.
Consider benzene, C6H6. -- Each carbon atom is ___ hybridized. -- This leaves... a 2p
orbital on each Cthat is oriented to plane of m’cule.
C
C
C
C
C C
H
H
H
H
H
H
-- 6 e– shared equally by 6 C atoms
sp2
Which of the following exhibit delocalized bonding?
H3C–C–CH3
O=
SO42–
SO2
sulfate ion
sulfur dioxide18 e–
32 e–
O–S=O...... ..
....
“YEP.”
HH–C–C–C–H
O=
.. ..
––H –H
H
–
“NOPE.”
“NOPE.”[ ]S–OO–....
..
....
....
....
O
O
.. ....2–
24 e–
propanone
(res.)
Molecules respond to themany wavelengths of light.The wavelengths that areabsorbed and then re-emitteddetermine an object’s color, while the wavelengths that areNOT re-emitted raise thetemperature of the object.
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.
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
atomicorbitals produces two
MOs.
1s 1s
H atomic orbitals
+
H2 molecular orbitals
s1s
(bonding MO)
(antibonding MO)s*1s
-- 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)
1s 1s
s1s
s*1s
H atom H2 m’cule
H atom
1s 1s
H atomic orbitals
+
H2 molecular orbitals
s1s
(bonding MO)
(antibonding MO)s*1s
Consider the energy-level diagram for thehypothetical He2 molecule…
1s 1s
s1s
s*1s
He atom He2 m’cule
He atom
2 bonding e–, 2 antibonding e–
No energy benefit to bonding.
He2 molecule won’t form.
no bond
bond order = ½ (# of bonding e– – # of antibonding e–)
-- the higher the bond order, the greater the bond stability
-- a bond order of... 0 = single bond1 =double bond2 =triple bond3 =
-- MO theory allows for fractional bond orders as well.
What is the bond order of H2+?
1 e–
total1 bonding e–,zero antibonding e– BO = ½ (1–0) = ½
James Bond7 =007 =
Second-Row Diatomic Molecules
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.
Li3
6.941 Be
4
9.012 B
5
10.811 C
6
12.011 N
7
14.007 O
8
15.999 F
9
18.998 Ne
10
20.180
Li2
Li Li s1s
s*1s
1s 1s
2s 2s
s*2s
s2s
Use MO theory topredict whether Li2and/or Be2 couldpossibly form.
BO = ½ (4–2) = 1 “YEP.”
Be2
Be Be
2s 2s
s*2s
s2s
BO = ½ (4–4) = 0 “NOPE.”
Bonding and antibonding e– cancel eachother out in core energy levels, so any
bonding is due to e– in bonding orbitals of outermost shell.
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
The other 2p orbitals (i.e., 2px and 2py) overlap insideways fashion, so the bonds are... p bonds.
-- the corresponding MOs are: p2p (two of these) and p*2p (two of these)
z
x
y
Rule 3 above suggests that, from low energy tohigh, the 2p MOs SHOULD follow the order:
s2p < p2p < p*2p < s*2pLOW
ENERGYHIGH
ENERGY
General energy-level diagrams for MOsof 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.)
Here, the interaction betweenthe 2s of one atom and the 2pof the other is strong. Theorbital energy distribution isaltered.
Here, the interaction is weak.The energy distribution isas expected.
(1s MOs are down here)
For B2, C2, and N2... For O2, F2, and “Ne2”...
s2s
s*2s
p2p
p*2p
s2p
s*2p
s2s
s*2s
s2p
s*2p
p*2p
p2p
same forboth
“Mr. B”(or Mr. C)(or Mr. N)
paramagnetism: describes the attraction of molecules with unpaired e– to a magnetic field
diamagnetism: describes substances with no unpaired e–
-- such substances are VERY weakly (almost unnoticeably) repelled by a magnetic field
Use the energy diagramsabove to tell if diatomicspecies are paramagneticor diamagnetic.paramagnetism of liquid oxygen
(“di-” = two; diamagnetic = “dielectron”)~
Paramagnetic or diamagnetic?
B2
C2
N2
O2
F2
O2+
O22–
C22–
(10)
(12)
(14)
(16)
(18)
(15)
(18)
(14)
s2s
s*2s
p2p
p*2p
s2p
s*2p
s2s
s*2s
s2p
s*2p
p*2p
p2p
s1s
s*1s
s1s
s*1s
P
D
D
P
D
P
D
D