1 VSEPR Theory. 2 Lewis structures are valuable because?? But what about the 3-D shape of molecules?...
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Transcript of 1 VSEPR Theory. 2 Lewis structures are valuable because?? But what about the 3-D shape of molecules?...
2
VSEPR Theory
•Lewis structures are valuable because??
•But what about the 3-D shape of molecules?
•They have 3-D shapes, but is it important?
• What about enzymes: is their 3-D shape important?
• What about DNA: is their 3-D shape important?
3
VSEPR Theory
•Just as the 3-D shape of proteins and enzymes are
crucial for their proper function, many chemical
reactions are very shape specific.
•Many rxns will only occur if a molecule of the “right
shape” comes along.
•So how do we predict the 3-D shape of molecules?
•VSEPR Theory: valence shell electron pair repulsion
theory.
4
VSEPR Theory
•The underlying principle behind VSEPR is that
electron pairs (whether bonding pairs or nonbonding
pairs) repel each other.
•So molecules want to maximize the distance between
electron pairs.
•But this means that there are “bond angles” between
electron pairs.
5
VSEPR Theory
•But bond angles means 3-D geometric shapes!•So if we can predict the bond angles between electron pairs, we can predict the geometry and thus the 3-D shape.
•We still use Lewis structures, as to find the bond angles you must first draw the Lewis structure.
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VSEPR Terminology
•Electron pairs: pairs of electrons which are within the electron cloud of an atom (they may be bonding or nonbonding).
•Nonbonding pairs (lone pairs): pairs of electrons which are not used in bonding; they are not “shared” between 2 atoms.
•Nonbonding electrons: electrons which are not used in bonding.
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VSEPR Terminology
•Electron Domain (electron charge clouds): the electron domain is a number between 2 and 6. It is:
ED = (#atoms bonded to central atom) + (#lone pairs on the central atom)*
*Although we start with finding the electron domain of central atoms, we can use the same technique to find the electron domain of any atom in a molecule.
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VSEPR Terminology
•VSEPR Geometry: based on the ED, the molecule is assigned a 3-D shape around the central atom (or atom of interest). There are 5 basic VSEPR geometries.
•Molecular Geometry (molecular shape): Based on the ED and the number of lone pairs on an atom, the actual 3-D shape of the molecule around the atom is assigned. There are 13 different molecular shapes.
•You need to know them all!
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VSEPR Prediction
To predict the 3-D shape:1. Draw the Lewis structure2. Determine how many lone pairs are on the atom of
interest.3. Determine the ED4. Determine the VSEPR geometry and the molecular
geometry around the atom of interest.
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Covalent Bonding and Orbital Overlap
Lewis structures and VSEPR do not explain why a bond forms.How do we account for shape in terms of quantum mechanics? What are the orbitals that are involved in bonding?We use Valence Bond Theory:
Bonds form when orbitals on atoms overlap.There are two electrons of opposite spin in the orbital overlap.The more the orbitals overlap, the stronger the bond.When p or d orbitals are involved, then the resulting bond has direction (somewhere in x, y, z axis).
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Covalent Bonding and Orbital OverlapThis works great for a few simple molecules, but it fails for most!
Example: CH4.
If you look at the orbital energy diagram for C, C only has 2 unpaired electrons. (the 2p2 orbital electrons)
How does C form 4 bonds??
So chemists had to develop a theory to explain how orbitals overlap to form covalent bonds.
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Hybrid Orbitalssp Hybrid OrbitalsConsider the BeF2 molecule (experimentally known to exist):
Be has a 1s22s2 electron configuration.There is no unpaired electron available for bonding.We conclude that the atomic orbitals are not adequate to describe orbitals in molecules.We know that the F-Be-F bond angle is 180˚ (VSEPR theory).We also know that one electron from Be is shared with each one of the unpaired electrons from F.
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Hybrid Orbitalssp Hybrid OrbitalsWe assume that the Be orbitals in the BeF bond are 180° apart.
We could promote and electron from the 2s orbital on Be to the 2p orbital to get two unpaired electrons for bonding.BUT the geometry is still not explained. Why?We can solve the problem by allowing the 2s and one 2p orbital on Be to mix or form a hybrid orbital (process called hybridization).The hybrid orbital comes from an s and a p orbital and is called an sp hybrid orbital.
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Hybrid Orbitalssp Hybrid Orbitals
•Since only one of the Be 2p orbitals has been used in hybridization, there are two unhybridized p orbitals remaining on Be.
•Note: If you start with x number of orbitals, you end up with x orbitals!
•Once the 2 sp hybrid orbitals are formed, they overlap with 2 F atoms to form the 2 covalent bonds.
•Also, if the Electron Domain of an atom is 2, then it is sp hybridized!
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Hybrid Orbitalssp2 Hybrid Orbitals• Important: when we mix n atomic orbitals we must get n hybrid orbitals.
• sp2 hybrid orbitals are formed with one s and two p orbitals. (Therefore, there is one unhybridized p orbital remaining.)
•The large lobes of sp2 hybrids lie in a trigonal plane.•All molecules with Electron Domain of 3 have sp2 orbitals on the central atom.
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Hybrid Orbitalssp3 Hybrid Orbitals• sp3 Hybrid orbitals are formed from one s and three p orbitals.
•Therefore, there are four large lobes.•Each lobe points towards the vertex of a tetrahedron.•The angle between the large lobes is 109.5°•All molecules with an Electron Domain of 4 are sp3 hybridized.
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Hybrid OrbitalsHybridization Involving d Orbitals•Since there are only three p-orbitals, trigonal bipyramidal and octahedral electron pair geometries must involve d-orbitals.
•Trigonal bipyramidal geometries require sp3d hybridization.
•Octahedral geometries require sp3d2 hybridization.•Note the Electron Domain from VSEPR theory determines the hybridization.
31
Hybrid OrbitalsSummaryTo assign hybridization:
1) draw a Lewis structure; 2) assign the Electron Domain using VSEPR theory; 3) from the Electron Domain, determine the hybridization; and 4) name the geometry by the positions of the atoms.
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Multiple Bonds• σ-Bonds: electron density lies on the axis between the nuclei. (Head-on overlap.)
•All single bonds are σ-bonds.•π-Bonds: electron density lies above and below the plane of the nuclei. (Side-by-side overlap.)
•A double bond consists of one σ-bond and one π-bond.•A triple bond has one σ-bond and two π-bonds.•Usually the p-orbitals involved in π-bonding come from unhybridized orbitals.
34
Multiple BondsEthylene, C2H4, has:
one σ- and one π-bond; both C atoms sp2 hybridized;both C atoms with trigonal planar electron pair and molecular geometries.
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Multiple BondsConsider acetylene, C2H2
the electron pair geometry of each C is linear;therefore, the C atoms are sp hybridized;the sp hybrid orbitals form the C-C and C-H σ-bonds;there are two unhybridized p-orbitals;both unhybridized p-orbitals form the two π-bonds;one π-bond is above and below the plane of the nuclei;one π-bond is in front and behind the plane of the nuclei.When triple bonds form (e.g. N2) one π-bond is always above and below and the other is in front and behind the plane of the nuclei.
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Multiple Bonds
Delocalized π BondingSo far all the bonds we have encountered are localized (or fixed) between two nuclei.In the case of benzene, C6H6,
there are 6 C-C σ bonds, 6 C-H σ bonds,each C atom is sp2 hybridized,there are 6 unhybridized p orbitals on each C atom.
40
Multiple Bonds
Delocalized π BondingIn benzene there are two options for the 3 π bonds
localized between C atoms ordelocalized over the entire ring (i.e. the π electrons are shared by all 6 C atoms).You can draw 2 resonance structures for benzene.Experimentally, all C-C bonds are the same length in benzene.Therefore, all C-C bonds are of the same type (recall single bonds are longer than double bonds).
42
Multiple Bonds
General Conclusions•Every two atoms share at least 2 electrons.•Two electrons between atoms on the same axis as the nuclei are σ bonds.•σ-Bonds are always localized.• If two atoms share more than one pair of electrons, the
second and third pair form π-bonds.•When resonance structures are possible, delocalization
is also possible.
43
Molecular Orbitals• Some aspects of bonding are not explained by
Lewis structures, VSEPR theory and hybridization. (E.g. why does O2 interact with a magnetic field?; Why are some molecules colored?)
•For these molecules, we use Molecular Orbital (MO) Theory.
44
Molecular Orbitals• In Valence Bond Theory, atomic orbitals were
hybridized, but they were still atomic orbitals: that is, these orbitals belong to the atom.
•Molecular Orbital Theory takes atomic orbitals and creates molecular orbitals, that is, orbitals which belong to the entire molecule.
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Molecular Orbitals•Molecular orbitals:• each contain a maximum of two electrons;• have definite energies;• can be visualized with contour diagrams;• a molecule has as many molecular orbitals as it
originally had atomic orbitals;• come in 3 types: bonding, nonbonding, and
antibonding;• are mathematical combinations of the atomic
orbitals;• are associated with an entire molecule.
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Molecular OrbitalsThe Hydrogen Molecule When two AOs overlap two MOs form.Therefore, 1s (H) + 1s (H) must result in two MOs for H2:• one has electron density between nuclei (bonding
MO);• one has little electron density between nuclei
(antibonding MO).MOs resulting from s orbitals are σ MOs.The σ Bonding MO is lower energy than σ* (antibonding) MO.
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Molecular OrbitalsThe Hydrogen Molecule • Energy level diagram or MO diagram shows the
energies and orbitals in an orbital.• The total number of electrons in all atoms are placed
in the MOs starting from lowest energy (σ1s) and ending when you run out of electrons.• Note that electrons in MOs have opposite spins.• H2 has two bonding electrons.
• He2 has two bonding electrons and two antibonding electrons.
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Molecular OrbitalsBond OrderBond Order = ½(bonding e- - antibonding e-).
Bond order = 1 for single bond.Bond order = 2 for double bond.Bond order = 3 for triple bond.Fractional bond orders are possible.Bond order for H2 = ½(bonding e- - antibonding e-) = ½(2 -0) = 1. Therefore, H2 has a single bond.
Bond order for He2 = ½(bonding e- - antibonding e-) = ½(2 - 2) = 0. Therefore He2 is not a stable molecule.
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Second-Row Diatomic Molecules
We look at homonuclear diatomic molecules (e.g. Li2, Be2, B2 etc.).AOs combine according to the following rules:
The number of MOs = number of AOs;AOs of similar energy combine (e.g. 1s + 1s rather than 1s + 2s);As overlap increases, the energy of the MO decreases;Pauli: each MO has at most two electrons;Hund: for degenerate orbitals, each MO is first occupied singly.
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Molecular Orbitals for Li2 and Be2
Each 1s orbital combines with another 1s orbital to give one σ1s and one σ*
1s orbital, both of which are occupied (since Li and Be have 1s2 electron configurations).Each 2s orbital combines with another 2s orbital two give one σ2s and one σ*
2s orbital.The energies of the 1s and 2s orbitals are sufficiently different so that there is no cross-mixing of orbitals (i.e. we do not get 1s + 2s).
Second-Row Diatomic Molecules
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Molecular Orbitals for Li2 and Be2
There are a total of 6 electrons in Li2:2 electrons in σ1s;2 electrons in σ*
1s;2 electrons in σ2s; and0 electrons in σ*
2s.
Therefore the bond order is ½(4 - 2) = 1.Since the 1s AOs are completely filled, the σ1s and σ*1s are filled. We generally ignore core electrons in MO diagrams.
Second-Row Diatomic Molecules
55
Molecular Orbitals for Li2 and Be2
There are a total of 8 electrons in Be2:2 electrons in σ1s;2 electrons in σ*
1s;2 electrons in σ2s; and2 electrons in σ2s.
Therefore, the bond order is ½(4 - 4) = 0.
Be2 does not exist.
Second-Row Diatomic Molecules
56
Molecular Orbitals from 2p Atomic OrbitalsThere are two ways in which two p orbitals overlap:
end on so that the resulting MO has electron density on the axis between nuclei (i.e. σ type orbital);sideways so that the resulting MO has electron density above and below the axis between nuclei (i.e. π type orbital).The six p-orbitals (two sets of 3) must give rise to 6 MOs:σ, σ*, π, π*, π, and π*.Therefore there is a maximum of 2 π bonds that can come from p-orbitals.The relative energies of these six orbitals can change.
Second-Row Diatomic Molecules
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Electron Configurations for B2 through Ne2
· 2s Orbitals are lower in energy than 2p orbitals so
σ2s orbitals are lower in energy than σ2p orbitals.
·There is greater overlap between 2pz orbitals (they
point directly towards one another) so the σ2p MO
is lower in energy than the π2p orbitals.
·Bonding orbitals are lower in energy than their
corresponding antibonding orbital.
·The π2p and π*2p orbitals are doubly degenerate.
Second-Row Diatomic Molecules
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Electron Configurations for B2 through Ne2
As the atomic radius increases, it becomes more likely
that a 2s orbital on one atom can interact with the 2p
orbital on the other (lower e- repulsions).As the 2s-2p interaction increases, the σ2s MO lowers in
energy and the σ2p orbital increases in energy.
For B2, C2 and N2 the σ2p orbital is higher in energy
than the π2p.
For O2, F2 and Ne2 the σ2p orbital is lower in energy
than the π2p.
Second-Row Diatomic Molecules
62
Electron Configurations for B2 through Ne2
Once the relative orbital energies are known, we add the required number of electrons to the MOs taking into account Pauli’s exclusion principle and Hund’s rule.
As bond order increases, bond length decreases.As bond order increases, bond energy increases.
Second-Row Diatomic Molecules
64
Electron Configurations and Molecular PropertiesTwo types of magnetic behavior:
paramagnetism (unpaired electrons in molecule): strong attraction between magnetic field and molecule;diamagnetism (no unpaired electrons in molecule): weak repulsion between magnetic field and molecule.Magnetic behavior is detected by determining the mass of a sample in the presence and absence of magnetic field:large increase in mass indicates paramagnetism,small decrease in mass indicates diamagnetism.
Second-Row Diatomic Molecules
66
Electron Configurations and Molecular PropertiesExperimentally O2 is paramagnetic.The Lewis structure for O2 shows no unpaired electrons.The MO diagram for O2 shows 2 unpaired electrons in the π*
2p orbital.Experimentally, O2 has a short bond length (1.21 Å) and high bond dissociation energy (495 kJ/mol). This suggests a double bond.The MO diagram for O2 predicts both paramagnetism and the double bond (bond order = 2).
Second-Row Diatomic Molecules