Chapter 9 Covalent Bonding: Orbitals

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Chapter 9Covalent Bonding Orbitals
2
Preview

• Hybridization of atomic orbitals.
• From an atomic orbital (AO) to a molecular
  orbital (MO).
• Bond orders.
• Paramagnetism and Diamagnetism from MO models.
• Localized and delocalized bondings.

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sp3 Hybridization
Chapter 9 Section 1

• The atom uses its valence orbitals to bond with
  other atoms.
• CH4 molecule.

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sp3 Hybridization
Chapter 9 Section 1

• Does the C atom bond to the H atoms using these
  orbitals (one s and three p orbitals)?
• From experiments, CH4 is known to have four
  similar orbitals (tetrahedral structure with
  109.5º bond angles).

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sp3 Hybridization
Chapter 9 Section 1

• The C atom in tetrahedral structure is assumed to
  have four equivalent orbitals.
• Hybridization
• The one s and three p orbitals in the free C
  atom are mixed to produce a new set of four
  equivalent orbitals.

Four sp3 orbitals
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sp3 Hybridization
Chapter 9 Section 1

• sp3 hybridized orbitals are identical in shape
  and energy and they enable the atom to have the
  tetrahedral geometry with 109.5º bond angles.
• This model can be generalized for other molecules
  that adopt the tetrahedral arrangement of its
  electron pairs.

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sp3 Hybridization in CH4 and NH3
Chapter 9 Section 1

• You should always think about how the electron
  pairs are arranged around the atom.

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sp2 Hybridization
Chapter 9 Section 1

• In ethylene (C2H4), the bond angles are 120º.
• One s and two p orbitals in the free C atom are
  mixed to produce a new set of three hybridized
  orbitals.
• This gives rise to the sp2 hybridization.

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sp2 Hybridization
Chapter 9 Section 1
Three sp2 orbitals

• For the sp2 hybridized orbitals, one p orbital
  is not being mixed.

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sp2 Hybridization
Chapter 9 Section 1

• The three sp2 orbitals are located on the same
  plane (xy-plane, for instance). The remaining
  non-hybridized p orbital is perpendicular to that
  plane (pz).

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sp2 Hybridization (Single and Double Bonds)
Chapter 9 Section 1

• The three sp2 hybridized orbitals will form three
  sigma (s) bonds with three 1s orbitals of the H
  atoms.
• The unchanged perpendicular 2p orbitals (one is
  located on each C atom) will share 2 electrons to
  form a pi (p) bond.

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Single and Double Bonds in Ethylene
Chapter 9 Section 1
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Single and Double Bonds in Ethylene
Chapter 9 Section 1

• A single bond consists of a s-bond.
• A double bond consists of a s-bond and a p-bond.

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sp2 and sp3 Hybridizations
Chapter 9 Section 1
? You should always think about how the electron
pairs are arranged around the atom.
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sp Hybridization
Chapter 9 Section 1

• The CO2 molecule has a linear structure with a
  bond angle of 180º.
• An sp hybridization (involving one s orbital and
  one p orbital that are combined) can provide a
  linear combination of C atomic orbitals.

Two sp orbitals
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sp Hybridization
Chapter 9 Section 1

• When one s orbital and one p orbital are
  hybridized, a set of two sp orbitals oriented at
  180 degrees results.

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The sp Hybridized Orbitals in CO2
Chapter 9 Section 1

• In CO2
• the C atoms have two sp hybridized orbitals, and
• the O atoms have three sp2 hybridized orbitals.

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sp Hybridization in CO2
Chapter 9 Section 1

• What is about the non-mixed p orbitals for the C
  atoms and O atoms?

p bond
s bond
p bond
The C atom in CO2
The O atom in CO2
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sp Hybridization in CO2
Chapter 9 Section 1

• The orbitals used to form the bonds in carbon
  dioxide. Note that the carbon-oxygen double bonds
  each consist of one s bond and one p bond.

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sp Hybridization in Acetylene (C2H2)
Chapter 9 Section 1

• C2H2 requires linear geometry which can be
  represented with sp hybridized orbitals.

sp
1s
1s
sp

• This account for the triple bond (one s and two
  p) in C2H2.

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sp Hybridization in N2
Chapter 9 Section 1

• N2 has the following Lewis structure

N
N
? You should always think about how the electron
pairs are arranged around the atom.
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sp Hybridization in N2
Chapter 9 Section 1
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sp, sp2 and sp3 Hybridizations
Chapter 9 Section 1
? You should always think about the geometry of
how the electron pairs are arranged around the
atom.
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Sample Exercise 9.5(b)
Chapter 9 Section 1

• Predict the hybridization for the BF4 ion, and
  describe the molecular structure.
• 1. Draw Lewis structure.
• 2. Determine electron pair arrangement using
  VSEPR model.
• 3. Specify the hybrid orbitals needed.

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s-Bonds and p-Bonds
Chapter 9 Section 1

• How many single and double bonds are in acetic
  acid?
• 7 s-bonds and 1 p-bond.

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dsp3 Hybridization
Chapter 9 Section 1

• PCl5 requires five atomic orbitals one s, three
  p, and one d orbitals.
• A set of five dsp3 orbitals can combine to have
  the trigonal bipyramidal structure.

PCl5
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dsp3 Hybridization in PCl5
Chapter 9 Section 1

• What electron arrangements, and hence what type
  of hybrid orbitals, do the Cl atom require?
• Three sp3 orbitals.

PCl5

• The five s bonds in PCl5 are formed by sharing
  electrons between the dsp3 orbitals on P atom and
  the sp3 orbitals on Cl atoms.

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d2sp3 Hybridization
Chapter 9 Section 1

• The d2sp3 hybridization mixes six orbitals that
  are arranged octahedrally.
• Example SF6

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Sample Exercise 9.4
Chapter 9 Section 1

• How is the Xe atom in the XeF4 hybridized?
• 1. Draw Lewis structure.
• 2. Determine electron pair arrangement using
  VSEPR model.
• 3. Specify the hybrid orbitals needed.

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Chapter 9 Section 1
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The Molecular Orbital (MO) Model
Chapter 9 Section 2

• The need for use MO model rather than the use of
  AO model
• The electron are delocalized in the real sense.
• AO model doesnt deal effectively with molecules
  containing lone pairs of electrons.
• Information about bond energies are not given
  correctly from AO model.
• The MOs can be approximated using the rules of
  quantum mechanics. The hydrogen atom is composed
  of a positively charged nucleus and an electron
  density (orbital) where the electron can be
  found, but its movement can not be accurately
  determined.

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The Molecular Orbital (MO) Model
Chapter 9 Section 2

• Two important similarities between an AO and a
  MO
• Both can hold two electron with opposite spins.
• The probability can be obtain by squaring the
  wave function associated to the given AO or MO.
• Combination of H atomic orbitals to produce MOs

Destructive interference
Constructive interference
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MOs for H2
Chapter 9 Section 2

• MO1 and MO2 in H2 molecule are known as sigma (s)
  molecular orbital because the probabilities of
  both MOs are centered along the line passing
  through the two nuclei.
• In the molecule, the atomic orbitals are no
  longer exist. The new type of MOs are where the
  electrons will occupy.

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MOs for H2
Chapter 9 Section 2

• The bonding MO available to the two electrons has
  lower energy than the AOs these electrons occupy
  in the separated atoms. This is known as
  probonding.
• The antibonding MO is higher in energy than the
  AOs of which it is composed, and the electrons in
  this case will favor the separated atoms, such as
  noble gas atoms.

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Bonding and Antibonding MOs
Chapter 9 Section 2
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MOs for H2
Chapter 9 Section 2

• In the bonding MO, the electrons have greater
  probabilities to be between the two nuclei. This
  leads to lower energy and to form a bond.
• In the antibonding MO, the electrons are outside
  the space between the nuclei, producing a higher
  energy and no bonding.

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MOs for H2
Chapter 9 Section 2

• We normally label the MOs as follow
• MO1 s1s
• MO2 s1s
• Each MO can hold two electron with opposite
  spins.
• The number of MOs and AOs is conserved.

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MOs for H2
Chapter 9 Section 2

• From the MO model, do we expect H2 to exist?
• Yes , we do.
• H2 is expected to be as twice stable as H2,
  since a net lowering of energy of one electron is
  predicted.

The Molecular orbital energy-level diagram for
the H2 molecule
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Bond Order
Chapter 9 Section 2

• Bond order
• Larger bond order means greater bond strength
• Bond order (H2) 1
• Bond order (H2) 0.5

no. of bonding electrons no. of antibonding
electrons
2
2 0
2
2 1
2
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MOs and Bond Order for He2
Chapter 9 Section 2

• Bond order (He2)
• 0
• It indicates that the He2 molecule is not stable
  compared with the free atoms.

2 2
2
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Bonding in Homonuclear Diatomic Molecules
Chapter 9 Section 3

• Li 1s2 2s1
• To participate in MOs, AOs must overlap in the
  space. These can only be the valence atomic
  orbitals.
• In general, in constructing MO diagram, we ignore
  the inner AOs.

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MOs in Lithium Molecule
Chapter 9 Section 3

• Li2 molecule has two 2s electrons.
• Bond order (Li2) (2 0) /2 1
• Lithium exists as solid having many Li atoms.

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Solid Lithium
Chapter 9 Section 3
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MOs in Beryllium Molecule
Chapter 9 Section 3

• Be2 molecule has four 2s electrons.

Bond order (2 2) /2 0
However, Beryllium metal exists.
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MOs in Boron Molecule
Chapter 9 Section 3

• B2 molecule has four 2s and two 2p electrons
  (B 1s22s22p1).
• Two pairs of parallel p orbitals can overlap
  side-on, and the third pair can overlap head-on.

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Head-On Overlap for p MOs in B2
Chapter 9 Section 3

• The two p orbitals, each on each boron atom, that
  overlap head-on produce two s MOs, one bonding,
  s2p ,(constructive interference) and one
  antibonding, s2p, (destructive interference).

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Side-On Overlap for p MOs in B2
Chapter 9 Section 3

• Two p orbitals that lie parallel overlap to
  produce two p MOs, one bonding , p2p
  ,(constructive interference) and one anitbonding,
  p2p (destructive interference).

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MO Energy Level Diagram for B2
Chapter 9 Section 3

• The s interaction is stronger than p interaction
  because the electrons are closest to the nuclei
  when the MOs are with head-on overlap.

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MO Energy Level Diagram for B2
Chapter 9 Section 3

• The 2s and 2p orbitals are assumed not to combine
  (no 2s-2p mixing).
• Bond order (B2)
• (4 2) / 2 1

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Paramagnetism and Diamagnetism
Chapter 9 Section 3

• Paramagnetism causes the substance to be
  attracted to a magnetic field.
• Diamagnetism causes the substance to be repelled
  from a magnetic field.
• Measuring the paramagnetism of a substance.

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Paramagnetism and Diamagnetism
Chapter 9 Section 3

• Paramagnetism is associated with unpaired
  electrons.
• Diamagnetism is associated with paired
  electrons.
• B2 molecule is diamagnetic.

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Paramagnetism of B2 Molecule
Chapter 9 Section 3

• In B2 molecule, experiments showed a net
  paramagnetism. This can be explained on the basis
  of that there is some mixing between the 2s and
  2p orbitals.
• The degree of 2s-2p mixing decreases as we go
  across the 2nd period from left to right.

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Paramagnetism and Diamagnetism
Chapter 9 Section 3

• The degree of 2s-2p mixing decreases as we go
  across the 2nd period from left to right.
• For O2 and F2 there is almost no mixing.

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Paramagnetism and Diamagnetism
Chapter 9 Section 3

• When the bond order increases, the bond strength
  increases (bond dissociation energy), and the
  bond length decreases.
• Bond order cant be associated with bond
  dissociation energy,

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Paramagnetism and Diamagnetism
Chapter 9 Section 3

• MO model predicts the O2 molecule to be
  paramagnetic.
• Liquid O2 remains between magnet poles until it
  totally evaporates, indicating its paramagnetism.

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Sample Exercise 9.6
Chapter 9 Section 3

• Give the electron configurations and bond orders
  for O2, O2, and O2.

O2
O2
O2
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Sample Exercise 9.7
Chapter 9 Section 3

• Predict the bond order and magnetism of Ne2.

• Bond order
• (8 8) / 2 0
• According to the MO theory, Ne2 doesnt exist.

Ne2
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Sample Exercise 9.7
Chapter 9 Section 3

• Predict the bond order and magnetism of P2.

• Bond order
• (8 2) / 2 3
• According to the MO theory, P2 exist and is
  diamagnetic.

P2
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Bonding in Heteronuclear Diatomic Molecules
Chapter 9 Section 4

• MO model can be expanded to diatomic molecules
  with two different nuclei whose electronic
  natures are not so different.
• NO is expected from MO theory to be paramagnetic
  and to have a bond order of
• (8 3) / 2 2.5

The MO energy-level diagram for the NO molecule
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Bonding in Heteronuclear Diatomic Molecules
Chapter 9 Section 4

• NO and CN molecules have 10 valence electrons.
• They are both diamagnetic and have a bond order
  of
• (8 2) / 2 3.

The MO energy-level diagram for the NO and CN
molecules
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Bonding in Heteronuclear Diatomic Molecules
Chapter 9 Section 4

• The HF molecule is an example of diatomic
  molecule that is composed of two very different
  nuclei.
• H 1s1.
• F 1s2 2s2 2p5.
• We assume the bonding is between the 1s orbital
  of the H atom and one of the 2p orbitals on the F
  atom.

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Bonding in Heteronuclear Diatomic Molecules
Chapter 9 Section 4

• The HF molecule is predicted to be stable because
  both electrons have lowered their energies.
• The 2p orbital of the F atom is lower in energy
  than the 1s orbital of the H atom because the 2p
  electron is more tightly bound to the nucleus of
  the F atom.

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Bonding in Heteronuclear Diatomic Molecules
Chapter 9 Section 4

• Electron probability distribution in the bonding
  MO of the HF molecule. There is a greater
  electron density closer to the F atom since its
  2p orbital is lower in energy than the 1s orbital
  on the H atom.

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Bonding in Heteronuclear Diatomic Molecules
Chapter 9 Section 4

• The MO theory accounts nicely for the bond
  polarity and the electronegativity differences.

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Combining the Localized Electron and MO Models
Chapter 9 Section 5

• Localized electron model assumes the bonding
  electron pair is being shared between two atoms
  (localized).
• In several cases, such as O3 and NO3, the
  bonding pairs are delocalized and are not present
  in one specific location in the molecule.

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Combining the Localized Electron and MO Models
Chapter 9 Section 5

• In molecule with resonance structures, s bonds
  can be viewed to be localized, and p bonds can be
  considered to be delocalized.
• The localized electron model can be used to
  describe s bonds and the MO model can be used to
  describe p bonds.

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Bonding in Benzene (C6H6)
Chapter 9 Section 5

• In benzene all the six C-C bonds are equivalent.
• Localized electron model can explain the
  structure of benzene in terms of resonance.
  However, our approach here is to use the
  localized electron model to describe the six s
  bonds and to use the MO model to describe the
  three p bonds.

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Bonding in Benzene (C6H6)
Chapter 9 Section 5

• Benzene is planar with trigonal planar
  electron-pair arrangement around the C atoms.
• This requires sp2 hybridization.

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Bonding in Benzene (C6H6)
Chapter 9 Section 5

• The MOs in benzene is formed by combining the six
  p orbtials from the six sp2 hybridized carbon
  atoms.
• The electrons in the resulting p MOs are
  delocalized over the entire ring of carbon atoms

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Bonding in Benzene (C6H6)
Chapter 9 Section 5

• The MOs in benzene is formed by combining the six
  p orbtials from the six sp2 hybridized carbon
  atoms.
• The electrons in the resulting p MOs are
  delocalized over the entire ring of carbon atoms

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Bonding in Nitrate Ion (NH3)
Chapter 9 Section 5

• The p orbitals is used to form the p bonding
  system in the NO3. The p electrons are said to
  be delocalized over the NO3ions.