Overlap and Bonding

1
Overlap and Bonding

• We think of covalent bonds forming through the
  sharing of electrons by adjacent atoms.
• In such an approach this can only occur when
  orbitals on the two atoms overlap.

2
Overlap and Bonding

• Increased overlap brings the electrons and nuclei
  closer together while simultaneously decreasing
  electron-electron repulsion.
• However, if atoms get too close, the internuclear
  repulsion greatly raises the energy.

3

• HYBRIDIZATION
• VALENCE BOND THEORY (s and p bonds)
• BOND ORDER

4
Hybrid Orbitals

• Its hard to imagine tetrahedral, trigonal
  bipyramidal, and other geometries arising from
  the atomic orbitals we recognize.

5
Hybrid Orbitals

• Atomic orbitals can mix or hybridize in order to
  adopt an appropriate geometry for bonding.
• Hybridization is determined by the electron
  domain geometry.
• sp Hybrid Orbitals
• Consider the BeF2 molecule (experimentally known
  to exist)

6
Hybrid Orbitals

• Consider beryllium
• In its ground electronic state, it would not be
  able to form bonds because it has no
  singly-occupied orbitals.

7
Hybrid Orbitals

• But if it absorbs the small amount of energy
  needed to promote an electron from the 2s to the
  2p orbital, it can form two bonds.

8
Hybrid Orbitals

• Mixing the s and p orbitals yields two degenerate
  orbitals that are hybrids of the two orbitals.
• These sp hybrid orbitals have two lobes like a p
  orbital.
• One of the lobes is larger and more rounded as is
  the s orbital.

9
Hybrid Orbitals (sp)

• These two degenerate orbitals would align
  themselves 180? from each other.
• This is consistent with the observed geometry of
  beryllium compounds linear.

10
Hybrid Orbitals

• With hybrid orbitals the orbital diagram for
  beryllium would look like this.
• The sp orbitals are higher in energy than the 1s
  orbital but lower than the 2p.

11
(No Transcript)
12

• Since only one of the Be 2p orbitals has been
  used in hybridization, there are two unhybridized
  p orbitals remaining on Be.

13

• sp2 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 trigonal planar electron pair
  geometries have sp2 orbitals on the central atom.

14
Hybrid Orbitals (sp2)

• For boron

15
Hybrid Orbitals

• three degenerate sp2 orbitals.

16
Hybrid Orbitals

• With carbon we get

17
Hybrid Orbitals

• four degenerate
• sp3 orbitals.

18

• sp2 and sp3 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 lobs is 109.5?.
• All molecules with tetrahedral electron pair
  geometries are sp3 hybridized.

19

• Hybridization Involving d Orbitals
• Since there are only three p-orbitals, trigonal
  bipyramidal and octahedral electron domain
  geometries must involve d-orbitals.
• Trigonal bipyramidal electron domain geometries
  require sp3d hybridization.
• Octahedral electron domain geometries require
  sp3d2 hybridization.
• Note the electron domain geometry from VSEPR
  theory determines the hybridization.

20
Sp3d and sp3d2 Hybrid Orbitals

• For geometries involving expanded octets on the
  central atom, we must use d orbitals in our
  hybrids.

21
Hybrid Orbitals

• This leads to five degenerate sp3d orbitals
• or six degenerate sp3d2 orbitals.

22
Hybrid Orbitals

• Summary
• Draw the Lewis structure.
• Determine the electron domain geometry with
  VSEPR.
• Specify the hybrid orbitals required for the
  electron pairs based on the electron domain
  geometry.

23
(No Transcript)
24
(No Transcript)
25
Hybrid Orbitals

• Once you know the electron-domain geometry, you
  know the hybridization state of the atom.

26

• Examples Determine the hybridization on the
  central atom of each
• NCl3
• CO2
• H2O
• SF4
• BF3
• XeF4

27

• Examples Determine the hybridization on the
  central atom of each
• NCl3 sp3
• CO2 sp
• H2O sp3
• SF4 sp3d
• BF3 sp2
• XeF4 sp3d2

28

• Examples Determine the hybridization on EACH
  atom

29

• Examples Determine the hybridization on EACH
  atom

sp2 sp
sp3
sp3
sp2
30
Valence Bond Theory

• Hybridization is a major player in this approach
  to bonding.
• There are two ways orbitals can overlap to form
  bonds between atoms.

31
Sigma (?) Bonds

• Sigma bonds are characterized by
• Head-to-head overlap.
• Cylindrical symmetry of electron density about
  the internuclear axis.

32
Pi (?) Bonds

• Pi bonds are characterized by
• Side-to-side overlap.
• Electron density above and below the internuclear
  axis.

33
Single Bonds

• Single bonds are always ? bonds, because ?
  overlap is greater, resulting in a stronger bond
  and more energy lowering.

34
Multiple Bonds

• In a multiple bond one of the bonds is a ? bond
  and the rest are ? bonds.

35
Multiple Bonds

• In a molecule like formaldehyde (shown at left)
  an sp2 orbital on carbon overlaps in ? fashion
  with the corresponding orbital on the oxygen.
• The unhybridized p orbitals overlap in ? fashion.

36
Multiple Bonds

• In triple bonds, as in acetylene, two sp orbitals
  form a ? bond between the carbons, and two pairs
  of p orbitals overlap in ? fashion to form the
  two ? bonds.

37
Delocalized Electrons Resonance

• When writing Lewis structures for species like
  the nitrate ion, we draw resonance structures to
  more accurately reflect the structure of the
  molecule or ion.

38
Delocalized Electrons Resonance

• In reality, each of the four atoms in the nitrate
  ion has a p orbital.
• The p orbitals on all three oxygens overlap with
  the p orbital on the central nitrogen.

39
Delocalized Electrons Resonance

• This means the ? electrons are not localized
  between the nitrogen and one of the oxygens, but
  rather are delocalized throughout the ion.

40
Resonance

• The organic molecule benzene has six ? bonds and
  a p orbital on each carbon atom.

41
Resonance

• In reality the ? electrons in benzene are not
  localized, but delocalized.
• The even distribution of the ?? electrons in
  benzene makes the molecule unusually stable.

42

• General Conclusions
• Every two atoms share at least 2 electrons.
• Two electrons between atoms on the same axis as
  the nuclei are ? bonds.
• ?-Bonds are always localized.
• If two atoms share more than one pair of
  electrons, the second and third pair form
  ?-bonds.
• When resonance structures are possible,
  delocalization is also possible.

43

• Bond Order
• Bond Order total number of covalent bonds
  between 2 atoms
• Bond order 1 for single bond.
• Bond order 2 for double bond.
• Bond order 3 for triple bond.
• If there is resonance then divide the bond order
  by the number of atoms that share the resonance
  structure.
• Fractional bond orders are possible (with
  resonance)

44

• Examples Draw Lewis Structures (including
  resonance). Determine the total number of s and
  p bonds in the molecule, and the bond order of
  each bond
• O3
• SO3
• CO2

45

• Examples Draw Lewis Structures (including
  resonance). Determine the total number of s and
  p bonds in the molecule, and the bond order of
  each bond
• O3
• SO3
• CO2

2s and 1p Each bond 1.5
3s and 1p each bond 1.33
2s and 2p each bond 2
46

• Electron Configurations and Molecular Properties
• Two types of magnetic behavior
• paramagnetism (unpaired electrons in atom or
  molecule) strong attraction between magnetic
  field and molecule
• diamagnetism (no unpaired electrons in atom or
  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

47
Second-Row Diatomic Molecules

• Electron Configurations and Molecular Properties
• large increase in mass indicates paramagnetism,
• small decrease in mass indicates diamagnetism.