Module 12: Valence Shell Electron Pair Repulsion Theory and Valence Bond Theory

1
Module 12 Valence Shell Electron Pair Repulsion
Theory and Valence Bond Theory

• By Alyssa Jean-Mary
• Source Modular Study Guide for First Semester
  Chemistry by Anthony J. Papaps and Marta E.
  Goicoechea-Pappas

2
Valence Shell Electron Pair Repulsion Theory
(VSEPR Theory)

• The VSEPR theory is useful in predicting
  geometries. It assumes
• Each valence shell electron pair around the
  central atom is significant
• Repulsions among valence shell electron pairs
  determine the shape of the molecule
• Valence shell electron pairs around the central
  atom are arranged so that repulsions among them
  are as small as possible.
• Order of repulsion LP/LP gtgt LP/BP gt BP/BP

3
Steps to Predicting Molecular Geometry and Bond
Angles

• Step 1 Draw the Lewis Structure for the species.
• Step 2 Count the number of bonding pairs and the
  number of lone pairs around the central element.
• A double bond or a triple bond is 1 bonding pair.
• Step 3 Use the table on the next slide to
  predict the geometry.
• Step 4 If lone pairs are present, bond angles
  are slightly less then those given in the table,
  except if it is linear, which is 180, or square
  planar, which is 90.
• If only 2 atoms are present in a species, it is
  not appropriate to speak of bond angles.

4
VSEPR VB Theory Molecular Geometries
5
Examples of Predicting Molecular Geometry and
Bond Angles

• Predict the molecular geometry and the bond
  angles for the following compounds
• CO2
• AlH3
• HCl
• NH2-
• IF4
• IF4-

6
Valence Bond Theory (VB Theory)

• The VB theory assumes
• Covalent bonds are formed when atomic orbitals on
  different atoms overlap and electrons are shared
• Each lone pair occupies a separate orbital
• Hybridization mixing of a set of valence atomic
  orbitals to form a new set of atomic orbitals
  with the same total number of electron capacity
  and with properties and energies intermediate
  between those of the original unhybridized
  orbitals
• Hybridization provides the necessary atomic
  orbitals that explain the experimentally observed
  geometry of some molecules
• Hybridization allows the central element to have
  orbitals that can overlap more effectively, which
  allows for a stronger bond to form, with the
  atoms surrounding it

7
VB Theory Valence Bond Picture of AB2 Molecules
with No Unshared Electrons on A Linear Molecules

• For example BeCl2
• Here, since Be bonds with 2 Cl, it must have two
  orbitals, each with 1 electron, to bond to each
  Cl electron that is unpaired (each Cl has only 1
  electron that is unpaired (in 3p)). But, since
  the last orbital in Be is full (2s), it has to
  promote 1 of these paired electrons to the next
  highest energy level (2p) in order to have two
  partially filled orbitals (a 2s and a 2p) that
  could then bond with the 2 Cl. But, these two
  orbitals (2s and 2p) cannot effectively and
  equally overlap with the 3p orbital of Cl, so the
  two Be-Cl bonds would be different. This does not
  agree with experimental observations, which show
  that both Be-Cl bonds are equivalent in bond
  length and bond strength. So, in addition to the
  promotion of an electron, hybridization of the 2s
  and one of the 2p orbitals occurs. Hybridization
  produces two new orbitals, which are called sp
  hybrid orbitals. Each of these sp hybrid
  orbitals contains 1 electron. Thus, each 3p Cl
  orbital with 1 electron will overlap with each of
  the two sp hybrid orbitals in Be. The orbitals
  will be arranged to that repulsions among them
  are as small as possible (the VSEPR theory),
  which results in the Be and the 2 Cl nuclei lying
  in a straight line, which agrees with
  experimental observations, which show the
  molecule as linear.

8
VB Theory Valence Bond Picture of AB3 Molecules
with No Unshared Electrons on A Trigonal Planar
Molecules

• For example AlH3
• Here, since Al bonds with 3 H, it must have three
  orbitals, each with 1 electron, to bond to each H
  electron that is unpaired (each H has only 1
  electron that is unpaired (in 1s)). But, since
  the last orbital in Al has only 1 unpaired
  electron (3p), it has to promote 1 of its paired
  electrons from 3s to 3p in order to have the
  three partially filled orbitals (a 3s and two 3p)
  it needs to bond to 3 H. This promotion leads to
  hybridization of the 3s and two of the 3p
  orbitals, which produces three new orbitals,
  which are called sp2 hybrid orbitals. Each of
  these three sp2 hybrid orbitals on Al can bond
  to each 1s orbital on the 3 H.

9
VB Theory Valence Bond Picture of AB4 Molecules
with No Unshared Electrons on A Tetrahedral
Molecules

• For example CH4
• Here, since C bonds with 4 H, it must have four
  orbitals, each with 1 electron, to bond to each H
  electron that is unpaired (each H has only 1
  electron that is unpaired (in 1s)). But, since
  the last orbital in C has only 2 unpaired
  electrons (2p), it has to promote 1 of its paired
  electrons from 2s to 2p in order to have the four
  partially filled orbitals (a 2s and three 2p) it
  needs to bond to 4 H. This promotion leads to
  hybridization of the 2s and the three 2p
  orbitals, which produces four new orbitals, which
  are called sp3 hybrid orbitals. Each of these
  four sp3 hybrid orbitals on C can bond to each
  1s orbital on the 4 H.

10
VB Theory Valence Bond Picture of AB3 Molecules
with 1 Unshared Pair of Electrons on A Trigonal
Pyramidal Molecules

• For example NH3
• Here, since N bonds with 3 H, it must have three
  orbitals, each with 1 electron, to bond to each H
  electron that is unpaired (each H has only 1
  electron that is unpaired (in 1s)). N does have 3
  orbitals, each with 1 electron, so it would seem
  that, in this case, hybridization is not needed.
  But, hybridization is needed to account for the
  experimental observations of NH3. So,
  hybridization of the 2s and the three 2p orbitals
  occurs, which produces four new orbitals, which
  are called sp3 hybrid orbitals. One of these
  sp3 hybrid orbitals on N has a lone pair
  present, and each of the other ones (three) sp3
  hybrid orbitals on N can bond to each 1s orbital
  on the 3 H.
• This shows that, on the central element, all
  valance orbitals, whether they have 1 electron or
  2 electrons preset, must be hybridized.

11
VB Theory Valence Bond Picture of AB5 Molecules
with No Unshared Electrons on A Trigonal
Bipyramidal Molecules

• For example PH5
• Here, since P bonds with 5 H, it must have five
  orbitals, each with 1 electron, to bond to each H
  electron that is unpaired (each H has only 1
  electron that is unpaired (in 1s)). But, since
  the last orbital in P has only 3 unpaired
  electrons (3p), it has to promote 1 of its paired
  electrons from 3s to 3d in order to have the five
  partially filled orbitals (a 3s, three 3p, and a
  3d) it needs to bond to 5 H. This promotion leads
  to hybridization of the 3s, the three 3p, and the
  3d orbitals, which produces five new orbitals,
  which are called sp3d hybrid orbitals. Each of
  these five sp3d hybrid orbitals on P can bond
  to each 1s orbital on the 5 H.

12
VB Theory Valence Bond Picture of AB6 Molecules
with No Unshared Electrons on A Octahedral
Molecules

• For example SH6
• Here, since S bonds with 6 H, it must have six
  orbitals, each with 1 electron, to bond to each H
  electron that is unpaired (each H has only 1
  electron that is unpaired (in 1s)). But, since
  the last orbital in S has only 2 unpaired
  electrons (3p), it has to promote 2 of its paired
  electrons, 1 from 3s to 3d and 1 from 3p to 3d in
  order to have the six partially filled orbitals
  (a 3s, three 3p, and two 3d) it needs to bond to
  6 H. This promotion leads to hybridization of the
  3s, the three 3p, and the two 3d orbitals, which
  produces six new orbitals, which are called
  sp3d2 hybrid orbitals. Each of these six sp3d2
  hybrid orbitals on S can bond to each 1s orbital
  on the 6 H.

13
VB Theory Valence Bond Picture of Compounds
Containing Double Bonds

• For example C2H4
• To bond everything singly, each C will have 3
  bonds (1 to the other C and 2 to each of 2 H), so
  they each need three orbitals, each with 1
  electron. But, since the last orbital in C has
  only 2 unpaired electrons (2p), it has to promote
  1 of its paired electrons from 2s to 2p in order
  to have the three partially filled orbitals (a 2s
  and two 2p) it needs to make the 3 bonds. This
  promotion leads to hybridization of the 2s and
  the two 2p orbitals, which produces three new
  orbitals, which are again called sp2 hybrid
  orbitals. Each of these three sp2 hybrid
  orbitals on C can bond to each 1s orbital on the
  2 H and to a sp2 hybrid orbital on the other C.
  But, when the electron was promoted from 2s to
  2p, four partially filled orbitals were actually
  created, even though only three were hybridized.
  The other orbital (2pz) remains unhybridized.
  Since having each C singly bonded to three
  elements (the other C and 2 H) does not complete
  each Cs octet (it only gives each of them 6
  electrons), another bond is needed between them.
  This additional bond is created by overlapping
  each Cs unhybridized 2pz orbitals sideways. This
  type of bond is referred to as a pi bond
  (p-bond), whereas the other bonds created are
  referred to as sigma bonds (s-bonds).

14
VB Theory Sigma Bonds (s-bonds) Pi Bonds
(p-bonds)

• A sigma bond (s-bond) is a bond that results when
  atomic orbitals overlap head. In a sigma bond,
  the region of electron sharing is along a
  cylindrically symmetrical imaginary line
  connecting the bonded atoms.
• A pi bond (p-bond) is a bond that results when
  atomic orbitals overlap side-on. In a pi bond,
  there are two regions of electron sharing. These
  regions are on opposite sides of the imaginary
  line connecting the bonded atoms and parallel to
  this line.
• p-bonds are weaker than s-bonds
• Double bonds have one s-bond and one p-bond. The
  p-bond is perpendicular to the s-bond.
• Triple bonds have one s-bond and two p-bonds. The
  two p-bonds are perpendicular to each other as
  well as perpendicular to the s-bond.
• The order for bond strength triple bond gt double
  bond gt single bond

15
VB Theory Valence Bond Picture of Compounds
Containing Triple Bonds

• For example C2H2
• To bond everything singly, each C will have 2
  bonds (1 to the other C and 1 to a H), so they
  each need two orbitals, each with 1 electron. C
  does have 2 orbitals, each with 1 electron, so it
  would seem that, again, in this case,
  hybridization is not needed. But, to account for
  the experimental observations of C2H2 and taking
  into account that bonding everything singly is
  not the correct structure for C2H2, hybridization
  is still needed. So, it has to promote 1 of its
  paired electrons from 2s to 2p, which leads to
  hybridization of the 2s and the 2p orbitals
  (since only two orbitals are needed to make the 2
  bonds), which produces two new orbitals, which
  are again called sp hybrid orbitals. Each of
  these two sp hybrid orbitals on C can bond to
  the 1s orbital on the H and to a sp hybrid
  orbital on the other C. But, when the electron
  was promoted from 2s to 2p, four partially filled
  orbitals were actually created, even though only
  two were hybridized. The other two orbitals (2py
  and 2pz) remain unhybridized. Since having each C
  singly bonded to two elements (the other C and a
  H) does not complete each Cs octet (it only
  gives each of them 4 electrons), two more bonds
  are needed between them. These additional bonds
  are created by overlapping each Cs unhybridized
  2py orbitals and each Cs unhybridized 2pz
  orbitals sideways, thus creating 2 p-bonds.

16
Hybridization on Carbon

• More than 3 million organic compounds (i.e.
  compounds that contain carbon) are known, whereas
  only about 300,000 inorganic compounds (i.e.
  compounds that contain atoms other than carbon)
  are known
• Organic compounds are special since carbon can
  form stable bonds with other carbon atoms in
  addition to bonds with hydrogen, oxygen,
  nitrogen, and the halogens
• Since carbon doesnt have any lone pairs under
  normal circumstances, the hybridization and
  molecular geometry for carbon depends on the
  number of bonding pairs around carbon
• 4 bonding pairs sp3 hybridized tetrahedral in
  shape
• 3 bonding pairs sp2 hybridized trigonal
  planar in shape
• 2 bonding pairs sp hybridized linear in shape

17
Molecular Polarity

• Even if the bonds in a molecule are polar, the
  molecule as a whole could be non-polar (i.e. it
  could have no permanent dipole moment) if it is
  symmetrical.
• To determine if a molecule is polar or non-polar
• Step 1 - Draw its Lewis Structure.
• Step 2 Determine if it is symmetrical. It is
  symmetrical if
• the same atoms are bonded to the central element
  and the central element doesnt have any lone
  pairs
• the central element contains 2 identical bond
  pairs and 3 lone pairs (i.e. it is sp3d
  hybridized and linear in shape)
• the central element contains 4 identical bond
  pairs and 2 lone pairs (i.e. it is sp3d2
  hybridized and square planar in shape)

18
Examples Involving Molecular Polarity
19
THE END