Covalent bonding

1. Structure 2. Models of bonding
and structure
Structure 2.2—The Covalent Model
Guiding question: What determines the
covalent nature and properties of a substance?
Syed Arshad Mushtaq

2. Resonance
Learning outcomes (S2.2.11)
After studying this topic, you should be able to:
Understand:
 Resonance structures occur when there is more than one possible
position for a double bond in a molecule.
 Benzene, C6H6, is an important example of a molecule which has
resonance.
Apply your knowledge to:
 Deduce resonance structures of molecules and ions.
 Discuss the structure of benzene from physical and chemical evidence.
( Will be discussed in S3.2)

3. RESONANCE
Resonance is a concept used to describe the structures
when there are multiple ways to depict the same
molecule.
If you can put a double bond in more than one
position, you will be expected to draw the
resonance structures.
The electrons are delocalized in the areas of the
double bonds and are spread out equally among all
bonding positions.
Bond strength and length are in between that of single
and double bonds.

4. Resonance structures allow us to depict all the possible
positions of the double bonds.
The true structure, however, is an intermediate form known
as a resonance hybrid.
Double arrows are placed between all resonance structures.
Ref: myweb.astate.edu

5. OZONE
www.chemwiki.ucdavis.edu

6. CARBONATE ION
www.archives.evergreen.edu

7. Delocalization
Resonance structures can be
described as a single structure
using the concept of
delocalization.
In covalent bonds, two atoms
share a pair of electrons, which
are localized between the two
atoms. Delocalization occurs
when electrons are shared by
more than two atoms in a
molecule or ion, as opposed to
being localized between a pair
of atoms.

8. Benzene and resonance
Learning outcomes (S2.2.12)
After studying this topic, you should be able to:
Understand:
Benzene, C6H6, is an important example of a molecule that has
resonance.
Apply your knowledge to:
Discuss the structure of benzene from physical and chemical
evidence.

9. BENZENE C6H6
www.pixmule.com

12. Expanded octets
Learning outcomes (S2.2.13)
After studying this topic, you should be able to:
Understand:
 some atoms can form molecules in which they have an expanded octet
of electrons.
Apply your knowledge to:
 represent Lewis formulas for species with five and six electron domains
around the central atom.
 predict the electron domain geometry and the molecular geometry for
these species

14. Drawing Lewis formulas for expanded octets
 Drawing Lewis structure for expended octet is same as for normal molecules, the only
difference is central atom have more than eight electrons

15. Five electron domains: trigonal bipyramidal geometry

17. Six electrons' domains: octahedral geometry

18. Formal charge
Learning outcomes (S2.2.14)
After studying this topic, you should be able to:
Understand:
 Formal charge values can be calculated for each atom in a species and
used to determine which of several possible Lewis formulas is preferred
Apply your knowledge to:
 Apply formal charge to determine a preferred Lewis formula from
different Lewis formulas for a species

19. FORMAL CHARGE
 Sometimes it is possible to come up with more than one
valid Lewis structure.
 This is most often seen in molecules that can have an
expanded octet.
 One such example is sulfur dioxide, SO2.
 Two resonance structures or the expanded octet.

20. FORMAL CHARGE
Which structure is the most stable?
The concept of formal charge treats covalent bonds as
if they were purely covalent with equal electron
distribution and ignores electronegativity.
You basically count how many electrons belong to
each atom in the Lewis structure and compare this
with the number of valence electrons in the non-
bonded atom.
The difference is the formal charge.

21. How to calculate formal charge
FC = V-(1/2B + L)
The lower the formal charge, the more stable the structure.

22. Back to SO2
 The formal charge for the resonance structure:
Sulfur has 6 valence electrons, 3 bonding pairs and 1 lone pair (2e-).
FC = 6 – (3 +2) = +1
The left oxygen has 6 valence e-, 2 bonding pairs and 2 lone pairs (4e-).
FC = 6 – (2+4) = 0
The right oxygen has 6 valence e-, 1 bonding pair and 3 lone pairs (6e-).
FC = 6 – (1 + 6) = -1

23.  The expanded octet structure:
Sulfur has 6 valence e-, 4 bonding pairs and 1 lone pair (2e-) so
FC = 6 – (4+2) = 0
Both oxygens – 6 valence e-, 2 bonding pairs and 2 lone pairs (4e-) so
FC = 6 – (2 + 4) = 0

24. 0 0
0
0
+1
-1
The more stable structure would be the expanded octet with the formal charge
closest to zero.
Formal charge does ignore electronegativity so the most stable structure should
have the lowest formal charge and the negative values for formal charge on the
most electronegative atom.

25. Example with N2O
 Figure out the best Lewis structure using FC.
https://www.youtube.com/watch?v=ZlHIQhJlWNs

26. Sigma bonds (Ω) and pi bonds (π)
Learning outcomes (S2.2.14)
After studying this topic, you should be able to:
Understand:
 Sigma bonds σ form by the head-on combination of atomic orbitals
where the electron density is concentrated along the bond axis.
 Pi bonds π form by the lateral combination of p-orbitals where the
electron density is concentrated on opposite sides of the bond axis.
Apply your knowledge to:
 Deduce the presence of sigma bonds and pi bonds in molecules and
ions.

27. Sigma bonds
 All single covalent bonds are sigma
bonds.
 Sigma bonds form by the overlap of
orbitals along the bond axis.
 The following form sigma bonds:
 s and s (H2)
 s and p (HCl)
 p and p (Cl2)
 Hybrid orbitals and s
 Hybrid orbitals and hybrid orbitals

28. Pi Bonds
 All double covalent bonds contain one pi and
one sigma bond.
 All triple covalent bonds contain two pi and one
sigma bond.
 The following form pi bonds:
 p and p sideways
 Pi bonds are weaker than sigma bonds since
their electron density is farther away from the
positive nucleus.
 The double bonds break more readily and are
more reactive than those with only sigma
bonds.

29. Sigma σ and Pi π bonds

30. Hybridization
Learning outcomes (S2.2.14)
After studying this topic, you should be able to:
Understand:
 Hybridization is the concept of mixing atomic orbitals to form new
hybrid orbitals for bonding..
Apply your knowledge to:
 Analyse the hybridization and bond formation in molecules and ions.
 Identify the relationships between Lewis formulas, electron domains,
molecular geometry and type of hybridization.
 Predict the geometry around an atom from its hybridization, and vice
versa.

31. Hybridization
Hybridization is the concept of mixing atomic orbitals to form new
hybrid orbitals for bonding

32. What about carbon?
 Carbon almost always forms four covalent
bonds.
 Its electron configuration though is
1s22s22px
12py
1 which would lead one to expect
carbon to only form two covalent bonds due to
the two singly occupied orbitals.
 Because carbon does form 4 and not 2 covalent
bonds, its ground state electron configuration
changes during bonding.
The process of “excitation” occurs so that one of the “2s” electrons is excited to the
unoccupied “p” orbital, thus giving it four singly occupied orbitals available for bonding.

33. Why hybridization?
Let’s look at methane – CH4
It has a tetrahedral shape with identical bonds and bond angles of 109.5
degrees.
We know that the bonding is happening between the valence electrons of
the carbon and four hydrogen atoms.
All of the hydrogen valence electrons come from the 1s orbitals. However,
the carbon atomic orbitals available for bonding are 1 – s and 3-p orbitals
after the process of excitation.
It would be expected that the energies from these bonds would be different
since they are being formed from different types of orbitals.

34. However, we know that the four C-H
bonds in methane have the same
energy so somehow the orbitals have
been changed and been made equal
during the bonding process.
When unequal atomic orbitals mix to
form new hybrid atomic orbitals which
are the same as each other but
different from the original orbitals,
hybridization has occurred.

35. Comments about hybrid orbitals
1. They do not exist in isolated atoms.
2. They are found only in covalent compounds.
3. They are equivalent in a compound.
4. The number of hybrid orbitals in a bonded atom is equal to the
number of atomic orbitals used to form the hybrid orbitals.
5. The type of hybrid orbitals depends upon the electron domain
geometry.
6. The atom can form stronger covalent bonds using hybrid
orbitals.

36. Hybridization animation
• Formation of Methane molecule
• https://www.chemtube3d.com/orbitalshybrid/
• Formation of Ethene molecule
• https://www.chemtube3d.com/orbitalsethene/
• Formation of Ethyne molecule
• https://www.chemtube3d.com/orbitalsacetylene/

37. sp3 Hybridization
The previous example which
explained the bonding in methane,
CH4, is a classic example of sp3
hybridization.
This type of hybridization occurs
when the three “p” orbitals and one
“s” orbital hybridize to form four
identical sigma bonds.
The shape is tetrahedral and has
bond angles of 109.5 degrees.

38. Methane molecule

39. sp2 Hybridization
When carbon forms a double bond as
in ethene, it undergoes sp2
hybridization.
This type of hybridization occurs
when the two “p” orbitals and one “s”
orbital hybridize to form three hybrid
orbitals and leaves one unhybridized
“p” orbital.
The shape is trigonal planar with bond
angles of 120 degrees.
The unhybridized “p” orbitals overlap
sideways forming a pi bond.

40. Ethene Molecule

41. sp Hybridization
When carbon forms a triple bond as
in ethyne, it undergoes sp
hybridization.
This type of hybridization occurs
when the one “p” orbitals and one
“s” orbital hybridize to form two
hybrid orbitals and leaves two
unhybridized “p” orbitals.
The shape is linear with bond angles
of 180 degrees.
The unhybridized “p” orbitals
overlap sideways forming two pi
bonds.

42. Ethyne molecule

43. Ref: jahschem.wikispaces.com

45. https://www.chemtube3d.com/orbitalsbenzene/

46. Citations
 Oxford Resource for IB Chemistry Course Companion 2023 Edition
 Brown, Catrin, and Mike Ford. Higher Level Chemistry. 2nd ed. N.p.: Pearson
Baccalaureate, 2014. Print.
 Most of the information found in this power point comes directly from these
textbooks.
 https://www.thinkib.net/chemistry
 International Baccalaureate Organization. Chemistry Guide, First assessment 2016.
Updated 2015
 The power point has been made to directly complement the Higher Level Chemistry
textbooks and is used for direct instructional purposes only.