My presentation for the HL content of T14.

1. T14- A closer look at covalent bonding & structure.

2. Lesson outline…..
• Lesson 1- Topic 4 review.
• Lesson 2- More shapes of molecules.
• Lesson 3- Deciding on the best resonance structure (F.C.)
• Lesson 4- O3- A case study in resonance, molecular polarity and FC.
• Lesson 5- Mixing it all up! (orbital hybridization)

3. 14.1- Further aspects of
covalent bonding & structure.

4. Lesson 1- Topic 4 review☺
Level 4: Recall the keyterms used to describe the
bonding in the four types of solid.
Level 7: Deduce the molecular shape and
polarity of covalently bonded molecules up to
4 electron domains.
Level 5/6: Explain the physical properties of
various compounds based on their structure
and bonding.

5. Starter-Try this quizlet, then get stuck into the
review questions on topic 4 Bonding & structure.
https://quizlet.com/189340897/ib-chemistry-topic-4-bonding-flash-cards/

6. Lesson 2- More shapes of molecules.
Level 4: State what V.S.E.P.R. theory is.
Level 7: Determine the polarity of a molecule
from its molecular shape.
Level 5/6: Deduce the shapes of compounds up
to and including 5&6 electron domains.

7. Starter
• In year 12, we looked at drawing Lewis structures and from these then determined the correct
molecular shape of a covalent compound. We worked up to compounds with 4 electron domains
around the central atom. Try these as a reminder:
CO2 ,BF3, *SO2, CH4, NH3, H2O, HCl
1. Draw the Lewis diagram/structure.
2. Determine the electron geometry around the central atom.
3. Determine the molecular geometry based on the number of bonding and
non-bonding electrons.
4. Deduce whether each molecule is polar or non-polar overall.

8. Molecular shapes up to 4 electron domains (‘clouds’)
Pearson pg 189

9. Expanded octet.
• Small atoms like H, He, Be and B form stable molecules with fewer
than 8 electrons.
• Atoms of elements in period 3 and below may expand their octets by
using d orbitals in their valence shell.
• This can allow for molecules with 5 or 6 electron domains around a
central atom…..

10. Molecules with5 electrondomains.
https://phet.colorado.edu/sims/html/molecule-shapes/latest/molecule-shapes_en.html

11. Moleculeswith 6 electron domains.
https://phet.colorado.edu/sims/html/molecule-shapes/latest/molecule-shapes_en.html

12. Polar or non-polar?
Always follow these steps:
1. Draw the correct molecular shape.
2. Identify any polar bonds.
3. Determine if the polar bonds are
arranged symmetrically or
asymmetrically.
4. Asymmetrically arranged polar
bonds will not cancel each other
out.
5. Symmetrically arranged polar bonds
will cancel each other out.

13. Check for learning…. (*print out- with answers)

14. Lesson 3- Deciding on the best resonance structure.
Level 4: Recall the equation for determining the
formal charge (F.C.) of a covalent molecule.
Level 7: Explain what a resonance hybrid is and
determine bond order in a hybrid structure.
Level 5/6: Determine the most stable of two
non-equivalent resonance structures using FC.

15. Starter- deciding on the ‘best’ resonance
structure…
If you draw all of the possible Lewis structures for the molecule SO2,
you should come up with 3. Try and draw them :
Q. How do we determine which of these is the most stable structure?

16. (Formal charge) FC = V – (1/2B+ L)
Note-thisis notin yourdatabook,so finda wayto memorisethis…
• All covalent bonds are treated as
pure covalent, i.e. that each
electron pair is shared evenly(so
ignore concept of electronegativity)
• The lower the formal charge on a
resonance structure, the more
stable it is considered to be.
• Only used to compare two non-
equivalent structures (no point in
comparing equivalent structures as
they will have the same F.C.!)
Vs.
Taken from Pearson pg 192

17. Your turn to try FC = V – (1/2B+ L)

18. Q. What if both resonance structures give the
same FC?!
N2O is an interesting molecule. Use formal charge to decide which structure
is the most stable:
*Key point; if two non-equivalent resonance structures result in the same
formal charge, then you must consider the electronegativity of each element
involved and put the –ive charges on the more electronegative element, so
structure (i) is preferred.

19. Let’s practice!

21. Resonance hybrids.
The different resonance structures for a molecule are sometimes represented as a resonance hybrid
structure. Some important examples to know would be:
SO3(l)
Benzene(l)

22. *Bond order in resonance hybrids- remember:
Double bond- ‘bond order’ of 2
Single bond- ‘bond order’ of 1.
Q. What would the bond order be for each bond in these two hybrid structures?
SO3 B.O. = 4 pairs of e- /3 positions= 1.3
Benzene B.O.= 9 pairs of e- / 6 positions= 1.5

23. Lesson 4- Ozone- A case study in resonance, molecular
polarity and FC.
Level 4: State the main structural differences
between O2 and O3.
Level 7: Explain the significance of ozone being
able to absorb UV-B and UV-C light.
Level 5/6: write general equations for the
catalytic destruction of ozone by NOx and Cl.

24. Starter- The molecular structure of O3.
White boards please☺
1. Draw the two possible Lewis structures for O3 (which obey the
octet rule).
2. Now draw the molecule with the correct molecular shape.
3. *Is Ozone a polar or non-polar molecule?
4. What is its formal charge?
5. What would the resonance hybrid look like?
6. What would the bond order be in this hybrid structure?
7. What is the bond order in the O2(g) molecule?
8. Why does it take radiation of lower intensity to break down O3?
*Great summary video here: https://www.youtube.com/watch?v=wSNMkeCht7c

25. What is ozone?!
https://www.youtube.com/watch
?v=0ZfBgjUnXIs
Watch to 3.48min

26. The natural ozone ‘cycle’
In this cycle you must remember that both O2 and O3 gas molecules are
split with UV radiation of the appropriate wavelengths (<242 and
<330nm respectively)
Both of these initial reactions will yield O
. radicals.
These free radicals can then further react with O3 to break it down into
2 molecules of O2. The cycle would stop here normally and is only
propagated/kept going by more UV light.

27. Discovering the CFC problem- see handout of this story(slide doesn’t work)

28. 1. Catalytic destruction of O3 by CFCs.
Two equations to write which should
add up to the nett equation for the
normal ozone cycle (see notes)
https://www.youtube.com/watch?v=
ppSf35Jvd4w

29. 2. Catalytic destruction of O3 by NOx gasses.
Two equations to write which
should add up to the nett
equation for the normal ozone
cycle (see notes)
Unpaired electron, highly
reactive (‘radical’!)

30. Reduction of these pollutants.
CFCs.
• Use alternative gasses in
aerosol cans and fridges!
• Hydro fluoro carbon gasses e.g.
CH2F2(g).
• Simple alkanes e.g. CH4(g).
• However, these are thought to
contribute to the greenhouse
effect!:/
NOx gasses
• Introduction of catalytic
convertor into all cars as
standard in the 70s.
Catalytic convertor:
https://www.youtube.com/watch
?v=rmtFp-SV0tY

31. Check for learning…

32. Answers…

33. 14.2- Hybridization
(of orbitals in covalent bonding)

34. Lesson 5- Mixing it all up!
Level 4: Draw electron box/orbital diagrams to
illustrate the process of hybridization.
Level 7: Explain the reason for the formation of
sp3, sp2 or sp hybrid orbitals in methane,
ethene and ethyne.
Level 5/6: Identify the orbital hybridization in
simple covalent molecules as sp3, sp2 or sp.

35. Starter- A focus on our old friend carbon.
Carbon atoms serve as the backbone of
the entire field of organic chemistry. In
all organic compounds that we draw
with carbon, we always show the
carbon atom with 4 covalent bonds.
If the electron configuration for a
carbon atom is: 1s2 2s2 (full) 2p2, how
can the 2nd energy level accommodate 4
equivalent covalent bonds!?

36. Step 1-The formation of covalent bonds starts with the
excitation of electrons in atoms!!
• Excitation of carbon showing
promotion of the paired s orbital
electron (in red) to the empty pz
orbital.
• This results in four singly
occupied orbitals being available
for bonding!

37. Step 2- The s and p orbitals (within the atom) then mix together.
• Although we now have 4 singly
occupied orbitals, they are clearly not
all of the same energy.
• They therefore hybridize/mix together to
form an intermediate hybrid orbital of in-
between energy.
• Hybrid orbitals have different energies,
shapes and orientation in space than their
‘parent’ orbitals.
From Pearson pg. 201- Great analogy!
Go to this link and watch the videos on hybridization:
http://glencoe.mheducation.com/sites/0003152012/st
udent_view0/chapter10/animations.html#

38. sp3 hybridization
(tetrahedral shape/4 single bonds)
*A ‘sigma’ bond σ is a bond between the s-orbital of another atom and a hybrid orbital. In this case, the
normal s orbital of a H atom and one of the sp3 hybrid orbitals of the carbon atom.
Now go to this link again and view the video on sigma and pi bonding for ethane to begin with:
http://glencoe.mheducation.com/sites/0003152012/student_view0/chapter10/animations.html#

39. sp2 hybridization
(trigonal planar shape 2 single bonds+ 1 double= 4 bonds)
A double bond is made up of a sigma bond (as before) and a pi bond (π) . Remember, the sigma bond is
formed between the sp2 hybridized orbital of the carbon atom, while the pi bond is formed between the
unhybridized p orbitals on adjacent carbons.
Now go and view ethene on sigma and pi bonding at this link:
http://glencoe.mheducation.com/sites/0003152012/student_view0/chapter10/animations.html#

40. sp hybridization
(linear shape- 1 single bond + 1 triple bond= 4 bonds)
A triple bond is made up of 1 sigma bond and two different pi bonds. This is a very electron
rich/dense area which is a real target for electrophilic addition (e.g. H+ ions)
Now go and view ethyne (‘acetylene’) on sigma and pi bonding at this link:
http://glencoe.mheducation.com/sites/0003152012/student_view0/chapter10/animations.html#

41. Benzene- a special case study….
• Look at the model of benzene.
Q. What sort of hybridization do the
carbon atoms have?
Q. Identify the orbital which is
represented by the purple and pink.
Q. What do you notice about these
orbitals on the model?
Q. What do we call this linking of
orbitals?
Q. Is benzene more/less reactive than
1,3,5- cyclohexatriene?

42. Hybridization and molecular shape
(up to 4 electron domains only).

44. The answers! (note- for more detail, see worked solutions from Pearson on my conference)
43. Wider spacing between the electrons in a pi bond. Closer spacing in a
sigma bond.
45. (a)1 single and a double bond = sp2
(b) 4 single bonds = sp3
(c) trigonal planar shape= sp2
(d) linear, two single bonds= sp
(e) 1 double and 2 singles = sp2
46. cyclohexane, all single bonds, sp3 hybridization, bond angle of 109.5
benzene, 1 single + 1 double (per carbon atom), sp2, bond angle of 120,
flat on the page☺