1. Explaining Molecular Formulas
Pg. 89 #5 (a-f), 6 (a-e)
Section 3.2 (pg )
Pg. 90 #1-4, 6

2. Explaining Molecular Formulas
Objectives:
Draw electron dot diagrams of atoms and molecules, writing structural formulas for molecular substances using Lewis structures to predict bonding in simple molecules
Illustrate, by drawing or building models, the structure of simple molecular substances
Explain why the formulas for molecular substances refer to the number of atoms of each constituent element
Section 3.2 (pg )

3. Molecular Elements
Many molecular elements are diatomic and some are polyatomic
You will need to memorize the formulas of the 9 molecular elements as they will not be given to you:
Name
Symbol
hydrogen
H2(g)
nitrogen
N2(g)
oxygen
O2(g)
fluorine
F2(g)
chlorine
Cl2(g)
iodine
I2(g)
bromine
Br2(g)
phosphorous
P4(g)
sulfur
S8(g)

4. In structural formulas, lone pairs are not shown
Why are they diatomic?
Remember fluorine has 7 valence e-’s and needs 1 more e- to be stable?
Well 2 fluorine atoms could obtain a stable octet of e-’s if they shared a pair with each other
F - F F2
In structural formulas, lone pairs are not shown
Remember: This is a simplified 2-D version, not where the electrons actually are

5. Diatomic Elements O = O N Ξ N What about oxygen and nitrogen?
Would sharing only 1 electron each work?
What about oxygen and nitrogen?
O = O
Each oxygen atom only has 6 valence electrons. So by sharing 2 electrons each, the two oxygen atoms can create a full octet. This creates a double bond
N Ξ N
Each nitrogen atom only has 5 valence electrons. So by sharing 3 electrons each, the two nitrogen atoms can create a full octet. This creates a triple bond

6. Molecular Compounds Background:
Molecular compounds have covalent bonds (shared electrons) between non-metals and non-metals
Can be solid, liquid or gas as SATP
May or may not be soluble in water (more later)
Don’t ever conduct electricity - even when (aq)
Generally have lower m.p. and b.p than ionic compounds
Review from Section 1.5 Notes

7. Molecular Compounds Background:
Empirical Formulas – show the simplest whole number ratios of atoms in a compound
Very useful for IONIC compounds
Formula Unit – the ratio of ions that repeats in a pattern within the crystal; the chemical formula of ionic compounds represents the formula unit
Not useful for MOLECULAR compounds
Na242Cl242  Na16Cl16  NaCl
CH  C2H2  acetylene
 C6H6  benzene
 C8H8  octene
All are extremely different compounds but the empirical formula would be the same

8. Molecular Compounds Background: C2H4O2  CH3COOH
Molecular Formulas – a molecular formula shows the actual number of atoms that are covalently bonded to make-up each molecule
We use this because chemical formulas for molecular compounds result from sharing electrons, therefore a variety of compounds are possible (which we determine empirically through experiments)
Often the symbols are written in a sequence that helps you determine how the atoms are bonded
C2H4O2  CH3COOH
Empirical formula: CH2O incorrect for molecular compounds

9. Review of Molecular Compound Formulas
See
Pg. 88

10. Q: How do we know how molecular compounds bond?
(Aka: How do we draw Lewis Formulas?)
Where do these come from???

11. Determining Lewis Formulas
Bonding Capacity: the maximum number of single covalent bonds that an atom can form
REMINDER: How many e-’s does an atom want in its valence energy level to be satisfied?
H = 2 e- C, N, O, F, P, S, Cl, etc. = 8 e-
REMINDER: What types of covalent bonds are possible?
F – F single = sharing one e- pair
O = O double = sharing two e- pairs
N Ξ N triple = sharing three e- pairs

12. Determining Lewis Formulas
So why do we care about bonding capacity?
If we know how many bonding e-’s an atom has, we can predict what structure a molecular compound will have
Atom
Number of
valence electrons
bonding electrons
Bonding capacity
carbon 4
nitrogen 5 3
oxygen 6 2
halogens 7 1
hydrogen H
I.e. Carbon can form 4 single bonds, 2 double bonds, 1 triple and 1 single, or 1 double and 2 singles

13. Lewis Formulas- Guided Ex. #1
Determine the Lewis formula and structural formula for sulfur trioxide, SO3(g)
1. Count the number of valence electrons there are in total? (If polyatomic ions are included, subtract or add electrons to account for the net charge)
Oxygen = 3 atoms x 6 valence e-’s each = 18 valence e-’s
Sulfur = 1 atom x 6 valence e-’s each = 6 valence e-’s
24 valence e-’s

14. Lewis Formulas – Guided Ex. #1
2. Choose your central atom
In our course, we will limit our formulas to ones with one central atom (unless extra info is provided)
So how do you know which is the central atom?
Usually the one in lesser quantity (SO3(g)) OR
The one with the higher bonding capacity
Carbon usually – because 4 is the highest bonding capacity
So which is the central atom?

15. Lewis Formulas – Guided Ex. #1
3. Arrange peripheral atoms around central atom and place one pair of valence e-’s between them 4. Place lone pairs on all peripheral atoms to complete their octet 5. Place any remaining valence e-’s on the central atom as lone pairs. For this example, all 24 have been assigned

16. Lewis Formulas – Guided Ex. #1
6. If the central atom’s octet is not complete, move a lone pair from a peripheral atom to a new position between the peripheral and central atom.
7. Show the structural formula but omit lone pairs and replace every bond with a line
(Count the electrons around each atom to confirm the octet rule. Each atom should have 8 e-’s around it; exception: H)

17. Lewis Formulas – Guided Ex. #2
Determine the Lewis formula & structural formula for the nitrate ion, NO3-
1. Count the valence electrons (*look for a net charge if an ion).
nitrogen = 1 x 5 valence e-’s = 5
oxygen = 3 x 6 valence e-’s = 18
(b/c net charge is -1) = 24
2. Which is the central atom? Nitrogen (in lesser quantity)
3. Arrange peripheral atoms around central atom and
place 1 pair of valence e-’s between them N

18. Lewis Formulas – Guided Ex. #2
4. Place lone pairs on all peripheral atoms to complete their octet
5. Place any remaining valence e-’s on the central atom as lone pairs.
6. If the central atom’s octet is not complete, move a lone pair from a peripheral atom to a new position between the peripheral and central atom.
7. If the entity is a polyatomic ion, place square brackets around the entire Lewis formula and then write the net charge outside the bracket on the upper left. N N N

19. Practice Pg. 89 #5 (a-f), 6 (a-e) Pg. 90 #1-4, 6
Watch 5 (f) there is an exception noted. The central atom does not follow the octet rule.
We will go through these answers as a class.
Pg. 90 #1-4, 6
The theory presented today is not absolute – there are exceptions. But rather than presenting a more detailed theory, your textbook will always note such exceptions.

20. Tomorrow... Molecular Model Investigation Thought Lab Investigation
Morse Code Assignment