Lecture 11 - Coulomb's Law, Lewis Structures I

Thursday, February 15, 2024

1:30 PM

"The beauty of chemistry is that I can design my own molecular world." - ﷟HYPERLINK "https://www.nobelprize.org/prizes/chemistry/1999/press-release/"Ahmed Zewail

Class notes can now be found here: https://bricejurban.github.io/CHEM111/
Assignments due this week:
﷟HYPERLINK "https://boisestatecanvas.instructure.com/courses/28699/assignments/990975"HW 5 - Quantum Part 2 (Aktiv Chemistry)(Sun 2/18)
New Activity: ﷟HYPERLINK "https://boisestatecanvas.instructure.com/courses/28699/assignments/993257"Emerging Tech in Chemistry
Post which technology you're interested in (Fri 2/16)
Reading for today's lecture 7.1-7.4, 7.8
Reading for next lecture 7.5-7.10
Office Hours: 
Friday 11-1 CIC 
﷟HYPERLINK "https://calendly.com/bricejurban/office-hours"By appointment

Today (2/15)
Coulomb's Law
Electronegativity
Covalent Bonding
Lewis Structures Part I
Formal Charge
Tuesday (2/20) 
Lewis Structures Part II
Coulomb's Law can be used to calculate the energy of an ion pair
Untitled picture.png QIQ2 
oulomb (231 aJ•pm) 
The energy of interaction between two ions is directly proportional to the product of their electrical charges and is inversely proportional to the distance between their centers.

The proportionality constant is 231 aJ•pm when:
The energy is expressed in attojoules (1 x 10-18 J)
The distance between the centers is in picometers (1 x 10-12 m) 
The charges (Q) are measured in units of the charge on a proton (+1, +2, -1, -3 etc)


What is the energy of attraction between a sodium ion and a fluoride ion?






Untitled picture.emf 
Charles-Augustin de
Coulomb

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What is the energy of attraction between a calcium ion and an oxide ion?











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Untitled picture.png TABLE 6.4 Crystallographic radii of ions in picometers (l pm 
Ion Radius 
Cations: 
Ion 
Cu24 
24 
Ni24 
Sr24 
Z 1124 
Radius 
135 
100 
95 
65 
73 
61 
72 
69 
118 
74 
Ion 
Gab* 
In 34 
La 34 
"I-13* 
Radius 
54 
23 
62 
55 
62 
80 
103 
89 
90 
Ion 
C e44 
Ti4+ 
44 
= 10-12 m) 
Radius 
87 
61 
89 
84 
Ion 
Radius 
Ion 
Tez- 
Radius 
184 
198 
221 
Ion 
Anions: 
Radius 
171 
212 
cuv 
115 
167 
77 
138 
76 
102 
152 
150 
196 
181 
154 
220 

Untitled picture.png TABLE 6.5 Electron affinities 
of the atoms of some reactive 
nonmetals 
Atom 
o 
s 
EA/aJ 
—0.12 
—0.545 
—0.580 
—0.540 
—0.490 
—0.234 
+1.30 (EA2) 
—0.332 
+0.980 (EA2) 

Untitled picture.png TABLE 4.1 Successive ionization energies of the elements hydrogen through sodium* 
Ionization energy/aJ 
Z 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
Element 
He 
Be 
c 
o 
2.18 
3.94 
0.86 
1.49 
1.33 
1.81 
2.32 
2.17 
2.78 
3.45 
0.83 
8.72 
12.1 
2.92 
4.04 
3.90 
4.75 
5.63 
5.60 
6.56 
7.57 
19.6 
24.7 
6.08 
7.67 
7.60 
8.80 
10.0 
10.2 
11.5 
34.9 
41.5 
10.3 
12.4 
12.4 
14.0 
15.6 
15.8 
54.5 
62.8 
15.7 
18.2 
18.3 
20.2 
22.2 
16 
78.5 
88.4 
22.7 
25.2 
25.3 
27.6 
107 
118 
29.7 
33.2 
33.4 
140 
153 
38.3 
42.3 
177 
192 
48.0 
218 
238 
264 
Useful tables
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Untitled picture.png TABLE 4.1 Successive ionization energies of the elements hydrogen through sodium* 
Ionization energy/aJ 
Z 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
Element 
He 
Be 
c 
o 
2.18 
3.94 
0.86 
1.49 
1.33 
1.81 
2.32 
2.17 
2.78 
3.45 
0.83 
8.72 
12.1 
2.92 
4.04 
3.90 
4.75 
5.63 
5.60 
6.56 
7.57 
19.6 
24.7 
6.08 
7.67 
7.60 
8.80 
10.0 
10.2 
11.5 
34.9 
41.5 
10.3 
12.4 
12.4 
14.0 
15.6 
15.8 
54.5 
62.8 
15.7 
18.2 
18.3 
20.2 
22.2 
16 
78.5 
88.4 
22.7 
25.2 
25.3 
27.6 
107 
118 
29.7 
33.2 
33.4 
140 
153 
38.3 
42.3 
177 
192 
48.0 
218 
238 
264 



Coulomb's law can be used in conjunction with ionization energies and electron affinities to estimate the energy gained in the formation of an ionic bond.
 Let's calculate the value of the energy change in aJ per ion pair for this reaction:   Na  +   H  → Na+H– 












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The lattice (crystal) energy is Coulomb's law applied to all of the neighboring ions




Bond Polarity
The difference in strength two atoms have on shared valence electrons
In an ionic bond, valence electrons are completely transferred to the anion. 
In a covalent bond, valence electrons are shared or mostly shared
Ionic Bond =  Unhappy baby
Untitled picture.png 
Covalent Bond = Tug of war
Untitled picture.png 
﷟HYPERLINK "https://ptable.com/#Properties/Electronegativity"Electronegativity 𝝌
How chemists quantify bond polarity
"The tendency of an atom to attract a shared pair of electrons towards itself".
Linus Pauling came up with electronegativity (𝜒) scale based on a geometric averaging of the ionization energy and electron affinities
Scale of 0 to 4
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What element has the highest tendency to attract electrons?

    Fluorine

Which elements have no EN?

Noble (Inert) gases
Which of these elements is the least (nonzero) EN?

Francium

Where does Hydrogen fall in terms of the other elements?

Right in the middle (2.1)



Untitled picture.png 2.1 
g 
0.98 
11 
0.93 
87 
0.7 
2 
He 
10 
4 
Be 
1.57 
12 
Mg 
1.31 
88 
0.9 
5 
2.04 2.55 
3.04 3.44 3.98 
89+ 
1.1-1.3 
2.20 
2.28 2.54 2.00 2.04 2.33 
2.02 2.0 
2.2 
Figure 7.13 Electronegativities 
of the elements as calculated 
by Linus Pauling. Note that the 

Untitled picture.png > 
3.98 
3.44 
Cl 
3.16 
> 
N 
3.04 
> 
S 
2.58 
> 
C 
2.55 
> 
P 
2.19 
> 
H 
2.1 
The most useful electronegativity values
Electronegativity differences Δ𝜒 determine bond polarity
Type of Bond
Example
Bonded Elements
Nonpolar Covalent Bond
Also called Pure Covalent

Cl-Cl
Δ𝜒<0.4
Strong NM† & Strong NM
or Weak NM & Weak NM
Polar Covalent Bond
H-Cl
0.4≤Δ𝜒≤﷐2.0﷮∗﷯
Weak NM & Strong NM
Ionic Bond
Na-Cl


Δ𝜒>﷐2.0﷮∗﷯
Metal & NM
Examples of calculating the polarity of a bond: 
C - N
Na - O
Mg - C
Br - O  (Br 𝜒 = 2.96)
N - N
S - O
P - F
Cr - Cl (Cr 𝜒 = 1.66)
 | 2.55 - 3.04 | = 0.49 Polar Covalent
 | 0.94 - 3.44 | = 2.50 Ionic
 | 1.31 - 2.55 | = 1.24 Polar Covalent
 | 2.96 - 3.44 | = 0.48 Polar Covalent
 | 3.04 - 3.04 | = 0.00 Nonpolar Covalent
 | 2.58 - 3.44 | = 0.86 Polar Covalent
 | 2.19 - 3.98 | = 1.79 Polar Covalent
 | 1.66 - 3.16 | = 1.50 ?
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Metallic Bond
Na-Na



Metal and Metal
† NM = Non-metal
* These are general guidelines. At the extremes, we tend to get a mix of ionic and covalent character (i.e. HF, AlI3)
Bond polarities can be indicated with partial charges
﷐𝛿﷮+﷯ means slightly positive.
﷐𝛿﷮−﷯ means slightly negative
Cr - Cl (Cr 𝜒 = 1.66)
Fe - H  (Fe 𝜒 = 1.83)
Ca - O (Ca 𝜒 = 1.0)
 | 1.66 - 3.16 | = 1.50 ?
 | 1.83 - 2.20 | = 0.37 Nonpolar Covalent
 | 1.00 - 3.44 | = 2.44 Ionic
Note that some metals and nonmetals have significant covalent character and these are just estimates of the bond character.
Drawing Covalent Molecular Structures (Lewis Structures)
Instead of the transfer of electrons to obtain a noble-gas electronic configuration, in covalent molecules there is a simultaneous sharing of electrons. For example the molecule Cl2:


Untitled picture.png 
Untitled picture.png 
Untitled picture.png 

Cl ([Ne]3s23p5) + Cl ([Ne]3s23p5) ⟶ Cl2 ([Ar][Ar]) 

Both atoms now have a noble gas outer electron configuration of eight electrons. G. N. Lewis generalized this result into what we call the octet rule: each element forms covalent bonds such that eight electrons occupy its outer shell. Unpaired electrons are known as lone pairs.

Algorithm for Drawing Lewis Structures

Arrange the symbols of the atoms that are bonded together in the molecule
next to one another.
Compute the total number of valence electrons in the molecule by adding
the number of valence electrons for all the atoms in the molecule. If the
species is an ion rather than a molecule, then you must take the charge of
the ion into account by adding electrons if it is a negative ion or subtracting
electrons if it is a positive ion
Represent a two-electron covalent bond by placing a line between the
Untitled picture.png 
 G. N. Lewis - Introduced Lewis structures and covalent bonds in the 1920s.

Tips: 
The central atom is usually the least electronegative unless hydrogen which is always terminal.
Some compounds will not have a central atom, but several.
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Represent a two-electron covalent bond by placing a line between the
atoms that are assumed to be bonded to each other
Arrange the remaining valence electrons as lone pairs about each atom so
that the octet rule is satisfied for each one
Check the formal charge on each atom to help with competing structures 
If there is not enough electrons, you may need to use double bonds, triple bonds or rings or the central atom(s) has a deficient octet or you have a radical
If there are too many electrons, you may need to put electron pairs on the central atom and/or form double bonds to reduce formal charges (this will not be assessed in CHEM 101)
Draw any valid resonance structures if requested



Some compounds will not have a central atom, but several.
Be, B, and Al have deficient octets
Elements in the 3rd row (S, P . . . )can expand their octet to have 10, 12, or more e-.
If the molecule is an ion, bracket the structure and put a charge in the corner
Molecules with an odd number of electrons will form a free radical, a reactive species.
Examples of Lewis Structures (Steps 1-4 only)
OF2














NH2–
PCl4+
 CHCl3 (chloroform)




N2H4 (hydrazine)
CH3OH (methanol)
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Assigning Formal Charges help with determining the best structure (Step 5)
 Sometimes more than one possible structure may be possible. In that case it is necessary to assign a formal charge to the atoms in the structure to help aid us in choosing the correct one. We assume each pair of shared electrons are shared equally and assign one of the electrons to each atom. Lone electron pairs are assigned to the atom they are located on. Use this equation:

Untitled picture.png formal charge 
on an atom In 
a Lewis formula 
total number of 
= valence electrons 
in the free atom 
total number 
of lone-pair 
electrons 
total number 
of shared 
2 
electrons 











NH4+ (ammonium)





H3O+ (hydronium)
OF2 (written as OFF)
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Which structure is preferable for hydroxylamine?
NH3O


Pick up here next Tuesday





NH2OH
Which structure is preferable for hydrogen peroxide?
H2OO



HOOH
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When there are not enough electrons (Step 6)
Some elements are electron deficient are do not require an octet when forming a covalent bond. 
This is restricted to H, Li, Mg, Be, B, and Al.
What is more likely is that you need to use double or triple bonds, especially if there is C or O present.
If the molecule is larger, you can also form ring structures to use up 2 electrons.
Lastly, if you have an odd number of electrons radical (unpaired electron) compounds are also possible.
H2 





Li2 




BeCl2





MgH2





BH3





AlBr3






C2H6 (ethane)


C2H4 (ethene)
C2H2 (ethyne)









C6H6 (benzene)








C2H3O2– (acetate)

CH3• 
(methyl radical)





O2•– 
(superoxide anion radical)
 (C6H5)3C•
Trityl radical 
 First organic radical discovered. Moses Gomberg in 1900 at University of Michigan (my alma mater)
(superoxide anion radical)









When there are too many electrons (Step 7)
Some elements allow for an expanded octet when forming covalent bonds. 
This is restricted to elements in the third shell or higher including the elements: P, S, Cl, As, Se, Te, Br, I, and even Xe.
Never will an element such as C, N, O, or F have an expanded octet
I will not test on this (unless it is an extra credit problem)

 SF6
 
 
 
 
 
 

 
 
XeF2
IF5

 

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