School of Chemistry and
Physics
Westville Campus, Durban
General Principles of
Chemistry - CHEM 110
Tutorial 9 – 23rd and 25th April 2014
Electronegativity and Polarity
1.
(a) Determine the polarity of the bonds given below using the electronegative values of
the elements in the table given below:
Table 1. Electronegativities of some elements
Element
Electronegative
value
i)
H-H
ii)
Li-F
iii)
H-Br
H
2.1
Li
1.0
F
4.0
Br
2.8
In (i):
∆EN = 2.1 - 2.1 = 0, therefore non polar
In (ii)
∆EN = 4.0 – 1.0 = 3.0, therefore ionic
In (iii)
∆EN = 2.8 - 2.1 = 0.7, therefore polar
Bromine (and the other halogens) are all more electronegative than hydrogen, and so all
the hydrogen halides have polar bonds with the hydrogen end slightly positive and the
halogen end slightly negative.
1
School of Chemistry and
Physics
Westville Campus, Durban
General Principles of
Chemistry - CHEM 110
b) Which is the more electronegative atom in each bond in (ii) and (iii) above?
In (ii) F is more electronegative and in (iii) Br is more electronegative
Lewis Structures
2.
Draw Lewis structures for the following compounds. All working must be shown.
a)
H2 O
No. of electrons:
O = 1 x 6 eH = 1 x 2 eTotal = 8 e
b)
PO43- (In your answer, make sure the octet rule is obeyed).
2
School of Chemistry and
Physics
Westville Campus, Durban
General Principles of
Chemistry - CHEM 110
Resonance and Formal Charge
3. a)
Draw two possible resonance Lewis structures for carbon dioxide, CO2.
No. of electrons:
C = 1 x 4 e- = 4 e
O = 2 x 6 e- = 12 e
Total = 16 e
This is a resonance Lewis structure for carbon dioxide.
b)
Determine the formal charges of the atoms in each structure. Show all working.


For each atom, count the electrons in lone pairs and half the electrons it shares
with other atoms.
Subtract that from the number of valence electrons for that atom: The difference
is its formal charge.
3
School of Chemistry and
Physics
Westville Campus, Durban
General Principles of
Chemistry - CHEM 110
Shown here for 1st structure: A
O
C
O
Valence electrons:
6
4
6
(Electrons assigned to atom):
6
4
6
Formal Charge:
0
0
0
O
C
O
Valence electrons:
6
4
6
(Electrons assigned to atom):
5
4
7
Formal Charge:
+1
0
-1
Shown here for 2nd structure: B
c)
Which Lewis structure is the preferred one? Give reasons for your answer.
Structure A
All the formal charges on Structure A equal zero. The formal charges on Structure B
shows one end is positively charged and the other is negatively charged. Since the overall
distribution of Structure A is zero, Structure A is the most correct Lewis structure for
CO2.
4.
Draw two plausible Lewis structures for methyl isocyanate, CH3NCO, and use formal
charges to determine which of the two structures is the more stable. The skeletal
structure of CH3NCO is:
H
H C N C O
H
4
General Principles of
Chemistry - CHEM 110
School of Chemistry and
Physics
Westville Campus, Durban
Answer:
3+4+5+4+6 = 22
22-12 = 10
Distributing 10 e-s around atoms gives,
H
H C N C O
H
Three possible Lewis structures are:
H
H
H C N C O
H
H C N C O
H
H
H
FC(H) = 1-(0+1) = 0
FC(CN) = 4-(0+4) = 0
FC(N) = 5-(0+4) = +1
FC(CO) = 4-(0+4) = 0
FC(O) = 6-(6+1) = -1
= 1-(0+1) = 0
= 4-(0+4) = 0
= 5-(2+3) = 0
= 4-(0+4) = 0
= 6-(4+2) = 0
0
0
+1
0
0
H
-1
H C N C O
(a)
= 1-(0+1) = 0
= 4-(0+4) = 0
= 5-(4+2) = -1
= 4-(0+4) = 0
= 6-(2+3) = +1
0
H
0
H C N C O
0
0
0
0
0
H C N C O
H
H
0
0
H
0
0
-1
0
+1
H C N C O
H
0
(b)
(c)
Since formal charges are zero for all atoms in structure (b), (b) is the more plausible
structure.
5