School of Chemistry and Physics, University of KwaZulu-Natal, Westville Campus, Durban
General Principles of Chemistry – CHEM110
Tutorial 9 – 3 May 2013
Electronegativity and Polarity
1.
(a) Which of the following bonds are polar:
i)
B-F
ii)
Se-O
iii)
H-I
The bonds in ii) and iii) are polar and the bond in i) is ionic.
b) Which is the more electronegative atom in each polar bond?
In (ii) O is more electronegative and in (iii) I is more electronegative and in the ionic
bond i), F is more electronegative.
2.
For each of the following bond pairs, decide which is the more polar and indicate the
partial negative and partial positive poles:
(a) B—Cl or C—Cl , (b) P—F or P—C
Use the following electronegativity values:
B = 2.0, F = 4.0, Cl = 3.0, C = 2.5, O = 3.5 and P = 2.1.
Answer:
a) From the electronegativity values,
∆ENBCl = 3.0 - 2.0 = 1.0
∆ENCCl = 3.0 – 2.5 = 0.5
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The difference in the electronegativities of chlorine and boron is 1.0; the difference between
chlorine and carbon is 0.5. Hence the B—Cl bond is more polar.


B Cl
b) ∆ENPF = 4.0 – 2.1 = 1.9
∆ENPC = 3.0 – 2.1 = 0.9
P-F is more polar
The difference in the electronegativities of chlorine and phosphorus is 0.9. The difference
between fluorine and phosphorus is 1.9. Hence the bond is more polar.
Lewis Structures
3.
Draw Lewis structures for the following compounds. All working must be shown.
a) CH2Cl2
C=4
Cl = 7 x 2 = 14
H=1x2=2
20 valence electrons
Carbon is the central atom.
2
Total = 20
–
b) BrO3 ion.
The total number of valence electrons is therefore 7 + (3 × 6) + 1 = 26.
4.
–
The following are three possible Lewis structures for the thiocyanate ion, NCS :
No. of electrons:
N= 5
C=4
S=6
Charge = 1
Total = 16
I
II
III
(a) 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.
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Shown here for 1st structure: I
N
C
S
Valence electrons:
5
4
6
(Electrons assigned to atom):
7
4
5
Formal Charge:
-2
0
+1
N
C
S
Valence electrons:
5
4
6
(Electrons assigned to atom):
6
4
6
Formal Charge:
-1
0
0
N
C
S
Valence electrons:
5
4
6
(Electrons assigned to atom):
5
4
7
Formal Charge:
0
0
-1
Shown here for 2nd structure: II
Shown here for 3rd structure: III
a)
Which Lewis structure is the preferred one? Give reasons for your answer.
N is more electronegative than C or S. Comparing N and C, N (in group 15, while C in group 14,
EN increases across a period, therefore N more EN than C. Comparing N and S (N is in period
2, and S is in Period 3: EN decrease down a group, therefore N more EN than S).Therefore, we
expect that any negative formal charge will reside on the N atom. Further, we usually choose the
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Lewis structure that produces the formal charges of smallest magnitude. For these two reasons,
–
the middle structure is the preferred Lewis structure of the NCS ion.
5.
2-
Draw equivalent resonance structures for the carbonate ion, CO3 . Show all working.
C=4
O=3x6e
Charge = 2 e
Total = 24 e
Expanded Octet
6.
–
Draw the Lewis structure for ICl4 .Show all working.
Iodine (group 7A) has seven valence electrons. Each chlorine (group 7A) also has seven. An
extra electron is added to account for the 1– charge of the ion. Therefore, the total number of
valence electrons is (7 + 28) + 1 = 36. The I atom is the central atom in the ion. Putting eight
electrons around each Cl atom (including a pair of electrons between I and each Cl to represent
the single bond between these atoms) requires 8 × 4 = 32 electrons.
We are thus left with 36 – 32 = 4 electrons to be placed on the larger iodine:
Iodine has 12 valence electrons around it, four more than needed for an octet.
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