Chemistry Reference
In-Depth Information
B. From Figure 4-1a, we find:
E.N. of H = 2.2
E.N. of O = 3.5
Difference in E.N. = 1.3 (3.5 - 2.2 = 1.3).
Because 1.3 is less than 1.7, this bond is covalent.
Because 1.3 is greater than 0.3, this bond is polar covalent.
C.
From Figure 4-1a, we find:
E.N. of H = 2.2
E.N. of Cl = 3.2
Difference in E.N. = 1.0 (3.2 - 2.2 = 1.0).
Because 1.0 is less than 1.7 this bond is covalent.
Because 1.0 is greater than 0.3, this bond is polar covalent.
D. From Figure 4-1a, we find:
E.N. of O = 3.5
E.N. of O = 3.5
Difference in E.N. = 0.0 (3.5 - 3.5 = 0.0).
Because 0.0 is less than 1.7 this bond is covalent.
Because 0.0 is less than 0.3, this bond is non-polar covalent.
Answers:
A. KCl
ionic bond
B. H
2
O
polar covalent bond
C. HCl
polar covalent bond
D. O
2
non-polar covalent bond
Before you try the practice problems on page 119, let me remind you of
two things. First, because we are looking for the difference between the
electronegativities, we always subtract the smaller number from the larger.
Second, remember not to use the subscripts that come with formulas as
multipliers for the electronegativities. For example, if you wanted to deter-
mine the type of bond that forms between hydrogen and carbon in the
compound CH
4
, you would not multiply the electronegativity of carbon by
a factor of four.
