Electronegativity & Polar Covalent Bonds:

Some elements tend to attract electrons more strongly than others.  This property is roughly described as "electronegativity."  If two atoms of differing electronegativity form a bond, the electrons spend more time on the more electronegative atom.  In the extreme, we have an ionic bond.

However, it should be noted that no bond can be 100% ionic--there is always some sharing of electrons between atoms.  For instance, NaCl is about 80% ionic and 20% covalent.  Conversely, polar covalent bonds can be thought of as paritially ionic (such as the bond in HCl which is often mentioned to be about 10% ionic and 90% covalent).

Ionic and covalent are useful concepts but it must be remembered that they are extremes and that many bonds lie somewhere in between.  Only a nonpolar covalent bond is close to 100% pure (and, even then, some valence bond structures of these can be partially ionic).

Problem 7.38:  What general trend in electronegativity occur in the periodic table?

What we shall do here is first show another modified periodic table.  This time electronegativities are shown for all the elements.  We have both a number for each one and a direct indication of trends as a whole.

Obviously the trends are now easily stated:

Problem 7.39:  Predict the electronegativity of the undiscovered element with Z = 119.
Again, we show the chart.  This has two purposes: Here is the chart again:  We note that element #119 would be directly below francium.

All we need do is look at the left-most column.  We see the following numbers:
 

Li
1.0
Na
0.9
K
0.8
Rb
0.8
Cs
0.7

Just extrapolating, and noting that the increase in atomic number, if plotted vs. this would look rather shallow (the atomic numbers go as 3, 11, 19, 37, 55).  Fr is 87 and we would probably predict its electronegativity as 0.7 or 0.6.  (Since we only go to +0.1 in the estimates, these are probably good numbers.)

The end result here is that we would probably set the E.N. of element #119 to be about 0.6.  This is quite low but one would expect it to be the least electronegative of ALL elements in any case.

Problem 7.42:  Which of the following substances are largely ionic and which are covalent?
With electronegativities, there is a fairly simple, albeit crude, rule:
Between two atoms: We apply these rules to the bonds given.  In order to facilitate things for the web students and to aggravate those who insist on having hard copy for everything, we again show the E. N. periodic table figure:

Now, we cruise forth and handle the bonds.  Just use the numbers above to get those below.

(a)  HF ENF = 4.0; ENH = 2.1:  DEN = 1.9 HF is polar covalent.
(b)  HI ENI = 2.5; ENH = 2.1:  DEN = 0.4 HI is polar covalent.
(c)  PdCl2 ENCl = 3.0; ENPd = 2.2:  DEN = 0.8 PdCl2 is polar covalent.
(d)  BBr3 ENBr = 2.8; ENB = 2.0:  DEN = 0.8 BBr3 is polar covalent.
(e)  NaOH Na+-OH is ionic!  So, we look at the
hydroxide ion:
ENO = 3.5; ENH = 2.1:  DEN = 1.4

OH- is polar covalent.
(f)  CH3Li Here, we know that the C-H bond is polar
covalent (with just a difference of 0.4).  So,
what about C-Li?  We look at this now.

ENC = 2.5; ENLi = 1.0:  DEN = 1.5

CH3Li is polar covalent.

There are lots and lots of bonds which are polar covalent.  Just a few are ionic!
Problem 7.43:  Use the electronegativity data in Figure 7.4 to predict which bond in each of the following pairs is more polar.
Just use the chart for these!  In fact, here it is again (to help you and to infuriate the folks who insist on printing out everything):

(a)  C-H or C-Cl Respective DEN's: 0.4 & 0.5 C-Cl is more polar
(b)  Si-Li or Si-Cl Respective DEN's: 0.8 & 1.2 Si-Cl is  more polar.
(c)  N-Cl or N-Mg Respective DEN's: 0.0 & 1.8 N-Mg is more polar.
Not really difficult, is it?  Note that, in the last line, N and Cl have about the same EN's.  Only O and F exceed these--but by quite a bit!