Question 5.5: Consider the following bonds N—Cl Ca—F C—O B—H N—O a. Ra......

First, the electronegativity difference for each of the bonds is calculated using the electronegativity values in Figure 5.11.
\quad\quad\quad\quadN—Cl:       3.0 – 3.0 = 0.0
\quad\quad\quad\quadCa—F:        4.0 – 1.0= 3.0
\quad\quad\quad\quadC—O:         3.5 – 2.5 = 1.0
\quad\quad\quad\quadB—H:         2.1 – 2.0 = 0.1
\quad\quad\quad\quadN—O:         3.5 – 3.0 = 0.5
a. Bond polarity increases as electronegativity difference increases. Using the mathematical symbol <, which means “is less than,” we can rank the bonds in terms of increasing bond polarity as follows:
\quad\quad\quad\quadN—Cl < B—H < N—O < C—O < Ca—F
\quad\quad\quad\quad   0.0         0.1         0.5         1.0         3.0
b. The direction of bond polarity is from the least electronegative atom to the most electronegative atom. The more electronegative atom bears the fractional negative charge (\delta^{-}).
\quad\quad N-Cl\quad\overset{⇸}{B-H}\quad\overset{⇸}{N-O}\quad\overset{⇸}{C-O}\quad\overset{⇸}{Ca-F}
c. Nonpolar covalent bonds require a difference in electronegativity of 0.4 or less, and an electronegativity difference of 2.0 or greater corresponds to ionic bonds. Bonds with an electronegativity difference greater than 0.4 but less than 1.5 are polar covalent. If the electronegativity difference is between 1.5 and 2.0, the bond is polar covalent if it involves two nonmetals, but is considered ionic if the bond involves a metal and a nonmetal. Using these guidelines the bond classifi cations are:
\quad\quad\quad\quadNonpolar covalent:           N—Cl, B—H
\quad\quad\quad\quad        Polar covalent:           N—O, C—O
\quad\quad\quad\quad                           Ionic:           Ca—F