How can the polarity of a covalent bond be determined?

1 Answer
Sep 8, 2014

The polarity of a covalent bond can be determined by determining the difference in electronegativity (#Delta"EN"#; where #Delta# means difference) between the two bonded atoms. Electronegativity is the tendency of an atom to attract electrons to itself. Larger, nonmetallic atoms tend to have higher electronegativities than metals. The following is a periodic table showing the electronegativities of the elements using the Pauling scale.

http://www.amazingrust.com/Experiments/background_knowledge/Redox.html

The difference in electronegativities (#Delta"EN"#)between the bonded atoms determines the character of the bond. The following rules apply:

  1. If the ΔEN is less than 0.5, then the bond is nonpolar covalent.
  2. If the ΔEN is between 0.5 and 1.6 (some use 1.7), the bond is considered polar covalent
  3. If the ΔEN is greater than 2.0, then the bond is ionic.
  4. If the ΔEN is between 1.6 and 2.0, and a metal is involved, then the bond is considered ionic. If only nonmetals are involved, the bond is considered polar covalent.

*Note that absolute values are used; there are no negative differences in electronegativity.

**A note about Rule 4: Some websites and textbooks consider a #Delta"EN"# greater than 1.6 as ionic. Others consider a #Delta"EN"# greater than 0.5 and less than 2.0 as polar covalent. Ask your teacher what he or she prefers.
(Source: http://www.chemteam.info/Bonding/Electroneg-Bond-Polarity.html)

Example
Determine the #Delta"EN"# and bond type for the following elements using the four rules listed above and the periodic table showing electronegativities.

  1. silicon (Si) and phosphorus (P): EN of Si = 1.90, EN of P = 2.19
    #Delta"EN"# = 1.90 - 2.19 = 0.29, therefore the bond is nonpolar covalent.

  2. carbon (C) and oxygen (O): EN of C = 2.55, EN of O = 3.44
    #Delta"EN"# = 2.55 - 3.44 = 0.89, therefore the bond is polar covalent.

  3. sodium (Na) and chlorine (Cl): EN of Na = 0.93, EN of Cl = 3.16
    #Delta"EN"# = 0.93 - 3.16 = 2.23, therefore the bond is ionic.