2
$\begingroup$

In introductory college chemistry, bonds between elements that have an electronegativity difference $\Delta EN > 0.4$ are generally consider to be polar (and thus have a dipole moment), while $\Delta EN < 0.4$ bonds are considered to be nonpolar. I think more advanced chemistry classes generally don't adhere strictly to this threshold, but I'm still curious if this 0.4 threshold has any meaning (or if it's just arbitrary).

For example, if there's an experiment plotting $\Delta EN$ vs. the intermolecular force strength, and there's a big jump on the graph at 0.4, that would be convincing evidence. Does anyone have an explanation along this line of reasoning?

(My question could also apply to the $\Delta EN > 1.7$ threshold for ionic vs. polar covalent. I saw a question on this before here, but I think it's worth further discussion.)

$\endgroup$
3
$\begingroup$

Both the $0.4$ threshold for nonpolar/polar bonds and the $1.7$ threshold for polar/ionic bonds are arbitrary choices. When defining nonpolar, polar and ionic bonds, it is important to allow $\ce{C-H}$ bonds ($\Delta \chi = 0.4$) to be nonpolar — it is the best description of alkanes’ low polarities and it facilitates discussions so much. Likewise, a few compounds commonly known as salts should fall just within the ionic boundary for ease of discussion. Teaching introductory levels is just so much easier if you can draw things black and white.

At higher levels, it should always be remembered that not only the thresholds are arbitrary but even electronegativity as a concept follows quite arbitrary definitions.

  • Except for one electronegativity scale that links electronegativity to ionisation enthalpy and electron affinity, all definitions and all defined reference points (typically fluorine having $\chi = 4.0$) are arbitrary.

  • Choosing a definite value for a specific element is also arbitrary since the physical implication of electronegativity depends on:

    • bonding partners
    • structure (i.e. hybridisation)
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.