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Which polar covalent bond of the following:

Cl-F

S-O

P-N

C-Cl

It's trivia - I guess. And I got it wrong.

Electronegativity list was given for the following: F, O, Cl, N, S, C, H, P

My wrong reasoning was: why would two highly electronegative atoms bond with each other. Even they did, given their high electronegativity as both pull the electron cloud to each side (tug of war, weak dipole moments) with much strength by forming a weak bond. I refused to consider the difference between electronegativity of the species. (Even I did, S-O would be the highest).

I applied the same to S-O and P-N as well. C-Cl was the pick and bad!

C-Cl has a acceptable dipole moment given C's less en 2.55 and Cl's high en. 3.16. The size of atoms wise, Cl-F is shorter compared to C-Cl. Which is questionable.

I checked few other related questions and answers, such as this peculiar one and this.

As I read ron's answer (the latter), the concept is not about a compound like CCl4 which negates the polarity between C-Cl bonds in the tetrahedral geometry. Thus C-Cl is a polar covalent bond. (isn't it?)

Where did I go wrong in reasoning, what did I miss? PS: Not homework but some theory revision questions. Trying to reinstate/expand the understanding of concepts.

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  • $\begingroup$ This is really difficult to answer. (1) There are multiple tables of electronegativity values. So which were you given? (2) Off the top of my head I'd assume that since you were given a table of electronegativity values that the pair with the greatest difference would be the most polar. (3) you say that it isn't S-O or C-Cl. So what was the right answer? What are we trying to rationalize for you? $\endgroup$
    – MaxW
    Commented Mar 19, 2017 at 17:05
  • $\begingroup$ Cl-F was the answer. I have no clue how it came about. Care to discuss and shoot an answer? $\endgroup$
    – bonCodigo
    Commented Mar 20, 2017 at 9:12
  • $\begingroup$ @MaxW revisiting this question again, and looking at your comment, why do we have multiple tables of electronegativity values for the same bonds from single periodic table? $\endgroup$
    – bonCodigo
    Commented Oct 7, 2021 at 0:27

1 Answer 1

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There are various electronegativity scales.

Electronegativity vales from Wikipedia

$$\newcommand{\d}[2]{#1.&\hspace{-1em}#2} \begin{array}{lrl} \hline \text{Element} & \text{Pauling} & \text{Allen}& \\ \hline \text{H} & 2.20 & 2.30 \\ \text{C} & 2.55 & 2.544 \\ \text{N} & 3.04 & 3.066 \\ \text{O} & 3.44 & 3.610 \\ \text{F} & 3.98 & 4.193 \\ \text{P} & 2.19 & 2.253 \\ \text{S} & 2.58 & 2.589\\ \text{Cl} & 3.16 & 2.896 \\ \hline \end{array} $$

The compounds are all binaries, so the polarity, $\mu$, is equal to the difference in charge, $\delta$, times the distance, $d$, between ions ($\mu = \delta \cdot d$). Only given a table of electronegativities, the first level of analysis would be to assume that all the bond lengths were equal and that that the polarity would simply be a function of the difference in electronegativity. The differences are thus:

$$\newcommand{\d}[2]{#1.&\hspace{-1em}#2} \begin{array}{lrl} \hline \text{Element} & \text{Pauling} & \text{Allen}& \\ \hline \Delta\text{Cl-F} & 0.82 & 1.297 \\ \Delta\text{S-O} & 0.86 & 1.021 \\ \Delta\text{P-N} & 0.85 & 0.813 \\ \Delta\text{C-Cl} & 0.61 & 0.352 \\ \hline \end{array} $$

So on the Pauling scale S-O should be the most polar, but on the Allen scale Cl-F should be the most polar.

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  • $\begingroup$ Are you implying the diff between electronegavity of the bond was the only reason? That feels too naive to conclude... because the values given were similar to Pauling scale and therefore answer can not be Cl-F. That's where my doubts began and to post this question. $\endgroup$
    – bonCodigo
    Commented Mar 21, 2017 at 22:51
  • $\begingroup$ I'm not saying that the difference in electronegativity is the only reason. The dipole moment is given by the different in charge times the bond length. However only given the tables of electronegativities, I'd assume the bond lengths for the four compounds were equal and choose an answer based on the electronegativity difference. Don't really know how else you're supposed to analyze the problem. // I edited the problem to reflect the reasoning better... $\endgroup$
    – MaxW
    Commented Mar 22, 2017 at 4:44
  • $\begingroup$ I need sometime to reply to you. I will have a bit more sub questions to discuss. $\endgroup$
    – bonCodigo
    Commented Mar 30, 2017 at 2:44

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