2
$\begingroup$

I was going through my chemistry textbook (IB Pearson), and it explicitly stated that the higher the absolute difference between the electronegativity of elements in a binary compound, the more 'ionic' that compound is, which does make intuitive sense given that a higher electronegativity difference leads to the compound being more 'polar.'

However, the book came up with the following example:

Which of the following pairs will form the most ionic compound?

1) Be and F (the electronegativity difference is $|4-1.6| = 2.4)$
2) Si and O (the electronegativity difference is $|3.4-1.9| = 1.5)$
3) N and Cl (the electronegativity difference is $|3.2-3| = 0.2)$
4) K and S (the electronegativity difference is $|2.6-0.8| = 1.8)$

$\ce{K2S}$ will be the most ionic compound as the difference in electronegativity is the greatest.

I don't get how this works, the difference in electronegativity between beryllium and fluorine is visible greater, so why does the book say that the binary compound formed by K and S is more ionic?

$\endgroup$
2

2 Answers 2

2
$\begingroup$

Tl, dr -- upon further review we can't really tell whether potassium sulfide or beryllium fluoride is more ionic without some quantitative details.

There is more involved in the balance between ionic and covalent bonding than electronegativity differential. You have to consider molecular orbital structure, too. One of the answers to this question describes how molecular orbital structure can drive highly ionic bonding when electronegativity differences are modest (as in salts with aromatic ions) or even the "wrong way" (as in alkalides).

With single atoms, as pointed out in the other question, you cannot expect 100% ionic bonds even when molecular orbitals favor it. Yet there can still be an impact which we can see by comparing organolithium compounds with organic fluorides. The former acts as a ready source of anionic carbon, while the latter are notoriously unreactive towards the displacement of fluoride ions, as if the carbon-lithium bond had much more ionic character. Yet the electronegativity differentials in the carbon-lithium and carbon fluorine bonds are about the same (oppositely directed with respect to carbon, of course). The difference comes from lithium having a diffuse valence orbital that overlaps poorly with carbon, whereas fluorine is more compact and overlaps well, thus promoting electron sharing between both atoms in the bonding orbital.

Similarly, potassium has a diffuse valence orbital, even more so than lithium, which overlaps poorly with most nonmetals. The ionicity of potassium sulfide, already considerable based on electronegativity, is thus enhanced even more. Beryllium and fluorine, with good covalent overlap, would have a more covalent interaction than what might be expected from their electronegativity difference.

Since potassium sulfide has poor overlap and beryllium fluoride has a somewhat higher electronegativity difference, the two effects oppose each other when the compounds are compared. We cannot tell from these qualitative arguments which one is more ionic.

$\endgroup$
3
  • 1
    $\begingroup$ "One of the answers" is a funky way of referring to an earlier answer of you. ;) [Btw. did you perhaps mean to link to it directly? I really needed to think twice before I got where you wanted to go with that paragraph.] // I generally agree with your conclusions; have a look at "Is beryllium difluoride covalent or ionic?" which suggests that $\ce{BeF2}$ is indeed very, very ionic. $\endgroup$ Oct 24, 2019 at 16:00
  • $\begingroup$ Yup, you got me. I had no better way to "hide" what I was referring to. :-). Woujd have been interesting to see what the referenced method would say about potassium sulfide, I suspect this would be found above 90% ionic too. $\endgroup$ Oct 24, 2019 at 16:03
  • 1
    $\begingroup$ Why hide it? I quite like your answer, and it is very fitting indeed. Most of us re-use our answers if they are related, there is nothing wrong with that; and I think it's better than repeating a lot of the same arguments. $\endgroup$ Oct 24, 2019 at 16:07
1
$\begingroup$

It seems the examples were thrown together without considering that someone would try to understand them.

The rule does make sense, but there are several factors involved here, and they are not mentioned. The problem is not even that you are comparing apples and oranges; it's more like fruit salad vs vegetable stew!

Let's simplify: Keep one of the elements the same and change the other. N and Cl have a difference of 0.2 and the compound is covalent (with lots of charge separation). If we look at KCl, the difference is 3.2 - 0.8 = 2.4, so the difference is larger and suggests a larger ionicity. That difference suggests why KCl completely ionizes in water.

That extra factor involved is theoretically comparing "degrees of ionicity" by calculations vs seeing the actual ions in water (well, observing the changes that we ascribe to ionization, like freezing point depression) from KCl. Comparing the calculated differences is a mental mastication, but if you just change one variable at a time, you can make some useful explanations or testable predictions.

$\endgroup$
0

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.