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As an exercise I have to predict the formula of the compounds that fluorine and tellurium can build using group properties.

My idea was to use the octet rule and therefore the formula would be $\ce{TeF2}$ so both atoms have eight valence electrons.

However, the correct solution given by the exercise is $\ce{TeF4}$ and $\ce{TeF6},$ and I have literally no clue how to get these solutions. Can you give me some advice on how to retrieve these solutions?

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    $\begingroup$ That is actually a very sensible answer, unfortunately Mother Nature has conspired against you here. As to what is the reason for the answer it is difficult to say without knowing how much you know already - what compounds of fluorine and sulphur do you know? $\endgroup$ – Ian Bush Jun 16 at 12:48
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    $\begingroup$ @IanBush I was going to write an answer about how it was due to the presence of d and f orbitals on elements of higher groups and how those could be involved in bonds even being higher in energy. Than I found out about diazomethane ($\ce{CH2N2}$) and it all went to space lol. Wikipedia article about Hypervalent Molecules might be an interesting read for the OP though. $\endgroup$ – IanC Jun 16 at 13:43
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    $\begingroup$ @JohnyDow I believe Ian was hinting at that Tellurium could/should form the same structures with Fluorine as Sulfur (or Selenium). $\endgroup$ – Tyberius Jun 16 at 18:06
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    $\begingroup$ From my research, the answer seems much more complicated than it could be conveyed in a simple answer. At first I thought it was due to the d and f orbitals being involved in bonding. But apparently this explanation (dated about 1950) was incorrect, since experiments showed the contribution of d and f orbitals is very low on those hypervalent molecules. It seems the more up to date explanation is the (3c-4e) bond model, which I'm still reading about. $\endgroup$ – IanC Jun 16 at 19:24
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    $\begingroup$ I guess the most simple answer that can be given to the OP without further complicating things is that elements in lower rows of the periodic table present this feature of not complying to the octet rule by having more bonds than it would be predicted. This almost never happens with elements with valence shells of quantum number 2 (the previously cited diazomethane being the only exception I found to this). $\endgroup$ – IanC Jun 16 at 19:27

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