-2
$\begingroup$

This question came in my exam to draw the Structure of $\ce{[IO_2F_5]^2-}$

I am confused. Here's what I reasoned:

Iodine will have hybridisation $\ce{sp^3d^3}$ so it will be pentagonal bi pyramidical in shape. I can't reason where will be the position of oxygen atoms... will they be in the axial position or in the pentagon. According to Bent's rule, more electronegative atom occupies the position which have more bond length , smaller bond angle so by this reasoning the fluorine should be on axial position and the oxygen atoms in the pentagon.

Is my reasoning correct or there is something more to it?

$\endgroup$
  • $\begingroup$ Aren't the oxygen atoms singly bonded with excess negative charge (that's why the 2- above the bracket)? The more electronegative atoms should be axial. There is a good MO explanation why, and I think that someone mentioned it in an answer. I will search for it. $\endgroup$ – RBW Nov 8 '16 at 16:48
  • 2
    $\begingroup$ Don’t use hybridisation involving d-orbitals for main group elements; it is wrong. $\endgroup$ – Jan Nov 8 '16 at 18:22
  • $\begingroup$ @NeeleshVij See this question and the ones linked therein. $\endgroup$ – Jan Nov 9 '16 at 19:36
-2
$\begingroup$

See, before answering the question, let's look at the orbital splitting that takes place. $$sp^3d^3\equiv sp + p^2d^3$$

Since double repulsions are much more than single bond. Thus, they occupy axial, more $s$ containing orbitals, the $sp$ orbitals.

Also note what Bent's rule says:

  1. The more electronegative atom prefers to stay in the orbital having less $s$ character, while the lone pair and double bond prefers to stay in the orbital having more $s$ character.

  2. The orbital occupying more space around that central atom will have more $s$ character. Thus justifying the answer.

Hope it's useful!

$\endgroup$
  • 1
    $\begingroup$ See my comments on the question on the topic of involving d orbitals in main group hybridisation. $\endgroup$ – Jan Nov 9 '16 at 20:32
  • 1
    $\begingroup$ Jan, I don't clearly see the relation between your comment and the answer to question that you mentioned. Please elaborate your opinion. Also note that I've taken these concepts from a standard book, which I believe should be right. Kindly enlighten me. $\endgroup$ – Kishore S Shenoy Nov 10 '16 at 0:33
  • 1
    $\begingroup$ Yes, standard books did, for quite some time, involve d-orbitals to explain hypercoordination. However, that concept has been proven wrong. Main group elements do not use their d-orbitals in bonding. Your answer suggests they do. Remember that iodine’s 5d orbitals are even higher in energy than its 6s orbitals, yet nobody suggests 6s should hybridise alongside with 5s and 5p. Think about it. $\endgroup$ – Jan Nov 10 '16 at 19:39

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.