This question already has an answer here:

This question was inspired from the following questioN:

enter image description here

I thought the answer would be either A (maybe accounting for larger sizes?) or B (Bent's rule) but apparently the answer is D. Can someone explain why this is so?

This question is #54 on the 2010 USNCO Part 1


marked as duplicate by Mithoron, Todd Minehardt, airhuff, Jon Custer, aventurin Mar 21 '18 at 19:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ A screenshot or picture of an exercise is not searchable. Please consider rewriting it, so that it can be of help for future visitors. $\endgroup$ – Martin - マーチン Mar 21 '18 at 4:17
  • $\begingroup$ I have edited it for the source $\endgroup$ – phi2k Mar 21 '18 at 4:21
  • $\begingroup$ The reason for (D) is pretty much given here, and in short, s-p separation is large in sulfur, therefore no hybrid orbitals. Which also excludes Bent's rule. While (A) is true, a longer bond would not affect the bond angle at all, so that cannot be the reason. (B) is a bit difficult to judge, because there are many factors affecting polarity, however, polarity does not necessarily influence the geometry that much. And (D) is just plain wrong, wrong, wrong, wrong. No d orbitals for sulfur. $\endgroup$ – Martin - マーチン Mar 21 '18 at 4:24
  • $\begingroup$ I mean, according to VSEPR theory sulfur can access d-orbitals for bonding. I know that this is false now but why is it false? Also isn't the s and p separation smaller in the n = 3 subshell than in the n = 2 subshell? $\endgroup$ – phi2k Mar 21 '18 at 5:01
  • 2
    $\begingroup$ VSEPR is taught more often wrong than it is taught right. "according to VSEPR theory sulfur can access d-orbitals for bonding" is such a common, but completely false statement. There have been multiple studies, showing that octet expansion is a wrong concept and that d-orbitals will not be used for bonding at all. This is due to the fact that energy levels become more spread out with higher n. Which also answers your last inquiry. $\endgroup$ – Martin - マーチン Mar 21 '18 at 5:22

Browse other questions tagged or ask your own question.