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I want to ask a question about the MO diagram of BN.

I was provided with the following question and answer and describe the situation occurring:

enter image description here

After performing the s-p mixing for BN, I found this diagram worked as the $\ce{1\pi}$ MO was the HOMO, but I thought using atomic structure knowledge that as B and N are in the same group, and N has a greater nuclear charge (and hence greater effective nuclear charge) its orbital size would be smaller, not bigger as the diagram shows above.

Am I right in saying that the label above for these orbitals are wrong, so it should be showing $\ce{N-B}$ from left to right?

The whole question is given here below

enter image description here

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  • $\begingroup$ If we knew the other answer options, we would be better equipped for making guesses as to what was the author's intention. $\endgroup$ – Ivan Neretin Jun 15 at 13:31
  • $\begingroup$ @IvanNeretin done! $\endgroup$ – vik1245 Jun 15 at 13:49
  • $\begingroup$ Great. Now what do they really mean by orbitals' size? You think it is the physical size of the corresponding orbitals. Maybe so, but there is another possibility. In many situations, people draw their AOs bigger or smaller to reflect the contribution of the corresponding AO to the MO. $\endgroup$ – Ivan Neretin Jun 15 at 14:05
  • $\begingroup$ @IvanNeretin So,how should I decide which AO contributes more? Please give me a few more hints to solve this problem $\endgroup$ – Yusuf Hasan Jun 15 at 16:45
  • $\begingroup$ @yusuf the lowest energy orbital with a given symmetry favors the atom whose own contributing orbital is itself lower (typically the more electronegative atom), in this case nitrogen as opposed to boron. Here we have that with the lowest energy pi orbital. $\endgroup$ – Oscar Lanzi Jun 16 at 14:08

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