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As I take sigma 2p antibonding molecular orbital as an example enter image description here

Would it have 3 or 1 nodal planes. The two supposed nodal planes pass through the centres of the two atoms and are same as in the original atomic p orbitals. By LCAO A and B, I suspect that there will be some electron density due to the wave function of B where A is zero. Is there a fault in my logic?

For pi there is no problem, there are 1 nodal planes for pi bonding mo and 3 for antibonding mo as there is a common nodal plane to both MO.

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  • $\begingroup$ just so I surely have it down, dyz dyz Pi bond bmo has 1 and abmo has 2 nodal planes? with z axis as internuclear axis. $\endgroup$ – Mrigank Jan 18 '17 at 14:09
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    $\begingroup$ No, should be more than that. Draw them with blue and red regions for different signs of $\psi$, then you'll see. $\endgroup$ – Ivan Neretin Jan 18 '17 at 14:18
  • $\begingroup$ m.imgur.com/xNFcJKf taking internuclear axis as z $\endgroup$ – Mrigank Jan 18 '17 at 14:20
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You are right in that node of one AO is not a node of other AO, and hence not a node of the MO (that is, unless we're talking about π bonding between two p-orbitals). In effect, there still would be three nodal planes, but two of them would not quite match those of individual AOs, and might even be not quite planar.

The same applies to any other type of orbitals.

Look at these crude approximations of $\pi$ orbitals formed by two d-type AOs.

Bonding:

Bonding MO

Antibonding:

Antibonding MO

See those vertical nodal "planes" which are not quite planar?

That's why, BTW, I'd rather not ask about nodal planes. By asking that, we get bogged down in a pointless and un-chemical quibbles concerning the definition of a plane. Ask about nodal surfaces, and you'll get a meaningful answer.

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  • $\begingroup$ Sorry, I accidentally deleted a comment. "Just as cones are not considered nodal planes, these won't either right?" $\endgroup$ – Mrigank Jan 18 '17 at 14:53
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    $\begingroup$ I started writing a comment, then noticed that your comment is gone, so I just added mine into my answer as the last paragraph. $\endgroup$ – Ivan Neretin Jan 18 '17 at 14:56
  • $\begingroup$ Just one last example, for delta bond with z as internuclear axis for dxy, dxy is the following correct? (3rd for abmo is z=0 plane) m.imgur.com/2wVkIDt $\endgroup$ – Mrigank Jan 18 '17 at 14:58
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    $\begingroup$ Yeah, looks about right. Then again, you might want to put (+) and (-) all over your drawings to denote the signs of $\psi$. $\endgroup$ – Ivan Neretin Jan 18 '17 at 14:59
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    $\begingroup$ Yes, should be that way. $\endgroup$ – Ivan Neretin Jan 18 '17 at 19:37
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AFAIK: The nodal planes are where the wavefunction sign inverses. If you drew the colors according to the sign value then you would see that for antibonding sigma MO there is a virtual nodal plane, since the sign changes between + and - between the two centre orbital lodes. Even if they by some magic grew larger, they can never touch, they interfere destructively. There will always be a plane between them where the wavefunction returns 0. (when the two spheres meet they will form a planar interface. Imagine soap bubbles.)

I found this, it is more descriptive; https://socratic.org/questions/how-can-i-draw-antibonding-orbitals

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  • $\begingroup$ That is the point I was making, they don't equally get larger, the wave function of B is negative everywhere near A then, Negative part of A grows more than positive side, resulting in nonzero wave function where the nodal plane of A used to be. $\endgroup$ – Mrigank Jan 18 '17 at 14:30
  • $\begingroup$ But then you are not considering the molecular orbital, you are considering two atomic orbitals. The molecular orbital is combined for both A and B, there is no "B part of orbital". Consider the sum of both A and B, if you will. $\endgroup$ – Stian Yttervik Jan 18 '17 at 15:11
  • $\begingroup$ I was following the LCAO model. I think Ivan has answered all of the queries. $\endgroup$ – Mrigank Jan 18 '17 at 15:13
  • $\begingroup$ You were right, but the nodal plane is not the same as that of an participating atom. I forgot that film between the soap bubbles is acutally curved to account for the excess pressure in case of different sized bubbles. :) $\endgroup$ – Mrigank Jan 18 '17 at 15:29

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