# Generating hybrid atomic orbitals for Boron

My professor was showing us how to form hybrid atomic orbitals for Boron by taking a linear combination of rotated 2p orbitals and 2s orbitals (of Boron). This was done in a particular manner such that the geometry of BH3 is correct. We were given BH3: He proposed 3 new hybrid atomic orbitals: 2Py(rot,1): a 2Py orbital rotated 30 degrees clockwise  -2Py(rot,2): take the negative of 2Py, which is then rotated 30 degrees counterclockwise

Can someone provide a visual (picture) showing the "mixing" of the atomic orbitals to create the three hybrid orbitals (while maintaining the same orientation of BH3 as in my first image)?

EDIT: Feodoran helped me clear my confusion regarding the first hybrid wavefunction. If possible, I would still be interested if someone could display the formation of the hybrid wavefunctions with some visuals.

• How do you define 2Py(rot,1) and 2Py(rot,2) ? Apr 17 '20 at 10:20
• @Maurice 2Py(rot,1) is a 2Py orbital rotated 30 degrees clockwise. -2Py(rot,2) is achieved by taking the NEGATIVE of 2Py which is then rotated 30 degrees counterclockwise. This is done to match the angles in BH3. The B-H bonds in 2nd and 3rd quadrant is 30 degrees from the y-axis. I have the specific equations describing each rotation, but didn't post them. Apr 17 '20 at 12:32
• @Oscar Lanzi Yes...ultimately those are linear combinations of unrotated Px and Py as I have shown in the equations. But my problem is not with chi2 or chi3 but with chi1. Apr 19 '20 at 17:11
• $\chi_1$ is oriented along the $x$ axis, on which one of the $\ce{B-H}$ bonds lies. What do you mean by "opposite direction"? Rotation of the $p_x$ orbital by 180° around the $z$ or $y$ axis results in the same orbital, just with a sign change. But the sign is arbitrary anyway. So there is not "wrong" direction, both are fine. Apr 19 '20 at 18:47
• You are only changing one orbitals sign. This will result in qualitatively different molecular orbitals that may not contribute to the bonding. Relative phase needs to be the same, so either you change all signs or none! $\chi_1=\phi_1-\phi_2$ is the "same" (except for the overall sign) as $-\chi_1=-\phi_i+\phi_2$, but something very different than $\chi_2=\phi_1+\phi_2$. Apr 21 '20 at 7:00