# Confusion about the choice of axis for Molecular Orbital Theory

This video from MIT OpenCourseWare says that pi bonds and anti-bonds are for 2px and 2py, i.e., $\pi_{2px}$, $\pi_{2py}$ and $\pi_{2px}^*$, $\pi_{2py}^*$ and sigma bonds and anti-bonds are for 2pz i.e., $\sigma_{2pz}$ and $\sigma_{2pz}^*$.

But in my reference books it says pi bonds and anti-bonds are formed for 2py and 2pz i.e., $\pi_{2py}$, $\pi_{2pz}$ and $\pi_{2py}^*$, $\pi_{2pz}^*$ and sigma bonds and anti-bonds are for 2pz i.e., $\sigma_{2px}$ and $\sigma_{2px}^*$.

Which is actually the right one? Please explain.

• It depends on the choice of the bond axis: I guess it was chosen to be $z$-axis in the MIT lecture, but $x$ in your book. Hence the difference. – Wildcat Aug 8 '16 at 17:23
• Representing it in either way is correct? If so, can we choose $y$-axis as the bond axis? Add your answer below so that I can mark it as the correct answer. – Siddharth Venu Aug 8 '16 at 17:25
• Right. In principle, you're free to choose any direction as the bond axis. But the convention is to choose $z$-axis for this purpose. – Wildcat Aug 8 '16 at 17:28
• Add your answer below so that I can mark it as the correct answer. – Siddharth Venu Aug 8 '16 at 17:31

Conventionally, and as it is (implicitly?) done in the MIT video lecture, it is chosen to be the $z$-axis, in which case $\pi$ bonds are indeed formed from $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals while $\mathrm{p}_z$ can take part in $\sigma$-bonding. If, however, the bond axis is chosen to be the $x$-axis, then $\pi$ bonds are formed from $\mathrm{p}_y$ and $\mathrm{p}_z$ orbitals while $\mathrm{p}_x$ takes part in $\sigma$-bonding.