= Solution
\b[End-on overlap produces $\sigma$ and $\sigma^*$ <molecular orbitals>; side-on overlap produces $\pi$ and $\pi^*$ <molecular orbitals>.] Matching signs of the facing <atomic orbital> lobes give constructive <interference> and a <bonding molecular orbital>. Opposite signs give destructive <interference> and an additional nodal plane between the <atomic nuclei> in the <antibonding molecular orbital>.
\Image[/past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025-p-orbital-combinations.png]
{title=Original signed combinations of p <atomic orbitals> for four p-derived <molecular orbitals>}
Take the internuclear direction as $x$, and define both atomic $p_x$ functions as positive on their own right-hand side. Up to <wavefunction normalization>, the combinations are
$$
\begin{array}{c|c}
\sigma&p_{x,A}-p_{x,B}\\
\sigma^*&p_{x,A}+p_{x,B}\\
\pi&p_{y,A}+p_{y,B}\\
\pi^*&p_{y,A}-p_{y,B}
\end{array}
$$
The minus sign in the bonding $p_x$ combination follows from this common-axis <quantum phase> convention: the facing lobes of the two unmodified $p_x$ functions have opposite signs. It does not make that combination antibonding. Every $\pi$ <molecular orbital> also has a nodal plane containing the internuclear axis; the additional midpoint node distinguishes $\pi^*$ from $\pi$.
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