Can a molecule be neither gerade nor ungerade?

Or can any molecular orbit always be written a linear combination of gerade and ungerade basis states?




1. g/u is not a property of a molecule; I assume you meant molecular orbital.
2. All functions can be written as a linear combination of even + odd functions: there is a short explanation on [Wikipedia](https://en.wikipedia.org/wiki/Even_and_odd_functions#Other_algebraic_properties). Basically, you have a function $f(x)$; now define 


$$\begin{align}
f\_\mathrm e(x) &= \frac{1}{2}[f(x) + f(-x)] \\
f\_\mathrm o(x) &= \frac{1}{2}[f(x) - f(-x)] \\
\end{align}$$


It is clear from the definition of parity that $f\_\mathrm e$ is even (since $f\_\mathrm e(x) = f\_\mathrm e(-x)$) and likewise that $f\_\mathrm o$ is odd. Now since $f(x) = f\_\mathrm e(x) + f\_\mathrm o(x)$, we have shown that any function $f(x)$ can be expressed as a linear combination of even and odd functions.


In three dimensions where you have $\psi(x,y,z)$ you simply need to define


$$\begin{align}
\psi\_\mathrm g(x) &= \frac{1}{2}[\psi(x,y,z) + \psi(-x,-y,-z)] \\
\psi\_\mathrm u(x) &= \frac{1}{2}[\psi(x,y,z) - \psi(-x,-y,-z)] \\
\end{align}$$


and analogously to above one can see that every function $\psi(x,y,z)$ can be expressed as the sum of a gerade component $\psi\_\mathrm g$ and an ungerade component $\psi\_\mathrm u$. These components don't have any physical meaning (unless $\psi$ itself is either g or u, in which case one component is simply $\psi$ and the other is zero), but that's somewhat beside the point.


If a molecule does not possess a centre of inversion then its (canonical) molecular orbitals will not possess g/u symmetry.


So the answer is yes, and yes. There is no "or".


 8