Molecular orbitals in an ionic diatomic compound

The wavefunction of a heteronuclear diatomic molecule, after the orbital approximation and the Born-Oppenheimer approximation, is


$$\psi = c\_A \chi\_A \pm c\_B \chi\_B $$


where $\chi$ are the starting AO wave functions. Calling


$$\gamma = \dfrac{c\_A}{c\_B}$$


the coefficient value can be calculated by


$$ c\_A = \sqrt{\dfrac{\gamma^2}{\gamma^2 + 1 + 2\gamma S}} $$


where $S$ is the overlap integral


$$ c\_B = \dfrac{c\_A}{\gamma} $$


In an ionic diatomic compound, $c\_A = 0$ and $c\_B = 1$: how can $c\_B = 1$ if


$$c\_B = \dfrac{0}{\gamma}$$




As


$$\gamma = \dfrac{c\_A}{c\_B}$$


when $c\_A=0$, $\gamma$ *must* also be zero. This follows as $c\_B$ can not be also be zero as the wavefunction is normalised to unity. Thus under such conditions it is invalid to divide through by $\gamma$. Instead simply use the normalisation condition and the fact that $c\_A$ is zero to show that $|c\_B|=1$ .


 3

Using the normalisation condition


$$c^2\_A + c\_B^2 + 2c\_A c\_B S = 1$$


while $c\_A = 0$


$$0 + c^2\_B + 2 \cdot 0 \cdot c\_B S = 1$$


$$ c^2\_B = 1 $$


$$ c\_B = \sqrt{1} = \pm 1$$


 3