Molecular orbitals in an ionic diatomic compound

Question

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

$$c_B = \dfrac{0}{\gamma}$$

Answer 1

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 $|c_B|=1$ .

Answer 2

Using the normalisation condition

$$c^2_A + c_B^2 + 2c_A c_B S = 1$$

while $c_A = 0$

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

$$ c^2_B = 1 $$

$$ c_B = \sqrt{1} = \pm 1$$