# 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

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

• Hint: CB can be 1 if gamma=0. It might be easier if you start by writing it as CA=gamma*CB rather than as a ratio, so avoiding confusing division by zero problems. Sep 4, 2021 at 20:31
• So, your advice is: $c_A = 0 \cdot c_b$ ? Sep 4, 2021 at 20:39
• Close. CA=0 implies gamma=0 from your first relation as CB can not be zero under such conditions, then calculate CB by the normalisation condition. Sorted. Sep 4, 2021 at 20:42

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 $$|c_B|=1$$ .

Using the normalisation condition

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

while $$c_A = 0$$

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

$$c^2_B = 1$$

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