Question

Consider a bonding electron in a diatomic molecule from the molecular orbital point of view. If the probabilities of finding the electron in atomic orbitals Wa and Wb are 0.25 and 0.75, respectively, what is the LCAOwavefunction for the electron (Neglect overlap)

Answer 1

Answer on Question #55173 - Chemistry - General chemistry

Question:

Consider a bonding electron in a diatomic molecule from the molecular orbital point of view. If the probabilities of finding the electron in atomic orbitals $\Psi_{\mathrm{A}}$ and $\Psi_{\mathrm{B}}$ are 0.25 and 0.75, respectively, what is the LCAO wave function for the electron (Neglect overlap).

Answer:

$$\Psi = C _ {1} \Psi_ {\mathrm {A}} + C _ {2} \Psi_ {\mathrm {B}}.$$

$\Psi_{\mathrm{A}}$ and $\Psi_{\mathrm{B}}$ are orthonormal which means:

$$\frac {1}{4} = \int C _ {1} \Psi_ {A} C _ {1} \Psi_ {A} d V = C _ {1} ^ {2}$$ $$C _ {1} = \pm \frac {1}{2}$$

Likewise:

$$C _ {2} ^ {2} = \frac {3}{4}$$ $$C _ {2} ^ {2} = \pm \frac {\sqrt {3}}{2}$$

We have to possible answer:

$$\Psi = - \frac {1}{2} \Psi_ {\mathrm {A}} + \frac {\sqrt {3}}{2} \Psi_ {\mathrm {B}}$$ $$\Psi = \frac {1}{2} \Psi_ {\mathrm {A}} - \frac {\sqrt {3}}{2} \Psi_ {\mathrm {B}}$$

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