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MOLECULAR ORBITAL THEORY VARIATION PRINCIPLE

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Illustration 2 Using the variation principle (2) To find the values of the coefficients c A and c B in the linear combination that corresponds to the energy E+ from Illustration 1, we use ebove (with α A =α B =α) to write Normalization

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For A = B =

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Proceeding in a similar way to find the coefficients in the linear combination that corresponds to the energy E , we write

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Two simple cases The second simple case is for a heteronuclear diatomic molecule but with S 0

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The solutions can be expressed in terms of the parameter (zeta), with

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and are When B A 2 and 2 / B A 1 arctan 2 / B A 2 / B A Don’t forget them For x 1, sin x = x, cos x = 1, tan x x, and arctan x tan -1 x x.

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It follows that tan ζ ≈ |β |/(α B −α A ) and β /|β | = −1,

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Example 11.3 Calculating the molecular orbitals of HF Calculate the wavefunctions and energies of the orbitals in the HF molecule,taking 1.0 eV and the following ionization energies: H1s: 13.6 eV, F2s: 40.2 eV, F2p: 17.4 eV. Answer Setting H 13.6 eV and F 17.4 eV gives tan 2 0.58; so 13.9°. Then E − = −13.4 eV ψ − = 0.97χH − 0.24χF E + = −17.6 eV ψ + = 0.24χH + 0.97χ F

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Dengan cara seperti contoh di atas tentukan E dan untuk molekul CO dan NO. Tugas I dan II

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Atomic QM to Molecular QM (16.4-16.6) Solution of SE for molecules is more complicated due to much larger number of electrons and multiple nuclei – SE.

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