Estados Eletrônicos e Constantes Moleculares

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114 Chapter 4 where is the energy of the jth excited electronic state relative to the ground state. Show that if we define the ground state to be the zero of energy, then = + + + The first and second excited electronic states of O(g) lie 158.2 cm⁻¹ and 226.5 cm⁻¹ above the ground electronic state. Given 8₀ = 5, g₁ = 3, and g₂ = 1, calculate the values of and (ignoring any higher states) for O(g) at 5000 K. Substituting the values given in the problem into the definition of gives = 0.69509 cm = 227.6 K = 0.69509 226.5 cm = 325.8 K We can write as (Equation 4.8) = = = 5 + = 8.803 4-23. Determine the symmetry numbers for H₂O, HOD, SF₆, C₂H₂, and C₂H₄. Symmetry numbers of selected molecules Molecule Symmetry Number 2 HOD 1 CH₄ 12 SF₆ 24 C₂H₂ 2 C₂H₄ 4 4-24. The HCN(g) molecule is a linear molecule, and the following constants determined spectro- scopically are I = 18.816 X 10⁻⁴⁷ = 2096.7 cm⁻¹ (HC-N stretch), = 713.46 cm⁻¹ (H-C-N bend, two-fold degeneracy), and v₃ = 3311.47 cm⁻¹ (H-C stretch). Calculate the values of and and at 3000 K. We can use the definitions of and to write = h² = 2.1405 K = 3016 K (HC-N stretch) = 1026 K (H-C-N bend) = 4765 K (H-C stretch)