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Not considering the electronic spin, the degenracy of the second excited state of `H^(-)` is |
Answer» Correct Answer - 3 In case of H-atom, the energies of the orbitals are in the order. `1s lt 2s = 2p lt 3s = 3p = 3d lt 4s = 4p = 4d = 4f lt`.... ,br. In case of multielectron atoms or ions (e.g. `H^(-)` viz `1s^(2)`), the order is `1s lt 2s lt 2p lt 3s lt 3p`.... (i.e., they follows `(n + l)` rule) Hence, in case of `H^(-) (1s^(2))`, the first excited state would be `1s^(1) 2s^(1)` Second excited state would be `1s^(1) 2s^(0) 2p^(1)`. Thus, degeneracy of the second excited state `(2p^(1))` is three viz. `2p_(x) 2p_(y) 2p_(z)`, i.e., number of degenerate orbitals = 3 [In case of H-atom `(1s^(1))`, 1st excited state would be 2s or 2p and 2nd excited state would be 3s, 3p or 3d, i.e., number of degenerate orbitals would be 1(3s), 2 (3p) and 5(3d), i.e., total = 9] |
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