The 2s and the 2p orbital energies of hydrogen atom are E2s(H) and E2p(H), respectively. The 2s and the 2p orbital energies of lithium atom are E2s(Li) and E2p(Li), respectively. The correct option(s) about the orbital energies is(are)
- E2s(Li) < E2p(Li)
- E2s(H) = E2p(H)
- E2p(H) < E2s(Li)
- E2s(H) > E2s(Li)
Answer
(1) E2s(Li) < E2p(Li), (2) E2s(H) = E2p(H), (4) E2s(H) > E2s(Li)
Solution
In a one-electron atom (H), the energy depends only on n, so E2s(H) = E2p(H) = -3.4 eV. (B) is correct.
In a multi-electron atom (Li), 2s penetrates closer to the nucleus than 2p, so E2s < E2p. (A) is correct.
Li 2s energy ≈ −(IE of Li) = −5.39 eV, which is lower than −3.4 eV. So E2s(H) > E2s(Li). (D) is correct.
For (C): E2p(H) = -3.4 eV is not lower than E2s(Li) = -5.39 eV, so (C) is wrong.
Answer: (A), (B), (D)