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Will the orbital energies for multielectron atoms depend on their angular momentum quantum number ℓ? (A) In the H atom, the orbital energies depend on.

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Presentation on theme: "Will the orbital energies for multielectron atoms depend on their angular momentum quantum number ℓ? (A) In the H atom, the orbital energies depend on."— Presentation transcript:

1 Will the orbital energies for multielectron atoms depend on their angular momentum quantum number ℓ? (A) In the H atom, the orbital energies depend on the principal quantum number n only, if there are no external fields present. This will be the same in multielectron atoms.  E depends only on n. (B) The repulsion between different electrons is different for different ℓ (e.g. s or p orbitals), because their spatial distributions are different. Therefore, E depends on n and ℓ.

2 Will the orbital energies for multielectron atoms depend on their angular momentum quantum number ℓ? (A) In the H atom, the orbital energies depend on the principal quantum number n only, if there are no external fields present. This will be the same in multielectron atoms.  E depends only on n. (B) The repulsion between different electrons is different for different ℓ (e.g. s or p orbitals), because their spatial distributions are different. Therefore, E depends on n and ℓ.

3 For constant n, how will the orbital energies for multielectron atoms depend on ℓ? (A) Higher ℓ → higher angular momentum → more “circular” behavior → more screening of nuclear charge → for the same n, E will increase with increasing ℓ (B) Higher ℓ → higher angular momentum → more “circular” behavior → less overlap with orbitals of “core” electrons → less Coulomb repulsion → for the same n, E will decrease with increasing ℓ

4 For constant n, how will the orbital energies for multielectron atoms depend on ℓ? (A) Higher ℓ → higher angular momentum → more “circular” behavior → more screening of nuclear charge → for the same n, E will increase with increasing ℓ (B) Higher ℓ → higher angular momentum → more “circular” behavior → less overlap with orbitals of “core” electrons → less Coulomb repulsion → for the same n, E will decrease with increasing ℓ

5 Consider the H atom in its ground state: H(1s). What is the term symbol? (A) 1 S 0 (B) 2 S 1/2 (C) 3 P 2 (D) 3 S 0 (E) 1 S 1/2 2S+1 L J

6 Consider the H atom in its ground state: H(1s). What is the term symbol? (A) 1 S 0 (B) 2 S 1/2 correct: S=1/2; L=0; J can only be 1/2 (C) 3 P 2 (D) 3 S 0 not possible, not with any electron configuration! (E) 1 S 1/2 not possible, not with any electron configuration! 2S+1 L J


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