Ch7 Lec6

(4/28/20, bes)

Greetings,

I wanted to post this last lecture well before exam 3 (Friday, May 1st) so that you do not feel too rushed...review these notes as time permits. This is the last lecture for Chapter 7.

Multi-electron Energy Levels

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The figure above was presented in yesterday's notes...building upon this...please notice:

The shift in the energy levels mixes up the order in which the orbitals are filled.

- the 3d orbitals are now above the 4s orbitals

- the 4d orbitals are now above the 5s orbitals (not shown in figure)

- the 5d orbitals are now above the 6s orbitals (not shown in figure)

For boron (B) (as previously discussed), it contains 5 electrons...with 2 electrons per orbital, then the electron configuration is: 1s² 2s² 2p¹.

(Note: in WA i would ask that you enter the answer as, 1s2 2s2 2p1 <-- no superscripts).

For carbon (C) (as previously discussed), it contains 6 electrons...with 2 electrons per orbital, then the electron configuration is: 1s² 2s² 2p².

(Note: in WA i would ask that you enter the answer as, 1s2 2s2 2p2 <-- no superscripts).

For neon (Ne), it contains 10 electrons...with 2 electrons per orbital, then the electron configuration is: 1s² 2s² 2p⁶.

(Note: in WA i would ask that you enter the answer as, 1s2 2s2 2p6 <-- no superscripts).

For argon (Ar), it contains 18 electrons...with 2 electrons per orbital, then the electron configuration is: 1s² 2s² 2p⁶ 3s² 3p⁶.

(Note: in WA i would ask that you enter the answer as, 1s2 2s2 2p6 3s2 3p6 <-- no superscripts).

For calcium (Ca), it contains 20 electrons...with 2 electrons per orbital, then the electron configuration is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s².

(Note: in WA i would ask that you enter the answer as, 1s2 2s2 2p6 3s2 3p6 4s2).