Understanding Atomic Electron Configurations and Inert Gases, Lecture notes of Molecular biology

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What are inert Gases again??

NOBLE GASES!!

Hebden Pg 156 #27 ( odds only)

1. This means “having the same number of electrons / the same electron configuration as a noble gas” but not the same number of protons or the same properties.
2. Recall …Mg vs. Mg 2+ Mg 1s 2 2s 2 2p 6 3s 2 Mg 2 +  1s 2 2s 2 2p 6 *loses two electrons to be isoelectronic with Neon (Ne) The e- in Mg 2 + are the same as Neon, and the configuration is the same, but they are still different elements – most elements want to do this - be Isoelectronic (look-alikes) with Noble gases.

Ne 1s

2

2s

2

2p

6

Titanium Ion…Ti

Ti

 [Ne] 3s

3p

You can’t just write [Ar]!!

1. Br 1 -
2. Sc 3 +
3. N 3 -
4. Sb 3+ Br 1 - :[Ar]3d 10 4s 2 4p 6 Mn 2 + : [Ne] 3s 2 3 p 6 N 3 - : [He]2s 2 2p 6 Sb 3+ : [Kr]4d 10 5 s 2

1. 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2
2. 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 1
3. 1s 2 2s 2 2p 4 3s 1
4. 1s 2 2s 2 2p 6 3s 1
5. 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 (20 e-) = Calcium GS
6. 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 1 (20e-) = Calcium ES!!!
7. 1s 2 2s 2 2p 4 3s 1 (9 e-) = Fluorine ES!!!
8. 1s 2 2s 2 2p 6 3s 1 (11e-) = Sodium GS

1. The most stable orbitals are completely full (all paired electrons) or half full (no paired electrons). a. These types of orbitals are lower in energy (more stable) than those which are partially filled. b. This is especially important for p, d, and f orbitals
2. Atoms will sometimes rearrange their electrons between orbitals close in energy in order to obtain one of these above conditions.

3. The transition metals (d-orbitals) will lose electrons from their outer orbitals (highest n value) first, and then the d-orbitals to form stable ions. (p before s before d ) e.g. Sn [Kr] 36 4d 10 5s 2 5p 2 Sn 2 + [Kr] 36 4d 10 5s 2 5p 0 Sn 4 + [Kr] 36 4d 10 5s 0 5p 0

“Those electrons in an atom, that are not in an

inert Gas formation core, or are not in a full d or f

subshell” ….

Al  [Ne] 3s

2

3p

1

Ga  [Ar] 3d

10

4s

2

4p

1

Pb  [Xe] 4f

14

5d

10

6s

2

6p

2

Xe  [Kr] 4d

105

s

2

5p

6

= 3 valence e-

= 3 valence e-

= 4 valence e-

= 0 valence e-

(noble gas)