Electron Configeration

Electronic Configeration = orbitals electron occupy + # of electrons in each orbital.
History: Bohr's proposal-electrons moves from one orbital to another when emit or absorb energy.
Energy level= amount of energy an electron can possess( n)
Ground State: all e- in lowest possible energy level
Excited State: 1 or more e- in energy level other than lowest available level

( orbital: region occupied by an electron in a specific energy level.-> S, P, D, F
( shell: set of all orbitals w same n-value. (n) = energy level
( subshell: set of orbitals of same type.

Comparing Energy level of Hydrogen to Polyelectronic atom


H (1S1)   1 e-
He ( 1S 2)  2 e-
Be (1S2 2S2)  4 e-

Orbital:
n=1 : only s-type possible
n=2: s, p-types are possible
n=3: s, p, d-types are possible
n=4: s, p, d, f-types are possible

Subshell:
s-subshell include 1 s-orbital
p-subshell include 3 p-orbitals
d-subshell include 5 d-orbitals
f-subshell include 7 f-orbitals



The diagram above shows how s, p, d, f are distrbuted on the periodic table for each element, and we can use this diagram to understand and write the electronic configeration.

TWO ESSENTIAL RULES:

1. electrons MUST be added to lowest energy orbital 1st.
2. maximum of 2 e- in each orbital:
 2 e- in s-type subshell, 6 e- in p-type subshell, 10 e- in d-type subshell, 14 e- in f-type

E-Configeration for Neutral atoms
like Na, P, Ca, Fe, ......

1.how many e-?
2.each e- has 1 upward and 1 downward arrow

ex. Silicon, Si has 14 e-.

Look at the diagram below and starts with the lowest energy level, then keep adding until it fulfills the all # of e-.

Si ( 1s2 2s2 2p6 3s2 3p2)

Exercise:

a)Mn b) Co c) Rb d) P

Answer:
a) 1s2 2s2 2p6 3s2 3p6 4s2 3d5
b) 1s2 2s2 2p6 3s2 3p6 4s2 3d7
c) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1
d) 1s2 2s2 2p6 3s2 3p3

Core notation: simplifies the work by replacing a set of configeration with the nearest noble gas with less atomic number.

try those with the same exercise above, did you get the answer?
a) [Ar] 4s2 3d5
b) [Ar] 4s2 3d7
c) [Kr] 5s1
d) [Ne] 3s2 3p3

E.Configeration for ions

negative: add e- (same as charge) to last UNFILLED subshell, where neutral atom left off.

Ex. P 3-
1s2 2s2 2p6 3s2 3p3 -> 1s2 2s2 2p6 3s2 3p6

positive: remove e- from largest n-value, remove p-electron before s-electron before d-electron.

Ex. Sn 2+
[Kr]5s2 4d10 5p2 -> [Kr] 5s2 4d10

I believe simple rule following like this won't preplex all of you!

Thanx for comin!