Why does the 4 s level in neutral atoms lie below the 3 d ?
The s orbital has a small fraction of its probability density close to the nucleus.
3 d orbitals do not have such inner regions, as they only have planar nodes
Hence an s electron from a higher shell will sometimes occur at lower energy than a d electron in a lower shell

The charge that a 2s or 2p electron feels is different due to the shielding from the electrons in the 1s orbital
From Li to Ne, nuclear charge increases from 3 to 10
-1.77
-1.42
-2.78
-3.15
-3.51
-3.87
2 s orbital penetrate into the 1s orbital and therefore are shielded less on average than d orbitals
Note: Shielding effect increases as the number of e’s increase
.
This is the result of additional shielding from the 2 s and 2 p e’s
Note: As Z* increases orbital shrink towards nucleus as e’s are held more tightly dues to stronger electronic interactions.

Consider the change in size of the atoms from Li to F
Decrease strongly
Consider the size change from Li to Rb
Increase significantly
Consider the size change from F to I
Increase gently
• Consider the size change from F to I

Atomic radius decreases along the period, and increases down the group
The radius of an anion is larger than its neutral atom.
Adding the extra electron increases shielding without changing the charge of the nucleus. i.e Z* is smaller.
The radius of a cation is smaller than its neutral atom
Removing the electron decreases shielding without changing the charge of the nucleus. i.e., Z* is larger.
Valence electrons of a cations are in a lower energy shell than in the neutral atom, decreasing the ionic radius.

Anions are larger than cations
This is always true across a period of the table
Ions in each group of the table get larger in size down the group
Isolectronic ions decrease in size across the period, as Z* increases dramatically.
Ex) N 3to F Na + to Al 3+

IE
1
 
E for E

E
  e

The energy that must be absorbed in order to remove a valence electron from a neutral atom in the gas phase

Li
s
2 1 s F s
2 2 s p
5 e -
3+ e e r = 152 pm
Z*= 1.28
EA
1
= 520 kJ/mol
>
<
< e e e e e -
9+ e e e e r = 71 pm
Z*= 5.13
EA
1
= 1681 kJ/mol

1
List of the IE
1 in kJ/mol for the elements
IE increases across the period
IE increases up the group
Z* increases
Shielding effect decreases i.e
Z* increases

Energy released when an element attracts an extra electron into the lowest-energy unoccupied orbital to form an anion
For large Z* e’s are held closely to the nucleus therefore e-n interactions will be stronger for an additional electron coming in.
Compare Li ( Z * = 1.28) with F ( Z * = 5.13 )
Li
 e
 
Li

Negative since energy is released
F
 e
 
F


H
EA increases in magnitude across period

H
EA decreases in magnitude down the group

&
c
i) How strongly does an element hold onto its own electrons ?
ii) How strongly is an element able to attract electrons from other elements?
A combination of ionization energy and enthalpy of electronic attraction.
Which element(s) should have the highest electronegativity?
Which element(s) should have the lowest electronegativity?
10

symbol atomic number
Carbon
6
C

“The properties of the elements are periodic functions of atomic number.”
Group
Similar chemical properties
Period
Repetition of properties
Nonmetals – insulators not ductile
Metalloids - Semiconductors
Ductile ?
Metals – Conducting, Ductile