Covalent radius .Metallic radius . Cationic Radii. van der Waals radius . Screening and Penetration.Slater's Rules.

1. The Sizes of Atoms and Ions
Dr. K. Shahzad Baig
Memorial University of Newfoundland
(MUN)
Canada
Petrucci, et al. 2011. General Chemistry: Principles and Modern Applications. Pearson Canada Inc., Toronto, Ontario.
Tro, N.J. 2010. Principles of Chemistry. : a molecular approach. Pearson Education, Inc..

2. The Sizes of Atoms and Ions
Physical properties such as density, melting point, and
boiling point are related to size of atoms.
The covalent radius is one-half the distance between
the nuclei of two identical atoms joined by a single
covalent bond
Metallic radius is one-half the distance between the
nuclei of two atoms in contact in the crystalline solid metal
The van der Waals radius is one-half the distance between the centers of neighboring
atoms when the electrostatic forces between them are balanced

3. The ionic radius is based on the distance between the nuclei
of ions joined by an ionic bond. Because the ions are not
identical in size, this distance must be properly apportioned
between the cation and anion.
The convention we have chosen to use is to assign O2- an ionic radius of 140 pm.
Starting with a radius of 140 pm for O2- the radius of Mg2+ can be obtained from
the internuclear distance in MgO.
the radius of Cl- from the internuclear distance in MgCl2 and the radius of Na+ from
the internuclear distance in NaCl
One way to apportion the electron density between the ions is to define the radius of
one ion and then infer the radius of the other ion.
The angstrom unit, Å, has long been used for atomic dimensions (1 Å = 10-10 m).
The angstrom, however, is not a recognized SI unit. The SI units are the nanometer
(nm) and picometer (pm)

4. Atomic radius

5. Screening and Penetration
Penetration is a gauge to measure how close an electron gets to the nucleus
When interpreting the radial probability distributions, it
was observed that s electrons, penetrate better than p
electrons, which in turn penetrate better than d electrons.
𝑟𝑛𝑙 =
𝑛2
𝑎0
𝑍 𝑒𝑓𝑓
1 +
1
2
1 −
𝑙(𝑙 + 1)
𝑛2
𝐸 𝑛 = 𝑅 𝐻
𝑍 𝑒𝑓𝑓
2
𝑛2
the Effective nuclear charge is replaced by (Zeff) so
that the orbital energy is approximated by
The average size of an orbital is taken to be the average
value of the distance, of the electron in that orbital from
the nucleus

6. let us assume that the core electrons completely cancel an equivalent charge on the nucleus.
In this way, the core electrons shield, or screen, the outer-shell electrons from the full attractive force
of the nucleus.
Let us also assume that the outer-shell electrons do not screen one another
Actually, neither assumption we made above full screening by inner-shell electrons and no screening
by outer-shell electrons is correct.
These assumptions ignore the fact that the electrons, both inner and outer, occupy orbitals with
different radial probability distributions and, consequently, different degrees of penetration . Thus,
an ‘s’ e-, with its greater penetration, will be screened by inner electrons less than will a ‘p’ e-.
Similarly a ‘p’ electron is shielded less than a ‘d’ electron, which has a much lower penetration.
In Na, the 10 core electrons cancel only about 8.5 units of charge; thus, is about not Also, outer-shell
electrons do screen one another somewhat because of penetration effects.
Each outer-shell electron is about 1/3 effective in screening the other outer-shell electrons.

7. Slater's Rules
the actual charge felt by an electron =
= (expected the charge to be from a certain number of P+ ) – (certain amount of charge from other e-)
Step 1: Write the electron configuration of the atom in the following form:
(1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) (5s, 5p) . . .
Step 2: Identify the electron of interest, and ignore all electrons
•Step 3: two cases: the shielding experienced by an s- or p- electron,
• electrons within same group shield 0.35, except the 1s which shield 0.30
• electrons within the n-1 group shield 0.85
• electrons within the n-2 or lower groups shield 1.00
•the shielding experienced by nd or nf valence electrons
• electrons within same group shield 0.35
• electrons within the lower groups shield 1.00
𝑆 = 𝑍 − 𝑍 𝑒𝑓𝑓

8. Penetration and Shielding
Slater rules: For a 1s electron, S = 0.3.
For electrons in an s or p orbital with n > 1, the screening constant is given by
S = 1.00·n2 + 0.85·n1 + 0.35·n0
n0 represents the number of other electrons in the same shell,
n1 represents the number of electrons in the next smaller shell (n-1), and
n2 is the number of electrons in other smaller shells (n-2 and smaller).
Zeff = Z - S
Zeff = Z – (1.00·n2 + 0.85·n1 + 0.35·n0)
The effective nuclear charge is

9. Problem statement
What is the shielding constant experienced by a 2p electron in the nitrogen atom?
N: 1s2 2s2 2p3
N: (1s2)(2s2,2p3)
S (x) 1.00·n2 + 0.85·n1 + 0.35·n0
S [2p] = 1.00(0) + 0.85(2) + 0.35(4) = 3.10
Problem statement
What is the shielding constant experienced by a 3d electron in the bromine atom?
Br: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5
Br: (1s2)(2s2,2p6)(3s2,3p6)(3d10)(4s2,4p5)
S (x) = 1.00·n2 + 0.85·n1 + 0.35·n0
S [3d] = 1.00(18) + 0.35(9) = 21.15

10. Problem statement
What is the effective nuclear charge experienced by a valence p- electron in boron?
B: 1s2 2s2 2p1
B: (1s2)(2s2,2p1)
Zeff = Z – (1.00·n2 + 0.85·n1 + 0.35·n0)
Zeff = 5 – (1.00 x 0 + 0.85 x 2 + 0.35 x 2)
Zeff = 5 – (2.40)
Zeff = 2.60

11. Cationic Radii
Cation
a positively charged ion, i.e. one that would be
attracted to the cathode in electrolysis.
Cations are smaller than the atoms from which
they are formed.
For isoelectronic cations, the more positive the
ionic charge, the smaller the ionic radius.

12. Anionic Radii Anions
Anions are larger than the atoms from which they are
formed. For isoelectronic anions, the more negative
the charge, the larger the ionic radius
Pure aluminum oxide Al2O3 is
colorless, but with this small
amount of chromium(III) ion,
Cr3+ it is a beautiful red color.
Ruby
When Cr3+ ions replace about 1% of the AL3+
ions in aluminum oxide Al2O3, This substitution
is possible because ions are only slightly larger
(by 9 pm) than ions.