2. ELECTRON SHIELDING
Electrons are attracted by the nucleus (+), but
repelled by other electrons (-).
Thus electrons ‘shield’ other electrons from the
attraction of the nucleus.
This shielding reduces the full nuclear charge to an
effective nuclear charge (Zeff), the nuclear charge
an electron actually experiences.
2p
2s
Forces of attraction are
1s reduced by repulsion
O
nucleus
3. ELECTRON SHIELDING
Effective nuclear charge can be calculated using the
following equation:
Zeff = Z – S
Zeff = effective nuclear charge
Z = atomic number
S = shielding constant
The higher the effective nuclear charge, the greater
the attraction between the electron and the nucleus,
and the lower the energy of the orbital in which that
electron is found.
4. ELECTRON SHIELDING
How to calculate S?
1) Write out the electronic configuration of the element in the following
order of groupings: (1s) (2s, 2p) (3s, 3p) (3d) (4s 4p) (4d) (4f) (5s 5p),
etc.
2) Electrons in any group to the right of the (ns, np) group contribute
nothing to the shielding constant
3) All of the other electrons in the (ns, np) group shield the valence electron
to an extent of 0.35 each
4) All electrons in the (n -1) shell shield to an extent of 0.85 each
5) All electrons (n-2) or lower shield completely; i.e., their contribution is
1.00.
When the electron being shielding is in an (nd) or (nf) group, rules 2
and 3 are the same but rules 4 and 5 become:
6) All electrons in groups lying to the left of the (nd) or (nf) group contribute
1.00
5. ELECTRON SHIELDING
1) Write out the electronic configuration of the element in the following order of groupings: (1s) (2s, 2p) (3s, 3p)
(3d) (4s 4p) (4d) (4f) (5s 5p), etc.
2) Electrons in any group to the right of the (ns, np) group contribute nothing to the shielding constant
3) All of the other electrons in the (ns, np) group shield the valence electron to an extent of 0.35 each
4) All electrons in the (n -1) shell shield to an extent of 0.85 each
5) All electrons (n-2) or lower shield completely; i.e., their contribution is 1.00.
When the electron being shielding is in an (nd) or (nf) group, rules 2 and 3 are the same but rules 4
and 5 become:
6) All electrons in groups lying to the left of the (nd) or (nf) group contribute 1.00
Note: A value of 1.00 as seen in rule 5 means that the electron is
fully capable of shielding an outer electron. However, values of 0.85
or 0.35 mean that the electron is only partially capable of shielding
an outer electron.
3p Electron of interest
3s Can only shield
slightly (0.35)
2p
Can only shield 2s
partially (0.85)
1s Can fully shield (1.00)
O
nucleus
7. ELECTRON SHIELDING
1) Write out the electronic configuration of the element in the following order of groupings: (1s) (2s, 2p) (3s, 3p)
(3d) (4s 4p) (4d) (4f) (5s 5p), etc.
2) Electrons in any group to the right of the (ns, np) group contribute nothing to the shielding constant
3) All of the other electrons in the (ns, np) group shield the valence electron to an extent of 0.35 each
4) All electrons in the (n -1) shell shield to an extent of 0.85 each
5) All electrons (n-2) or lower shield completely; i.e., their contribution is 1.00.
When the electron being shielding is in an (nd) or (nf) group, rules 2 and 3 are the same but rules 4
and 5 become:
6) All electrons in groups lying to the left of the (nd) or (nf) group contribute 1.00
Example #1:
Consider the effective nuclear charge of the valence electron of
potassium: K = 1s22s22p63s23p64s1
Rule 1: (1s2) (2s22p6) (3s23p6) (4s1)
Rule 2: does not apply
S = (8 x 0.85) + (10 x 1.00)
Rule 3: does not apply = 16.80
Rule 4: All (3s23p6) = 8 x 0.85 Zeff = Z – S
= 19.00 – 16.80
Rule 5: (1s2) (2s22p6) = 10 x 1.00 = 2.20
.: Zeff = 2.20
8. ELECTRON SHIELDING
1) Write out the electronic configuration of the element in the following order of groupings: (1s) (2s, 2p) (3s, 3p)
(3d) (4s 4p) (4d) (4f) (5s 5p), etc.
2) Electrons in any group to the right of the (ns, np) group contribute nothing to the shielding constant
3) All of the other electrons in the (ns, np) group shield the valence electron to an extent of 0.35 each
4) All electrons in the (n -1) shell shield to an extent of 0.85 each
5) All electrons (n-2) or lower shield completely; i.e., their contribution is 1.00.
When the electron being shielding is in an (nd) or (nf) group, rules 2 and 3 are the same but rules 4
and 5 become:
6) All electrons in groups lying to the left of the (nd) or (nf) group contribute 1.00
Example #2:
Consider the Zeff for the valence electron of scandium:
Sc = 1s22s22p63s23p64s23d1
Rule 1: (1s2) (2s22p6) (3s23p6) (3d1)(4s2)
Rule 2: does not apply S = (1 x 0.35) + (9 x 0.85) +
(10 x 1.00)
Rule 3: the other 4s electron = 1 x 0.35 = 18.00
Rule 4: (3s23p6)(3d1) = 9 x 0.85 Zeff = Z – S
= 21.00 – 18.00
Rule 5: (1s2) (2s22p6) = 10 x 1.00 = 3.00
.: Zeff = 3.00
9. ELECTRON SHIELDING
Example #3:
Consider the Zeff for
a)the valence electron of titanium
b)the effective nuclear charge that the electron in the 3d orbital
experiences for titanium
Ti = 1s22s22p63s23p64s23d2
a) Rule 1: (1s2) (2s22p6) (3s23p6) (3d2)(4s2)
Rule 2: does not apply
Rule 3: the other 4s electron = 1 x 0.35
Rule 4: (3s23p6)(3d2) = 10 x 0.85
Rule 5: (1s2) (2s22p6) = 10 x 1.00
S = (1 x 0.35) + (10 x 0.85) + (10 x 1.00) = 18.85
Zeff= = 22.00 – 18.85 = 3.15 .: Zeff = 3.15
10. ELECTRON SHIELDING
Example #3:
Consider the Zeff for
a)the valence electron of titanium
b)the effective nuclear charge that the electron in the 3d orbital
experiences for titanium
Ti = 1s22s22p63s23p64s23d2
b) Rule 1: (1s2) (2s22p6) (3s23p6) (3d2)(4s2)
Rule 2: 4s2 does not affect shielding
Rule 3: the other 3d electron = 1 x 0.35
Rule 6: 18 x 1.00
S = (1x 0.35) + (1.00 x 18) = 18.35
Zeff= = 22.00 – 18.35 = 3.65
.: Zeff = 3.65
11. ELECTRON SHIELDING
Example #4:
Consider the Zeff for the valence electron of Chromium
a)as an exception
b)if Cr is [Ar]4s23d4
Cr = 1s22s22p63s23p64s13d5 vs. Cr = 1s22s22p63s23p64s23d4
a) Rule 1: (1s2) (2s22p6) (3s23p6) (3d5)(4s1)
Rule 2: does not apply
Rule 3: does not apply
Rule 4: (3s23p6)(3d5) = 13 x 0.85
Rule 5: (1s2) (2s22p6) = 10 x 1.00
S = (13 x 0.85) + (10 x 1.00) = 21.05
Zeff= = 24.00 – 21.05 = 2.95 .: Zeff = 2.95
12. ELECTRON SHIELDING
Example #4:
Consider the Zeff for the valence electron of Chromium
a)as an exception
b)if Cr is [Ar]4s23d4
Cr = 1s22s22p63s23p64s13d5 vs. Cr = 1s22s22p63s23p64s23d4
b) Rule 1: (1s2) (2s22p6) (3s23p6) (3d4)(4s2)
Rule 2: does not apply
Rule 3: 4s1 = 1 x 0.35
Rule 4: (3s23p6)(3d4) = 12 x 0.85
Rule 5: (1s2) (2s22p6) = 10 x 1.00
S = (1 x 0.35) + (12 x 0.85) + (10 x 1.00) = 20.55
Zeff= = 24.00 – 21.05 = 3.45 .: Zeff = 3.45
