introduction:
Periodic table is design in such a way so that we can understand all the physical and chemical properties in a proper order . Elements having similar properties are placed either in same group or same period as well as.
Periodicity :
The repetition of physical and chemical properties after the regular time interval is defined as periodicity of periodic table. Basically it depends on many factor i.e screening effect,electron affinity and electronegativity etc.
Screening effect or Shielding effect
The phenomena of decrement in the force of attraction between the outer electron and nucleus because the nucleus and the valence shell electron due to inner electron is defined as screening effect or shielding effect.
It is defined by slater
effective nuclear charge |
It is defined by slater
Z* = Z - σ
here Z* is effective nucleus charge , Z is nuclear charge and σ is screening const or slater const .
The reduce force of attraction by the nucleus on the valence electron (see in above figure) is defined as effective nuclear force. It is discuss in term of charge so it is named as effective nuclear charge.
Calculation of screening effect of any element through following slater rule:
Let's take some example to more understanding
Ex. 1. Li ➝ 1s2 2s1
σ = 2 x 0.85 = 1.7 so, Z*= Z- σ = 3-1.7=1.3
here we are finding screening const of last electron so we have not consider nth electron.
Same things have done in upcoming example .
2. Na ➝ 1s2 2s2 2p6 3s1
↑ ↑ ↑
(n-2) (n-1) n
σ = 8 x 0.85 + 2 x 1 = 8.8
so Z*= 11-8.8 = 2.2
3. K ➝ 1s2 2s2 2p6 3s2 3p6 4s1
↑ ↑
inner (n-1)
σ = 8 x 0.85 + 2 x 1 + 1 x 0.35 = 9.15
so Z*= 12-9.15 = 2.85
4. Mg ➝ 1s2 2s2 2p6 3s2
↑ ↑ ↑
(n-2) (n-1) n
σ = 8 x 0.85 + 2 x 1 + 1 x 0.35 = 9.15
so Z*= 12-9.15 = 2.85
5. Ca ➝ 1s2 2s2 2p6 3s2 3p6 4s2
↑ ↑ ↑ ↑
(n-3) (n-2) (n-1) n
σ = 8 x 0.85 + 10 x 1 + 2 x 0.35 = 17.15
so Z*= 20-17.15 = 2.85
let's try it
Find the screening constant and then effective nuclear charge for the last for the last e- of :
(a) Zn (b) B (c) N
Check solution for more clear :
https://drive.google.com/open?id=1DX-f57Vg1l7FZStN7xwYK2SEWpKpwnYX
As you can see in above example of Na , K and Ca ( which belongs to same group i.e 1st group ).
The Z* (Effective nuclear charge ) of all the elements are almost same .So we conclude a important note that there is no change in Z* in group .
Now take some example in the same period elements.
1. B = 1s2 2s2 2p1
↑ ↑
n-1 n
σ = 2 x 0.85 + 2 x 0.35 = 2.40
Z* = 5 - 2.40 = 2.6
2. C = 1s2 2s2 2p2
↑ ↑
n-1 n
σ = 2 x 0.85 + 3 x 0.35 = 2.75
Z* = 6 - 2.75 = 3.25
3. N = 1s2 2s2 2p3
↑ ↑
n-1 n
σ = 2 x 0.85 + 4 x 0.35 = 3.1
Z* = 7- 4 = 3.9
B -- C -- N -- O here consecutive difference = 0.65
↑ ↑ ↑ ↑
2.60 3.25 3.90 4.55
so we get another important point
Note : In period ,Z* increases by 0.65 for each movement in period.
here Z* is effective nucleus charge , Z is nuclear charge and σ is screening const or slater const .
The reduce force of attraction by the nucleus on the valence electron (see in above figure) is defined as effective nuclear force. It is discuss in term of charge so it is named as effective nuclear charge.
Calculation of screening effect of any element through following slater rule:
- Arrange the orbital in the given form
(1s) (2s 2p) (3s 3p) (3d) (4s 4p) (4d 4f)
- The amount of extent from which the force of attraction decreasing is named as screening constant (σ ) .
- The value of screening constant for the different shell are : -
Shell
| (n-4)th | (n-3)th | (n-2)th | (n-1)th | nth |
---|---|---|---|---|---|
σ
|
1
|
1
|
1
|
0.85
|
0.35
|
Let's take some example to more understanding
Ex. 1. Li ➝ 1s2 2s1
σ = 2 x 0.85 = 1.7 so, Z*= Z- σ = 3-1.7=1.3
here we are finding screening const of last electron so we have not consider nth electron.
Same things have done in upcoming example .
2. Na ➝ 1s2 2s2 2p6 3s1
↑ ↑ ↑
(n-2) (n-1) n
σ = 8 x 0.85 + 2 x 1 = 8.8
so Z*= 11-8.8 = 2.2
3. K ➝ 1s2 2s2 2p6 3s2 3p6 4s1
↑ ↑
inner (n-1)
σ = 8 x 0.85 + 2 x 1 + 1 x 0.35 = 9.15
so Z*= 12-9.15 = 2.85
4. Mg ➝ 1s2 2s2 2p6 3s2
↑ ↑ ↑
(n-2) (n-1) n
σ = 8 x 0.85 + 2 x 1 + 1 x 0.35 = 9.15
so Z*= 12-9.15 = 2.85
5. Ca ➝ 1s2 2s2 2p6 3s2 3p6 4s2
↑ ↑ ↑ ↑
(n-3) (n-2) (n-1) n
σ = 8 x 0.85 + 10 x 1 + 2 x 0.35 = 17.15
so Z*= 20-17.15 = 2.85
let's try it
Find the screening constant and then effective nuclear charge for the last for the last e- of :
(a) Zn (b) B (c) N
Check solution for more clear :
https://drive.google.com/open?id=1DX-f57Vg1l7FZStN7xwYK2SEWpKpwnYX
As you can see in above example of Na , K and Ca ( which belongs to same group i.e 1st group ).
The Z* (Effective nuclear charge ) of all the elements are almost same .So we conclude a important note that there is no change in Z* in group .
Now take some example in the same period elements.
1. B = 1s2 2s2 2p1
↑ ↑
n-1 n
σ = 2 x 0.85 + 2 x 0.35 = 2.40
Z* = 5 - 2.40 = 2.6
2. C = 1s2 2s2 2p2
↑ ↑
n-1 n
σ = 2 x 0.85 + 3 x 0.35 = 2.75
Z* = 6 - 2.75 = 3.25
3. N = 1s2 2s2 2p3
↑ ↑
n-1 n
σ = 2 x 0.85 + 4 x 0.35 = 3.1
Z* = 7- 4 = 3.9
B -- C -- N -- O here consecutive difference = 0.65
↑ ↑ ↑ ↑
2.60 3.25 3.90 4.55
so we get another important point
Note : In period ,Z* increases by 0.65 for each movement in period.
So we have seen how to find screening constant and effective nuclear charge for a particular elements . These concept are very important to understanding other periodic properties in the table.
Some of the important properties like lanthanide construction , we will see in upcoming post.
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