Answer:
Sn < Pb < Te < S < Cl
Step-by-step explanation:
The ionization energy increases across a period but decreases down a group.
This means that the elements with the lowest ionization energies would be in the bottom left-hand corner of the periodic table.
The change in ionization energies is also bigger going down the periodic table than going across the periodic table.
P b is the element that is in the lowest period at 6 (and lowest group at 14) in the periodic table; it should have the smallest ionization energy.
In period 5 we have the elements Sn and Te.
Since ionization energy increases across a period, Sn will have a smaller ionization energy than Te.
The third period, where we have S and C l .
Since S is before C l . So it has a lower ionization energy than Cl
The order ( increasing) is: Pb < Sn <Te <S <Cl
BUT
The electron configuration of Sn is 5s 2 4d 10 5p 2
The electron configuration of Pb is 6s 2 4f 14 5d 10 6p 2 .
The 4f electrons in Pb are poor at shielding the outermost electrons.
So, the outer electrons experience a greater effective nuclear charge, what makes it more difficult to remove them.
So this means Pb has a higher ionization than Sn
So the correct order is Sn < Pb < Te < S < Cl
.