How would you arrange the following elements in order of increasing ionization energy: te, pb, cl, s, sn?

Sn < Pb < Te < S < Cl

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

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