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
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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