How do you arrange ionization energy in increasing order?

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 (by change within a group) than going across the periodic table (by change within a period).

So let's start from the bottom of the periodic table:
#Pb# is the element that is in the lowest period at 6 (and lowest group at 14) in the periodic table; it's the smallest ionization energy.

The period above (5) has two of the elements: Sn and Te. Well, since ionization energy increases across a period, Sn will have a smaller ionization energy than Te.
#Pb, Sn, Te#

Now, let's go to the third period, where #S# and #Cl# are. Since #S# is before #Cl,##S# has a lower ionization energy than #Cl#.
#Pb, Sn, Te, S, Cl#

The order is #"Sn < Pb < Te < S < Cl"#.

You have learned that ionization energy increases from top to bottom and from left to right in the Periodic Table.

You probably saw a diagram something like this.

Here's the portion of the Periodic Table that includes the elements in this question.

You would naturally predict the order to be

#"Pb < Sn < Te < S< Cl"#

This is almost correct, but the correct order is #"Sn < Pb"#, as shown in the image below.

Why is this so?

The electron configuration of #"Sn"# is #"[Kr] 5s"^2 "4d"^10 "5p"^2#.

The electron configuration of #"Pb"# is #"[Xe] 6s"^2 "4f"^14 "5d"^10 "6p"^2#.

The #"4f"# electrons in #"Pb"# are poor at shielding the outermost electrons.

Thus the outer electrons experience a greater effective nuclear charge, and it is more difficult to remove them.

Hence #"Pb"# has a higher ionization than #"Sn"#, and the correct order is #"Sn < Pb < Te < S < Cl"#.

I hope that your instructor told you about this phenomenon before asking you to make a prediction.

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Solution

The correct option is C Po,Te,Se,S,O
Down the group ionization energies decreases
So, the increasing order of ionization energies in the 16th group is Po,Te,Se,S,O.

Answer

Verified

Hint: Ionisation energy decreases as the nuclear effective charge decreases on the atom. Nuclear effective charge is also called as ${{Z}_{eff}}$. Size of the atom is inversely proportional to the Nuclear effective charge.

Complete step by step answer:
Ionisation energy is the quantity of energy that an isolated, gaseous atom in the group electronic state must absorb to discharge an electron, resulting in a cation. When considering an initial neutral atom, expelling the first electron will require less energy than expelling the second, the second will require less energy than the third, and so on. This is due to the reason that after removal of one electron the atom acquires a positive charge, therefore, the rest of the electrons present on the ion, will experience more nuclear effective force. As we move from top to bottom in the group the ionization energy will be decreased, because the atomic radius will be increased so removal of electrons will be easy.
Li, Na, K, Rb, Cs are IA group elements. Ionisation energy decreases from Li to Rb. So the correct order of ionization energy will be (A) Cs < Rb < K < Na < Li . So, the correct answer is “Option A”.

Note: Ionisation energy is usually expressed in kJ/mol, or the amount of energy it takes for all the atoms in a mole to lose one electron each. The more electrons that are lost, the more positive this ion will be, the harder it is to separate the electrons from the atom, more is the ionization energy.