Question

Several consecutive ionization energies for a certain element are given below, in units of kJ/mol.

[I1 is the first ionization energy, I2 is the second ionization energy and so on.]

Which element matches this pattern?

I1 = 1060                  (M → e- + M+)           

I2 = 1890                (M+ → e- + M2+)

I3 = 2905                (M2+ → e- + M3+)

I4 = 4950                (M3+ → e- + M4+)

I5 = 6270                (M4+ → e- + M5+)

I6 = 21,200             (M5+ → e- + M6+)

A) Na

B) Al

C) P

D) Si

E) Mg

F) S

G) Ar

H) Cl

Answer 1

Ans: C) P (Phosphorus)

Electronic configuration of Phosphorus is 1s² 2s² 2p⁶ 3s² 3p³

The first three ionization energies increases regularly as the effective charge on the outermost electron increases with loss of electrons.

After 3rd ionization, the electronic configuration is 1s² 2s² 2p⁶ 3s²

I4 increase considerably because eletron has to be removed from fully filled 3s² orbital. I5 increases slightly following the usual trend where effective charge increased even further.

After 5th ionization, the electronic configuration is: 1s² 2s² 2p⁶

I6 has very high value as it would be very difficult to remove electron form stable nobel gas configuration.