Answer to Question #85832 in Inorganic Chemistry for SK IQBAL HOSSAIN

Question #85832
Explain the variation in ionization energy across the elements of period 2 in the periodic table.
1
Expert's answer
2019-03-08T05:01:23-0500

First ionisation energies of the elements of second period

IE1(Li) = 520 kJ/mol

IE1(Be) = 899 kJ/mol

IE1(B) = 801 kJ/mol

IE1(C) = 1086 kJ/mol

IE1(N) = 1400 kJ/mol

IE1(O) = 1314 kJ/mol

IE1(F) = 1680 kJ/mol

IE1(Ne) = 2081 kJ/mol



Reference for picture: https://www.grandinetti.org/ionization-energy-trends


As we move across the second period from left to right, in general, the first ionisation energy increases. But we can see two drops: between Be and B and between N and O. The explanation lies in the electronic structures of atoms.

Li [He] 2s1

Be [He] 2s2

B [He] 2s22p1

We can see in atom of B one more electron is added but it is added to the p orbital. P orbital is higher in energy than s orbital  and electron on this orbital is further from the nucleus. The first ionization energy decreases 1) due to the screening effect of 2S2 electrons that reduces the pull from the nucleus and 2) due to the increased distance between

2p electron and nuclei that reduces attraction between them.

C [He] 2s22p2

N [He] 2s22p3

O [He] 2s2 2p4

We can see that three p orbitals in N atom are half- filled (px1py1pz1) whereas in O atom one p orbital is fully filled (px2py1pz1). The screening effect is the same for all p electrons, we remove an electron from the same p orbital (px) but for the pair of electrons repulsion exists therefore it is easier to remove electron from the pair. For the next atoms (F and Ne)addition of one more proton has more influence on p electrons than repulsion between electrons of a pair.


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