Elements of group 6A, compared to 5A, require less ionization energy due to the paired electrons of 6A.

But a question arises: why does group 7A, compared to group 6A, require more ionization energy in spite of the paired electrons in group 7a too.

Regardless of the general trend that the ionization energy increases as we go from left to right in periods, why didn't we apply the same procedure as the first comparison?


2 Answers 2


It's not just adding one more electron. Another proton is also added to the nucleus. The overall trend across a row is driven by shielding effects.

In general, as you go across a row, the shielding effect is fairly small compared to the effect of increasing nuclear charge.

As you note, there is a slight deviation in this trend between 5A (i.e., $\ce{N}$) and 6A (i.e., $\ce{O}$) because the extra electron in oxygen is paired. This increases the electron-electron repulsion, so the ionization energy decreases slightly.

As you note, $\ce{F}$ (7A) resumes the overall trend of increasing ionization energy driven by the attractive effects of the higher effective nuclear charge $Z_{eff}$.


Elements of group 6A (also called group 16) require less ionization energy than group 5A (group 15) elements because group 5A elements have a half filled p-orbital.

Group 7A (group 17) DOES NOT have paired electrons. Ionization energy of group 7A elements is more than that of group 6A elements due to higher effective nuclear charge on group 7A elements.

  • 1
    $\begingroup$ but Group 7A (17) has paired electrons in the first and the second orbitals and an unpaired electron in the last orbital, of the three orbitals of the subshell p !!! $\endgroup$
    – user14840
    Commented Mar 16, 2015 at 14:31
  • $\begingroup$ @user14840 Elements are said to have paired electrons when all the electrons are paired. In this case 1 electron is still unpaired. $\endgroup$ Commented Mar 16, 2015 at 15:50
  • $\begingroup$ I am not aware of such a definition of paired electrons and I think it is a poor choice. There would be no way to describe complexes between high and low spin. $\endgroup$ Commented Mar 17, 2015 at 8:46
  • $\begingroup$ @Martin I agree with you on the issue of distinguishing between high spin and low spin complexes. However in this case we're talking about native elements and their stability and ionization energies, and I feel in this case while referring to paired electrons, it is generally meant all electrons are paired. $\endgroup$ Commented Mar 17, 2015 at 10:29

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