Why 1st Ionization energy of oxygen is less than that of nitrogen?

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26
Oct 2, 2015

In Nitrogen, 2p subshell is half filled, thus making it slightly stable, while in Oxygen, in 2p subshell one electron gets paired up, so it's easy to remove electron from that orbital due to repulsion.

Explanation:

Due to slight repulsion in one of the orbitals(since there are paired electrons) in Oxygen, it's easy to remove one electron from it, so the 1st ionization energy in oxygen is slightly more than nitrogen, which is slightly stable due to Hund's half-filled rule.

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11
Oct 6, 2016

Due to electron-electron repulsion.

Explanation:

When comparing the ionization energies of two elements we should consider four main factors:

1. Distance from nucleus.
2. Effective nuclear charge.
3. Shielding effect.
4. Electron-electron repulsion.

Consider the electron configuration for:
""_7N: 1s^(2)2s^(2)2p^(3) $\text{ " " " " " " " " }$""_8O: 1s^(2)2s^(2)2p^(4)
$\text{ " " " " }$

When looking at the electron configurations of $N$ and $O$ we can notice that the first ionization energy of oxygen is less than that of Nitrogen for the following reasons:

1. Distance from nucleus: The electron removed from both Nitrogen and Oxygen are at the same energy level $n = 2$ and therefore, this is not the reason (same distance).
2. Effective nuclear charge : Even though Oxygen's nuclear charge is $+ 8$ and that of Nitrogen is $+ 7$; with a single proton difference the nuclear charge effect is very minimal in this case.
3. Shielding effect: The electron removed from both Nitrogen and Oxygen is shielded by the same number of electrons which is two from the inner shell $1 {s}^{2}$.
4. Electron-electron repulsion: The electron repulsion effect is the dominant effect in this case, since there is no electron-electron repulsion in the case of Nitrogen, and it is present in the case of oxygen, and therefore, this is the main reason.

Periodic Trends | Ionization Energy.

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anor277 Share
Oct 2, 2015

This is possibly a manifestation of Hund's rule of maximum multiplicity with respect to electron spins. Note that I deal with the atomic and not the molecular species.

Explanation:

Valence configuration of atomic $N$ is $2 {s}^{2} 2 {p}^{3}$, the $2 p$ orbitals are conceived to be each singly occupied and all the electrons are spin-aligned; valence configuration of atomic $O$ is $2 {s}^{2} 2 {p}^{4}$. The ${O}^{+}$ species has available an energetically favourable electron spin state; when the $N$ atom is ionized, we have to change the spin state from its most stable one. (When I say energetically favourable, of course energy has to be expended to ionize each atom!)

As you realize, other things being equal (in the absence of Hund's rule!), atomic oxygen should hold its valence electrons more tightly in that it has greater nuclear charge.

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