Question

Why does nitrogen have a higher ionization energy than oxygen?

Answer 1

Well, nitrogen atom has a first ionization energy of `"14.53 eV"`, so it would be `(b)`.

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But of course, you would not know that off-hand. To rationalize it, consider their electron configurations:

`"N": [He] 2s^2 2p^3`

`underbrace(ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr)))`
`" "" "" "" "2p`

`ul(uarr darr)`
`2s`

`"O": [He] 2s^2 2p^4`

`underbrace(ul(uarr darr)" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr)))`
`" "" "" "" "2p`

`ul(uarr darr)`
`2s`

Since oxygen atom has a paired electron, it repels the other `2p` electrons, which makes it easier to remove the first electron from oxygen atom.

How does that lead to the first ionization energy of oxygen atom being less than for nitrogen atom?

Answer 2

Well, let us first look at the data....

Explanation:

We gots...

`N(g) +DeltararrN^(+)(g) + e^-`...`Delta_N=1402.3*kJ*mol^-1`

`O(g) +DeltararrO^(+)(g) + e^-`...`Delta_O=1313.9*kJ*mol^-1`

The data are from this site...

Facilely, we might predict that oxygen should have a greater ionization energy than nitrogen given that we got `Z=7` for nitrogen, versus `Z=8` for oxygen, so another issue must dominate...

And this is probably attributable to Hund's rule of maximum multiplicity; and for nitrogen the spin quantum number may be maximized if the electrons SINGLY occupy the `"p-orbitals"`. On the other hand, the addition of an EXTRA electron in the oxygen atom, detracts from the spin quantum number by the necessary pairing of one `"p-orbital"`. Anyway, consult the relevant chapter in your text for a fuller explanation....