Exploring Electron Configurations: Why I.E. Increases Across the Second Period

In summary, the conversation discusses the first ionization energy of Nitrogen and Oxygen, with Oxygen having a lower value despite having more protons. This is attributed to the repulsion of the electron pair in the 2px suborbital. However, the trend of increasing ionization energy in Fluorine and Neon is not affected by the additional electron pairs in their orbitals. This is due to the stronger nuclear-electronic attraction in these elements. It is also noted that none of the three p orbitals in Oxygen are singled out, but rather one is doubly occupied and the other two are singly occupied.
  • #1
Hrnmhmm
2
0
My level of knowledge on Chemistry is first year. I'm teaching myself about electron configurations using Google to find and compare sources.

In this webpage - http://www.chemguide.co.uk/atoms/properties/ies.htm [Broken]
Jim Clark said:
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N ... 1s2, 2s2, 2px1, 2py1, 2pz1 ... 1st I.E. = 1400 kJ/mol
O ... 1s2, 2s2, 2px2, 2py1, 2pz1 ... 1st I.E. = 1310 kJ/mol

It's demonstrated that Oxygen has a lower first ionization energy than Nitrogen despite having more protons. The explanation given is that the 2px suborbital in Oxygen (2px2) contains one more electron than in Nitrogen (2px1), forming an electron pair, and their repulsion negates some positive charge.

If the only difference between Nitrogen and Oxygen is in the electron pair 2px, why do Fluorine and Neon continue the second period's upward trend of ionization energy as normal? Shouldn't the repulsion of the additional electron pairs 2py2 and 2pz2 also affect ionization energy in these elements, reversing the trend?
 
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  • #2


It seems half-filled orbital subshells are particularly stable.
 
  • #3


Hrnmhmm said:
If the only difference between Nitrogen and Oxygen is in the electron pair 2px, why do Fluorine and Neon continue the second period's upward trend of ionization energy as normal?
No, it's not the only difference. With increasing nuclear charge, when going from left to right in a period, the orbitals contract. This contraction is due to the stronger attraction of the electrons by the nucleus and it will increase ionization energy. Comparing Nitrogen and Oxygen, by Hund's rule,
in Nitrogen the three p electrons will each occupy a separate p orbital. When going t O, the new electron has to go to an orbital already filled whence it experiences a stronger repulsion although the nuclear attraction has increased somewhat, too.
The effect of ionization in fluorine is also that of going from a doubly occupied orbital to a singly occupied one; the repulsion of the electrons being approximately comparable. Hence the only effect which makes a difference is the stronger nuclear-electronic attraction in F as compared to O.

Btw, none of the three p orbitals is singled out, it makes no sense to say e.g. in Oxygen the p_x orbital is doubly occupied while the p_y and p_z are singly occupied .
You can only say that one p orbital is doubly occupied and 2 only singly occupied.
 

1. What is electron configuration?

Electron configuration refers to the arrangement of electrons in an atom's energy levels and sublevels.

2. Why does ionization energy (I.E.) increase across the second period?

The ionization energy increases across the second period because the atoms in this period have an increasing number of protons in their nucleus, which results in a stronger attraction for their valence electrons. This makes it harder to remove an electron and increases the ionization energy.

3. How does the number of valence electrons affect ionization energy?

The number of valence electrons directly affects ionization energy. The more valence electrons an atom has, the higher its ionization energy will be. This is because these electrons are farther from the nucleus and are therefore easier to remove.

4. What is the relationship between electron configuration and ionization energy?

The electron configuration of an atom determines its ionization energy. Atoms with a half-filled or fully filled valence shell have lower ionization energy because their electrons are more stable and harder to remove.

5. How does the trend in ionization energy across the second period relate to other periodic trends?

The trend in ionization energy across the second period is similar to the trend in atomic radius and electronegativity. As the number of protons increases across the period, the atomic radius decreases and electronegativity increases. This is due to the increasing nuclear charge, which affects all three of these properties.

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