4
$\begingroup$

I want to address two exceptions in the trend of ionization energy across the period, that are causing me problems:

Taking the second period as an example :Consider only the second period

The two exceptions from the general trend are the ionization energies of B lesser than Be and that of O less than N. My teacher told me the reason to both was that half filled and fully filled orbitals of N and Be are more stable and hence require more energy to pull off an electron. He also said that this affects only the next element in the period for a change in trend( he never said why though). This place and my textbook tells me that the reasons are:

1) Be,B :

The electron is to be removed from the 2p orbital in B and the 2s orbital in Be. The 2s orbital penetrates more into the nucleus and hence it is more difficult to remove an electron form Be than from B.

Question: In that case why doesn't this apply to C? Shouldn't it be easier to remove one of the electrons from C's 2p orbital than Be's 2s orbital just like the B's 2p orbital? And so shouldn't the ionization energy of Be must be more than C too?

2) N,O:

N 1s2 2s2 2px1 2py1 2pz1
O 1s2 2s2 2px2 2py1 2pz1

The reason I have understood here is that there are two electrons in the 2px orbital of O and because of electron-electron repulsion, it is easier to remove an electron from here as opposed to the 2px orbital of N which had only one electron.

Question: Why doesn't this apply to F?

F 1s2 2s2 2px2 2py2 2pz1

F has paired electrons too like O in the px and py subshells. Wouldn't there be repulsion here too? And hence why isn't it easier to remove electrons from F than is it to from N which doesn't have any paired electrons at all? Why is the ionization energy of F more than N instead?

The same problems can be seen in the next period.I hope they can be sorted out with the same logic that i am missing for this period.

Finally, was what my teacher said just a simplification of everything I found in the link and the textbook? Was he trying to convey the same thing? Or are they two complete different explanations?

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ As you said, the ionisation energies of all the atoms after B, such as C, N, etc. are lowered from what they would be if we were to take out a 2s electron from them. However you're missing the big picture. If you ignore the small dips going from B to C and from N to O, you get an overall increase going from Li to Ne. Why is that so? $\endgroup$
    – orthocresol
    Jun 2, 2015 at 18:19
  • $\begingroup$ Also, chemguide is correct. The explanation involving half- and fully-filled orbitals involves something called exchange energy, but these effects are not as important as the 2s -> 2p change going from Be to B and the pairing of the electrons (leading to greater repulsion) going from N to O. Exchange energy is more commonly used to rationalise the ground-state electronic configurations of Cr and Cu. Also my previous comment should of course have said Be to B, not B to C. I am very sleepy today, it seems. Sorry. $\endgroup$
    – orthocresol
    Jun 2, 2015 at 18:27
  • $\begingroup$ @orthocresol The overall increase is because the atomic radius decreases across the period. But however small the dips are , they are dips. Why does it affect only the immediate to the right elements in the period? $\endgroup$
    – SMcCK
    Jun 3, 2015 at 3:14
  • 1
    $\begingroup$ Well, it's a simple case of balancing two factors. Going from Be to B, 1) IE increases due to a decrease in atomic radius 2) IE decreases because you are going from 2s->2p. In this case factor no. 2 wins out. Going from Be to C, you again have the same two factors, but factor 1 wins out. And that is because the decrease in atomic radius is larger going from Be to C than it is going from Be to B, because you are adding two protons instead of just one. $\endgroup$
    – orthocresol
    Jun 3, 2015 at 11:56
  • 1
    $\begingroup$ To put it another way, going from B to C there should definitely be an increase because the nuclear charge increases (leading to a smaller atomic radius). This increase is large enough to counteract the decrease going from Be to B, such that going from Be to C the IE increases. $\endgroup$
    – orthocresol
    Jun 3, 2015 at 12:00

1 Answer 1

Reset to default
5
$\begingroup$

OK. Lets get a few things cleared up first. The main factor affecting ionizing energy (IE) is the attraction between the nucleus and the electrons. The more the attraction, the more energy is needed to remove an electron from it's orbital and hence higher the IE. The other factors modify this main factor and thus change the expected IE values to the anomalous observed values. Now coming to your questions:

  1. As we go to the right of a row, size of the atoms get smaller. $\ce{C}$ is smaller than $\ce{Be}$ and $\ce{B}$ and hence has higher IE.

  2. The same thing happens in case of $\ce{F}$. The size of $\ce{F}$ is smaller compared to $\ce{N}$ and this factor over powers the repulsion between the paired $p$ orbital electrons.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.