0
$\begingroup$

There was a question which was asking about the order of ionization energies of K,Li,Ca,S. The electronic configuration of all the elements were as follows

  • Li:${1s^2 2s^1}$
  • K :${1s^2 2s^2 2p^6 3s^2 3p^6 4s^1}$
  • Ca :${1s^2 2s^2 2p^6 3s^2 3p^6 4s^2}$
  • S :${1s^2 2s^2 2p^6 3s^2 3p^4}$

From the electronic configurations:

  • Potassium should have lowest ionization energy as it is very easy to remove electron from 4th shell.

  • Then it should be followed by Calcium as its nuclear charge increases from going across the period

  • Then it should be followed by sulphur as it is more easier to remove the electron from 3rd shell than from 2nd shell.

  • Then it should be followed by Lithium as it has lowest principle quantum number(n) so it is more difficult to remove an electron from 2nd shell than from 3rd and 4th shell.

So according to my calculation the order should be K < Ca < S < Li. But the correct answer in my textbook is K < Li < Ca < S. If I am wrong anywhere then what factor would help in determining the Ionization energies of the following elements, and what would be the exact order of I.E for these elements.

$\endgroup$
  • 1
    $\begingroup$ frankly I hate these sort of questions. Typically they call for some rationalization which only can be justified by some hand waving. $\endgroup$ – MaxW Jul 30 '18 at 14:58
1
$\begingroup$

Looked up ionization energies (kJ/mol) from Wikipedia

  • K = 418.8
  • Li = 520.2
  • Ca = 589.8
  • S = 999.6

so the book answer is correct...


Ok, here is best hand waving explanation that I can give. The general trend...

enter image description here

Li and K are both alkali metals. As the atomic number goes up the shells get further away from the nucleus and the inner electrons shield the valence electron. So in order of increasing ionization energy

Li > Na > K > Rb > Cs > Fr

So to start we easily get Li > K.

Ca and K are in the same period in the periodic table. Generally as you go up in atomic number in a period the ionization energy will increase. So Ca > K.

enter image description here

Sulfur is a non-metal in the upper right hand of the periodic table so it should have the highest ionization energy.

So we're down to either S > Ca > Li > K or S > Li > Ca > K.

To me giving either of these answers wouldn't be embarrassing.

If you go to the Bohr model of the hydrogen atom then energy of the $n$th shell is

$\mathrm{E}_n \propto \dfrac{Z^2}{n^2} $

If we assume that the 2s and 3s electrons are perfectly shielded by all the electrons in the lower shells, then

$ \mathrm{E_{Li}} \propto \dfrac{1}{2^2} = 0.25$

The two Ca 3s electrons being equal the charge is 2

$ \mathrm{E_{Ca}} \propto \dfrac{2^2}{4^2} = 0.25$

So I would have guessed S > Li > Ca > K

Now I know that is wrong. I know that the "problem" is with the shielding of the electrons. Below is a section of the periodic table with ionization energies listed.

I just don't know how to explain the "right" answer without hand waving about shielding.

I don't know how you'd rationalize all the relationships between the group 1 and 2 elements.

enter image description here

If you assume the Bohr model as noted above then you end up with this which correctly predicts the trend down a group but overestimates shielding of the inner electrons.

enter image description here

$\endgroup$
  • $\begingroup$ I can't understand what are you trying to say about hand waving? $\endgroup$ – pranjal verma Jul 30 '18 at 16:59
  • 1
    $\begingroup$ Hand waving - a pejorative label for attempting to be seen as effective – in word, reasoning, or deed – while actually doing nothing effective or substantial. It is most often applied to debate techniques that involve fallacies, misdirection and the glossing over of details. $\endgroup$ – MaxW Jul 30 '18 at 20:47
  • $\begingroup$ As good an answer as one could come up with. $\endgroup$ – Jon Custer Jul 31 '18 at 0:44
  • $\begingroup$ I think you are trying to say that respective ionization energies of elements is just a matter of fact and we should see it just as a number, but anyway thanks for the answer. $\endgroup$ – pranjal verma Jul 31 '18 at 9:26
  • $\begingroup$ @pranjalverma - Some of the tends in the ionization energies you can deduce from the position of the elements in the periodic chart. However some you just can't. // The answer is somewhat of a rant. These kinds of problems are posed many times to new students and then the answer is "justified" by what I believe to be some deceptive logic. You can invoke shielding to explain Li/Ca, which is true, but I think you need a very detail calculation to prove it rather than some hand waving. In other words the hand waving for Li/Ca would also have to explain Na/Sr where the relative values reverse. $\endgroup$ – MaxW Jul 31 '18 at 15:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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