X, Y and Z are three unknown elements whose first 5 ionization energies are given below. Which of the 3 is the most electronegative?$$ \begin{array}{|c|c|c|c|c|c|}\hline &\text{IE}_1&\text{IE}_2&\text{IE}_3&\text{IE}_4&\text{IE}_5\\\hline \ce{X}&577&1815&2740&11600&15000\\\hline \ce{Y}&800&2427&3660&25026&32827\\\hline \ce{Z}&738&1451&7733&10543&13630\\\hline \end{array} $$

Because of the IE ratios, I concluded that X and Y are in the third column and Z is in the second column. I know that EN increases from left to right for elements in the same row. Is there anything else I should know in order to solve this?


Let's look at what amounts of energy we get from IEs.

$$\ce{X}: 577 \rightarrow 1815 \rightarrow 2740 \rightarrow 11600 \rightarrow 15000$$

After taking out three electrons, we see a first jump in IEs. Thus, there are three electrons in the valence shell.

$$\ce{Y}: 800 \rightarrow 2427 \rightarrow 3660 \rightarrow 25026 \rightarrow 32827$$

Again, after three electrons out, we have the first jump. The result is the same as X: Three electrons in the valence shell.

$$\ce{Z}: 738 \rightarrow 1451 \rightarrow 7733 \rightarrow 10543 \rightarrow 13630$$

You see the first bump in IEs in between IE2 and IE3. So, the third electron comes out of a lower shell and thus, two electrons in the valence shell.

The more the electrons, the more shells they will fill and the farther they'll be from the nucleus (among other lesser important factors). That means X has more electrons than Y because of relatively lower IEs, and therefore Y is more electronegative as it's closer to the fluorine.

You're correct that from left to right we get more electronegativity for the atoms, but note that
though with exceptions, such as Al and Ga, from top to bottom you get lesser electronegativities.

Even in the case of Al and Ga, the one that's more electronegative will get relatively more IE to lose electrons. So that's not what you should worry about. So let's conclude:

If you see two elements that are in the same group, the one with a generally higher IE is more electronegative.

  • $\begingroup$ As a side note, a really nice way to compare electronegativities is through atomic radii. $\endgroup$ – M.A.R. Feb 11 '15 at 14:25

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.