4
$\begingroup$

I'm taking a first-year Inorganic Chemistry course in college, and stumbled upon this problem:

For $\ce{Mn}$ and $\ce{P}$, calculate the first ionization energy and explain the difference between the obtained values.

Now, my course materials (written by our professor) apparently say that the first ionization energy can be computed using this formula. I say apparently because the "course materials" are poorly written PowerPoint slides with loads of unexplained terminology and abbreviations, so I may be misunderstanding.

$$ \text{IE} = \frac{Z_\mathrm{eff}^2}{n_\mathrm{eff}^2} \cdot 13.6~\mathrm{eV} $$

where $Z_\mathrm{eff}$ is the atomic number minus the differentiating electron's shielding constant (calculated using Slater's rules) and $n_\mathrm{eff}$ is pulled out of a table of values (see here).

Unfortunately, it only seems to produce (somewhat) numerically correct results for $\ce{H}$ and $\ce{Li}$, and stops being useful for comparing them at $Z=5$. I wrote an implementation of the algorithm in Mathematica and compared the results it produces to the first ionization energies returned by ElementData (suitably converted from molar to regular), and it's a very rough approximation.

Much too rough, in fact, to be usable for the problem above: it returns $47.38$ instead of $7.43$ for $\ce{Mn}$ and $45.71$ instead of $10.48$ for $\ce{P}$. It sorts them the wrong way around, in other words.

My question is, what am I missing? As far as I can tell from Wikipedia, there's no way to simply calculate ionization energies - they have to be measured. Is there some obvious detail I haven't spotted? Another way to interpret the problem text? Some elementary knowledge I don't have?

$\endgroup$
  • 1
    $\begingroup$ The formula I know uses $(Z_\mathrm{eff}/n)^2$. The idea of using either an effective nuclear charge, $Z_\mathrm{eff}$, or an effective principal quantum number $n_\mathrm{eff}$, is that you want to account for the electron-electron repulsions. However, if you use both of them, you're accounting for e-e repulsions twice. Can you try and run your script again just using the unadjusted integer values for $n$? $\endgroup$ – orthocresol Jan 22 '16 at 21:43
  • $\begingroup$ Actually, scrap that. The difference is way too much - a factor of 4 to 9 - so I don't feel like what I said would make a difference. Wikipedia does also write that Slater was attempting to find one-electron wavefunctions that had both $Z$ and $n$ adjusted. Maybe check your code, make sure it's outputting the correct values of $Z_\mathrm{eff}$. I would help you do it but I'm not familiar with Mathematica syntax at all, sorry. $\endgroup$ – orthocresol Jan 22 '16 at 21:46
  • $\begingroup$ @orthocresol: tried it anyway - changing from $n_{eff}$ to just $n$ actually makes the numeric differences worse. $\endgroup$ – Andy C. Jan 22 '16 at 21:52
  • $\begingroup$ Qualitatively, that doesn't seem like it should be the case - since $n_\mathrm{eff} < n$, if you divide by $n^2$ you should get a smaller value for the ionisation energy, which is what you want $\endgroup$ – orthocresol Jan 22 '16 at 21:55
  • $\begingroup$ @orthocresol: my bad, it was just Mathematica redering the graph differently for some reason. It does give smaller values, but not too much smaller, and of course only for atoms that have the differentiating electron in $n>3$. $\endgroup$ – Andy C. Jan 22 '16 at 22:05
3
$\begingroup$

I tried to find out empirically whether your formula makes sense. The data I used was effective nuclear charges from this Wikipedia page (always the minimal value, Pt was discarded), and ionization energies already gathered for the answer to this question on this site. The principal quantum number was also manually added to the data set.

I explored the data in many ways and tried to find a simple expression which could give an estimate for where the ionization energy should lie.

First, I looked at what your formula would give:

Proposed formula from question.

It is quite obvious that your formula describes two atoms perfectly and is absolutely terrible for the overwhelming majority of elements in the periodic table.

I present to you the best guess so far, based on simple manual adjustments and some exploration of the data:

$$ E_\text{ion} / \mathrm{eV} = 6 \left(\frac{Z_\text{eff}}{n^{1.5}} \right)^2 + 4$$

Below is the graphical representation:

Plot demonstrating the accuracy of the formula presented above.

The error will still be large for many atoms, I estimate the median error to be about 2 eV. But if you want to use a simple formula to calculate the first ionization barrier, this would be the one I would use.

The data exploration was done in Python and is recorded in this iPython notebook. Feel free to have a look at it and even come up with a better model.

$\endgroup$

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.