# Why does a Fluoride ion only have a -1 charge and not a -2 charge or more?

Transition elements can form ions with different charges. Why can't elements other than transition elements form ions with different charges? If it is a Fluoride ion, why does it have to be an anion of charge -1? Why can't it have a -2 charge, or more?

The octet rule is usually used to explain the charges that atoms in groups I, II, VI, and VII have in ionic compounds, by saying something like "noble gases are stable, therefore having 8 valence electrons must be stable for some reason." This is true, but understanding why noble gas configurations are stable is important, and without knowing that, it does seem like magic that some atoms don't adopt a noble gas configuration in ionic compounds.

The octet rule is a useful way to quickly predict ionic charges and write Lewis structures, but it's not a magical law that atoms follow. They don't "want" or "try" to obtain specific electronic structures, although we often talk like they do. What is really important is the total energy of the electron configuration. The electron configuration with the lowest energy is the most stable one.

"Stability" really means "lowest energy"

The total energy of the electron configuration is governed by two things:

1. The energy of the electron orbitals
2. The effective nuclear charge that is "seen" by the electrons

The orbital energies determine the order in which atomic orbitals will be populated by electrons in the ground state - in other words, which orbitals the electrons will exist in for a given atom. We know from empirical evidence and from quantum mechanical calculations that the order goes like this for most elements (not for some transition metals and heavier elements):

Electrons will occupy these orbitals from the "ground up" (called the Aufbau principle). We know, therefore, that for 4th-period elements, the $3d$ sub-shell will fill before the $4p$ sub-shell will.

The atomic orbital diagram indirectly tells us something else. The principle quantum number, $n$ (which for the valence shell is equal to the row or period number), roughly corresponds to the distance from the nucleus of the orbital's highest electron density region. In other words, any sub-shell with $n=4$ will have most of its electron density further away from the nucleus than a sub-shell with $n=3$. This means that $3d$ electrons are closer to the nucleus than $4s$ electrons, even though $3d$ electrons initially have higher energy.

The distance of the electron from the nucleus is roughly proportional to $n$

Since the electrons have a negative charge, and the nucleus is positively charged, there is an electrostatic potential energy between the electrons and the nucleus. If the electrons were uniformly distributed throughout the atom, we would expect that the nuclear charge "felt" by each electron would be equal to the number of protons - the atomic number. However, because of the "rules" of quantum mechanics (the allowed solutions to Schrodinger's equation), electrons are not uniformly distributed - they occupy orbitals in a specific arrangement. Since some electrons are closer to the nucleus (lower $n$) than others, this means that the nuclear charge is partially screened by the inner electrons, and the outer electrons experience an effective nuclear charge that is somewhat less than the full nuclear charge. This diagram from wikipedia demonstrates the idea:

And this one shows the idea using the Bohr model (note that this is a schematic - not an accurate picture of what electrons "look" like in orbitals - only the relative distances from the nucleus is right)

When the valence electrons experience a large effective nuclear charge, they are harder to remove (the energy cost for removing them increases), and when the effective nuclear charge is lower, they are easier to remove. The actual effective nuclear charge is very difficult to accurately calculate, but it is very easy to estimate - all you need to do is take the number of protons, then subtract the number of "screening" electrons. Basically, we are assuming that each electron between the valence shell and the nucleus cancels out one proton:

$$Z_{eff} = Z - S$$

where $Z_{eff}$ is the effective nuclear charge, $Z$ is the actual nuclear charge, and $S$ is the screening constant, here assumed to equal the number of "core" electrons.

To find $S$, all you have to do is count the number of electrons with $n$ less than the valence shell (highest $n$).

Taking fluorine as an example, the electron configuration is:

$$1s^22s^22p^5$$

We see that $Z=9$ and $S=2$, giving an effective nuclear charge of +7. Compare that to lithium: $3-2=+1$. The valence electrons on fluorine are roughly 7 times harder to remove than the valence electron on lithium. In fact, any element to the left of fluorine has a lower effective nuclear charge for the valence shell, and will give up electrons more easily than fluorine will. In other words, it's not so much that fluorine won't give up electrons, it's that nothing can take them from it.

What if we add an electron? For $\ce{F-}$, the electron configuration becomes:

$$1s^22s^22p^6$$

$Z_{eff}$ is still +7, since the number of core screening electrons hasn't changed. This means that it is energetically favorable for fluorine to pick up an electron, since the additional electron would still "see" a large positive nuclear charge.

What if we add another, to make $\ce{F^{2-}}$? Now the configuration looks like this:

$$1s^22s^22p^63s^1$$

$Z_{eff}$ is now $9 - 10 = -1$ - note that the number of screening electrons jumped to 10 because the "new" valence shell has $n=3$. In other words, because of the allowed orbital configurations the new electron would "see" a negative charge on the nucleus - it would require energy to keep it.

If you work through this process for all of the "fixed charge" elements, you find similar things - the effective nuclear charge is large for elements that gain electrons, until they gain too many (an octet), at which point it becomes negative. For fixed-charge metals, something else happens. Let's look at magnesium.

$\ce{Mg}$: $12-10=+2$

$\ce{Mg^1+}$: $12-10=+2$

$\ce{Mg^2+}$: $12-2=+10$

Once you remove two electrons, the effective nuclear charge jumps to +10 for the remaining valence electrons. The same is true for all the group I and II metals - once you get to the noble gas "core" - the octect - the effective nuclear charge seen by the valence electrons jumps way up.

This explains why "fixed charge" elements always have the same charge in ionic compounds, but what about transition metals?

Let's look at chromium. The electron configuration looks like this:

$$1s^22s^22p^63s^23p^64s^23d^4$$

Which electrons are the valence electrons? The ones with the highest $n$ - the 4s electrons. The effective nuclear charge they "see" is $24 - 22 = +2$ - note that the 4 electrons in the $3d$ sub-shell are counted as screening, since they have $n=3$ which is less than 4, putting them closer to the nucleus than the valence shell, despite their initially higher energy.

If we remove an electron, we get:

$$1s^22s^22p^63s^23p^64s^13d^4$$ (This isn't exactly correct, there would really be no $4s$ electron and 5 $3d$ electrons, but the reason is complicated and it doesn't matter to this explanation)

Note that we removed a valence electron - a $4s$ electron. Now $Z_{eff} = 24 - 22 = +2$ - it hasn't changed. We can remove one more to make $\ce{Cr^{2+}}$ and get:

$$1s^22s^22p^63s^23p^63d^4$$

Now the definition of "valence" becomes a little less clear - we know that the highest $n$ value corresponds to the valence shell, but what about $l$, the orbital shape? It turns out that d orbitals are larger than s and p orbitals, and so the s and p electrons end up screening the nucleus from the d electrons. This means that d electrons would be removed next, and the effective nuclear charge "seen" by them is $24 - 18 = +6$. This is a larger charge, but not so large that other elements (notably oxygen and the halogens) can't still pull the d electrons away. This gives us a range of chromium charges (oxidation states) from +1 to +6. For $Cr^{6+}$, the electron configuration would be:

$$1s^22s^22p^63s^23p^6$$

Now the effective nuclear charge is $24 - 10 = +14$ - much too strong for anything else to "win" against.

As you can see, the rules of quantum mechanics determine the relative size of atomic orbitals and the order in which they are filled, which in turn determines the effective nuclear charge that valence electrons "see" for any given neutral atom or ion. The combination of the two leads to constraints on the charges that atomic ions can have, with the practical result being that group I, II, VI, and VII elements adopt a "filled octet" or "noble gas" configuration, while the other groups have more flexibility.

• Nice usage of nuclear effective charge, and a good answer in general. I never considered that in extreme situations the calculation could actually result in a negative value, strongly suggesting at least one valence electron is unbound from the nucleus. However, I should point out to other readers that your calculations are very qualitative, much more so than even those based on Slater's rules. Your argument still stands, of course, but I would not recommend anyone take the magnitude of the calculated effective charges too seriously. – Nicolau Saker Neto Jul 21 '14 at 11:15
• @Nicolau - that's why I tried to make it clear that it was an estimate, not the actual effective charge. – thomij Jul 21 '14 at 13:22
• @thomij I'm wondering about your statement, "This means that 3d electrons are closer to the nucleus than 4s electrons, even though 3d electrons have higher energy." Is that true, why would an electron closer to the nucleus have higher energy? – ron Jul 21 '14 at 17:08
• @ron - The argument commonly used is that the local electron density of the d shells is higher than the next higher s shell, both because of "crowding" in the d sub-shell and because of the proximity to the 3s and 3p orbitals. The real reason is more complicated, but in short, the 4s electrons have density peaks closer to the nucleus, even though the outermost peak is further out. The energy levels actually move around as the orbitals fill, and by the time the 3d orbitals have started to fill, they may or may not have higher energy. I will edit my answer to make it more correct. – thomij Jul 21 '14 at 17:37

The electronic structure of a fluorine atom is $\ce{1s^2 2s^2 2p^5}$. There is a strong driving force for atoms to attain an octet (achieve an inert gas configuration) due to the extra stability associated with a filled shell of electrons. In order to complete its octet and achieve the neon inert gas configuration ($\ce{1s^2 2s^2 2p^6}$), fluorine must gain 1 electron and become the fluoride anion ($\ce{F^{-}}$). If it gains two electrons and becomes $\ce{F^{-2}}$, or loses 1 electron and becomes $\ce{F^{+}}$, it won't have an octet in its outer shell - it won't be anywhere near as stable as if it just gains one electron. The same reasoning explains why sodium, for example, prefers to lose only one electron. Transition elements have larger electronic configurations and may need to gain or lose 3 or 4 electrons to achieve an inert gas configuration. It requires a lot of energy to add or remove 3 or 4 electrons, so adding or removing 1 or 2 electrons, while not achieving an inert gas configuration, becomes an acceptable alternative because of the lower energy required to remove fewer electrons. In these cases, oxidation states of (plus or minus) 1, 2, 3 or 4 become feasible.

Some main group elements do form multiple ions, however, analogues to the transition elements are less common.

Examples:

• Hydrogen - Hydrogen forms both the $\ce{H+}$ ion (the proton) and the $\ce{H-}$ ion (the hydride)
• Tin forms the $\ce{Sn^{2+}}$ and $\ce{Sn^{4+}}$ cations

The fluoride anion can have a negative two charge. It's not impossible to add two electrons to the fluoride anion. But at that point it would be rather unstable as the fluoride anion with a negative 1 formal charge is isoelectronic with a noble gas - neon.

In addition, consider the high charge density of the fluoride anion. Tacking on an extra electron to fluorine is OK, but not as favorable relative to the less charge dense chlorine atom. Fluorine actually has a lower electron affinity than chlorine despite fluorine's higher electronegativity. Tacking on two extra electrons would be unfavorable.

Here's an easy argument to complement the others, which doesn't directly involve octets and electron shells. Compare the isoelectronic species $\ce{Na^0}$, $\ce{Ne^{-}}$ and $\ce{F^{2-}}$, which all have the same $1s^22s^22p^63s^1$ fundamental configuration. Notice that going from left to right in the list, you are gradually removing a proton from the nucleus, making the valence electrons more poorly bound to the atom (less positive charge in the nucleus to pull the electrons inwards). Elemental metallic sodium is already very reactive because it has a comparatively loose valence electron (a low ionization energy). Removing two protons from the nucleus to generate $\ce{F^{2-}}$ would create a species much more reactive than even metallic sodium(!), and it would be expected to immediately lose at least one electron in any conditions.

In truth, $\ce{F^{2-}}$ is too unstable to be formed in the first place. Again, comparing the isoelectronic species in the list above, we look to $\ce{Ne^{-}}$. Neon actually already has an electron affinity that is practically zero or even endergonic. In other words, the reaction $\ce{Ne^0_{(g)} + e^{-} -> Ne^{-}_{(g)}}$ already has very little drive to the right, if not actually being driven to the left. This suggests the reaction $\ce{F^{-}_{(g)} + e^{-} -> F^{2-}_{(g)}}$ (fluorine's second electron affinity) is strongly endergonic, and will not ever happen to a significant extent.