15

Fluorine, though higher than chlorine in the periodic table, has a very small atomic size. This makes the fluoride anion so formed unstable (highly reactive) due to a very high charge/mass ratio. Also, fluorine has no d-orbitals, which limits its atomic size. As a result, fluorine has an electron affinity less than that of chlorine. See this.


11

The electron being gained by fluorine would be taken in to a much smaller 2p orbital and requires more electron coupling energy than that of much larger 3p orbital of chlorine. Therefore, energy released during the electron gaining process of fluorine is less than that of chlorine.


8

It totally does not work like that. You can't say that $\ce{2F^-}$ is less stable (or more stable) than $\ce{F2}$; you can't compare them at all. People would say "B is more stable than A" when there is a spontaneous reaction $\ce{A\to B}$, but this is not the case here. A reaction must be balanced in charge and in all elements, otherwise it is just a bunch ...


8

Okay, here's what I did (and I highly encourage everybody to do this along with this tutorial-style answer, just so you get the feel for prepping and looking at data from the world wild web). Data Acquisition First, I went and copied the data from the tables at Wikipedia: This one for the ionization energies and this one for the electron affinities. Next ...


7

Sure, why not. It's just the question of your bombarding electrons having the right energy. At low energies, a bombarding electron will likely be captured by one of the iodine atoms: $$ \ce{I2 + e- -> I- + I} \, . $$ But somewhere around $8.62 \pm 0.06 \, \mathrm{eV}$ according to this study, you'll indeed have the ion pair formation: $$ \ce{I2 + e- -> ...


7

Let's take a look at a table of standard electrode potentials and find the relevant half-reactions: Reductions (there is also a half-reaction for water reduction that doesn't involve a proton at -0.83 V, but for the sake of simplicity, we can ignore it): $$\begin{align*} &\ce{Na+ + e- <=> Na} & &E°=-2.71\ \mathrm{V}\\ &\ce{2H+ + 2e- &...


7

Neither K nor Ca "wants" to ionize. Both ionization energies are endothermic. This question refers to eelctron affinity, which is gaining an electron. K has a more exothermic electron affinity because the electron gained fills the 4s orbital. In Ca the electron goes to 3d. The energy gap between 4s and 3d is enough to make the process barely exothermic. ...


6

ETA: I'm unclear about your question. Are you asking why nitrogen isn't the central atom or why sodium isn't? If it's the former, the explanation is below. If it is the latter, sodium, then it's because this is an ionic compound, and we build the lewis structures of the cation and anion separately. First and foremost, carbon (EN=2.5) is less ...


6

Two sign conventions: (1) the more common one states that a positive electron affinity value represents energy release when an electron is added to an atom; (2) the other states that a negative electron affinity represents a release of energy. http://www.rose-hulman.edu/~brandt/Chem251/GenChem_Review_v3.pdf Recommended values for these electron ...


5

This is sadly, one of those matters of "convention" that chemistry is so plagued by. For most neutral atoms, the magnitude of $\Delta H$ for the following reaction is negative (i.e. energy is released upon the addition of an electron). $$\ce{A (g) + e- (g) -> A- (g)} \tag{1}$$ The value of $\Delta U$ for the reaction is, in some places, called the "...


5

To quote chemguide: The first ionisation energy is the energy required to remove the most loosely held electron from one mole of gaseous atoms to produce 1 mole of gaseous ions each with a charge of 1+. That means, the ionization energy of fluorine is the energy of the following reaction: $$\ce{F -> F+ + e-}$$ The ionization energy of chlorine is ...


5

So consider a vessel wherein which we have Fluorine atoms (F) in gaseous state. Now, they have seven outer shell electrons and would really love to get the 8th one for an octet. What would have to happen is: $$\ce{2F -> F+ +F-}$$ It's not possible for both to become fluoride ions because there would be no charge balance. The total change in energy is ...


5

Energy of electron in vacuum is zero. If it can attach (however weakly), the energy is gained and process is therefore exotermic. It is not an atom which releases energy, it is the whole system. Exactly opposite of previous. The atoms you mentioned gladly release electrons to other atoms, but not to vacuum (which is how ionization energy is defined). In some ...


4

The amount of screening is the same in the two groups but the effective positive charge that the incoming electron feels is stronger in Group 7. For instance, oxygen has the electronic structure of $$\ce{1s^{2} 2s^{2} 2p_{x}^{2} 2p_{y}^{1} 2p_{z}^{1}}$$ and 8 protons. The effective positive charge that pulls the incoming electron (which approaches the ...


4

Looking at the comment thread above, I believe a bit more elaboration can help, so here it is Why is electron gain generally exothermic? Think of a unit positive sphere and a unit negative sphere. When is their system most stable? It is when the spheres are closest to each other. Why is that position stable? Because if we slightly disturb those spheres to ...


4

Chart of Electron Affinities The Electron Affinity can be thought of as the "electrical advantage" given by adding an electron to an atom. So, if you have a halogen that gains an electron, it becomes more stable because now it has its octet, orbitals filled, etc. This is why the 1st EA is large. But if you want to add an electron, it needs to go to a ...


4

This is likely due to second period elements' being quite small, so electron-electron repulsion is much more significant than in a third period element. The general trend is that EA is more positive as you move down the periodic table since effective nuclear charge is the same, but the electrons are farther away from the nucleus. Maybe it's easier to ...


4

The presence of $d$ orbitals can indeed provide a means for otherwise closed-shell atoms of alkaline-earth metals to accept electrons. Wu et al. [1] describe carbonyl complexes of Ca, Sr, and Ba in which the valence $d$ orbitals of these metal atoms are indeed engaged in (covalent) bonding. In principle, all atoms actually have $d$ orbitals. Getting them to ...


3

They all have 2 electrons in their highest occupied s sublevel and 10 electrons in their highest occupied d sublevel (both sublevels are full). As a result, they can not complete any orbitals by adding an electron and they can not put an electron in a sublevel that is occupied. All of the other transition metals will either add an electron to complete the ...


3

I think you mean to ask about the electron affinity of the alkali metals. That is, the enthalpy gain or lost when adding an electron to an atom. While the electron affinity for alkali metals are much smaller than the halogens, they are slightly positive. That is, quoting from Wikipedia: Electron capture for almost all non-noble gas atoms involves the ...


3

F-center defects in crystals are due to unpaired electrons and can be caused by ionizing radiation; these defects can accumulate over thousands of years. This can be used to find the age of minerals with retroactive inclusions, or to intentionally color gems. An electron gun, as in a cathode ray tube (CRT), will also do the job. Older televisions had CRT's, ...


3

Actually, not all atoms are stable, at least not in the gas phase, which is where we measure the electron affinity. You should first read Why do atoms "want" to have a full outer shell? for detailed elaboration. Now, electron affinity involves the gain of an electron to an isolated gaseous atom, but not all atoms actually want to gain an electron, ...


3

In summary: IE and EA are just partial terms in the formation enthalpy of solid ionic compounds, that includes also LE and atomization enthalpy of elements. Their values alone can give a hint, but do not provide the final word. Notice that the text says nothing explicit about ion pairs. IE + EA relate to free gaseous unbound ions. Like: $$\ce{M(g) -> M+(...


3

A simple, intuitive answer based solely on high-school chemistry and available emperical data; 1s < 2s < 2p < 3s < 3p < 4s < 3d For Be, all the orbitals up to 2s are filled, so the "new" elctron has to go into the 2p orbital. Since there is a large gap between the 2s/2p energy levels of Be(it is one of the only elements in the 2nd ...


2

After rediscovering my own question, I'm posting it as an answer for anyone else to reference or to correct. Electron affinity (EA) refers to "the energy released when an atom in the gas phase accepts an electron." (Burdge, Atoms First, p.118) Lower EA means that the process of accepting an electron is not energetically favorable. In other words, lower EA =...


2

The most important fact here is that the electron added to $\ce{S-}$ goes into an $n=3$ orbital ($3p$), whereas the one added to $\ce{O-}$ goes into an $n=2$ orbital ($2p$). The larger $n=3$ orbitals are further from the nucleus and hence experience a weaker attraction to the nucleus and are much higher in energy. This dependence with quantum number ...


2

Yes, the electron affinity for group 17 elements (group VII by older IUPAC nomenclature) is generally exothermic, so the uptake of an electron should be favourable. But as Ivan mentioned in a comment, there is no such thing as a free or lonesome electron. Each electron must first be displaced from another atom — ionised; the corresponding thermodynamic ...


2

A picture is worth a thousand words. In the picture above, start with a mole of solid sodium and half a mole of Cl$_2$ gas, in the ground state. Then gasify the sodium with 107.3 kJ of heat, then ionize it with 496 kJ of energy, then dissociate Cl$_2$ to Cl radicals with 121.7 kJ, then allow the electron from the sodium to combine with the chlorine radical. ...


2

When you talk about the ionization energy of $\ce{Mg^2+}$, it is the third ionization of $\ce{Mg}$, which is equal to $\pu{7734 kJ/mol}$. When you talk about the ionization energy of $\ce{Ne}$, it is the first ionization of $\ce{Ne}$, which is $\pu{2081 kJ/mol}$. Thus, actually, $IE_\ce{Mg^2+} \gt IE_\ce{Ne}$. Keep in mind that the first and second ...


2

The Ionisation Energy of an atom is defined as the minimum amount of energy required to remove the most loosely bound electron of an isolated neutral gaseous atom or molecule. In this case, since $E_1$ amount of energy is used to ionise $N_0/2$ atoms, the ionisation energy of each atom is $\frac{E_1}{N_0/2} = \frac{2E_1}{N_0}$. The Electron Affinity is ...


Only top voted, non community-wiki answers of a minimum length are eligible