Creating KF Molecule from Neutral Atoms: A Process of Ionization and Binding

In summary, the energy required to create a KF molecule from neutral atoms involves ionization of K and F, with K losing an electron and F gaining one. The first ionization energy of K is 418 kJ/mol and for F is 1681 kJ/mol, while the electron affinity for F is 328 kJ/mol. This results in a total energy requirement of 90 kJ/mol. However, the process of creating KF also releases 640 kJ/mol, resulting in a net energy release of 550 kJ/mol. The discrepancy between this expected value and the experimental measure of 498 kJ/mol for the energy required to dissociate KF into neutral atoms can be explained by taking into account the
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zenterix
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Homework Statement
I am a bit confused by the following concepts when we analyze them together:

- electron affinity
- electronegativity
- ionization energy

I will illustrate my doubts with the following problem

Consider the KF molecule, which has an ionic bond. The bond length is ##2.17\cdot 10^{-10}##m.

(a) Calculate the energy required to dissociate the KF molecule into the ions ##K^+## and ##F^-##.

(b) The energy required to dissociate KF into neutral atoms is 498 kJ/mol. Given that the first ionization energy for K is 418 kJ/mol, calculate the electron affinity (in kJ/mol) for F.
Relevant Equations
It seems that one way to calculate the dissociation energy of a KF molecule is to use classical electromagnetism. We consider each nucleus as a point charge and calculate the potential energy of the system composed of these two charges..

$$U(r)=k_e\frac{Z_1Z_2e^2}{r}=\frac{(+1)\cdot(-1)\cdot 1.602\cdot 10^{-19}}{2.17\cdot 10^{-10}} J$$
$$-1.06\cdot 10^{-18} J$$
$$=-640 kJ/mol$$

What this means is that to bring together these two ions it takes ##1.06\cdot 10^{-18} J## of work by the electrostatic forces involved. Similarly, to bring them infinitely far apart, the electrostatic forces involved do the negative of this amount of work, and whatever force is bringing the charges apart does ##1.06\cdot 10^{-18} J## of work.

Now, apparently, this all means that when these ions are brought together, energy is released. In fact, I do see that the final potential energy is lower than the initial potential energy, so difference had to go somewhere (though this still feels very imprecise at this point).

Hence, the process of breaking the ionic bond is the opposite of creating the bond by bringing the ions together, and thus requires 640 kJ/mol.
Let's think now about the energy to create a KF molecule from neutral atoms.

First we need to ionize both K and F: K loses an electron and F gains an electron. Then we have to bring the ions together.

The first ionization energy of K is 418 kJ/mol and for F is 1681 kJ/mol. The electron affinity for F is 328 kJ/mol.

Thus, ##K(g)\to K^+(g)+e## and 418 kJ is required per mol of K for this to happen.

Also, ##F^-(g)+e\to F(g)## and 328 kJ is released per mol of F.

Hence, just to ionize a mol of each of these elements we need to supply 418-328=90 kJ/mol of energy.

Next, as we've calculated previously, to bring the two ions together releases 640 kJ/mol of KF.

Thus, the entire process of creating KF from neutral atoms requires (-640+90)=-550 kJ/mol. That is, energy is released in the formation of KF.

My first question is if the calculations above are correct?

Next, apparently the energy required to dissociate KF into neutral atoms is 498 kJ/mol. I assume this is the experimental measure, which differs from our expected value of 550 kJ/mol due to the fact that we treated the atoms as point charges and ignored quantum mechanics.

Is this explanation for the discrepancy correct?

To get from KF to neutral atoms we have the following equations

$$KF \to K^+ +F^-$$

$$K^++e\to K$$

$$F^-\to F+e$$

In the first equation 640 kJ/mol is provided to the system. In the second equation, I assume we have the reverse operation implicit in the concept of "ionization energy". That is, if it takes 418 kJ/mol to ionize K, then this amount is released from a system of ##K^++e## when we form the neutral K.

Finally, in the third equation I again assume we are dealing with the opposite of the concept implicity in electron affinity of F. That is, a certain energy ##EA_F## is released when we go from ##F+e## to ##F^-##, so I assume we have to provide ##EA_F## to ##F^-## to obtain the neutral F.

Thus we have that ##498=640-418+EA_F##

and thus

##EA_F=276 kJ/mol##

What I am looking for are pointers about how to think about these concepts. I believe the final answers above are correct (because I can see the final solutions on MIT OCW), but it is the process of obtaining these answers that has me thinking a lot about if I actually understand what is happening.
 
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  • #2
zenterix said:
- electron affinity
- electronegativity
- ionization energy
It would help if one would provide definitions, or equations, for each term, and cite the source of the problem. For example,

https://ocw.mit.edu/courses/5-111sc...unit-ii-chemical-bonding-structure/lecture-8/
https://ocw.mit.edu/courses/5-111sc...unit-ii-chemical-bonding-structure/lecture-9/

As it indicates in the notes in the MIT OCW lectures, and elsewhere, ionization energy usually refers to the 1st ionization potential (unless otherwise stated), and it refers to the energy needed to remove an electron from the neutral atom.

Electron affinity is simply the energy released when an atom, e.g., F, binds with an electron. In the case of F, the extra electron fills the outer shell/orbital, and the atom becomes a negative ion. Oxygen normally attracts two electrons and becomes a doubly negatively charged ion.

Lecture 8 has discusses ionization energy (IE) and electron affinity (EA). In Lecture 9, please review the notes, II. IONIC BONDS.
 
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  • #3
The equation for F should read ##F+e^-\to F^-## but other than that, it seems like you have the basic idea.
 

FAQ: Creating KF Molecule from Neutral Atoms: A Process of Ionization and Binding

What is the process of ionization in the context of creating a KF molecule?

Ionization is the process of removing or adding electrons to an atom to create ions. In the context of creating a KF molecule, ionization involves removing an electron from a potassium (K) atom to form a K+ ion and adding an electron to a fluorine (F) atom to form an F- ion. These oppositely charged ions then attract each other to form the ionic bond in potassium fluoride (KF).

Why is potassium (K) more likely to lose an electron while fluorine (F) is more likely to gain one?

Potassium has one electron in its outermost shell, which it can lose easily to achieve a stable, noble gas electron configuration. Fluorine, on the other hand, has seven electrons in its outer shell and needs just one more to complete its octet and achieve stability. This makes potassium more likely to lose an electron and fluorine more likely to gain an electron during the ionization process.

What type of bond is formed between potassium (K) and fluorine (F) in a KF molecule?

The bond formed between potassium (K) and fluorine (F) in a KF molecule is an ionic bond. This type of bond occurs due to the electrostatic attraction between the positively charged potassium ion (K+) and the negatively charged fluoride ion (F-).

What are the energy changes involved in the formation of a KF molecule?

The formation of a KF molecule involves several energy changes: the ionization energy required to remove an electron from potassium, the electron affinity gained when fluorine accepts an electron, and the lattice energy released when the K+ and F- ions come together to form the ionic lattice of KF. The overall process is exothermic, meaning it releases energy.

How does the lattice structure of KF contribute to its stability?

The lattice structure of KF contributes to its stability by maximizing the electrostatic attraction between the K+ and F- ions. In the solid state, these ions arrange themselves in a regular, repeating pattern that minimizes the potential energy of the system, leading to a stable and strong ionic compound.

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