# Calculating the bonding energy between two ions

• Chemistry
• jisbon
In summary, I seem to be having problems with part two in general, but here's my solution.I understood that to find the bonding energy, I need to find the condition where the attractive force + repulsive force = 0 and find out the radius from there. However, I can't seem to find the radius. I attempted to find the condition where the attractive force + repulsive force = 0, but I can't seem to find it. I think I may have been doing something wrong.
jisbon
Homework Statement
The repulsive potential energy of 2 ions ##A^{2+}## and ##B^{2-}## is given as ##3.6*10^{-2} r^{-5}## eV.
i) Derive the attractive force between the two ions, and subsequently the attractive potential energy.

ii) Calculate the bonding energy of the 2 ions
Relevant Equations
-
I seem to be having problems with part two in general, but here's my solution.

Attractive force:
##F_{A}=\dfrac {k\left( 2\right) \left( -2\right) }{r^{2}}=\dfrac {-4k}{r^{2}} ##
##E_{A}=-\int ^{r}_{\infty }F_{A}\cdot dr ##
##\begin{aligned}E_{A}=\int ^{r}_{\infty }\dfrac {4k}{r^{2}}dr\\ =\left[ \dfrac {4k}{r}\right] ^{r}_{\infty }=\dfrac {4k}{r}\end{aligned} ##

(ii) This is where I started having trouble. I understand that to find the bonding energy, I need to find the condition where the attractive force + repulsive force = 0 and find out the radius from there. However, I can't seem to find the radius.
This is what I attempted,

Bonding energy when F =0,

##\begin{aligned}E_{r}=\dfrac {3.6\times 10^{-2}}{r^{5}}\\ F_{r}=\int E_{r}dr\end{aligned} ##
##\begin{aligned}F_{r}=-\dfrac {0\cdot 18}{r^{6}}\\ F_{0}=0=-\dfrac {0.18}{r^{6}}+\dfrac {-4k}{r^{2}}\end{aligned} ##Can anyone show me where I did wrong? Thanks

##
F_{r}=\int E_{r}dr
## Isn't right (And it isn't what you do to get from ##E## to ##F## either... ) what should it have been ?

jisbon
You're not being careful with signs. The integral of 1/r2 is -1/r. F = -dE/dr, so Fr is +0.18/r6.

jisbon
Don't give it all away ! That robs @jisbon from the exercise !

jisbon said:
The repulsive potential energy of 2 ions ##A^{2+}## and ##B^{2-}## is given as ##3.6*10^{-2} r^{-5}## eV.

Attractive force:
##F_{A}=\dfrac {k\left( 2\right) \left( -2\right) }{r^{2}}=\dfrac {-4k}{r^{2}} ##
What units are you using for the distance ##r##? What is the value of ##k## in these units?

BvU said:
##
F_{r}=\int E_{r}dr
## Isn't right (And it isn't what you do to get from ##E## to ##F## either... ) what should it have been ?
Oops, differntating the potential energy should give force instead, and there should be a negative sign infront. I assume when integrating from force to potential energy, there should also be a negative sign in front?

TSny said:
What units are you using for the distance ##r##? What is the value of ##k## in these units?
r is in nano meters. k is simply the constant which is

In this case, I should convert my constant to nm?

Thanks

mjc123 said:
eV is in joules. So.. what would that actually imply? Sorry, abit confused here

eV is in eV. That is a unit, equal to 1.602 x 10-19 J. You have to convert 3.6 x 10-2 eV into Joules. (By the way, you're given that figure to 2 sig figs, so there's no justification for taking k to 17 sig figs!)

mjc123 said:
eV is in eV. That is a unit, equal to 1.602 x 10-19 J. You have to convert 3.6 x 10-2 eV into Joules. (By the way, you're given that figure to 2 sig figs, so there's no justification for taking k to 17 sig figs!)
So it will be...

##E_{r}=\dfrac {\left( 3.6\times 10^{-2}\right) \left( 1.602\times 10^{-19}\right) }{r^{5}}##
##\begin{aligned}F_{r}=-\dfrac {d}{dr}E_{r}\\ =\dfrac {2.8836\times 10^{-20}}{r^{6}}\end{aligned} ##
?

That should be 10-21

mjc123 said:
That should be 10-21
Because of the unit of distance?

Sorry, my mistake. I was multiplying 10-2 by 10-19, and forgot about the factor of 5 when you differentiate.

Because the exercise uses eV for energy, I would stick to eV for the answer as well ...

BvU said:
Because the exercise uses eV for energy, I would stick to eV for the answer as well ...
Which part are you referring to? Sorry :/

mjc123 said:
Sorry, my mistake. I was multiplying 10-2 by 10-19, and forgot about the factor of 5 when you differentiate.
Oh okay.
Since
##\begin{aligned}F_{r}=-\dfrac {d}{dr}E_{r}\\ =\dfrac {2.8836\times 10^{-20}}{r^{6}}\end{aligned} ##
##\dfrac {2.8836\times 10^{-20}}{r^{6}}= \dfrac{4k}{r}##
##2.88\times 10^{-20}=4kr^{5} ##
where k =9.0*10^9
r=0.000000956m?

Isn't it a bit too short? Or is this actually correct?

You're using the wrong expression. FA = -4k/r2.

jisbon said:
Which part are you referring to?
I am referring to
jisbon said:
is given as ##3.6*10^{-2} r^{-5}## eV.

jisbon said:
Isn't it a bit too short?
Actually, 1 ##\mu##m is huge on an atomic scale

BvU said:
I am referring toActually, 1 ##\mu##m is huge on an atomic scale
Oh.. I realized, it may a bit too large? I recalled seeing my notes that states the avg is around 0.3nm. In this case, is 1nm still correct?Also regarding the eV units, if I don't convert it to joules, won't I be able to proceed?

jisbon said:
Also regarding the eV units, if I don't convert it to joules, won't I be able to proceed?
Yes, but you'd have to convert everything else (e.g. k) into consistent units. I find it easier to convert everything to SI, then convert back at the end if necessary.

BTW, when you said above "r is in nanometres", did you mean that the question says that Er = 3.6 x10-2/r5 eV when r is in nm? If so, what units are you using for r in the expression FA = -4k/r2, if you are using a k value in SI units? And what about the charge units? You've got to make sure everything's consistent, whichever units system you're working in.

jisbon said:
r is in nano meters
How do you know this ? Can you render the full and complete problem statement, litterally ?

mjc123 said:
Yes, but you'd have to convert everything else (e.g. k) into consistent units. I find it easier to convert everything to SI, then convert back at the end if necessary.

BTW, when you said above "r is in nanometres", did you mean that the question says that Er = 3.6 x10-2/r5 eV when r is in nm? If so, what units are you using for r in the expression FA = -4k/r2, if you are using a k value in SI units? And what about the charge units? You've got to make sure everything's consistent, whichever units system you're working in.
BvU said:
How do you know this ? Can you render the full and complete problem statement, litterally ?

Regarding:
BTW, when you said above "r is in nanometres", did you mean that the question says that Er = 3.6 x10-2/r5 eV when r is in nm?
Yes, it is.

Getting back into the question, and decided to do it in terms of metre and joules first (SI units)

Since ##E_{r}=\dfrac {3.6\times 10^{-2}}{r^{5}\left[ nm\right] }eV ##
##E_{r}=3.6\times 10^{-2}\times \left( 1.6\times 10^{-19}\right) \times \left( 10^{9}\right) ^{5} ##
to convert eV to joules and r to m. Since it is in the denominator it will be positive power.

##E_{r}=\dfrac {5.76\times 10^{24}}{r^{5}\left[ m\right] }J ##

Hence differentiating it with a negative sign will give me the repulsive force:

##\begin{aligned}F_{R}=\dfrac {2.88\times 10^{25}}{r^{6}}N\\ \dfrac {-4\left( 1.6\times 10^{-19}\right) }{4\pi \varepsilon _{0}r^{2}}+\dfrac {2.88\times 10^{25}}{r^{6}}=0\end{aligned} ##

where r is in metres. Is this now correct? Or did I make something wrong again?

The product of the charges is -4*(1.6x10-19)2. Otherwise it looks correct.

mjc123 said:
The product of the charges is -4*(1.6x10-19)2. Otherwise it looks correct.
Oh yes, forgot to put that in. Thanks for checking

Just wanted to bump this and apparently the interatomic distance seems to be VERY big? Why is this so?

In your expression for Er, you should have multiplied by (10-9)5. I missed that first time.

mjc123 said:
In your expression for Er, you should have multiplied by (10-9)5. I missed that first time.
I did so. Refer to post #21

That's 10-9, not 109.

mjc123 said:
That's 10-9, not 109.
oops gotacha. It should be -9 because it is in the denominator right?

I've edited my equation accordingly and now am facing this problem.

Why is this so?

EDIT: nvm, online calc seem to be having problems haha

Last edited:

## What is bonding energy between two ions?

Bonding energy between two ions is the amount of energy required to separate the ions and break the bond between them. It is a measure of the strength of the bond between the two ions.

## How is bonding energy calculated?

Bonding energy is calculated by using Coulomb's law, which takes into account the charges and distance between the ions.

## What factors affect the bonding energy between two ions?

The bonding energy between two ions is affected by the charges of the ions, the distance between them, and the size of the ions.

## Why is calculating bonding energy important?

Calculating bonding energy can help us understand the stability of compounds and the strength of chemical bonds. It is also useful in predicting the properties and reactivity of substances.

## Can bonding energy be negative?

Yes, bonding energy can be negative if the ions have opposite charges and are attracted to each other. This indicates a stable bond between the ions.

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