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AS Chemistry : Lattice Energy

What are the answers for the questions below?

http://prntscr.com/m31teb
CaO
Reply 2
Original post by BobbJo
CaO


Why it's not D? Isn't sulfur having a larger atomic radius? And polarization effect is maximised when the cation is having the highest charge density and the anion having the highest possible atomic radius?
Original post by OmarEdExcel
Why it's not D? Isn't sulfur having a larger atomic radius? And polarization effect is maximised when the cation is having the highest charge density and the anion having the highest possible atomic radius?

Lattice energy is higher if charges are higher, and if ionic radii are smaller

Since O2- is smaller and S2-, the lattice energy is greater in CaO.

The polarisation effect does increase L.E, but effect of ionic radius is greater. S2- has a larger ionic radius, so lattice energy is smaller

See AgCl, for example,
Experimental lattice enthalpy = 916 kJ
Theoretical Lattice enthalpy = 769 kJ
The discrepancy is explained by saying that AgCl has covalent character.

See here:
http://www.wiredchemist.com/chemistry/data/lattice-energies
E.g NaCl > NaBr > NaI in magnitudes of lattice energy
It is due to the ionic radius of the ions getting larger as we go from Cl to I. So the ionic bonding is weaker as charge density is smaller. A weaker lattice is formed (less close packing)
(edited 5 years ago)
Reply 4
So, the smaller the anion of an ionic compound, the more exothermic the lattice energy would be?

Regarding the polsarisation example of AgCl, why would it have a lower lattice energy then the calculated theoretical value?
Original post by OmarEdExcel
So, the smaller the anion of an ionic compound, the more exothermic the lattice energy would be?

Regarding the polsarisation example of AgCl, why would it have a lower lattice energy then the calculated theoretical value?

Yes since charge density is higher

It has a more negative L.E due to covalency. The effect is additive; there are ionic bonds and a degree of covalency. So total L.E is higher since more energy is released when the lattice forms, or more energy is required to break the bonds and separate the ions
Original post by OmarEdExcel
So, the smaller the anion of an ionic compound, the more exothermic the lattice energy would be?

Regarding the polsarisation example of AgCl, why would it have a lower lattice energy then the calculated theoretical value?

Here are some values:

experimental (kJ mol-1) v/s theoretical (kJ mol-1)
AgF +967 +953
AgCl +915 +864
AgBr +904 +830
AgI +889 +808


As you can see, the difference between experimental and theoretical LE is increasing.

For AgF, the values are close, so AgF is almost purely ionic.

For AgCl to AgI, the difference is larger. The bonding cannot be purely ionic bonding. There must be some covalency.
The electronegativity of halogen decreases from F to I. So there is more covalent character. (less like electron transfer, more like sharing of electrons)
The ions are larger and are more polarisable.

However, the trend is a decreasing lattice energy. Lattice energy depends on charge and ionic radii.

So if charge is higher, lattice energy is higher.
If atomic radii is lower, lattice energy is higher.

I hope this helps
I hope this is clearer
(edited 5 years ago)
Reply 7
It is indeed BobbJo, these are perfect replies.

I am aware that the greater the difference between the Born haber values and theoretical values indicate some sort of covalent character ( the polarisation effect ) in the ionic compound. The more connected or related the values are, the more strong and rigid the structure is and the more ionic character it would have.

However, if the option CaF was there, it would obviously be the one with the most lattice energy ( even if CaCl2 is mentioned as an option since F is smaller which would increase (total) charge density hence increase the strength of the ionic bond hence increase the lattice energy ). Kindly confirm this.

Consider the next question.


http://prntscr.com/m32rtu

The answer is D.

Taking note that the question wants us to identify a compound that has a large difference between experimental ( Born-haber value ) and theoretical value, that means that we should seek a compound that has a clear covalent character. And for this to be ensured, we have to find a compound that would have a high polarisation effect, meanwhile - high charge density ( and small ionic radius of cation ) and high ionic radius of anion ( and low charge density ). Is this true? Sulfur was picked for the fact of having the highest radius here. Confirm this.
(edited 5 years ago)
Original post by OmarEdExcel
It is indeed BobbJo, these are perfect replies.

I am aware that the greater the difference between the Born haber values and theoretical values indicate some sort of covalent character ( the polarisation effect ) in the ionic compound. The more connected or related the values are, the more strong and rigid the structure is and the more ionic character it would have.

However, if the option CaF was there, it would obviously be the one with the most lattice energy ( even if CaCl2 is mentioned as an option since F is smaller which would increase (total) charge density hence increase the strength of the ionic bond hence increase the lattice energy ). Kindly confirm this.

Consider the next question.


http://prntscr.com/m32rtu

The answer is D.

Taking note that the question wants us to identify a compound that has a large difference between experimental ( Born-haber value ) and theoretical value, that means that we should seek a compound that has a clear covalent character. And for this to be ensured, we have to find a compound that would have a high polarisation effect, meanwhile - high charge density ( and small ionic radius of cation ) and high ionic radius of anion ( and low charge density ). Is this true? Sulfur was picked for the fact of having the highest radius here. Confirm this.

Do you mean CaF2? Ca forms Ca2+ ions and F forms F- ions.
I do not agree.
CaO will still be more exothermic. The reason is that the charge on the ion is higher. Despite having higher ionic radius, the greater charge offsets this and the charge density on O2- is higher.
I have looked up values: CaO is -3464 kJ/mol, CaF2 is -2611 kJ/mol.
The experimental values agree.

CaF would be even less exothermic, if that is what you meant. Ca would form only a Ca+ ion. Since lattice energy is proportional to product of charges, the L.E is likely to be about 4 times less than CaF2. You may have learnt this relation, that LE proportional to product of charges, and inversely proportional to sum of ionic radii.
For comparison: lattice enthalpies:
MgCl -753 kJ/mol
MgCl2 -2526 kJ/mol
MgCl3 -5440 kJ/mol

You should make use of ionic radii given.
Recall Fajan's rules:
1. Cation with high charge is highly polarising
2. Small cation is highly polarising
3. Large anion is highly polarisable

Consider charge densities of cations:
Ca 2+ has a higher charge density than Li+.

so Ca2+ is more polarising
so options C and D
S2- has larger ionic radius, so more polarisable
hence greater covalent character
hence D.

Make sure you understand it fully
(edited 5 years ago)
Reply 9
I actually meant CaF2. I apologise.

And yes, that's true, I should've reviewed my reply - CaO should have a higher lattice energy for the sake that O2- is of higher charge density than F-.

So, as a summary :

Lattice energy increases by the highest possible charge density of ions involved in the ionic compound.

For example : CaS would have a higher lattice energy than CaF for the sake that the S2- ion is of higher charge density thus increasing strength of the ionic compound.

Difference in the BH and Theoretical values :

These increase when the compound shows a covalent character - highest possible polarisation, possibly the following combinations :

High(est) charge density cation ( and a small ionic radius ) and low(est) charge density anion ( and a high ionic radius )

For example :

CaF2 has a higher polarisation effect than NaF ( for the sake that Ca is Ca2+ and Na is only Na+, smaller charge density )
(edited 5 years ago)
Original post by OmarEdExcel
I actually meant CaF2. I apologise.

And yes, that's true, I should've reviewed my reply - CaO should have a higher lattice energy for the sake that O2- is of higher charge density than F-.

So, as a summary :

Lattice energy increases by the highest possible charge density of ions involved in the ionic compound.

Difference in the BH and Theoretical values :

These increase when the compound shows a covalent character - highest possible polarisation, possibly the following combinations :

High(est) charge density cation ( and a small ionic radius ) and low(est) charge density anion ( and a high ionic radius ).

Confirm this.

Yes that is basically it

Read more here:
https://en.wikipedia.org/wiki/Fajans%27_rules
Alright, what about this question :

http://prntscr.com/m33b3p

A : The term polar means a compound with electrical poles ( both + and - on opposite sides ) , such as NH3, H2O, and H2S. These compounds have the S+ (delta-positive ) slightly positively charged and this is because of the bonding electrons attracted to the nucleus of the atom the hydrogen is attached to. In this question, ethene would form a carbonium atom, making it positively charged, does this make it polar? If it is not polar, is this because it doesn't have a negative pole? - because we know that the hydrogen attached to the top of the carbonium ion would have a pair of shared electrons with the carbonium ion and these pair would make the hydrogen atoms themselves S+ ( delta, slightly positive ) charged because the pair would be attracted to the nucleus of the carbonium ion? Please answer this.

B : I do confuse the between electrophiles and nucleophiles. What's the difference between them?

H+ would act as an electrophile when it seeks the electron-dense areas in an ethene. Correct me if I am wrong. What then about the nucleophile? It should be the opposite, a negative ion seeking a positively charged ion?

C : Why it is not?

D : Does Ethene even donate here? The Br- attracted to the carbonium ion already has a lone pair of electrons ( 0 0 ) Br -, why does it need a donation?


I appreciate your efforts BobbJo.
Original post by OmarEdExcel
Alright, what about this question :

http://prntscr.com/m33b3p

A : The term polar means a compound with electrical poles ( both + and - on opposite sides ) , such as NH3, H2O, and H2S. These compounds have the S+ (delta-positive ) slightly positively charged and this is because of the bonding electrons attracted to the nucleus of the atom the hydrogen is attached to. In this question, ethene would form a carbonium atom, making it positively charged, does this make it polar? If it is not polar, is this because it doesn't have a negative pole? - because we know that the hydrogen attached to the top of the carbonium ion would have a pair of shared electrons with the carbonium ion and these pair would make the hydrogen atoms themselves S+ ( delta, slightly positive ) charged because the pair would be attracted to the nucleus of the carbonium ion? Please answer this.

B : I do confuse the between electrophiles and nucleophiles. What's the difference between them?

H+ would act as an electrophile when it seeks the electron-dense areas in an ethene. Correct me if I am wrong. What then about the nucleophile? It should be the opposite, a negative ion seeking a positively charged ion?

C : Why it is not?

D : Does Ethene even donate here? The Br- attracted to the carbonium ion already has a lone pair of electrons ( 0 0 ) Br -, why does it need a donation?


I appreciate your efforts BobbJo.

Other thing about polarisation is ok now?

A: It is saying that ethene is polar, which is clearly false.

A carbonium ion is positively charged. You do not talk about dipole or polarity since there is no delta negative charge. I quote the following from wikipedia: "A polar molecule has a net dipole as a result of the opposing charges (i.e. having partial positive and partial negative charges) from polar bonds arranged asymmetrically.". They talk of molecules. I quote from wikipedia again: "A molecule is an electrically neutral group of two or more atoms held together by chemical bonds.". Hence since an ion is clearly not a molecule, discussing whether it's polar is not very fruitful. A polar molecule should have a delta negative end and a delta positive end (both required)

I think that it is just that you may not have read it correctly or interpreted it the way it is meant to be.

B: A nucleophile is an electron pair donor. An electrophile is an electron pair acceptor. Nucleophile means nucleus (from nucleo) loving (from phile). That means that it is attracted to the nucleus, i.e, something positive. This does not mean that it must be negatively charged. Ethene is a nucleophile since the pi electrons are donated to the Br+ electrophile. Water is a nucleophile but uncharged.

Following the same logic, an electrophile is attracted to electrons, i.e sites of negative charge.

The Br2 reagent behaves as an electrophile because it is polarised by proximity to the pi bond of ethene. It undergoes heterolysis to give Br+, which reacts with ethene.

C: It is a nucleophile since it donates an electron pair.

D: It donates an electron pair, not a single electron.
About polarisation, as long as what I summarised was correct, then I'm fine with it as I took notes of it in my personal notes collection.

Regarding the reaction mechanism, say propene and bromine. The Br- attracted to the carbonium ion, is negative although it is an electrophile ( electron-loving, meaning it is supposed to be positive ). Why, why is it negative?

Ow wait! Let me guess, it has gained those electrons from it's bond with the other Br atom it was attached to before the formation of the carbonium ion, confirm this.

Therefore, by it's own nature, it is positive.
(edited 5 years ago)
Original post by OmarEdExcel
About polarisation, as long as what I summarised was correct, then I'm fine with it as I took notes of it in my personal notes collection.

Regarding the reaction mechanism, say propene and bromine. The Br- attracted to the carbonium ion, is negative although it is an electrophile ( electron-loving, meaning it is supposed to be positive ). Why, why is it negative?

Br- is a nucleophile. Refer to the definition. It is an electron pair donor.
Original post by BobbJo
=

B: A nucleophile is an electron pair donor. An electrophile is an electron pair acceptor. Nucleophile means nucleus (from nucleo) loving (from phile). That means that it is attracted to the nucleus, i.e, something positive. This does not mean that it must be negatively charged. Ethene is a nucleophile since the pi electrons are donated to the Br+ electrophile. Water is a nucleophile but uncharged.


Quoting this part specifically, you said that :

A nucleophile : An electron-pair donor that donates electrons. In the question context, it donates them to the Br- in the reaction mechanism. ( Doesn't necessarily mean it is negative as long as it donates electrons )
An electrophile : An electron-pair acceptor that receives a pair of electrons. ( Doesn't necessarily mean it is positive - it is an electrophile as long as it receives a pair of electrons )

So, a nucleophile or an electrophile, both are identified by function and not symbol.

The key question here would be, how would I know which one donates/receives a pair of electron? ( So that I can identify which is which )
(edited 5 years ago)
Original post by OmarEdExcel
Quoting this part specifically, you said that :

A nucleophile : An electron-pair donor that donates electrons. In the question context, it donates them to the Br- in the reaction mechanism. ( Doesn't necessarily mean it is negative as long as it donates electrons )
An electrophile : An electron-pair acceptor that receives a pair of electrons. ( Doesn't necessarily mean it is positive - it is an electrophile as long as it receives a pair of electrons )

So, a nucleophile or an electrophile, both are identified by function and not symbol.

The key question here would be, how would I know which one donates/receives a pair of electron? ( So that I can identify which is which )

I don't know what symbol you are talking about.

Have a read here:
https://en.wikipedia.org/wiki/Nucleophile
https://en.wikipedia.org/wiki/Electrophile

It will probably answer the question
Original post by OmarEdExcel
What are the answers for the questions below?

http://prntscr.com/m31teb


The discussion you've already had is very good and useful. In terms of considering lattice enthalpy, it's quite hard because they dont tend to teach you or show you the Kapustinskii equation at A-level. This equation tells you that lattice enthalpy is proportional to the number of ions in the formula unit, proportion to the charge on each of the contsituent ions and roughly inversely porpotional to the sum of the two ionic radii.

So lattice enthalpy goes down as you increase radius, and increases as you increase the number of ions in the formula unit or the charge on any either of the ions

Kapustinskii equation:
https://en.wikipedia.org/wiki/Kapustinskii_equation
(edited 5 years ago)
Original post by OmarEdExcel
Quoting this part specifically, you said that :

A nucleophile : An electron-pair donor that donates electrons. In the question context, it donates them to the Br- in the reaction mechanism. ( Doesn't necessarily mean it is negative as long as it donates electrons )
An electrophile : An electron-pair acceptor that receives a pair of electrons. ( Doesn't necessarily mean it is positive - it is an electrophile as long as it receives a pair of electrons )

So, a nucleophile or an electrophile, both are identified by function and not symbol.

The key question here would be, how would I know which one donates/receives a pair of electron? ( So that I can identify which is which )


The mechanism for the bromination that you learn at a level has two steps: https://www.chemguide.co.uk/mechanisms/eladd/symbr2.html

see 'the simplified version of the mechanism' on this link.

In the first step, the alkene is the nucleophile and Br2 is the electrophile, in the second step the organic cation is the electrophile and Br- is the nucleophile.

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