# factors affecting lattice enthalpy?Watch

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#1
question1b https://pmt.physicsandmathstutor.com...%20A-level.pdf

this question realllyyyy confused me. how do i answer it?

having a more negative LE means the ionic bond forms easier right?

therefore the metal part of the salt needs to be further down the group (it gives its e- to the halogen more) and the halogen needs to be further up the group??

when comparing NaCl and NaBr, the one which will have more negative LE will be the one further up the group right??? (NaCl)

but for the metal part, the one which will have a more negative LE will be the one further down the group (larger ionic radius) so btwn NaCl and KCl, the KCl will have a more negative LE??
Last edited by anonymoussse; 1 month ago
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#2
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#3
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1 month ago
#4
(Original post by ggxsywes)
question1b https://pmt.physicsandmathstutor.com...%20A-level.pdf

this question realllyyyy confused me. how do i answer it?

having a more negative LE means the ionic bond breaks easier right?

therefore the metal part of the salt needs to be further down the group (it gives its e- to the halogen more) and the halogen needs to be further up the group??

when comparing NaCl and NaBr, the one which will have more negative LE will be the one further up the group right??? (NaCl)

but for the metal part, the one which will have a more negative LE will be the one further down the group (larger ionic radius) so btwn NaCl and KCl, the KCl will have a more negative LE??
Lattice enthapy may be positive (breaking the lattice) or negative (forming the lattice). Different specifications use different definitions. Lattice enthalpy may be derived from first principles, or by experimentation. The latter is the "real" value.

The MAGNITUDE of the lattice enthalpy is dependent on four factors:

The magnitude of the charges on the ions (most important)
The radius of the ions (less important than magnitude of charge)
The structure/packing of the lattice, i.e if its 6,6 coordination, Hexagonal close packed, cubic close packed etc.
The influence of a degree of covalency can also attenuate the experimental lattice enthalpy

The first two factors are those usually discussed, as the last two only attentuate the magnitude.
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#5
Ah yes I meant the magnitude. When u talk abut the ionic radii the metal ion needs to larger right? And the halide smaller? This means the bond breaks easier and the energy released is greater??
(Original post by charco)
Lattice enthapy may be positive (breaking the lattice) or negative (forming the lattice). Different specifications use different definitions. Lattice enthalpy may be derived from first principles, or by experimentation. The latter is the "real" value.

The MAGNITUDE of the lattice enthalpy is dependent on four factors:

The magnitude of the charges on the ions (most important)
The radius of the ions (less important than magnitude of charge)
The structure/packing of the lattice, i.e if its 6,6 coordination, Hexagonal close packed, cubic close packed etc.
The influence of a degree of covalency can also attenuate the experimental lattice enthalpy

The first two factors are those usually discussed, as the last two only attentuate the magnitude.
0
1 month ago
#6
(Original post by ggxsywes)
Ah yes I meant the magnitude. When u talk abut the ionic radii the metal ion needs to larger right? And the halide smaller? This means the bond breaks easier and the energy released is greater??

In terms of ionic radii (factor 3)
The strongest lattice is produced when the ions are similar in size, but you must take factors 1 adn 2 into account as they are more important.

Your sentence in bold, above, is completely wrong. Energy is released when the lattice is formed. This is an exothermic process.
Breaking the lattice requires energy to overcome the forces of attraction between oppositely charged ions. This is endothermic.
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#7
ah oops i probs look stupid i muddled enthalpy of solution and lattice enthalpy

u said if ions are smaller in size, the strongest lattice is produced (i thought LE measures the energy given OUT when the gaseous ions make bonds so shldnt the statement be "if ions are smaller in size, the most energy is given out"?

i bolded "ions" because is it not depending on whether ure looking at the + or - ion? the smaller the halide, the more it's gonna attract the metal ion. the bigger the metal ion, the more it's gonna readily give out its electrons to the electronegative halide... is this right or wrong? i dont get how both the + and - ions need to be smaller...if the metal ion is smaller it's going to have a greater attraction to it's outermost e-...the one it's given away. then the ionic attraction is weak.

this is the bottom line of my thread btw lol
(Original post by charco)

In terms of ionic radii (factor 3)
The strongest lattice is produced when the ions are similar in size, but you must take factors 1 adn 2 into account as they are more important.

Your sentence in bold, above, is completely wrong. Energy is released when the lattice is formed. This is an exothermic process.
Breaking the lattice requires energy to overcome the forces of attraction between oppositely charged ions. This is endothermic.
Last edited by anonymoussse; 1 month ago
0
1 month ago
#8
(Original post by ggxsywes)
ah oops i probs look stupid i muddled enthalpy of solution and lattice enthalpy

u said if ions are smaller in size, the strongest lattice is produced (i thought LE measures the energy given OUT when the gaseous ions make bonds so shldnt the statement be "if ions are smaller in size, the most energy is given out"?

i bolded "ions" because is it not depending on whether ure looking at the + or - ion? the smaller the halide, the more it's gonna attract the metal ion. the bigger the metal ion, the more it's gonna readily give out its electrons to the electronegative halide... is this right or wrong? i dont get how both the + and - ions need to be smaller...if the metal ion is smaller it's going to have a greater attraction to it's outermost e-...the one it's given away. then the ionic attraction is weak.

this is the bottom line of my thread btw lol
Metal ions do not "give away" electrons to anything.

You are getting confused between the actual lattice itself and the process of reaction between metals and non-metals.

Lattice enthapy may be positive (breaking the lattice) or negative (forming the lattice). Different specifications use different definitions. Lattice enthalpy may be derived from first principles, or by experimentation. The latter is the "real" value.

The MAGNITUDE of the lattice enthalpy is dependent on four factors:

The magnitude of the charges on the ions (most important)
The radius of the ions (less important than magnitude of charge)
The structure/packing of the lattice, i.e if its 6,6 coordination, Hexagonal close packed, cubic close packed etc. This is affected by the relative size of the ions.
The influence of a degree of covalency can also attenuate the experimental lattice enthalpy

The first two factors are (most important) those usually discussed, as the last two only attentuate the magnitude.
0
#9
1. I know different specs use different LE definitions
2. I know LE magnitude is dependent on 4 factors
3. My confusion is: the radius factor. Please explain the radius factor. I don’t understand why halide ions and metal ions BOTH need to be small, when different charged ions do different things (lose / gain e-). U didn’t answer it in the previous answer I thought metal ions need to be bigger not smaller. This will make the attraction between the metal outermost electron and the Metal nucleus small so the e- is closer to the halide. This makes the reaction more vigorous and the LE magnitude is greater. The question in the link confuses me too. Please explain it if U can.
Thank you.
(Original post by charco)
Metal ions do not "give away" electrons to anything.

You are getting confused between the actual lattice itself and the process of reaction between metals and non-metals.

Lattice enthapy may be positive (breaking the lattice) or negative (forming the lattice). Different specifications use different definitions. Lattice enthalpy may be derived from first principles, or by experimentation. The latter is the "real" value.

The MAGNITUDE of the lattice enthalpy is dependent on four factors:

The magnitude of the charges on the ions (most important)
The radius of the ions (less important than magnitude of charge)
The structure/packing of the lattice, i.e if its 6,6 coordination, Hexagonal close packed, cubic close packed etc. This is affected by the relative size of the ions.
The influence of a degree of covalency can also attenuate the experimental lattice enthalpy

The first two factors are (most important) those usually discussed, as the last two only attentuate the magnitude.
Last edited by anonymoussse; 4 weeks ago
0
4 weeks ago
#10
(Original post by ggxsywes)
1. I know different specs use different LE definitions

(Original post by ggxsywes)
2. I know LE magnitude is dependent on 4 factors

(Original post by ggxsywes)
The smaller the radius the stronger the force of interionic attraction and the larger the magnitude of the lattice enthalpy

(Original post by ggxsywes)
I don’t understand why halide ions and metal ions BOTH need to be small, when different charged ions do different things (lose / gain e-).
Ions DO NOT GAIN OR LOSE ELECTRONS

(Original post by ggxsywes)
U didn’t answer it in the previous answer I thought metal ions need to be bigger not smaller.
I did answer it - the ions need to be smaller.

(Original post by ggxsywes)
This will make the attraction between the metal outermost electron and the Metal nucleus small so the e- is closer to the halide. This makes the reaction more vigorous and the LE magnitude is greater. The question in the link confuses me too. Please explain it if U can.
Thank you.
You are confusing metallic bonding with ionic bonding. AND you are still taliking about reactions! There are no reactions involved in the lattice enthalpy ...
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