TBR Thermochem Sublimation Question (Ch 8 #15)

[grapepopsicle]

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The passage is about the Born-Haber cycle involving the formation of Lithium Fluoride
15. If sodium metal were used instead of lithium metal, what values would be affected?
I. The sublimation energy would increase
II. The ionization energy would increase
III. The lattice energy would increase

The answer key says choice I. is the only correct answer. It says that b/c sodium metal is heavier than lithium metal, it takes more energy to sublime sodium. This makes no sense to me. I would think that sodium metal, a larger atom, would pack less closely in its metal lattice than lithium, and therefore would require less energy to be sublimed (since during sublimation you're in essence disrupting the lattice structure interactions of the sodium metal). I would argue that none of the three options are correct.

This website agrees with me:
http://chemistry.bd.psu.edu/jircitano/sublime.html

Anyone have any ideas?

 

[MedPR]

Wasn't there a list of data and an explanation in the passage for this question?

 

[Chemdude]

The passage is about the Born-Haber cycle involving the formation of Lithium Fluoride
15. If sodium metal were used instead of lithium metal, what values would be affected?
I. The sublimation energy would increase
II. The ionization energy would increase
III. The lattice energy would increase

The answer key says choice I. is the only correct answer. It says that b/c sodium metal is heavier than lithium metal, it takes more energy to sublime sodium. This makes no sense to me. I would think that sodium metal, a larger atom, would pack less closely in its metal lattice than lithium, and therefore would require less energy to be sublimed (since during sublimation you're in essence disrupting the lattice structure interactions of the sodium metal). I would argue that none of the three options are correct.

This website agrees with me:
http://chemistry.bd.psu.edu/jircitano/sublime.html

Anyone have any ideas?

The website that you gave lists the sublimation enthalpy of Alkali Metals...The properties of Na are not the same as the properties of NaF. Although the sublimation energy of NaF and LiF are going to be very similar, NaF will have a slightly larger sublimation energy. NaF is a slightly larger molecule that LiF, so it is expected to have more London Dispersion Forces. More LDF=increased intermolecular forces=higher sublimation energy...I'm 90% sure that's the correct answer.

 

[grapepopsicle]

The website that you gave lists the sublimation enthalpy of Alkali Metals...The properties of Na are not the same as the properties of NaF. Although the sublimation energy of NaF and LiF are going to be very similar, NaF will have a slightly larger sublimation energy. NaF is a slightly larger molecule that LiF, so it is expected to have more London Dispersion Forces. More LDF=increased intermolecular forces=higher sublimation energy...I'm 90% sure that's the correct answer.

In a Born-Haber analysis we're talking about this sublimation reaction:

Na(s) + heat .→ Na(g)

You can't sublime an ionic solid. You sublime a solid by adding heat to it; if you add heat to an ionic solid the first thing that is going to happen is the lattice structure will be disrupted, and the ions will go into the gas phase:
NaF(s) + heat ..→ ..Na+(g).. + F-(g)

This doesn't happen: NaF(s) + heat ..→ NaF(g)

There is not a sublimation energy for NaF because ionic compounds can't be sublimed.

Dispersion forces are not at work here, we are talking about metallic bonding. The lattice structure of sodium metal vs. lithium metal is all we care about, and the strength of the lattice structure interactions is a function of how close the metal atoms pack. Sodium is larger than lithium. Therefore sodium atoms should pack less closely than lithium atoms, and have a lower sublimation energy.

I'm 100% sure this is just a typo in the book, but thanks for your help.
.