Stankovic123's chem q's

[zvezda]

Oxidation only occurs if it can get rid of the electrons. If it's covered in aluminium hydroxide, the aluminium can't really oxidise.

Dont metals have a metallic lattice structure? So if the electrons are delocalised, they just flow through the anode itself, meaning that the aluminium hydroxide has no substantial effect?

[lzxnl]

It's covered by aluminium hydroxide, which consists of ions that are packed together and can't really move.

[zvezda]

But the aluminium hydroxide isnt covering the anode where it is connected by a wire to the cathode?

Also, is the ionisation of acids endothermic or exothermic? I was thinking endothermic merely basing it off of the self-ionisation of water.
Thanks

[lzxnl]

If you reduce the effective surface area of aluminium that can be reduced, you reduce the reaction rate. That's what the gel does; it stops the electrons from leaving.

As for the ionisation, it could be either. Sulfuric acid, nitric acid and hydrogen chloride ionise exothermically, but I'm sure other acids dissociate endothermically (can't think of an example immediately right now).
It depends on the acid, the bonds it forms to itself and how well the ions are solvated by water. In the case of the auto-ionisation of water, water itself is pretty stable; there is a negative charge on the oxygen which allows water to abstract protons from other water molecules, but then you're left with a hydroxide ion, a small negative charge, which is a rather strong base. There's a rather large energy cost to do so.

[zvezda]

If you reduce the effective surface area of aluminium that can be reduced, you reduce the reaction rate. That's what the gel does; it stops the electrons from leaving.

As for the ionisation, it could be either. Sulfuric acid, nitric acid and hydrogen chloride ionise exothermically, but I'm sure other acids dissociate endothermically (can't think of an example immediately right now).
It depends on the acid, the bonds it forms to itself and how well the ions are solvated by water. In the case of the auto-ionisation of water, water itself is pretty stable; there is a negative charge on the oxygen which allows water to abstract protons from other water molecules, but then you're left with a hydroxide ion, a small negative charge, which is a rather strong base. There's a rather large energy cost to do so.

All right fair enough.
I basically asked that endo/exo question because there was a q in a neap paper which asked for whether the value for the heat of neutralisation of acids and bases would be different if you used hydrochloric acid and sodium hydroxide or ethanoic acid and sodium hydroxide. It basically said that the heat of neutralisation would be comparatively lower if one used ethanoic acid instead of hydrochloric acid, but how can you deduce this without prioir knowledge that the ionisation of ethanoic acid is endothermic?

Also, for q18 from vcaa 2011 exam 1, it says that the atomic masses of covalently bonded atoms affect the wavenumbers and the amount of infra-red radiation absorbed by covalent bonds. But how?
Isnt the bending and stretching of covalent bonds dependent on the dipole moment of the bond, which in effect is dependent on the electronegativities of the atoms in question? The text book doesnt give a very good explanation and neither does the assessor's report.
Cheers

[zvezda]

Ok, this has come to frustrate me.
2 statements from two separate vcaa exam 1s:
I: a UV spectrum is a result of electrons falling back from higher to lower electronic energy levels
II: the flame in the AAS when analysing the amount of copper in a copper sulphate solution is green due to electron transitions from a higher energy state to a lower energy state.

Apparently, I is an incorrect statement and II is correct. Please explain?

[lzxnl]

All right fair enough.
I basically asked that endo/exo question because there was a q in a neap paper which asked for whether the value for the heat of neutralisation of acids and bases would be different if you used hydrochloric acid and sodium hydroxide or ethanoic acid and sodium hydroxide. It basically said that the heat of neutralisation would be comparatively lower if one used ethanoic acid instead of hydrochloric acid, but how can you deduce this without prioir knowledge that the ionisation of ethanoic acid is endothermic?

Also, for q18 from vcaa 2011 exam 1, it says that the atomic masses of covalently bonded atoms affect the wavenumbers and the amount of infra-red radiation absorbed by covalent bonds. But how?
Isnt the bending and stretching of covalent bonds dependent on the dipole moment of the bond, which in effect is dependent on the electronegativities of the atoms in question? The text book doesnt give a very good explanation and neither does the assessor's report.
Cheers

Ah. I see the issue.
With acids that dissociate exothermically, that is generally a result of the enthalpy change when they dissolve, not when they dissociate. There are two steps involved. The dissociation itself would be endothermic.
But if you already have dissolved ethanoic acid, it's going to take energy to convert that to ethanoate ions and hydrogen ions.


Your second question is one I still disagree with. The answer given is acceptable. The atoms are like springs; higher mass with the same force means they won't oscillate as much. Same with the infrared absorption; larger mass of atoms means it won't oscillate as much for an identical bond strength. The dipole moment isn't the only thing that matters.

The second problem, though, is the atomic radius. That SHOULD also be a factor, because a smaller atomic radius means the atoms can bond closer, which creates stronger bonds. Sigh.

Ok, this has come to frustrate me.
2 statements from two separate vcaa exam 1s:
I: a UV spectrum is a result of electrons falling back from higher to lower electronic energy levels
II: the flame in the AAS when analysing the amount of copper in a copper sulphate solution is green due to electron transitions from a higher energy state to a lower energy state.

Apparently, I is an incorrect statement and II is correct. Please explain?

UV spectra are absorption spectra; we measure how much light is absorbed on its way through the solution.
However, the flame colour is a direct result of the electronic transition from higher to lower energy levels; that's how the light is emitted.

[zvezda]

Ah. I see the issue.
With acids that dissociate exothermically, that is generally a result of the enthalpy change when they dissolve, not when they dissociate. There are two steps involved. The dissociation itself would be endothermic.
But if you already have dissolved ethanoic acid, it's going to take energy to convert that to ethanoate ions and hydrogen ions.


Your second question is one I still disagree with. The answer given is acceptable. The atoms are like springs; higher mass with the same force means they won't oscillate as much. Same with the infrared absorption; larger mass of atoms means it won't oscillate as much for an identical bond strength. The dipole moment isn't the only thing that matters.

The second problem, though, is the atomic radius. That SHOULD also be a factor, because a smaller atomic radius means the atoms can bond closer, which creates stronger bonds. Sigh.

UV spectra are absorption spectra; we measure how much light is absorbed on its way through the solution.
However, the flame colour is a direct result of the electronic transition from higher to lower energy levels; that's how the light is emitted.

With the IR question;
But if there was no electronegativity difference, there would be no oscillation either; isnt that why oxygen and nitrogen for example do not absorb infra red radiation?
That also brings me to another point. The C-O bond has a lower wavenumber than the C-H bond. If an increase in mass did mean that less vibration would occur, why does C-O have a lower wavenumber? The C-H has a lower electronegativity difference and a smaller atomic radius?

Also, why doesnt the C-H on most IR spectra show up? Some spectra have little absorption around its wavenumber?

And cheers for clearing up the other problems

[lzxnl]

With the IR question;
But if there was no electronegativity difference, there would be no oscillation either; isnt that why oxygen and nitrogen for example do not absorb infra red radiation?
That also brings me to another point. The C-O bond has a lower wavenumber than the C-H bond. If an increase in mass did mean that less vibration would occur, why does C-O have a lower wavenumber? The C-H has a lower electronegativity difference and a smaller atomic radius?

Also, why doesnt the C-H on most IR spectra show up? Some spectra have little absorption around its wavenumber?

And cheers for clearing up the other problems

It's not that; IR only picks up oscillations in which there is a change in the dipole moment. For diatomic oxygen and nitrogen, they absorb infra red radiation, but we can't see it on an IR spectrum. You're right; electronegativity differences are one thing, but those really just strengthen the bonds due to the polarity that now exists.

C-O is polar, so there's that, but it's also heaps larger than C-H. C-H has a higher wavenumber both because of hydrogen's little mass and because of the size of hydrogen; stronger bond. Electronegativity isn't everything.

C-H does generally show up; it's at around 3000 cm^-1 and I've always seen something there.

[zvezda]

It's not that; IR only picks up oscillations in which there is a change in the dipole moment. For diatomic oxygen and nitrogen, they absorb infra red radiation, but we can't see it on an IR spectrum. You're right; electronegativity differences are one thing, but those really just strengthen the bonds due to the polarity that now exists.

C-O is polar, so there's that, but it's also heaps larger than C-H. C-H has a higher wavenumber both because of hydrogen's little mass and because of the size of hydrogen; stronger bond. Electronegativity isn't everything.

C-H does generally show up; it's at around 3000 cm^-1 and I've always seen something there.

But isnt the change in the dipole moment made easier by a higher difference in electronegativities?  And hang on. Are you saying a higher difference in electronegativity means a steonger covalent bond??

So i think im right in saying that electronegativity, molar mass and atomic radius size influence the vibrations of covalent bonds. If so, how do we quantiy their influences given this question that vcaa has put in an exam?

Also with regards to C-H on the IR spectra; the peaks they cause are not as pronounced as say the O-H yet theyre more abundant than O-H. Take the methyl ethanoate IR spectrum for example.

[Aurelian]

For diatomic oxygen and nitrogen, they absorb infra red radiation, but we can't see it on an IR spectrum.

Nah, O₂ and N₂ actually can't directly absorb IR radiation in the first place. You can get at their single vibrational modes via Raman spectroscopy though (I presume).

stankovic, you might like to have a look at this post I made a little while back (scroll down to the IR spectroscopy part).

The second problem, though, is the atomic radius. That SHOULD also be a factor, because a smaller atomic radius means the atoms can bond closer, which creates stronger bonds. Sigh.

Smaller atomic radii don't necessarily result in stronger bonds though (for example, the H-H bond in hydrogen is weaker than the O=O bond in oxygen). Moreover, there are a lot of other factors to consider when determining bond strength, not just atomic radius. So, I think it's pretty fair to say that the atomic mass is more significant factor than atomic radius. Since the question asked for the "best" explanation, I don't think there's anything especially controversial going on.

[lzxnl]

Nah, O₂ and N₂ actually can't directly absorb IR radiation in the first place. You can get at their single vibrational modes via Raman spectroscopy though (I presume).

stankovic, you might like to have a look at this post I made a little while back (scroll down to the IR spectroscopy part).

Smaller atomic radii don't necessarily result in stronger bonds though (for example, the H-H bond in hydrogen is weaker than the O=O bond in oxygen). Moreover, there are a lot of other factors to consider when determining bond strength, not just atomic radius. So, I think it's pretty fair to say that the atomic mass is more significant factor than atomic radius. Since the question asked for the "best" explanation, I don't think there's anything especially controversial going on.

For your second point, H-H is weaker than O=O because of the double bond I would have thought. Even so, O=O isn't much stronger than H-H according to some sites I see on the internet. O-O, for instance, is VERY weak.

What other factors would there be? There would be influence due to surrounding atoms, but VCE wouldn't consider that.
Is atomic mass really more significant? From my limited physics knowledge, the frequency of a spring's vibration depends on the inverse of the square root of the mass.
To be fair, I'm not sure how to estimate the force and its dependency on distance, but I'm sure it would be more than inverse square root of distance. I mean, http://www.cem.msu.edu/~reusch/OrgPage/bndenrgy.htm says that F-F is stronger than O-O, which supports the fact that distance does play a part in determining bond strengths. What else is there? I'm curious.

Although as for your first point, I know little to nothing about IR radiation xP

Edit: I just realised the dependency on perhaps polarities of the atoms and whether they're bonded to other electron-donating or withdrawing groups

[Aurelian]

For your second point, H-H is weaker than O=O because of the double bond I would have thought. Even so, O=O isn't much stronger than H-H according to some sites I see on the internet. O-O, for instance, is VERY weak.

What other factors would there be? There would be influence due to surrounding atoms, but VCE wouldn't consider that.
Is atomic mass really more significant? From my limited physics knowledge, the frequency of a spring's vibration depends on the inverse of the square root of the mass.
To be fair, I'm not sure how to estimate the force and its dependency on distance, but I'm sure it would be more than inverse square root of distance. I mean, http://www.cem.msu.edu/~reusch/OrgPage/bndenrgy.htm says that F-F is stronger than O-O, which supports the fact that distance does play a part in determining bond strengths. What else is there? I'm curious.

In the context of vibrational frequencies, the bond strengths just get lumped to together in a constant, k, called the "force constant" (analogous to the stiffness of a spring), so any distinction between discrete single vs. double bonds get lost, which is why I considered H-H vs. O=O to be a fair comparison. Moreover, the distinction between discrete single and double bonds doesn't really work anymore once you start doing molecular orbital (MO) analysis.

Now, to be clear, I never meant to say that atomic radius wasn't a factor at all! I was just trying to argue against the idea that it's as equally good an explanation in the context of that question as atomic mass. Atomic radius certainly does play a role in bond strength. However, don't forget that it's bond strengths which have the direct affect on the frequency of vibration. Now, a C-H bond has a bond energy of around 98 kcal/mol whereas the C-O bond in methanol has a bond energy of around 91 kcal/mol (unless Wikipedia is lying to me). These values aren't massively different, suggesting that the bond strengths also aren't that different. In contrast, an oxygen atom is 16 times more massive than a hydrogen atom. As such, bond strength (and hence atomic radius) isn't going to be the biggest influence on vibrational frequency here - atomic mass is.

The other sorts of considerations I was thinking of arise when you start doing MO theory; you have to consider things like orbital overlap and symmetry (i.e. lots of scary integrals and algebra). That said, when I think about it, I suppose that's really just the MO equivalent of thinking about atomic radius... Other than that, electronegativity difference and EDG/EWG substituent stuff are a very important factors, but you've already talked about that.

As an aside, be careful with talking about "bond length/distance" as a factor (as opposed to atomic radius) because, in my opinion, it's difficult to establish the line of causality clearly (is the bond short because the atoms are tightly bound together or are they tightly bound together because the bond is short?).

PS: Sorry for us semi-hijacking your thread, stankovic!

[lzxnl]

I wanted to have a discussion without molecular orbital theory. Sigh. That theory complicates things so much.
Yeah, I know what you mean, but you can still talk of double bonds like O=O consisting of two bonds as there are four more bonding electrons than antibonding electrons...right?

The thing about your point with C-O and C-H is that the two factors. atomic radius and electronegativity, have evidently just cancelled each other. Mass is certainly an important aspect, but I'm just saying that it's not the only reason.
I think numbers are the only real way of doing such questions to be honest.

Yeah, I only know of orbital overlap in the diagrammatical sense; if I get two atoms, I can sort of determine which orbitals will combine...but with integrals? Uh oh. I'm not looking forward to chemistry next year if I DO get into second year chemistry. After a LOT of studying after my last exam.

I think it's fair to say that large atoms will have larger bond lengths in general.

[Aurelian]

I wanted to have a discussion without molecular orbital theory. Sigh. That theory complicates things so much.
Yeah, I know what you mean, but you can still talk of double bonds like O=O consisting of two bonds as there are four more bonding electrons than antibonding electrons...right?

Yeah I guess that's fair enough =)

Mass is certainly an important aspect, but I'm just saying that it's not the only reason.

Yes I don't disagree with you ^.^

Yeah, I only know of orbital overlap in the diagrammatical sense; if I get two atoms, I can sort of determine which orbitals will combine...but with integrals? Uh oh. I'm not looking forward to chemistry next year if I DO get into second year chemistry. After a LOT of studying after my last exam.

If it makes you feel better, it's actually so amazing when you start doing stuff with symmetry - really eye-opening (though pretty hard to get your head around if you don't have the mathematical background...).

I think it's fair to say that large atoms will have larger bond lengths in general.

Haha yes so do I, I meant with respect to establishing causality in the connection between bond strength and bond length, rather than in the connection between atomic radius and bond length (sorry, that probably wasn't clear).