VCE Chemistry Question Thread

[keltingmeith]

when a base such as O² and it becomes OH- what happens to the other oxygen ion.

This question makes no sense, sorry man. Potentially may be worth pushing the preview button beforehand, so you can make sure things are formatted correctly if you're not used to posting on here yet

On the assumption you meant for it to be O₂, then it depends entirely on the situation where that other oxygen goes, and I'd have to see the context of the question. The most likely explanation, however, is that they both turn into hydroxide, and the equation just isn't balanced.

If you meant O^2-, then remembered that this is actually only one ion, not two - it's just that the one ion is doubly charged.

EDIT: lol even I'm not immune to bad formatting, fixed. Also beaten by Chocolatepistachio, but leaving for the first paragraph explanation

[Coolgalbornin03Lo]

I was just thinking about it and when a calorimeters insulation is not sufficient and heat is then lost to the environment does this heat heat up the water making the temperature difference measured larger OR does this heat just get lost to the environment?

Thanks!! I’m trying to think of how insulation can make an error

[kavithag]

I was just thinking about it and when a calorimeters insulation is not sufficient and heat is then lost to the environment does this heat heat up the water making the temperature difference measured larger OR does this heat just get lost to the environment?

Thanks!! I’m trying to think of how insulation can make an error

Hi. When insulation is insufficient, heat is lost to the surrounding environment. This makes for a calibration factor that is lower than expected.

Hope that helps!

[Corey King]

Hey guys,
Year 11 student here.
I am reading about Metallic Bonding but there is something I can't find a sufficient explanation for. The textbook teaches how the valence electrons in most metals tend to be easily taken from said metals, as the atoms of said metal seek stability like found in the elemental state of noble gases.
What I don't understand is how in a solid mass of metal, made up of one metal element, the valence electrons get stripped from the localized state in each atom. What forces make this happen?
Many thanks,
Corey

[Chocolatepistachio]

For beryllium fluoride how would you draw the Lewis dot structure and is there supposed to be 2 double bonds and why does beryllium not follow the octet rule

[sweetiepi]

For beryllium fluoride how would you draw the Lewis dot structure and is there supposed to be 2 double bonds and why does beryllium not follow the octet rule

The double bonded BeF2 is valid, but the fluorine will end up having a formal charge of +1. (Formal charge is calculated by: valance -non bonding - (bonding electrons)/2).

If you only have 1 bond between the Be and F, Fluorine is much happier and has a formal charge of 0, whilst retaining 16 valence electrons total.

Beryllium isn't strong enough (electronegatively speaking) to pull electrons away from Fluorine (which is very electronegative), and is happy not to have a whole set of eight valence electrons.

Hope this helps!

[keltingmeith]

Hey guys,
Year 11 student here.
I am reading about Metallic Bonding but there is something I can't find a sufficient explanation for. The textbook teaches how the valence electrons in most metals tend to be easily taken from said metals, as the atoms of said metal seek stability like found in the elemental state of noble gases.
What I don't understand is how in a solid mass of metal, made up of one metal element, the valence electrons get stripped from the localized state in each atom. What forces make this happen?
Many thanks,
Corey

Yeah, so this is mainly a confusion between ionic and metallic bonding. Yes - in the presence of a suitable atom (called an oxidant, but that terminology won't be relevant until later), the electrons in a metal will be ripped away from the metal and it will form an ion. This ion can then participate in ionic bonding.

You're right - in the presence of another, identical (and sometimes not identical) metal, why should a metal give up its electrons? This iron atom over here doesn't want any more electrons, so why would it take them from that iron atom over there? In fact, neither of them want to hold onto their valence electrons, so what's a metal to do?? Well, the problem is, when we're talking about small numbers of metals, there's no real solution to think of. No matter what way you try to arrange those extra electrons for (say) sodium - there'll always be x extra electrons for every x atom that tries to participate in bonding. But what if we didn't just have one atom? Or two? Or even 50? What if we had a number so big, that you couldn't count the atoms even if you tried?

First, a side-note: Have you ever noticed how choirs will often sound so good they could sell out concerts, even if no individual person in the choir could perform by themselves and sell out that concert? (/think to your school choir, who probably sound okay, even though there are people in that choir you know cannot sing to save their lives. Which I say as one of those people lmfao) It's a similar effect - things behave differently in large enough groups.

Well, that's what the metals do. The atoms cluster in groups so large, that there's just moles and moles of atoms (if you don't know what a mole is yet, just think of it as a really, really, really big number). Okay, so how does that fix the problem? Because now we just have moles and moles of extra electrons! Well, because they're in a group so big, those electrons can just move around. And they move around a lot. In fact, they move around so fast, that at any moment, each atom has 3 electrons around it, but over a period of time, the atom doesn't think they exist. The explanation for that, is, much more complicated, and I can't think of a real-world equivalent to get your head around it, sorry. But, the point is that the electrons are still attached to those metals - they still have a charge of 0 instead of the +1 that a sodium ion would have - however, the electrons are now dispersed in such a way that the metal can't feel them, and it /thinks/ it's in a +1 state. So as you can see, there is no force stripping the electrons from those metals - the metals technically have 3 electrons around it at any instant, the metal just tricks itself into thinking it doesn't.

Metallic bonding is hella complicated, which is why it's glossed over in high school - and tbh, the answer I gave isn't even technically correct, but it hopefully helps you understand the "sea of electrons" that is usually the explanation given in high school (and what I remember being in the study design). The answer as to why this actually works has to do with a thing called band theory - which is still studied at third year university physics level. That's right - physics, not chemistry, because the understanding of metallic bonding is so rooted in quantum mechanics that most chemists don't have the tools to understand it. Hell - I'd be lying if I said I understood it all perfectly myself.

Hopefully this isn't too disheartening to you, but that's the nature of the beast - macroscopic science is usually quite simple, but trying to explain what happens within and between atoms? Usually very complicated, and requiring a quantum mechanical explanation. Because, you're right - the idea that a lump of metals can share electrons when there's no driving force to pull those electrons away from the metal seems just a little bit off. In fact, chemistry is famous for each year in university being told, "so remember how we told you this? Yeah, that's wrong, bonding actually works THIS way", to then being told, "oh yeah, remember how we said that? Yeah, that's also wrong it's actually this", or, "remember how we said that? Yeah, that doesn't work for these types of molecules, so instead it's actually this". It's actually a little hilarious how often it happens.

[Corey King]

Yeah, so this is mainly a confusion between ionic and metallic bonding. Yes - in the presence of a suitable atom (called an oxidant, but that terminology won't be relevant until later), the electrons in a metal will be ripped away from the metal and it will form an ion. This ion can then participate in ionic bonding.

You're right - in the presence of another, identical (and sometimes not identical) metal, why should a metal give up its electrons? This iron atom over here doesn't want any more electrons, so why would it take them from that iron atom over there? In fact, neither of them want to hold onto their valence electrons, so what's a metal to do?? Well, the problem is, when we're talking about small numbers of metals, there's no real solution to think of. No matter what way you try to arrange those extra electrons for (say) sodium - there'll always be x extra electrons for every x atom that tries to participate in bonding. But what if we didn't just have one atom? Or two? Or even 50? What if we had a number so big, that you couldn't count the atoms even if you tried?

First, a side-note: Have you ever noticed how choirs will often sound so good they could sell out concerts, even if no individual person in the choir could perform by themselves and sell out that concert? (/think to your school choir, who probably sound okay, even though there are people in that choir you know cannot sing to save their lives. Which I say as one of those people lmfao) It's a similar effect - things behave differently in large enough groups.

Well, that's what the metals do. The atoms cluster in groups so large, that there's just moles and moles of atoms (if you don't know what a mole is yet, just think of it as a really, really, really big number). Okay, so how does that fix the problem? Because now we just have moles and moles of extra electrons! Well, because they're in a group so big, those electrons can just move around. And they move around a lot. In fact, they move around so fast, that at any moment, each atom has 3 electrons around it, but over a period of time, the atom doesn't think they exist. The explanation for that, is, much more complicated, and I can't think of a real-world equivalent to get your head around it, sorry. But, the point is that the electrons are still attached to those metals - they still have a charge of 0 instead of the +1 that a sodium ion would have - however, the electrons are now dispersed in such a way that the metal can't feel them, and it /thinks/ it's in a +1 state. So as you can see, there is no force stripping the electrons from those metals - the metals technically have 3 electrons around it at any instant, the metal just tricks itself into thinking it doesn't.

Metallic bonding is hella complicated, which is why it's glossed over in high school - and tbh, the answer I gave isn't even technically correct, but it hopefully helps you understand the "sea of electrons" that is usually the explanation given in high school (and what I remember being in the study design). The answer as to why this actually works has to do with a thing called band theory - which is still studied at third year university physics level. That's right - physics, not chemistry, because the understanding of metallic bonding is so rooted in quantum mechanics that most chemists don't have the tools to understand it. Hell - I'd be lying if I said I understood it all perfectly myself.

Hopefully this isn't too disheartening to you, but that's the nature of the beast - macroscopic science is usually quite simple, but trying to explain what happens within and between atoms? Usually very complicated, and requiring a quantum mechanical explanation. Because, you're right - the idea that a lump of metals can share electrons when there's no driving force to pull those electrons away from the metal seems just a little bit off. In fact, chemistry is famous for each year in university being told, "so remember how we told you this? Yeah, that's wrong, bonding actually works THIS way", to then being told, "oh yeah, remember how we said that? Yeah, that's also wrong it's actually this", or, "remember how we said that? Yeah, that doesn't work for these types of molecules, so instead it's actually this". It's actually a little hilarious how often it happens.

Wow, thanks for the great response keltingmeith

It is a bit dissapointing to know that Im learning things about a field that the field itself believes is false. Still, one step at a time. You did give me a feel for how the 'sea' model works

[ArtyDreams]

Does anyone know into what extent we need to know about amylose and amylopectin in the 2020 study design? Only because I've been seeing different things so now I'm confused

[keltingmeith]

Wow, thanks for the great response keltingmeith

It is a bit dissapointing to know that Im learning things about a field that the field itself believes is false. Still, one step at a time. You did give me a feel for how the 'sea' model works

I mean, let me at least cover VCE's arse like this: technically ALL models are inherently false, anyway. They're based on observations we have made at the time, and are designed to cover most of those observations and explain them. This is why the modelling for COVID in NSW and VIC are different - because the virus has behaved differently in those states, and so the models are different.

The point of a model is to predict properties, and all the models you are learning from are useful in their own ways. For example, the "sea of electrons" models very accurately predicts the thermal and conductive properties of metals, as well as why they're so malleable when compared to ionic compounds. Similarly, VSEPR is really good at predicting the geometries of simple organic compounds and small, main-group polyatomic ions. It breaks down particularly for larger molecules and transition metals particularly, but that doesn't make the models you're learning about useless.

So tbh, whenever you learn a model, you're technically learning something the field "believes is false", but it's also something that the field acknowledges is useful - the question is just how useful they find it, and that depends on what type of chemistry you're doing. I know chemists who currently do work where pretty much every model learned in high school is good enough for them, and I know some where a few of the high school ones are fine, and just one of the upper uni level ones are required - and then there's research like mine were most of the stuff in high school breaks down and isn't useful for me. It's just the way the cookie crumbles, but remember - we wouldn't be wasting your time with it in high school if it wasn't useful to know or learn.

[Corey King]

I mean, let me at least cover VCE's arse like this: technically ALL models are inherently false, anyway. They're based on observations we have made at the time, and are designed to cover most of those observations and explain them. This is why the modelling for COVID in NSW and VIC are different - because the virus has behaved differently in those states, and so the models are different.

The point of a model is to predict properties, and all the models you are learning from are useful in their own ways. For example, the "sea of electrons" models very accurately predicts the thermal and conductive properties of metals, as well as why they're so malleable when compared to ionic compounds. Similarly, VSEPR is really good at predicting the geometries of simple organic compounds and small, main-group polyatomic ions. It breaks down particularly for larger molecules and transition metals particularly, but that doesn't make the models you're learning about useless.

So tbh, whenever you learn a model, you're technically learning something the field "believes is false", but it's also something that the field acknowledges is useful - the question is just how useful they find it, and that depends on what type of chemistry you're doing. I know chemists who currently do work where pretty much every model learned in high school is good enough for them, and I know some where a few of the high school ones are fine, and just one of the upper uni level ones are required - and then there's research like mine were most of the stuff in high school breaks down and isn't useful for me. It's just the way the cookie crumbles, but remember - we wouldn't be wasting your time with it in high school if it wasn't useful to know or learn.

Some great points keltingmeith, as usual

[ArtyDreams]

Just a quick question because I've been trying to figure this out for much too long - when determining the melting point of a fat do we ALWAYS look at chain length first - and then look at the double bonds if the chain length is the same?

In one of my textbooks I had to look at the boiling point of these acids and put them in increasing order.
This is what I got:
Mystric, linoleic, arachidonic, arachidic

But my textbook says:
arachidonic, linoleic, myristic, arachidic

Help would be appreciated!

[Evolio]

Hey Arty.

I asked my teacher a similar question and we look at the number of double bonds and then the carbon chain, if they are different. Since double bonds cause kinks or 'bends' in the molecule, it actually changes the shape of the molecule and so this has a higher impact on melting point. However, the length of the carbon chain doesn't change the shape. Both the length of the C chain and double bonds impact the strength of dispersion forces but only double bonds change the shape. This helped me with understanding why double bonds sort of have a 'higher priority' in determining melting points.Although, as you said if one of the features is the same, then you look at the other as the deciding factor, determining which has a higher melting point.

For example, arachidonic has 4 double bonds, which is the greatest and so it has the highest melting point.

[Coolgalbornin03Lo]

Can you use Q=It and n(e-) = Q/F only with the cathode reaction? I’ve always thought that was the case but can you use it with the anode reaction? Or is your best bet to use th cathode reaction to find the mood of electrons and then begin working on the anode reaction?

[Corey King]

Hey guys,
A question regarding electron shells and orbitals.
I didn't know the answer to the attached question [ https://gyazo.com/2a62a2263e170759a10996e40340fbab ] so I just checked a random answer.
When I went to calculate it, I assumed there would be some d-block electrons in the relevant d shell, but they weren't counted as valence electrons.
Would there be no 6- energy level d-block electrons in a neutral Bi atom?
(I am able to correctly assign electrons to their orbitals, up to Ca.)
Many thanks,
Corey