H2 Chem Questions

last SG · 2012-09-08T17:17:47+00:00

I'll only help out with the non-calculation questions. Explain how you would expect the numerical magnitude of the LE of sodium chlorate (I) to compare with...

gohby SG

In future, please post a few questions at a time. It'll help your chances of getting a reply sooner. For instance, if I've only got a few mins to spare when dropping by this forum, I would be more inclined to immediately reply a post asking about 2 or 3 qns, but might decide not to reply (if busy, perhaps for a couple days or even more) a post asking about 8 qns.

Q1. It's nothing more than a typo error in the question.

Q2. AlCl3 doesn't conduct in the molten state because it remains covalent, but undergoes hydrolysis in aqueous state to generate ions. Topic : both chemical bonding and periodicity.

Q3. Option 1 only. Alcohols are capable of both donating and accepting H bonds, but ethers can only accept H bonds.

Q4. Only propane has two different possible mono-halogenated products. The rest only have 1 possible mono-halogenated product.

Q5. It's a trick question. It's not the S atoms that accept electrons in peroxodisulfate, it is the peroxo O atoms that accept electrons.

Q6. Based on NH3 generated, both N2O and N2O4 are possible. Based on H2 used up to generate H2O, only N2O4 is possible.

Q7. As pressure tends towards zero, a real gas tends towards ideal gas behaviour (due to reduced interactive forces between gaseous particles), where PV=nRT holds true, and hence = PV/RT = n =1.

Q8. Correct. But if the question is made tougher by having a Kc value closer to 1, then a mathematical proof (using algebra and ICE table) is required to discern the best answer.

My apologies, UltimaOnline. Thank you for your help and I will take note of that!

Q3, is there an efficient schematic way of ensuring that all different functional isomers have been drawn, before one compares the intermolecular forces of attraction to compare boiling points?

Q4: Doesn't all the compound have the ability to form a mono-substituted compound with HCl too?

Q5: Could you write down the equation to illustrate?

For Q6, the answer is N2O, what's the full equation, and where do I find the general equation for the reaction between nitrogen oxide and hydrogen?

UltimaOnline SG

No problem at all, gohby.

Q4. Yes, all of them except one of the options, can only generate one particular monohalogenated product. The singled out option, can generate two different monohalogenated products. This is due to, for instance, an internal plane of symmetry, such that no matter which H atom you substitute, you end up with exactly the same (ie. has the same structure and same IUPAC name) monohalogenated product. The phrasing "a mixture of monohalogenated products" in the qn implies that the correct answer can form more than one monohalogenated product.

Q5. The reduction half equation can be balanced easily, that's not the trick. The trick is in the OS (OS = formal charge + electronegativity consideration) of the peroxo / peroxy group O atoms, which have OS of -1, and thus will gain electrons to have an OS of -2. The sulfur atoms are neither oxidized nor reduced, in redox reactions involving the peroxodisulfate(VI) ion S2O8 2-.

Q6. I didn't actually work out the answer (so ignore my earlier post about the answer being N2O4 vs 2N2O). The underlying concept is this : once you've got the moles of NH3 generated, you can find the moles of N atoms present per mole of the oxide. Presumably this would be 2. So the answer is hence N2O or N2O4. Once you got the moles of H2 in excess, you then know the moles of H2 used up, and you can then determine how many moles of O atoms are present, per mole of the oxide. Presumably this would be 1, hence the answer is N2O. If the value is 4 (ie. 4 moles of O atoms present to combine with the H2 gas reacted), then the answer is N2O4.

The candidate is expected to deduce that hydrogenating oxides will ultimately generate H2O as the only stable oxygen & hydrogen containing product, and hydrogenating inorganic nitrogen compounds will produce NH3 as the only stable nitrogen & hydrogen containing product.

My apologies, UltimaOnline. Thank you for your help and I will take note of that!

Q3, is there an efficient schematic way of ensuring that all different functional isomers have been drawn, before one compares the intermolecular forces of attraction to compare boiling points?

Q4: Doesn't all the compound have the ability to form a mono-substituted compound with HCl too?

Q5: Could you write down the equation to illustrate?

For Q6, the answer is N2O, what's the full equation, and where do I find the general equation for the reaction between nitrogen oxide and hydrogen?

gohby SG

Thank you very much for your lucid explanation, UltimaOnline, they're of great help.

For Q3, in order to analyse if CH3CH2CH2CHO has a higher bp than its functional group isomer, I believe we are looking between the bp of aldehyde and ketone right? I've read up that ketone has a higher bp than its corresponding aldehyde as it is more polar (since the presence of 2 electron-donating alkyl groups (instead of 1 in aldehydes) will intensify the δ- charge on the C=O bond). However, why doesn't the electron-donating alkyl groups lessen the δ+ charge on the C instead, thereby lessening the polarity of the C=O bond in ketones (as compared to aldehydes)?

Another qn:

Beryllium forms covalent compounds which are said to be electorn deficient. In many ways, beryllium resembles aluminium.

BeF2 reacts with NAF

Since BeF2 only has 4 valence electrons, it needs 2 dative bonds right? How can the ionic compound of NaF form dative bonds with BeF2?

Once again, thank you! :)

UltimaOnline SG

Thank you very much for your lucid explanation, UltimaOnline, they're of great help.

Another qn:

Beryllium forms covalent compounds which are said to be electorn deficient. In many ways, beryllium resembles aluminium.

BeF2 reacts with NAF

Since BeF2 only has 4 valence electrons, it needs 2 dative bonds right? How can the ionic compound of NaF form dative bonds with BeF2?

Once again, thank you! :)

Ketones are less reactive than aldehydes due to both electronics (as you mentioned) and sterics. For 'A' levels, aldehydes and ketones with the same number of electrons and same molecular size are considered to have similar boiling points, which are only due to permanent dipole - permanent dipole Keesom van der Waals (and no hydrogen bonds involved).

The solution to the riddle is : because two moles of NaF react with one mole of BeF2.

gohby SG

The solution to the riddle is : because two moles of NaF react with one mole of BeF2.

Thank you for your reply, so for an MCQ question like

Q3: Which substances have a higher boiling point than its functional group isomer?
1.
2.
3. CH3CH2CH2CHO,

how do students know if option 3 is correct/wrong?

As for the BeF2 question, how does the NaF bond with BeF2?

UltimaOnline SG

Thank you for your reply, so for an MCQ question like

Q3: Which substances have a higher boiling point than its functional group isomer?
1.
2.
3. CH3CH2CH2CHO,

how do students know if option 3 is correct/wrong?

As for the BeF2 question, how does the NaF bond with BeF2?

Either Cambridge won't ask such an unfair question, and/or they will set the options such that the candidate can reasonably eliminate an unknown 1/2/3 option based on the options A/B/C/D available.

Each of the two F- (ion) Lewis bases donates one dative bond to the B (atom) Lewis acid.

gohby SG

Thanks!! :)

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