Help! Chemistry?!

Hello,
Please help me on these questions, i cant work out how why the answers are what they are?! My exams on tuesday and i'm starting to panic! :eek:

Thank You

Q1
In What species does Vandium have the highest oxidation state?
a,VOF3
b,V(OH)2
c,VCL4
d,VSO4.7H20 (why is it not this one??)

Q2
Phosphoric acid undergoes partial dissociation. The extent of dissociation at equilibrium could be increased by the addition of
a, sodium hydroxide
b, sulphuric acid
c, a catalyst
d, sodium dihydrogenphosphate

Q3
A white crystalline compound soluble in water, was found to react with both dilute hydrochloric acid and sodium hydroxide solution. Which of the following might have it been?
a, aminobenzene
b, aminoethanoic acid
c, ethylamine
d, ethylammonium chloride

Q4
A compound X, reduced hot copper(ii)oxide and the organic product dissolves in water forming a neutral solution
X could be
a, an aldehyde
b, a ketone
c, a secondary alcohol
d, a tertiary alcohol

Reply 1

Revenged

Sheilz001

Hello,
Please help me on these questions, i cant work out how why the answers are what they are?! My exams on tuesday and i'm starting to panic! :eek:

Thank You

Q1
In What species does Vandium have the highest oxidation state?
a,VOF3
b,V(OH)2
c,VCL4
d,VSO4.7H20 (why is it not this one??)

1d) isn't the answer because H20 has no oxidation state.

Therefore the oxidation state of vanadium is +2 because the oxidation state of the sulphate ion is -2

The highest oxidation state is a) where vanadium is +5. In b) the oxidation state is +2 and in c) the oxidation state is +4

Reply 2

CrazyChemist

Okay, it's a long post (understatement!) but it should help.

Q1

oxidation states: Vanadium is

a) +5
b) +2
c) +4 (btw, VCl4 plz - VCL4 is a little confusing. Just a little detail)
d) +2

To work these out, remember that the overall molecule has zero charge. You could get a question on ions with charges, in which case that charge has to be taken into account. So for example, Q1.a) you have Vanadium, Oxygen and 3 Fluorines. Oxygen has an oxidation state of -2 and Fluorine of -1. This makes an overall charge of -5, which must be balanced by the Vanadium oxidation state, so it must be +5.

With an ion, the vanadium (or other element - usually a transition metal) must balance the overall charge so that it equals the charge of the ion. So if the ion VOF₂^3- were given (don't know if it exists, but doesn't matter), the charge on the vanadium would be +1.

In d), you can ignore the .H₂O - this is the water of hydration I think it's called, and doesn't affect the oxidation state, so you only have to worry about the SO₄^2-, so Vanadium ends up as +2

A note on oxidation states: you always write the sign then the magnitude: -1, +2, -3 etc NOT 1-, 2+, 3-. This will lose you marks (lost me a few once anyway).

Q2

This is talking about the equilibrium HA [reversibly gives] H⁺ + A^-
where H⁺ is hydrogen and A^- is the rest of the acid (the anion)

Dissociating is talking about how much the acid goes into solution, and releases H⁺. Think of Le Chatelier's principle - to get the equilibrium to move further to the right (further dissociate) you must either take away products or add more reactants. The only option here that takes away products is to add alkali - this reacts with the H⁺ ions and takes them out of solution, so the equilibrium is driven to the right and the acid further dissociates.

Adding an acid would put more H⁺ into the solution, so drives the equilibrium back towards the reactants. A catalyst may cause a system to reach equilibrium FASTER, but it DOES NOT change where that equilibrium lies - important point. I’m not sure what option d) is - probably a red herring.

Q3

The answer here is c).

Aminoethanoic acid has an NH₂ group, which would react with the acid (the H⁺ from the acid bonds datively to the free pair of electrons on the N), and the COOH group would become COO^-Na⁺ (standard neutralisation reaction - the H⁺ is donated to make water with the OH^- from the alkali, leaving the salt)

The amino benzene will react with acid (same mechanism as before) but not alkali.

Ethylamine is just C-C-NH₂, so it will react with an acid and not an alkali - the same as amino benzene, and with the same mechanism.

Ethylammonium chloride is, I think, C-C-NHCl. I'm not sure what properties the Cl bestows on the molecule, but I think it would still react with an acid by the same mechanism. Would it react with an alkali? Hmmm...I suspect not because I don't think an OH^- ion would displace a Cl^- ion. This is a bit I’m gonna have to revise again.

Q4

Right, here we have a compound that reduces another, so it is itself oxidised. The only compounds here that can be oxidised are the aldehyde and the secondary alcohol. This is because neither of the others have a carbon that will lose a hydrogen in oxidation. Why? I would have to draw structures, and I'm not sure I can do that here.

The organic product means the 'main' product - they specify organic because there's often water or HCl or another inorganic product, which they want you to ignore.

If you oxidise a secondary alcohol, you get a ketone, which i'm pretty sure is not soluble in water, and if you oxidise an aldehyde you get an acid, which IS soluble in water, so the answer is c).

So there you go - I'm pretty sure about this lot, and good luck in the exam :smile:

Reply 3

Sheilz001

Thank You!!! :biggrin:

I think i will need all the luck i can get :redface:

Reply 4

glance

Good answer, CrazyChemist. Just to say, the .H20 is waters of crystallisation, I think. Q4 has to be an aldehyde as a ketone won't reduce the Benedicts reagents. Good luck!

Reply 5

glance

I'm not sure you can oxidise a secondary alcohol by adding Copper (II) oxide, can you? I'm pretty sure you have to perform an oxidation reaction involving sodium dichromate and high temperatures?

Reply 6

corkskrew

sodium not neccessary. acidified dichromate ions under high temp. refluxed. colour change is orange to green.

Reply 7

Grafter

Good answer to 1 and 2 cc, but im not so sure about the others, so the answers to 3 and 4;

A3) b

Aminoethanoic acid has both an acid and a base functionality, so it can react with both acids and bases. It also forms a zwitterion (cf. amino acids), which allows it to form the white crystalline, water soluble solid.

A4) a

This is referring to the fehlings/benedicts test for aldehydes. As the copper (II) is reduced, the aldehyde is oxidised to the corresponding carboxylic acid salt, and as such is water-soluble.

Good luck with exams.

Grafter

Reply 8

CrazyChemist

rats - another stupid mistake. in answer to Q3 i said aminoethanoic acid, but put c) as the answer, which is wrong...that's me in a nutshell that is...

just noticed that actually i've done the same for Q4 as well - i said the answer was an aldehyde, but said that the answer was c) :redface:

how dumb can you get?? oh dear...