Hardest Chemistry MCQs ever.

[A*(a*)]

For q37, I think 3 is incorrect because, look at this this way if you just replace the D with H:

NH4Cl + NaOH > H2O + NaCl + NH3, that would be the balanced equation. However, since you have NDH3Cl + NaOD, and you already know the molar rations are all 1 to 1 for everything, how can you get the gas ND3 if you have only 2 D atoms in the reactants?

Can you please explain 18 and 39?

 

[aalmuhannadi]

Can you please explain 18 and 39?

q18: A is wrong because CH3 is bonded to NH3 (let's ignore the iodine for now as it is not directly bonded to this since it's an ionic compound). Nitrogen has 5 valence outer-shell electrons as it is in group V. 3 of those electrons are used to bond with the 3 hydrogens, the 4th one is used to bond with the carbon in the CH3 to the left of it, and the final one has been transferred over to the iodine as CH3NH3+ is a cation with a 1+ charge, meaning that 1 electron has been lost (i.e. the 5th electron from the nitrogen), meaning that all of its 5 valence electrons have been used.

B is more obviously wrong because NH4Cl is an ionic compound: NH4+ Cl- ... four of nitrogen's 5 valence electrons are used to bond with the 4 hydrogens and the final electron is transferred to the chlorine to form a chloride ion and ammonium since we know NH4+ has a 1+ charge, meaning it has to have lost 1 electron (which is it's final outer shell electron, meaning that all 5 have been used now).

I am not sure about the explanation behind D because complex ions aren't part of the AS syllabus I'm pretty sure, but I know that a dative/co-oridnate bond forms between the Ag and the NH3 so automatically you should assume that the lone pairs are used.

This leaves C: which is an ionic compound Na+ NH2 - the nitrogen has 2 electrons bonded to 2 hydrogens, and one additional electron from the sodium ion (getting it back up to a total of 4 electrons now), so the initial bonding pairs haven't moved or changed in any way.

I hope that wasn't overdoing it for an explanation

 

[aalmuhannadi]

Can you please explain 18 and 39?

q39: just a matter of recalling the organic reactions:

CH3CH2Cl + NaOH (dilute) >>> CH3CH2OH + NaCl - no ethanoic acid made

CH3CO2CH3 + NaOH (dilute) >>> CH3CO2Na + CH3OH >>> after distilling off the alcohol, 'acidification' > add dilute HCl and the carboxylic acid is formed (CH3CO2H - ethanoic acid in this case) which is then distilled off

CH3CN + NaOH (dilute) + H2O >>> CH3CO2Na + NH3 >>> same as above, after 'acidification' > addition of HCl, ethanoic acid would be made which is then distilled off.

 

[Deedee55]

I solved whole past papers book from 01 to 12 and found these questions seriously hard!
If anyone good in chemistry help me solving them out, so that my past papers could come to an end.

Some of them are not really that hard. If you practiced the whole lot from 01 to 12 you are covered for chemistry MCQ's

 

[shafayat]

View attachment 24352

C.. I thinkso its C

 

[A*(a*)]

q18: A is wrong because CH3 is bonded to NH3 (let's ignore the iodine for now as it is not directly bonded to this since it's an ionic compound). Nitrogen has 5 valence outer-shell electrons as it is in group V. 3 of those electrons are used to bond with the 3 hydrogens, the 4th one is used to bond with the carbon in the CH3 to the left of it, and the final one has been transferred over to the iodine as CH3NH3+ is a cation with a 1+ charge, meaning that 1 electron has been lost (i.e. the 5th electron from the nitrogen), meaning that all of its 5 valence electrons have been used.

B is more obviously wrong because NH4Cl is an ionic compound: NH4+ Cl- ... four of nitrogen's 5 valence electrons are used to bond with the 4 hydrogens and the final electron is transferred to the chlorine to form a chloride ion and ammonium since we know NH4+ has a 1+ charge, meaning it has to have lost 1 electron (which is it's final outer shell electron, meaning that all 5 have been used now).

I am not sure about the explanation behind D because complex ions aren't part of the AS syllabus I'm pretty sure, but I know that a dative/co-oridnate bond forms between the Ag and the NH3 so automatically you should assume that the lone pairs are used.

This leaves C: which is an ionic compound Na+ NH2 - the nitrogen has 2 electrons bonded to 2 hydrogens, and one additional electron from the sodium ion (getting it back up to a total of 4 electrons now), so the initial bonding pairs haven't moved or changed in any way.

I hope that wasn't overdoing it for an explanation

thank youu so mucch May God bless you!
this is all what i thought but isn't in all the options except C the Nitrogen atom doesn't involve the non-bonding pair(lone pair) of electrons? Or Am i not getting the question right? :/

 

[A*(a*)]

q39: just a matter of recalling the organic reactions:

CH3CH2Cl + NaOH (dilute) >>> CH3CH2OH + NaCl - no ethanoic acid made

CH3CO2CH3 + NaOH (dilute) >>> CH3CO2Na + CH3OH >>> after distilling off the alcohol, 'acidification' > add dilute HCl and the carboxylic acid is formed (CH3CO2H - ethanoic acid in this case) which is then distilled off

CH3CN + NaOH (dilute) + H2O >>> CH3CO2Na + NH3 >>> same as above, after 'acidification' > addition of HCl, ethanoic acid would be made which is then distilled off.

So you are saying that Na+ can be replaced by H+ in acidification? Which reaction is this?

 

[A*(a*)]

Some of them are not really that hard. If you practiced the whole lot from 01 to 12 you are covered for chemistry MCQ's

I've already asked my teacher about rest of the confustions in the slot, I am clearing the rest ones, some of these MCQs were, in real, not that hard, I agree. But I had to write this to draw attention of people

 

[aalmuhannadi]

So you are saying that Na+ can be replaced by H+ in acidification? Which reaction is this?

Yes in q18 the answer is C you're right, I was explaining how C is the only one where the lone pair isn't used. In q39, the replacement of Na+ with H+ isn't really a reaction, it's just a step in the process of isolating the carboxylic acid if a base is used to convert ester > acid + alcohol or nitrile > acid + ammonia for example, I don't think you'd be asked for the reaction in that step of the process

 

[A*(a*)]

Yes in q18 the answer is C you're right, I was explaining how C is the only one where the lone pair isn't used. In q39, the replacement of Na+ with H+ isn't really a reaction, it's just a step in the process of isolating the carboxylic acid if a base is used to convert ester > acid + alcohol or nitrile > acid + ammonia for example, I don't think you'd be asked for the reaction in that step of the process

everything cleareed!! I can't thankyou enough
SO the thing to remember is that salt can be converted to acid by acidification. I didn't see any such thing in coursebook, thanks for tellin me

 

[aalmuhannadi]

everything cleareed!! I can't thankyou enough
SO the thing to remember is that salt can be converted to acid by acidification. I didn't see any such thing in coursebook, thanks for tellin me

No problem happy to help yeah I didn't get it from the cambridge coursebook I study from the revision guide and from my own notes taken from the chemguide website, so that's where you could find the acidification part hopefully mentioned

 

[NabzJ]

Q9: from the question it says 50cm^3 of 0.1moldm^-3 of a metallic salt react exactly with 25cm^3 of 0.1moldm^-3 sodium sulfite

The moles of each is therefore: n = cv = 0.05 * 0.1 = 0.005 mol of the salt and 0.025 * 0.1 = 0.0025 mol of sulfite ions, therefore from this we can now say that 1 mol of the sulfite ions react 'exactly' with 2 mol of the metallic salt. Therefore, if, from the given equation, 2 electrons are lost form the oxidation of the sulfite ions, then it is only logical that half the amount of electrons is gained by the metal (due to the molar ratio I just talked about), so if sulfur is oxidised from +4 to +6, then the metal should be reduced by half the amount, which is from +3 to +2. Hope that helps and please correct me if I'm wrong, or ask me to explain it if you don't get it

Hey,why would youhalf the amount , shouldnt you double it ? Ratio 1:2...

 

[aalmuhannadi]

Hey,why would youhalf the amount , shouldnt you double it ? Ratio 1:2...

No because if there are twice as many metal ions, then if the 2 electrons lost from the sulfite are transferred to the metal ions, each metal ion would only gain 1 electron (since there are two of the metal ions), and hence the oxidation number of each metal ion decreases by one