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  • hoay's Avatar
    161 posts since Jul '11
    • http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s13_qp_11.pdf

       

      white ppt formed in first reaction that may be AgCl or AgBr or AgI or AgAt.

      ppt dissolve in concentrated aqueous ammonia means it is either Cl- or Br-. 

      But on addition of HNO3 the ppt dissolves....Can you explain what happens here? and why Cl- is present here not Br-? Has this something to do with dynamic equilibrium?

  • Moderator
    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
    UltimaOnline's Avatar
    11,434 posts since May '05
    • Originally posted by hoay:

      http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s13_qp_11.pdf

       

      white ppt formed in first reaction that may be AgCl or AgBr or AgI or AgAt.

      ppt dissolve in concentrated aqueous ammonia means it is either Cl- or Br-. 

      But on addition of HNO3 the ppt dissolves....Can you explain what happens here? and why Cl- is present here not Br-? Has this something to do with dynamic equilibrium?


      For Q17, since the ppt dissolved in dilute NH3(aq), the ppt must have been AgCl, which re-precipitated out upon addition of acid, because position of equilibrium shifts right : NH3 ligands <~> NH3 (aq) <~> NH4+(aq)

      To address your question directly, it's Cl- not Br-, because the ppt dissolved in *DILUTE* NH3(aq). If the question used concentrated aqueous ammonia, then both Cl- and Br- could be possible (ie. then the MCQ would be flawed and cannot be answered definitively, unless the color of the ppt was specified).

      Edited by UltimaOnline 03 Apr `17, 4:10AM
  • hoay's Avatar
    161 posts since Jul '11
  • Moderator
    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
    UltimaOnline's Avatar
    11,434 posts since May '05
    • Originally posted by hoay:

      thank you so much....this question disturbed me for so long.


      No prob, hoay! :)

  • hoay's Avatar
    161 posts since Jul '11
    • http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s15_qp_13.pdf

      Q.8 Why C is not the answer? As the [methanol] will be become constant when the reaction is complete.

       

      Q.28 What is X in this diagram? 

  • Moderator
    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
    UltimaOnline's Avatar
    11,434 posts since May '05
    • Originally posted by hoay:

      http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s15_qp_13.pdf

      Q.8 Why C is not the answer? As the [methanol] will be become constant when the reaction is complete.

       

      Q.28 What is X in this diagram? 


      Q8. Option C assumes that the basic hydrolysis of ester is a 0 order reaction, but it is in fact, a 2nd order reaction. Hence the correct answer is option A. Option B is characteristic of an auto-catalytic reaction, but neither product of the basic hydrolysis catalyzes the hydrolysis. Option D is just nonsense.

      Q28. X is the transition state as the C-Br bond is being broken, ie. a C-Br partial bond, a partial +ve charge on the C atom, and a partial -ve charge on the Br atom.

      Edited by UltimaOnline 04 Apr `17, 1:18AM
  • hoay's Avatar
    161 posts since Jul '11
    • Order of the reaction is not taught at first year of A-level in CIE. any other explanation?

    • Another query is from June 2016 /MCQ. I am afraid that this paper is not available in any site on the web. 

       

      Methyl isocyanate , CH3NCO, is a toxic liquid.

      What is the approximate angle between the bonds formed by the N atoms in a a molecule of methyl isocyanate?

       

      H3C- N=C=O      120 is the answer.

       

      The electron pairs around N are 3 ; there are 2 bond pairsaround N; one is lone pairs. 321....so the bond angle must deviate from the ideal situation 330...120. It would be less than 120, around 109. 

       

       

  • Moderator
    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
    UltimaOnline's Avatar
    11,434 posts since May '05
    • Originally posted by hoay:

      Order of the reaction is not taught at first year of A-level in CIE. any other explanation?


      As reaction progresses, [ester] decreases, hence rate of reaction decreases, hence you obtain the decreasing rate curve in option A.

    • Originally posted by hoay:

      Another query is from June 2016 /MCQ. I am afraid that this paper is not available in any site on the web. 

       

      Methyl isocyanate , CH3NCO, is a toxic liquid.

      What is the approximate angle between the bonds formed by the N atoms in a a molecule of methyl isocyanate?

       

      H3C- N=C=O      120 is the answer.

       

      The electron pairs around N are 3 ; there are 2 bond pairsaround N; one is lone pairs. 321....so the bond angle must deviate from the ideal situation 330...120. It would be less than 120, around 109. 

       

       


      Excellent question. There are actually 5 levels to this at-1st-glance straightforward question.

      image
      Methyl isocyanate

      At the simplest level, the N atom has 3 electron regions or charge clouds : a single bond, a double bond, and a lone pair. Hence electron geometry is expected to be trigonal planar, ie. bond angle close to 120 degrees.

      On a deeper level, the lone pair is expected to have slightly more repelling power, hence the bond angle may now be expected to decrease slightly below 120 degrees.

      On an even deeper level, the double bond is expected to have slightly more repelling power, hence the bond angle may now be expected to increase slightly back to approximately 120 degrees.

      On a way deeper level, O has the greatest electronegativity, hence the O atom withdraws strongly by induction (ie. through the sigma bond) as well as by resonance (ie. through the pi bond) from the C atom, which consequently itself becomes somewhat electron-withdrawing vis-à-vis the N atom, despite N's greater electronegativity vis-à-vis the C atom. The effect of which is to reduce the electron density of the N=C double bond, hence the bond angle may now be expected to decrease back to slightly less than 120 degrees.

      On a way, way, way deeper level, you should be able to figure out that there is a minor resonance contributor in which the lone pair on the N atom forms a 2nd pi bond with the C atom, and the pi bond between the C atom and the O atom becomes a lone pair on the O atom, ie. +ve formal charge on the N atom, -ve formal charge on the O atom. This is a minor resonance contributor, valid only because of the high electronegativity of O. As such, the lone pair resides in an orbital which is mostly sp2, but has a little unhybridized 'p' character, in order for the slight overlap of unhybridized p orbitals (of N and C) to occur and for slight delocalization of the lone pair to form a partial 2nd pi bond with the C atom, ie. the N=C bond has partial triple bond character in the resonance hybrid. As such, the bond angle may now be expected to increase to above 120 degrees, ie. the N atom is mostly sp2 hybridized, with a little sp hybridized character.

      Indeed, experimental evidence has proven the C-N-C bond angle to be approximately 125 degrees, thus concurring with my 5 points of consideration above.

      Of course, Cambridge is aware that A level students are not able to consider all these factors to elucidate as deeply as I just did, and hence if it were a non-MCQ question, Cambridge will be reasonable and accept reasonable answers on condition of reasonable rationales. For a MCQ question, Cambridge will (as they did in this MCQ) only provide clear-cut options such as A) 120 degrees B) 109.5 degrees C) 90 degrees D) 180 degrees, in which the required answer is obvious, because the A level student is expected to choose the option which gives the closest answer.

  • hoay's Avatar
    161 posts since Jul '11
  • Moderator
    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
    UltimaOnline's Avatar
    11,434 posts since May '05
    • Originally posted by hoay:

      tahnk you for Excellent explanation


      No prob, Hoay! :)

    • For the interest of those of you following this thread on CIE papers, one of the toughest questions Cambridge has ever set, is Q6 f ii & iii of the CIE A Level 2013 Nov Question Paper 43. Go check it out and have fun! ;D

    • Another delightfully tricky IQ-type question that Cambridge can be proud of, would be Q5a of the CIE A Level 2009 Nov Question Paper 42.

      I recall someone (was it Hoay?) asking me about this question on the forum some time back.

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