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BedokFunland JC's A Level H2 Chemistry Qns (Part 1)

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    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
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    • 'A' Level Qn.

      by Dr Rod Beavon (http://home.clara.net/rod.beavon)



      Organic Chemistry Problem7



      A, C2H4, reacts with hydrogen bromide in the gas phase to give B, C2H5Br. This reacts with aqueous sodium hydroxide solution to give C, C2H6O, which can be oxidised by acidified potassium dichromate solution to give successively D, C2H4O, and E, C2H4O2. D reacts with both 2,4-dinitrophenylhydrazine and ammoniacal silver nitrate.

      With magnesium in dry ether B gives F, C2H5MgBr; this reagent will react with D to give G, C4H10O. Oxidation of this compound with acidified potassium dichromate solution gives H, C4H8O. This will also react with F, giving I, C6H14O. I is not oxidised by acidified potassium dichromate solution.

      If A is reacted with bromine in the presence of sodium chloride, two compounds are obtained; J, C2H4Br2, and K, C2H4BrCl. No C2H4Cl2 is formed.

      (a) Identify each of the compounds A – K, and explain all the observations as fully as you can.

      (b) Why is no C2H4Cl2 formed in the reaction given in the last paragraph?


      Solution :


      Edited by UltimaOnline 02 Feb `09, 1:02PM
    • 'A' Level Qn.

      by Dr Rod Beavon (http://home.clara.net/rod.beavon)



      Organic Chemistry Problem8



      Compound A, C4H8O, reacts with a solution of iodine in sodium hydroxide to give B, C3H5O2Na, and CHI3. A gives a precipitate with 2,4-dinitrophenylhydrazine but no reaction with ammoniacal silver nitrate.

      A can be reduced with lithium aluminium hydride in dry ether to give C, C4H10O, which is chiral. With sodium chloride in 50% sulphuric acid C gives D, C4H9Cl. With magnesium in dry ether D gives G, C4H9MgCl; G will react with solid carbon dioxide to give an intermediate which when treated with dilute hydrochloric acid gives H, C5H10O2.

      If D is heated with potassium hydroxide dissolved in ethanol, three compounds are formed; E is the major product and has two stereoisomers, and F is the minor product. Both varieties of E, and F, are C4H8.

      C reacts with H in the presence of sulphuric acid catalyst to give I, C9H18O2.



      Solution :


      Edited by UltimaOnline 02 Feb `09, 1:05PM
    • 'A' Level Qn.

      by Dr Rod Beavon (http://home.clara.net/rod.beavon)



      Organic Chemistry Problem9



      Compound A, a solid, C9H8O2 effervesces when added to a solution of sodium hydrogen carbonate (bicarbonate). It reacts immediately with bromine water, decolourising it, with the product being B, C9H8O2Br2. With aqueous sodium hydroxide solution B gives the sodium salt of C, C9H10O4; this compound can be oxidised by acidified potassium dichromate solution to D C9H6O4. D reacts with 2,4-dinitrophenylhydrazine but not with ammoniacal silver nitrate solution. D gives steamy fumes with phosphorus pentachloride, the organic product being E, C9H5O3Cl.

      With ammonia E gives F, C9H7O3N.

      Both G and A are oxidised by hot aqueous potassium manganate(VII) solution to the sodium salt of G, C7H6O2.

      Identify compounds A to G, explaining all the observations as fully as possible.



      Solution :


      Edited by UltimaOnline 05 Feb `09, 12:51AM
    • 'A' Level Qn.


      Q1) A student tried to synthesize para-methylphenylamine a.k.a 1-amino-4-methyl-benzene from phenylamine, chloromethane, and aluminium chloride. He ended up with another product instead. By drawing the mechanism, identify the end product obtained, and explain why his attempt failed.


      Q2) The same recalcitrant student (bless his soul) also attempted to synthesize propylbenzene from benzene, 1-bromobutane, and iron(III) bromide. He ended up with 1-methylethyl-1-benzene a.k.a. phenyl-1-methylethane. By drawing the mechanism, explain why his attempt failed.



      Solutions (partial; hints given; mechanisms not shown) :


      Q1) The amino group of phenylamine is more nucleophilic/basic than the benzene ring. There are 2 possible products

      (i) one from the Lewis acid-base reaction between AlCl3 and phenylamine, and

      (ii) the other from the Lewis acid-base reaction between phenylamine and AlCl4- (from a precursor Lewis acid-base reaction between chloromethane and AlCl3)


      Q2) Hydride shift carbocation rearrangement had occurred in the carbocation electrophile; this is due to the fact that a secondary carbocation is significantly more stable than a primary carbocation.  


      Edited by UltimaOnline 18 Mar `09, 11:50AM
    • 'A' Level Qn.


      Q1a) Why are ionic oxides basic, and covalent oxides acidic?

      Q1b) Illustrate your answer in (a) by drawing the mechanism for the hydrolysis of the oxide anion to produce the hydroxide ion.

      Q1c) Illustrate your answer in (a) by drawing the mechanism for the hydrolysis of sulfur trioxide to produce sulfuric(VI) acid.


      Solution :


      Ionic oxides contain the O2- ion. This is a strongly basic ion which reacts with water to produce hydroxide ions:

      O2-(aq) + H2O(l) à 2OH-(aq)


      Thus all ionic oxides are BASIC.


      Covalent oxides do not contain ions, but have a strongly positive dipole on the atom which is not oxygen. This attracts the lone pair on water molecules, releasing H ions:


      MO(s) + H2O(l) à MO(OH)-(aq)+ H+(aq)


      Thus all covalent oxides are ACIDIC.


      Intermediate oxides can react in either of the above ways, depending on the conditions. They can thus behave as either acids or bases and are thus AMPHOTERIC.


      (Acknowledgement : http://www.a-levelchemistry.co.uk)




      Mechanism for hydrolysis of oxide anion to produce hydroxide anion :


      1) Draw the Kekule structure of dinegative oxide anion and water molecule.

      2) Oxide anion acts as a base to abstract a proton from the water molecule (draw curved arrow from lone pair of oxide anion to H of water molecule).

      3) Check formal charges. Negative formal charge on the oxygens of both uninegative anions.


      Mechanism for hydrolysis of sulfur trioxide to produce sulfuric(VI) acid :

      1) Draw the Kekule structure of sulfur trioxide. Check your formal charges.

      2) Depending on your choice of resonance contributor, the sulfur atom is either positively formal-charged (+ or 2+), or strongly positively partial-charged (due to bonding with 3 electronegative O atoms).

      3) This invites nucleophilic attack by the available lone pair on the oxgen of a water molecule.

      4) Proton transfer occurs intramolecularly, to eliminate positive formal charge on the oxygen of the water molecule (which attacked in step #3 above).

      5) We end up with a sulfuric(VI) acid molecule (of which the 1st proton is strongly acidic and the second proton is weakly acidic). Check formal charges (overall charge of protonated conjugate acid form is zero).




      Edited by UltimaOnline 06 Feb `09, 4:48PM
    • 'A' Level Qn.



      Qn. Draw the Kekule structures of the following sulfur compounds, and work out the oxidation states (O.S.) of the sulfur atoms within. Explain the relationship between formal charges, electronegativity and oxidation states.


      sulfur dioxide hydrolyzed to sulfate(IV) ion SO3 2-

      sulfur trioxide hydrolyzed to sulfate(VI) ion SO4 2-

      thiosulfate ion S2O3 2-

      tetrathionate ion S4O6 2-

      peroxydisulfate(VI) ion S2O8 2-



      Solution (partial) :


      In the 'O' levels, using simple algebra, you merely obtain the average O.S. of an element in a specified compound. Eg. +2.5 for O.S. of sulfur in tetrathionate ion.


      In the 'A' levels, by drawing the Kekule structures of the ions, you see that the O.S. actually takes into consideration the formal charge as well as electronegativity (ie. polarity of covalent bonds), of the individual atoms. Eg. the O.S. of the 4 sulfur atoms in tetrathionate ions are +5, 0, 0, +5; which gives you an average of +2.5 (the 'O' level answer).



      O.S. of sulfur in the following sulfur containing compounds are :


      sulfur dioxide hydrolyzed to sulfate(IV) ion SO3 2- ----> +4


      sulfur trioxide hydrolyzed to sulfate(VI) ion SO4 2- ----> +6


      thiosulfate ion S2O3 2- ----> +2 (average of), +4 and 0.


      tetrathionate ion S4O6 2- ----> +2.5 (average of), +5, 0, 0, +5.


      peroxydisulfate(VI) ion S2O8 2- ----> +6

      (6 oxygen atoms have O.S. of -2 each, the 2 oxygen atoms of the peroxy group have O.S. of -1 each).


      If you're unsure of your Kekule structures (including resonance contributors and formal charges) for the ions above, google them out to verify.

      Edited by UltimaOnline 08 Feb `09, 2:07AM
    • 'A' Level Qn.


      Elucidate the structure of a compound C6H12 that can be oxidized to give 2 isomeric products with formula C3H6O.



      Solution (Hint) :

      The two (functional group) structural isomers with molecular formula C3H6O, are propanone and propanal. So it's obvious what the alkene is.




      Edited by UltimaOnline 10 Feb `09, 7:56PM
    • 'A' Level O.C. (Organic Chem) Qn.



      Qn. If you have 3-methylbenzoic acid reacting with the conc. nitric acid and conc. sulphuric acid, where will the NO2 group be substituted?



      Solution :


      The carboyxl group is deactivating and meta directing, due to a partial positive charge on its C atom. The methyl group is electron donating by induction and activating, and is hence ortho and para directing.

      Activating directors outweigh (ie. are more influential than) deactivating directors.

      Therefore, in this case, the nitronium cation, a sufficiently electrophilic species,will substitute away a proton on a position ortho or para to the methyl group.

      Considering that the carboxyl group presents a greater steric hinderance as compared to the methyl group, the major product will be the nitro group substituting ortho to the methyl group (ie. 4th C on the benzene), and the minor product will be the nitro group substituting para to the methyl group (ie. 6th C on the benzene).

      Edited by UltimaOnline 15 Feb `09, 3:37PM
    • _

      Edited by UltimaOnline 11 Oct `17, 11:29PM
    • 'A' levels Qn ('O' levels optional)


      Q1) Draw the Kekule and/or dot-&-cross structure to represent sodium nitrate(V).

      Q2) Draw the Kekule and/or dot-&-cross structure to represent carbon monoxide.



      Solution :


      For 'O' level students, you have a choice - either bo hiu / bo chap such questions, and don't bother about drawing structures of compounds that you won't be asked to draw at 'O' levels (assuming you *are* at 'O' levels... are you?), or if you want to learn them, then self-educate (this is the age of the internet - google be thy sword, wikipedia be thy shield) yourself about Kekule structures, Lewis structures, lone pairs of electrons, bond pairs of electrons, dative covalent bonding, electronegativity, electron orbital hybridization, etc.


      You can also easily google the Kekule and/or dot-and-cross structures of any compound you want. But it makes more sense, if you learn and understand the underlying principles, so you can make better sense of any structure you google out.


      You may or may not understand the following (its ok not to understand if you're 'O' levels (since it's not required at 'O' levels), but if you're 'A' levels, then you definitely must know how to draw the Kekule and dot-and-cross structures of polyatomic ions like nitrate(V) ion), which is the reasoning and thought process that 'A' level students have to do go through, in drawing structures like the polyatomic nitrate(V) anion.


      To draw the nitrate(V) or (NO3)- ion (it's recommended the student draw the Kekule structure first, then translate to dot-&-cross notation) :

      1) First draw the central atom, N.

      2) Then draw 3 O atoms, single bonded to it.

      3) The fact that the anion is uninegative, means at least one of the O atoms must be negatively formal charged.

      4) In other words, that O atom must have 3 lone pairs and one bond pair.

      5) Hence, try drawing 1 single bonded O, and the 2 other O atoms be double bonded to N.

      6) But in which case, N has violated its octet. Being in period 2, it does not have empty 3d orbitals with which to expand its octet.

      7) Consequently, you realize that 2 of the O atoms must be single bonded (ie. each with 3 lone pairs and 1 bond pair).

      8) Consequently in turn, you realize that the N atom must have a postive formal charge, because

      8ai) you have 4 bond pairs for N, ie. 4 valence electrons near the nucleus of N, and indeed, N is in Grp V (ie. 5 valence electrons for no charge); so the N atom should indeed have a +ve formal charge.

      8aii) the entire anion is uninegative, and since 2 of the O atoms have a -ve charge; so to cancel out the 2nd -ve charge, the N atom should indeed have a +ve formal charge.

      8b) Since (ai) and (aii) agree with each other, this supports the validity of your Kekule structure.

      9) Your N atom has trigonal planar electron geometry with sp2 electron orbital hybridization (the unhybridized p orbital of N overlaps with a p orbital from the double bonded O atom, to form the pi bond).


      Once you've drawn your NO3- anion, with individual formal charges inside the brackets, and a overall uninegative charge outside the square brackets; draw the unipositive Na+ cation, also with square brackets. The electrostatic attraction between the cation and anion represent ionic bonding. So obviously, sodium nitrate(V) has ionic bonding between the cation and anion, but covalent bonding within the anion.




      For carbon monoxide.


      The dative (ie. "donated") covalent bond is from oxygen to carbon. Meaning, one lone pair from oxygen becomes a bond pair between oxygen and carbon.


      Your Kekule or dot-&-cross structure should show :

      Carbon having 1 lone pair and 3 bond pairs.

      Oxygen having 1 lone pair and 3 bond pairs.

      Hence, carbon has a -ve formal charge, and oxygen has a +ve formal charge.

      This is due to the dative (ie. "donated") covalent electron bond pair; oxygen has lost an electron, carbon has gained an electron.


      The reason why the dative bond occurs or exists, is to allow both C and O to have a stable octet.


      For physical chemistry (but not organic chemistry*), reflect the nature of the dative bond with an arrow-headed line (as opposed to a normal line for a normal or non-dative covalent bond), for the Kekule structure.


      For the dot-&-cross structure, the dative bond should be represented as xx or oo (where x or o represent electrons from oxygen), as opposed to xo or ox for a normal or non-dative covalent bond.


      (* In O.C., curved arrows represent electron flow mechanism. And once the dative bond is formed, it is no different from any other covalent bond. Hence in O.C., you only show the formation of the dative bond using curved arrows; thereafter, normal lines to represent the bonds.

      In P.C., the arrow-headed line to represent the dative bond, is actually the mechanism for the formation of the dative bond. If in the exam, you're unsure which diagram to submit (ie. you're unsure whether it's an O.C. or P.C. type question), do what the intelligent exam candidate would do - draw both alternative answers or forms, as long as both are correct, you'll get your marks.)




      Hope you had fun. Everyone loves Chemistry! angel.png

      Edited by UltimaOnline 16 Feb `09, 11:05PM
    • 'A' Levels Qn (originally 'O' Level Qn) 


      >>> When chlorine is bubbled through a solution of iodine in hot aqueous sodium hydroxide, the two halogens react in the Cl2 : I2 ratio of 7 : 1, forming a white ppt A and a solution of sodium chloride.

      A has the following composition by mass:

      Na: 16.9%   ;   H: 1.1%   ;   I: 46.7%   ;   O: 35.3%

      Calculate the empirical formula of A and thus deduce the balanced equation for the reaction. <<<



      UltimaOnline's Solution and Enlightening Comments :


      The original question, if followed strictly, is a simple 'O' levels question.


      However, for those of you who are intrigued, fascinated and intellectually excited by this wonderfully mysterious chemical reaction, and are eager to more deeply understand the process proper, I'll describe what actually happens :


      The iodine undergoes disproportionation with hot aqueous sodium hydroxide, forming iodate(V) ions.


      Only participating ions :

      6l2 + 12OH- --> 10l- + 2lO3- + 6H2O


      Including spectator ions :

      6l2 + 12NaOH --> 10Nal + 2NalO3 + 6H2O



      The excess iodide ions are oxidized by chlorine.


      Only participating ions :

      10I- + 5Cl2 --> 10Cl- + 5I2


      Including spectator ions :

      10NaI + 5Cl2 --> 10NaCl + 5I2



      The iodate(V) ions are subsequently and simultaneously hydrolyzed (ie. nucleophilically attacked) by hydroxide ions and oxidized by chlorine (transfer of electrons), into a partially protonated form of the iodate(VII) ion.


      Watch carefully what happens, many students may find the following mechanism somewhat tricky. The iodate(V) ion has a central I atom, 2 double bonded O atoms, 1 single bonded O atom with a -1 formal charge. The central I atom has 5 bond pairs and 1 lone pair (no formal charge). When 3 hydroxide nucleophiles attack, the I atom now has 3 additional electrons (hence 3- formal charge). To be precise, it now has 2 double bonds and 4 single bonds. One of the pi bond pairs, becomes a lone pair on the O atom, giving it a -1 formal charge. All in all, we have a 2- formal charge on I (7 bond pairs, 1 lone pair) and 2 O atoms each with a -1 formal charge. Here's where the chlorine comes in. It glady accepts (reduction potential of +1.36V) the lone pair on the I atom, thereby removing its -2 formal charge, and is itself reduced to Cl- ion. Summarily, we obtain the partially protonated form of the iodate(VII) ion.


      Only participating ions :

      2IO3- + 6OH- + 2Cl2 -> 2(H3IO6)2- + 4Cl-


      Including spectator ions :

      2NaIO3 + 6NaOH + 2Cl2 -> 2Na2H3IO6 + 4NaCl



      Additional information on the fully protonated iodate(VII) ion, that may also be obtained from iodic(VII) acid :

      H5IO6 = HIO4.2H2O <---- iodic(VII) acid (central I atom double bonded to 3 O atoms, single bonded to OH) hydrated with 2 water molecules that nucleophilically attack and dative bond to the partial positively charged I atom; 2 proton transfers ensue, from the 2 water molecules to 2 double bonded O atoms; end product molecule (with central I atom) has 5 acidic protons at the end of 5 single bonded O atoms, and 1 double bonded O atom without acidic proton. Summarily, HIO4.2H2O becomes H5IO6, known as ortho-per-iodic acid, or hexa-oxo-iodic(VII) acid, or penta-hydro-oxido-iodine.



      Therefore, the correct balanced equation is obtained by combining the following equations that occur stepwise :


      6l2 + 12NaOH --> 10Nal + 2NalO3 + 6H2O

      10NaI + 5Cl2 --> 10NaCl + 5I2

      2NaIO3 + 6NaOH + 2Cl2 -> 2Na2H3IO6 + 4NaCl


      We obtain a summary balanced equation as follows :


      (6-5)I2 + (12+6)NaOH + (5+2)Cl2 --> 2Na2H3IO6 + (10+4)NaCl + 6H2O


      I2 + 18NaOH + 7Cl2 --> 2Na2H3IO6 + 14NaCl + 6H2O 




      angel.png We Love Chemistry!!! angel.png

    • 'A' Level Qn (Supplementary; not required in H2/H3/H1 syllabuses).



      Qn. Draw the mechanisms to explain the following disproportionation reactions :


      a) cold aqueous NaOH at 15 deg C

      Cl2 + 2NaOH --> NaCl + NaClO + H2O


      3Cl2 + 6NaOH --> 3NaCl + 3NaClO + 3H2O


      b) hot aqueous NaOH at 70 deg C

      Substitute 3ClO- --> 2Cl- + ClO3- into

      3Cl2 + 6NaOH --> 3NaCl + 3NaClO + 3H2O to obtain

      3Cl2 + 6NaOH --> 5NaCl + NaClO3 + 3H2O



      Solution :


      The mechanisms are as follows :


      a) cold aqueous NaOH at 15 deg C

      Cl2 + 2NaOH --> NaCl + NaClO + H2O


      3Cl2 + 6NaOH --> 3NaCl + 3NaClO + 3H2O


      1) A hydroxide ion nucleophile attacks Cl-Cl


      2) Heterolytic cleavage or fission occurs - bond pair between O and Cl, becomes lone pair on Cl leaving group to produce Cl- ion


      3) Proton transfer (acid-base reaction) from Cl-O-H to hydroxide ion, forming the chlorate(I) ion (aka hypochlorite ion) and water.



      b) hot aqueous NaOH at 70 deg C

      Substitute 3ClO- --> 2Cl- + ClO3- into

      3Cl2 + 6NaOH --> 3NaCl + 3NaClO + 3H2O to obtain

      3Cl2 + 6NaOH --> 5NaCl + NaClO3 + 3H2O


      The following steps occur simultaneously (not consecutively; oxygen has no empty d orbitals and cannot violate its octet) when in close proximity and with sufficient activation energy (hence high temperature needed) to overcome repulsion between the anions :


      1) Double (and non-dative) covalent bonds form between the O atoms of 2 chlorate(I) ions and the Cl atom of the 3rd chlorate(I) ion, to achieve stable octets for all atoms. (To represent formation of non-dative double bonds, draw 2 pairs of 2 single-hook arrows coming from opposite directions meeting in the middle; similar to 2 free radicals bonding in a termination step)


      2) To avoid violating oxygens' octets, heterolytic cleavage or fission occurs on the 2 chlorate(I) ions - the bond pairs between O and Cl, become lone pairs on the 2 Cl leaving groups, to produce 2Cl- ions.


      (Side note : Cl- ions are stable (hence excellent leaving groups; stability of conjugate base explains why HCl is a strong acid) because chlorine is relatively highly electronegative, and Cl- ion has a relatively large ionic radius to stablize the negative charge.)

      Edited by UltimaOnline 18 Feb `09, 11:18AM
    • _

      Edited by UltimaOnline 11 Oct `17, 11:30PM
    • 'O' Level & 'A' Level Qn.


      You have two bottles of acid, one containing concentrated hydrochloric acid, the other concentrated nitric(V) acid. You take two out coins, both made of the same copper alloy, and proceed to pour some acid, from the 1st bottle onto the 1st coin, and from the 2nd bottle onto the 2nd coin. For the 1st coin, no visible reaction was observed. For the 2nd coin, a blue-green solution is formed, together with some red-brown fumes. Identify the 1st bottle acid, the 2nd bottle acid, the blue-green solution, and the red-brown fumes.


      Solution :

      1st bottle acid is hydrochloric acid.

      2nd bottle acid is nitric acid.

      Copper, being unreactive, does not participate in a redox reaction with acidic protons (H+ ions). Instead it participates in a redox reaction with nitrate(V) ions, a powerful oxidizing agent. (Note that chloride ions from hydrochloric acid, is not an oxidizing agent, it is a very weak reducing agent, and copper metal is already in its reduced state.)

      Copper atoms are oxidized to copper(II) ions, hence the blue-green solution is copper(II) nitrate. Nitrate(V) ions are reduced to nitrogen dioxide gas, the red-brown fumes observed.




      'A' Level Qn.


      Gold is an unreactive metal. Concentrated nitric acid does not react visibly with gold. Concentrated hydrochloric acid does not react visibly with gold. But a 1:3 volumetric mixture of concentrated nitric acid and concentrated hydrochloric acid, reacts visibly with gold to dissolve it. Explain.


      Solution :

      Nitric acid provides nitrate(V) ions to oxidize gold (Au) to aurate(III) ions (Au3+) in an equilibrium reaction; while hydrochloric acid provides chloride ligands to form the soluble tetrachloroaurate(III) complex ion. The 2nd (ligation/coordination) reaction causes the position of equilibrium for the 1st (redox) reaction to shift towards the right; hence gold dissolves. For the full details, visit :


      Edited by UltimaOnline 24 Feb `09, 11:35PM
    • 'A' Level Qn.


      Given that experimental evidence has shown magnesium chloride to be soluble at all temperatures, explain this in terms of Gibbs Free Energy.


      Solution :


      This question was included in a top JC prelim paper, in which the mark scheme solution given was erroneous in its overgeneralized sweeping statement, "change in entropy of solution/dissolving is (*always) positive as there is a change from a more ordered state to a more disordered state for the ionic compound (*and water), from solid (*and liquid) to aqueous states." (*implied)


      If this were true, then magnesium carbonate would be soluble (since it has a favourably exothermic solution enthalpy), but it is not! (ie. it's insoluble).


      While the erroneous aspect of the mark scheme does not invalidate the question, or change the fact that magnesium chloride is soluble at all temperatures, the sentence is still erroneous and misleading, and I would encourage students to want to seek to understand the truth better for themselves, rather than blindly accept and blindly memorize on blind faith.


      Let's consider magnesium carbonate, which is known to be (almost) insoluble at all temperatures, even though the solution enthalpy (ie. lattice dissociation enthalpy + hydration enthalpy) is overall exothermic (-25.3kJ/mol), which means the solution/dissolving process should be favourable, based on enthalpy argument.


      But the fact that magensium carbonate is insoluble, means that the overall solution entropy change (from solid to aqueous) must be negative, or unfavourable, and that it outweighs the favourable solution enthalpy.


      Why does this happen? Many students have the misconception that solution entropy change is always positive or favourable, when a solid dissolves in a solvent (eg. water), into the aqueous state.


      The increase in disorderliness or entropy is obvious when you consider the orderly solid lattice ionic structure being broken apart into the more disorderly aqueous state, as far as the ions are concerned. However, do not forget that the water molecules now become more ordered or orderly (cations attracting partial -ve oxygens of water, anions attracting partial +ve hydrogens of water), due to ion-dipole interactions with the ions. Mg2+ is dipositive while CO3 2- is dinegative, and as such, both ions have strong ordering power over surrounding water molecules in aqueous state.


      Hence, overall entropy is negative (ie. decrease in entropy of solvent water molecules, outweighs increase in entropy of this particular ionic compound from solid to aqueous). And the fact that magnesium carbonate is (almost) insoluble at all temperatures, demonstrates or proves that the unfavourable decrease in entropy for magnesium carbonate must outweigh the favourable exothermic enthalpy for the solution/dissolving process.


      If entropy change (for a solid to aqueous state) was always positive for all ionic compounds (as implied by the errorneous mark scheme), then magnesium carbonate would be soluble (since it has a favourably exothermic solution enthalpy), but it is not! (ie. it's insoluble).


      Now, back to magnesium chloride.


      By Hess Law, the overall solution enthalpy is the sum of the (endothermic) lattice dissociation enthalpy and the (exothermic) hydration enthalpy. Experimental evidence has shown that the (exothermic) hydration enthalpy outweighs the (endothermic) lattice dissociation enthalpy, to give a overall favourable (exothermic) solution enthalpy of approximately -150kJ/mol.


      The overall entropy change of Mg(s) to Mg(g) to Mg2+(g) to Mg2+(aq), is unfavourably negative (ie. more orderly), due to the high charge density of the dipositive Mg2+ ion, resulting in a high ordering power over water molecules in (stronger) ion-dipole interactions.


      The overall entropy change of Cl2(g) to 2Cl(g) to 2Cl-(g) to 2Cl-(aq), is favourably positive (ie. more disorderly), due to the (relatively) lower charge density of the uninegative Cl- ion, resulting in a lower ordering power over water molecules in (weaker) ion-dipole interactions.


      Because (the question specifies that) experimental evidence has shown magnesium chloride to be soluble at all temperatures, hence the student / exam candidate is expected to be able to deduce that the increase in entropy for the solution of the chloride ion outweighs the decrease in entropy for the solution of the magnesium ion, such that overall entropy change is favourably positive.


      And furthermore, because solution enthalpy is also favourably exothermic, hence magnesium chloride is soluble at all temperatures, based on the Gibbs Free Energy concept and formula.


      Again, the point of this article, is that if you come across something that does not feel right to you, don't blindly accept it (eg. just because it came from a top JC's prelim paper mark scheme), but (hopefully you would want to) seek to learn the (**relative leading edge) truth for yourself.



      **All truth is relative. Just ask Einstein. Yesterday the Earth was flat. Today the Earth is round (spheroid). Tomorrow the Earth might be Hollow.



      Acknowledgement & Reference :

      Jim Clark's ChemGuide.co.uk : http://www.chemguide.co.uk/inorganic/group2/problems.html

      Edited by UltimaOnline 03 Mar `09, 2:11PM
    • 'A' Level Qn.


      Explain why alcohols with the CH3CH(OH)- group, give similar positive tri-iodomethane (iodoform) test results, as compared to carbonyl compounds with the CH3CO- group.


      Solution :


      They give similar results because their mechanism is similar, and in fact, identical from after the initial 2 steps.


      In the mechanism for the alcohols with the CH3CH(OH)- group, the first step is the oxidation of the alcohol to an aldehyde or ketone (depending on the 'R' group).


      (Although one might think of potassium iodide as an electron donor (rather than a hydride donor like LiAlH4 in dry ether), and hence iodine merely accepts electrons in carrying out oxidation; but in the presence of hydroxide ions, the combination of hydroxide ions and iodine can oxidize by removing hydrogen atoms. Hydroxide ions remove protons, while the excess electrons are removed by the iodine.)


      First, an enol (ie. alkene alcohol) intermediate is formed when both the alpha and beta carbons (ie. the 1st and 2nd carbon atoms from the hydroxy functional group) are deprotonated, and the excess (pair of) electrons are removed by an iodine molecule (consequently reduced to a pair of iodide ions).


      A lone pair on the (hydroxy group) oxygen shifts down to become a pi bond with the alpha carbon, forming the carbonyl group.


      To avoid violating the carbon's octet, the alkene pi bond nucleophilically attacks a (partially positively charged, due to an instantaneous dipole) iodine atom (of an iodine molecule).


      The formal charge on the (hydroxy group) oxygen is removed by the deprotonation of the hydroxy group.


      Alternatively, you can summarize the above mechanism by a redox equation that sees the alcohol (with the CH3CH(OH)- group) oxidized to a carbonyl compound (with the CH3CO- group), written as follows :


      2NaOH(aq) + I2(aq) + CH3CH(OH)-R(aq) --> 2NaI(aq) + CH3CO-R(aq) + 2H2O(l)


      (Acknowledgements and Reference : ChemistryDaily.com's http://www.chemistrydaily.com/chemistry/Iodoform_reaction )


      From this point on (step 3), the mechanism is exactly identical to that of the mechanism for tri-iodomethane (iodoform) test for carbonyl compounds with the CH3CO- group. Proceed from step 3 below :

      Edited by UltimaOnline 11 Oct `17, 11:31PM
    • 'A' Level Qn.


      Explain why cyclohexene has no geometric isomers.


      Solution :


      Notice that the two sp2 hybridized carbons (of the alkene) have trigonal planar molecular geometry, and bond angles of 120 deg C.


      Which means to say that the two H atoms bonded to these two sp2 (alkene) carbons, have no choice but to be both located outside the cycloalkane ring. You could regard this as the 'cis-isomer'.


      The 'trans-isomer' does not exist, because if it did, the cycloalkane ring structure would be broken up and destroyed.

      Edited by UltimaOnline 03 Mar `09, 7:03PM
    • 'A' Level Qn.


      For an equation A + B --> C + D,

      given the molarities of all 4 species, and given the Kc value, how can you determine

      (i) relative rates of forward versus backward reactions

      (ii) whether the LHS "A & B" or the RHS "C & D" gets the "-x" or "+x", for the "Change" in the ICE table.


      Solution :

      Calculate Qc value. Qc has the same formula as Kc, but is at initial rather than at equilibrium.

      If Qc < Kc, position of equilibrium lies to the right; rate of forward reaction exceeds rate of backward reaction until Qc = Kc where equilibrium is reached; "Change" has "-x" to "A & B" and "+x" to "C & D"

      If Qc > Kc, position of equilibrium lies to the left; rate of backward reaction exceeds rate of forward reaction until Qc = Kc where equilibrium is reached; "Change" has "+x" to "A & B" and "-x" to "C & D"

      If Qc = Kc, position of equilibrium has already been reached. There is no need for "Change" within the ICE table; initial = equilibrium.

      Edited by UltimaOnline 05 Mar `09, 12:03AM
    • (Challenging) 'O' Level Qn.

      (Standard) 'A' Level Qn.



      Titanium(IV) oxide is heated in hydrogen gas to give water and a new titanium oxide, TixOy. If 1.598g of titanium(IV) oxide produced 1.438g of TixOy, what is the formula of the new oxide?



      Solution :


      No. of mol of Ti before redox = no. of mol of Ti after redox

      0.02  =  (1.438/(47.9x+16y)) x

      x / y  =  2 / 3


      Hence, formula of new oxide is Ti2O3.

    • 'A' Level and 'O' Level Qn.


      The Haber Process Riddle



      The haber process is an exothermic process. Which means heat may be regarded as a product. Which means that when we increase temperature, the position of equilibrium shifts to the left. Which means that the rate of backward reaction exceeds the rate of forward reaction, until equilibrium is reached (upon which, the rates of forward and backward reactions will be the same). Which means that we get less ammonia, and more hydrogen and nitrogen.


      Now, if that is the case that in increasing temperature, the rate of backward reaction exceeds the rate of forward reaction, and we use up ammonia to produce more hydrogen and nitrogen, why the heck would we want to do that? How does that give us more ammonia? It seems we're losing ammonia, doesn't it? (Remember, we in the Haber process industry sell ammonia, not hydrogen and nitrogen!) 



      Solution :


      Increasing temperature does shift the position of equilibrium to the left (ie. decreasing yield of ammonia at equilibrium), meaning that (assuming we begun with the state of equilibrium already reached!) the rate of backward reaction increases much faster than the increase in the rate of the forward reaction.


      Note that assuming you begun with equilibrium already reached, whenever you raise temperature, the rates of BOTH the forward and backward reactions, will increase. But which increases faster (compared to the other), depends on whether the forward reaction is endothermic (ie. heat regarded as a reactant) or exothermic (ie. heat regarded as a product).


      The trick to solving this riddle (which confuses many students batch after batch), is to realize, that at initial conditions, you do NOT have equilibrium conditions! In fact, you have lots of nitrogen and hydrogen gases, but ZERO or NIL of ammonia.


      Hence, the Change (of your Initial, Change, Equilibrium table) will still use up nitrogen and hydrogen (ie. -x and -3x), to produce ammonia (ie. +2x). In other words, Qc = 0, which surely is less than Kc (regardless of temperature). Hence position of equilibrium still lies to the right, and rate of forward reaction will exceed rate of backward reaction, until Qc = Kc (at whatever temperature it may be), and equilibrium has been reached.



      For the sake of explaining the concept, let us consider the following scenarios.


      At a low temperature, you will get 90% yield of ammonia (at equilibrium), and 10% reactants (N2 and H2). But it takes a thousand years to reach equilibrium.


      At a high temperature, you will get 10% yield of ammonia (at equilibrium), and 90% reactants (N2 and H2). And it takes only an hour to reach equilibrium. 


      As the CEO of your ammonia manufacturing Haber Process industry, which temperature would you order your technicians to set? Remember, time is money.


      Every hour, you would carry out fractional distillation to separate the ammonia produced. Doing this, you can see how using a fairly high temperature to speed up rate of reaction, can actually produce far more ammonia in practice per day (or per thousand years!), even though the "yield at equilibrium" is less. Because equilibrium is reached much more quickly, and fractional distillation comes to the rescue.


      For more information on The Haber Process, including industrial reaction (temperature and pressure) conditions commonly utilized, see :


      Edited by UltimaOnline 11 Mar `09, 10:07PM
    • 'O' level (electrochem) qn.



      If you have an electric cell (not electrolytic cell) setup that consists of a zinc electrode, a copper electrode, sodium chloride solution; describe what happens, and name the compounds (both solid and aqueous) present in the electrolyte as the reaction proceeds.


      Solution :

      Zn(s) is oxidized to Zn2+(aq); oxidation at zinc anode.

      H+(aq) (from water) is reduced to H2(g); reduction at copper cathode.


      Ions present as the reaction proceeds :

      Zn2+, Na+, OH-, Cl-. 


      Hence, compounds present in electrolyte as the reaction proceeds are :

      Sodium chloride solution, sodium hydroxide solution, zinc chloride solution, zinc hydroxide precipitate.

      Edited by UltimaOnline 16 Mar `09, 10:50PM
    • On Photochemical Smog

      ('O' levels and 'A' levels)


      An 'O' level student asked :


      >>> This is not part of the question but I just want to know what are the products when unburnt hydrocarbons react with NO2? What exactly are the substances that cause the tearing of the eye? <<<


      Glad you asked. In your upcoming 'A' levels, you will learn about free-radical substitution mechanism (you already came across it in 'O' levels, when you mix a halogen gas with alkanes, under UV light; you just didn't know it was called free-radical substitution).


      Photochemical smog formation involves free radical substitution (initiation, propagation, termination). This is why it's called PHOTOchemical smog, because it involves UV radiation from sunlight allowing homolytic cleavage/fission, to generate highly reactive free-radicals.


      Details on the chemistry of photochemical smog :


      About photochemical smog as a pollutant :

      Edited by UltimaOnline 17 Mar `09, 10:52PM
    • 'A' Level Qn.


      Draw the mechanism for the formation of nitric(V) acid from the reaction between nitrogen dioxide and water.


      Solution :


      2NO2(g) + H2O(l) --> HNO3(aq) + HNO2(aq)

      This is a hydrolysis reaction, ie. water reacting with a species to form another species.


      In this case, begin by drawing the Kekule structure of nitrogen dioxide (this is near impossible to draw for 'O' level students, and still challenging for many 'A' level students... a JC tuition student of mine from said that she was the only one in the entire class who could successfully draw the structure required in a class test, thanks to my tuition).


      Note that because nitrogen is in period 2 and does not have empty d-orbitals with which to expand its octet, the structure of nitrogen dioxide has a single bonded O with a -ve formal charge, a double bonded O with no charge, and an unpaired lone electron on N, which has a +ve formal charge. The molecular geometry is V shaped or bent or non-linear.


      In the 1st step of the mechanism, two NO2 free-radical molecules form a (non-dative) covalent bond with their unpaired electrons, generating N2O4 (2 -ve formal charges on 2 singled bonded O atoms, 2 +ve formal charges on the 2 N atoms). Notice that this step confers upon both nitrogen atoms, a stable octet.


      In the 2nd step, because oxygen is more electronegative than nitrogen (which would give nitrogen a partial +ve charge), and moreover because the nitrogen already has a positive formal charge, water acts as a nucleophile to attack the +ve formal charged nitrogen. A lone pair on the O (of the water molecule) forms a dative bond pair with the +ve formal charged nitrogen.


      Simultaneously, to avoid violating the nitrogen's octet, the pi bond on that nitrogen becomes a lone pair on the oxygen, giving it a -ve formal charge.


      The resultant +ve charge on the oxygen is removed by the loss of the proton (eg. to another water molecule to form a hydroxonium ion), such that the proton's bond pair becomes a lone pair on the oxygen.


      In the 3rd step, a lone pair on one of the two single bonded -ve formal charged oxygens (that are bonded to the nitrogen that has just been attacked by water), forms a pi bond with the nitrogen. (This is the precursor for HNO3 nitric acid aka nitric(V) acid, with the loss of the leaving group described below).


      Simultaneously, in order to prevent violation of its octet, the sigma bond between the two N atoms, cleaves to become a lone pair on the leaving group NO2-, which abstracts a proton (eg. from a hydroxonium ion; recall that one was formed in an earlier step) to form HNO2 nitrous acid aka nitric(III) acid.


      Hence, in the final step, we obtain both acids HNO3(aq) and HNO2(aq).

      Edited by UltimaOnline 18 Mar `09, 11:54AM
    • (Challenging) 'O' Level Qn

      (Standard) 'A' Level Qn


      The metallic ion X^n+ is oxidised to the uninegative anion XO3- By MnO4- in acidic medium. If 2.68 X 10^-3 moles of X^n+ require 1.61 X 10^-3 moles of MnO4- for oxidation, what is the value of n? Construct a balanced redox equation for the reaction.



      Solution :


      1.61 x 10^-3 mol of MnO4- implies 5 x 1.61 x 10^-3 mol of electrons removed from 2.68 x 10^-3 mol of X^n.


      This implies 3 mol of electrons removed per mole of X^n.


      O.S. of X in XO3- = +5


      Therefore, n = O.S. of X before oxidation = +2


      [Reduction] MnO4– + 8H+ + 5e– ---> Mn2+ + 4H2O

      [Oxidation]  X2+ + 3H2O --> XO3- + 6H+ + 3e-

      [Balanced Redox] 3MnO4- + 5X2+ + 3H2O --> 3Mn2+ + 5XO3- + 6H+






      Edited by UltimaOnline 20 Mar `09, 12:31AM
    • 'O' Levels and 'A' Levels both.

      Electrolysis (Electrochemistry) notes.




      Red Cat riding An Ox.
      REDuction occurs at the CAThode; at the ANode, OXidation occurs.




      In electrolytic cells,
      cathode is the negative electrode.
      anode is the positive electrode.


      In electric / galvanic / voltaic cells,
      anode is the negative electrode.
      cathode is the positive electrode.


      How to remember? Here's how :


      In electrolytic cells, the battery provides the electrons, which flow from the negative terminal of the battery in order to carry out reduction, which by definition, occurs at the cathode.

      Hence, in an electrolytic cell, the cathode is the negative electrode, and the anode is the positive electrode.


      In electric / galvanic / voltaic cells, the entire setup IS the battery. The more reactive metal electrode gets oxidized (remember : reactive metals are eager to oxidize themselves into cationic oxidized form; while non-reactive metals lazily prefer to remain in atomic reduced form), and since by definition oxidation occurs at the anode, so the more reactive metal becomes the anode and the less reactive metal has no choice but to be the cathode. When the more reactive metal oxidizes itself at the anode, it releases electrons (eg. Zn ---> Zn2+ + 2e-), making the anode (which is the source of electrons) negative. Remember that in a battery (and this electric cell setup IS the battery!), electrons flow from the negative terminal.

      Hence, in an electric / galvanic / voltaic cell, the anode is the negative electrode, and the cathode is the positive electrode.


      Notice that in all cases above (ie. for both electrolytic and electric cells), the negative electrode is the source of electrons. That's why it's negative.




      Within the electrolyte, current is the (rate of) flow of charged ions. Within the external wire, current is the (rate of) flow of electrons.




      If the electric / galvanic / voltaic cell is going to be a successful one, you must ensure that the electrodes do not react directly with the electrolyte in a way that prevents electrons from flowing in the external wire (which is your objective, if you're designing an electric / galvanic / voltaic / cell).

      An example to illustrate this problem would be using sulfuric(VI) acid, with copper and zinc electrodes. Although you might want the electrons to flow from the anode (oxidation of Zn to Zn2+ + 2e-) to the cathode (where you might want the protons H+ from sulfuric(VI) acid to accept the electrons and be reduced to H2 gas), but in practice, no current will flow in the external wire.

      Why? Because the zinc metal would be involved in a redox reaction directly with the sulfuric(VI) acid electrolyte. In other words, the electrons would be transfered from the zinc directly to the protons H+ from the acid electrolyte, and no electrons would flow in the external wire. The setup has failed as an electric / galvanic / voltaic cell.

      The best way to work around this problem, is to use two different electrolytes in two separate compartments for the two electrodes, and connecting them using a salt-bridge.

      An important example of an electric / galvanic / voltaic cell setup that uses a salt-bridge, is the Daniel Cell.

      Edited by UltimaOnline 21 Mar `09, 9:59AM
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