h2 chem

• Z een

  8 Nov 09, 15:59

  1) Why do Group II metals form oxidation number of +2 and not +1?

  I understand why they form +2.  The ans scheme states that they do not exhibit +1 oxidation state due to the low lattice energy of the compounds formed. Do they mean that these compunds formed would be less stableand hence not favourable as compared to having a +2 oxidation state?

  2) when 2 -OH groups are bonded to a benzene ring (not sure what this compound name is called) and this compund reacts with dilute nitric acid, will there be 2 substitutions or 1 substitution with the nitronium ion?

   

  3) Do primary amines and phenol react with NaOH (aq) , followed by CH3I and heat? I thought this is the tri-iodomethane ppt obtained from the test? (I know that phenol react with NaOH to give phenoxide ion)

   

  4) I suck at T.M and am not sure how to determine the products when they give me a complex ion and the reactants. Like for example, the dichromate ion reacts with sulphur dioxide to give A. And they want us to give the formulae of the chromium-containing species for A. The ans given was [Cr(H2O)6]3+. How do you determine that this is the product?

• Z een

  8 Nov 09, 16:20

  Oh oh, another qns.

   

  5) Write an equation to illustrate how con. sulfuric acid behaves as an oxidizing agent in the reaction with solid NaBr.

  I wrote that 1 mole sulfate ion + 2 mole of bromide ion + 4 moles of H+ will give 1 mol of sulphur dioxide with 2 mol of water and 2 mol of bromine. But i realise that it might not feasible bcos the E value= 0.17-1.07= -0.9, which means this is not feasible?

• UltimaOnline

  8 Nov 09, 17:29

  1) It's more worth it (more 'wu hua' instead of 'bo hua' in hokkien) for the universe to expand the 1st and 2nd endothermic ionization energies required to generate dipositive Grp II cations, because the resulting exothermic lattice energy compensates adequately and even profitably (ie. 'wu hua'). If the universe were to expand less energy ('cheapskate' efforts/investments yield 'cheapskate' results/profits) to ionize only to generate unipositive Grp II cations, the much smaller exothermic lattice energy that results either doesn't adequately compensate the endothermic ionization energy expanded in absolute values (ie. profits don't cover costs, therefore net loss), or doesn't compensate as much (as in the case of dipositive Grp II cations), (ie. why invest only $10 to make only $20 profit, when we can invest $20 and make $80 profit!). 2) Dinitration would occur, in the two positions para to the phenolic groups. Each phenol group donates strongly by resonance, so even if the nitro group withdraws electrons (by both induction and resonance) and hence deactivates the benzene ring (ie. making it a weaker nucleophile), but two phenol groups activates the benzene ring (ie. making the it a stronger nucleophile) sufficiently for dinitration to occur. Para positions (instead of ortho) because of steric hinderance. Not meta because phenol donates by resonance (further explanation lookup my Collection of Chem qns thread; not required at 'A' levels). 3) Amines (primary amines have less steric hinderance and thus are more effective nucleophiles) and phenols are both nucleophilic. Iodine is a good leaving group because iodide ions are very stable (due to electronegativity and large ionic radius to stabilize the negative formal/ionic charge). Hence SN2 nucleophilic substitution occurs (not SN1 because methyl carbocation is too unstable). 4) Sulfur dioxide reduces dichromate(VI) to chromium(III), and is itself oxidized to sulfate(VI). No other available ligands, so water functions as the ligands to generate the hexaaquochromium(III) ion. 5) You're using standard redox potentials. Obviously we're not gonna use standard molarities to carry out these redox reactions. So yes, bromide ions are oxidized to molecular bromine, and are able to reduce sulfate(VI) ions to sulfur dioxide (aka sulfur(IV) oxide). Notice the change in O.S. of sulfur from +6 to +4. In contrast, with iodide ions (larger ionic radius results in greater reducing power), the O.S. of sulfur is reduced from +6 all the way to -2 in hydrogen sulfide gas.