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BedokFunland JC's A Level H2 Chemistry Qns (Part 2)

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    • Originally posted by MightyBiscuits:

      May I seek help for this question UltimaOnline? Here is the picture for the qns. Answer for this mcq question is B, I don't have the solution though.

      https://www.instagram.com/p/BG6O0ZbDnn7/


      Find OS of Fe in the reactant in terms of n. Find moles of electrons transferred per mole of anion. Hence find new OS of Fe in terms of n, in the product. Equate to actual OS of Fe in product. Hence solve for n.

    • A BedokFunland JC Original H2 Chemistry Qn.

      Methanoyl chloride is highly unstable, and readily decomposes (under standard conditions) into 2 covalent gases, one acidic and one neutral.

      Draw the curved-arrow electron-flow reaction mechanism.

      By considering bond enthalpies of the reactants and products, explain whether the thermodynamic driving force for the reaction is enthalpy or entropy or both.

      Edited by UltimaOnline 24 Jun `16, 2:34AM
    • Originally posted by Flying grenade:

      when do we use miscible and soluble? 

      is NH3 miscible and/or soluble with/in water?


      If a solute is miscible with a solvent, then the solute is considered to be soluble in that solvent.

    • Originally posted by Flying grenade:

       

      then is it right that all(thermodynamically feasible) chemical reactions are undergoing and happening , even if the reaction conditions are inadequate to provide the Ea required for the reaction to occur?

      means that all the reactants are indeed forming the products even when Ea of reaction is not achieved (albeit over thousands/millions of years) , instead of remaining as reactants as it is?


      Don't forget, all reactions are equilibrium reactions (it's a matter of where, and how far, the position of equilibrium lies, under what conditions, and conditions are changing all the time). What you call 'reactants' are actually products for the opposite reaction. And every species that exist in the Universe is part of a multiple web of different chemical reactions simultaneously. It's beautifully complex : infinite diversity in infinite creation.

      One more thing about your statement "even when Ea of reaction is not achieved", you've apparently forgotten the Maxwell-Boltzmann distribution : it's about the % of molecules with energy equals or exceeding Ea, not a binary 1 or 0, all or nothing. Even at low temperatures, a small % of molecules will have the required Ea for the reaction.

    • Originally posted by LavXuan:

      Is it possible to remove 2 carbon atoms at one go from the oxidation of an alkene? How


      Yes of course, if using KMnO4 (but not K2Cr2O7).

      The =CH2CH2= group will be oxidized to ethandial, then ethandioic acid, then 2 mol of carbonic(IV) acid, which decomposes into 2 mol of CO2(g) + H2O(l), with thermodynamically favorable positive entropy change (since CO2(g) is gaseous), and as predicted by Le Chatelier's principle, since CO2(g) leaves the reaction solution, pulling the position of equilibrium over to the RHS.

      Edited by UltimaOnline 25 Jun `16, 9:05PM
    • Originally posted by LavXuan:

      Does elimination reaction still occur for alcohol to form alkene when alcohol is REFLUXED with conc H2SO4 instead of carrying out the reaction at 170 degree celsius?


      It's not "heat" versus "reflux". You're supposed to heat under reflux at 170 deg C.

    • Originally posted by Christabellw:

       " Comparing HF and NH3 F is more electronegative than N, thus H-F bond is mote polar than N-H bond" What does it mean when a bond is more polar than the other?


      It means the dipole moment (in Debye units) has a larger magnitude. The larger the magnitude of electronegativity difference between the 2 atoms of different elements, the more polar the bond, the larger its dipole moment.

      Eg. the dipole moment in H-F is 1.91 D, which is larger than the dipole moment in H-Cl which is 1.08 D, which is larger than the dipole moment in H-Br which is 0.80 D.

    • Originally posted by Fxwhy:

      https://twitter.com/doublehalfarrow/status/746629255522443264                                 How does A undergo intramolecular nuclophilic substitution between its amine and alkyl halide to form D?

       

       


      What do you mean, "how does it undergo intramolecular nuclophilic substitution?". It undergoes because of the mechanism, that's how (by definition)!

      If you mean "how do you know" then it's because the no. of C atoms in A and D are the same, and the degree of unsaturation increases by 1 (due to formation of cyclic ring), that's how you know.

    • Originally posted by Christabellw:

      Explain why gaseous ethanoic acid, Ch3COOH has an apparent molecular mass of 120 instead of 60

      Ans: this is because the the gas particles consist of hydrogen bonded dimers

       

      Is it because organic acid in gaseous state have molecular mass twice of what it is expected??? If yes, why so??? So what if it consist of hydrogen bonded dimers?

       

       


      Not for all organic acids, but especially for small carboxylic acids. And not just gaseous state, but in the liquid state as well, and in non-polar solvents as well.

      For such questions, you're required to draw (and label!) the following hydrogen-bonded dimer :

      image

      http://www.chemguide.co.uk/organicprops/acids/background.html

      Edited by UltimaOnline 25 Jun `16, 9:04PM
    • Originally posted by nicolemantou:

      https://twitter.com/doublehalfarrow/status/7466…
      Why is [OH-] divided by 2 in the answer?


      Because adding equal volumes of both reactant solutions, doubles the volumes, hence halves the molarities of the aqueous ions.

    • Originally posted by ACA-Milkshake:

      For the bond angle of cl-c-cl bond of cocl2, even with the presence of the double bond with O, the bond angle won't be affected and remain at 120 degrees? 


      For A level purposes, the expected answer remains at 120 degrees, ie. Cambridge will award full credit (of 1 mark) for 120 degrees. In actuality, or at Uni levels, the bond angle in phosgene is 112 degrees, but Cambridge won't require this value for A levels.

      image

      https://en.wikipedia.org/wiki/Phosgene

      At most, Cambridge may ask, "Suggest a reason for the slight deviation from the 120 degrees as predicted by VSEPR theory", to which there are several contributing reasons (beyond the A level syllabus), one of which is as you mentioned (the greater repulsion of the double-bond).

      If the A level candidate wishes to incorporate such contributing factors that result in a deviation in the basic bond angles as predicted by VSEPR theory, the candidate (if he is exam-smart and wishes to safeguard his marks) must do so with a written explanation, first giving the basic bond angle of 120 degrees (with brief explanation), then make mention of factors which cause deviation, then suggest a concordantly modified bond angle.

    • Originally posted by ACA-Milkshake:

      h2 chemistry A Level TYS

      2007 Paper 3 qns 2(e)ii: My answer for the reaction I is base hydrolysis but the ans is neutralisation. Why is it neutralisation? 

      2007 Paper 3 qns 3(d)i: How do the product CH3CHO form from the reaction? 

      2007 Paper 3 qns 3(d)ii: Can CH2O decolourise purple KMnO4 and liberate CO2 gas? Cos i thought that ketones can't do so... 

       


      The amide group isn't hydrolyzed in the reaction, rather the N-H group is being deprotonated, ie. removal of H+, leaving behind a uninegative formal charged N atom as illustrated in the question.

      The mechanism is significantly beyond A levels. For A level purposes, use simple pattern recognition and mathematics. Based on the original reaction equation given, and based on the new reactants of ethanoate and methanoate ions in a 1:1 ratio, let A = H atom, B = methyl group, we mathematically obtain carbonyl molecules with AA, AB, BA and BB groups, hence we obtain a 1:2:1 ratio of methanal, ethanal and propanone. Notice that Cambridge isn't asking for an explanation, but merely "suggest the products formed and the ratio they're obtained".

      KMnO4 oxidizes methanal to methanoic acid to carbonic(IV) acid, which decomposes into CO2(g) + H2O(l), with thermodynamically favorable positive entropy change (since CO2(g) is gaseous), and as predicted by Le Chatelier's principle, since CO2(g) leaves the reaction solution, pulling the position of equilibrium over to the RHS.

    • Originally posted by Shinyphua:

      Why is the C-O-H bond slightly larger than the C-S-H bond angle? 

      For this question, the answer given is that O and S contains 2 lone pairs and 2 bond pairs, as O is more electronegative than S, the bond pair of electrons are held more closely towards O, hence, a stronger repulsion.

      My question is, can i answer in terms of the size of the molecule? O is smaller than S, and thus the C and H atoms are of closer proximity to the molecule than when its a large molecule like S, therefore, the lone pair - bond pair repulsion is of greater magnitude.


      Singapore JCs often use electronegativity to explain the deviation from the basic bond angles of period 2 elements as predicted by VSEPR theory, when discussing period 3 elements. Other than cases involving electronegative F (eg. why NF3 bond angles deviate from NH3 bond angles), using electronegativity to explain the deviation is actually incorrect, or rather, only a secondary factor (Chemistry, being a microcosm of real-life, is multi-factorial).

      The more important reason, is due to the significantly lesser extent of orbital hybridization (which requires energy, and thus only thermodynamically justified for less diffused valence orbitals of period 2 elements), due to the significantly lesser extent of electron-pair repulsions, due to the significantly more diffused valence orbitals (which the lone pairs reside in, or the bond pairs are formed from the overlap of) of period 3 elements.

      Consequently, the atoms of period 3 elements use their mostly unhybridized orbitals for their lone pairs to reside in, and for overlapping head-on or end-on to form sigma bonds. In the case of C-S-H, the S atom uses it's mostly unhybridized p orbitals to overlap head-on or end-on to form its sigma bonds with the C and H atoms, as well as for 1 of the lone pairs to reside in, with the remaining lone pair residing in the mostly unhybridized s orbital. This results in the C-S-H bond angle (with 2 orthogonally oriented p orbitals used for head-on or end-on overlap with the C and H atoms) being close to the unhybridized p orbital bond angle of 90 degrees.

      So you're partially right that it's about the size, though you'll need to elaborate (as I did) to secure your marks. Your school (and Singapore JCs in general) aren't entirely wrong, it's just that electronegativity isn't the most important contributing factor here (unlike in cases such as NF3 vs NH3).

      However, all this discussion may be moot, because as far as A levels are concerned, Cambridge won't require the student to give any answer other than the basic bond angles as predicted by VSEPR theory for period 2 elements, ie. just state 104.5 degrees as if the S atom were an O atom, and include the usual explanation about the S atom having 2 lone pairs and 2 bond pairs, and that lone pair - lone pair repulsions > lone pair - bond pair repulsions > bond pair - bond pair repulsions, that would gain you full marks.

      After writing the basic answer above, if you like, if you have the time, or if (unexpectedly, though I'd see it as a pleasant surprise) Cambridge specifically asks you to explain why the C-S-H bond angle deviates from the C-O-H bond angle, then you can elaborate further to why the C-S-H bond angle is actually less than the C-O-H bond angle, giving either my (more correct) explanation, or your school's (less correct) explanation, or both. But again, because asking this would be beyond the A level syllabus, it is rather unlikely for Cambridge to ask why the C-S-H bond angle deviates from the C-O-H bond angle for the actual A level exam (ie. your question, which you presumably took from your school's mid-year revision package).

      Edited by UltimaOnline 28 Jun `16, 3:14PM
    • Originally posted by Flying grenade:

      why is the first I.E. of Al and Ga the same value,  +577kj/mol?


      Because of d-block contraction.

    • Originally posted by Flying grenade:

      given Acid color of methyl orange is red, phenolphtalein is colorless , Alkaline color is, yellow and red , respectively

      Wouldn't methyl orange stay Red in color when the end pt of a SA-WB titration, is at a acidic pH?? Why will methyl orange turn Yellow only at the end pt , that occured at a acidic pH(for pg 157 SA-WB titration example, end pt pH≈5.5)? (given pH range methyl orange 3.1-4.4)


      If it stays the same color from beginning to end point, how the hell would u know you've to stop adding the strong acid from the burette?!? Then the indicator would be utterly useless, wouldn't it?!?

      And u seem to suffer from a ridiculously basic (pun intended) conceptual error, that "alkaline color" for the indicator means the solution must actually be alkaline (ie. >7). Nonsense! The "alkaline color" of the indicator could still be in acidic pH, eg. 4.5 for yellow colored methyl orange. It's "alkaline" only relative to the "acidic color" for which the pH is lagi acidic, eg. 3.0 for red colored methyl orange.

      The underlying idea, is that the suitable indicator must change color sharply over the vertical portion of the titration curve (the sharp change is due to the fact that at equivalence point, it is no longer a buffer region, so just a single drop of excess strong acid added will change the pH sharply, resulting in the sharp color change), ie. the pKa of the indicator must be as close as possible to the pH of the solution at equivalence point.

      So if u started using with a weak base in the conical flask (and stop ur acronym nonsense, Mr lazy WA SA BI), the methyl orange color would begin as yellow, until just enough strong acid is added from the burette to go past the buffer region into the equivalence point, thereby sharply changing the pH of the solution upon addition of 1 excess drop of strong acid, thereby sharply changing the methyl orange color to red, which indicates the change in pH of the solution has just transited past the pKa of the indicator (which if correctly chosen should be close to the equivalence point pH for this titration), which in turn indicates that the equivalence point has been reached.

      Lastly, don't forget that end point merely approximates equivalence point. But as long as you repeat the titration until you obtain 2 concordant or consistent titers, you're good to go, that's already as accurate as is humanly possible.

      And you prolly won't fully understand what I've said above, coz you (like most Singapore JC students) prolly have a terribly weak understanding of this topic. I've already given u an adequate response. If u still dun fully understand, go ask ur school teacher or private tutor. Move on to ur next qn on the next topic.

    • Originally posted by Flying grenade:

      my school cher say that benzene is not a functional group??


      It is.

      I already told u, don't trust Singapore JC teachers (who routinely disagree with each other, both across and within JCs).

      From the Chemistry department of the University of California, Los Angeles :

      http://www.chem.ucla.edu/~harding/notes/FG_01.pdf

      And here is the most important evidence why you should trust BedokFunland JC > Singapore JCs : to an actual recent A level Chemistry exam question which asked, "Name this functional group", the official Cambridge A level Mark Scheme answer is : "benzene / arene / aryl / phenyl"

      Which shows you that unlike the anally dogmatic marking of Prelim papers by Singapore JC teachers, Cambridge will reasonably accept a variety of reasonable answers, otherwise with the totally different phrasing of mark scheme answers of different Singapore JC Prelim papers, even Singapore JC teachers themselves would fail if they sat for each other's Prelim papers. Fortunately for you guys, Cambridge is more reasonable than anal.

      The only thing you need to be careful is, if groups like OH or NH2 or CONHR group are directly bonded to the benzene ring, instead of giving "benzene ring" and "alcohol / amine / amide" separately, you will need to say "phenol" or "phenylamine" or "benzamide", etc.

      Edited by UltimaOnline 01 Jul `16, 1:04PM
    • The video title is a Chemistry joke, geddit? ;Þ

      https://www.youtube.com/watch?v=8SlZ6n4iRZM

    • Originally posted by Flying grenade:

      page 51 cstoh advanced guide

      to form H bonds, a criteria is "a lp of e- on a very electro-ve atom such as F,O,N"

      is this not very true?

      is this right : e.g. H2S, the H with partial +ve charge in H2S can act as a H bond acceptor, and consequently H2S can form H bonds(maybe, e.g. with H2O) ? 

      or is it just intermolecular permanent dipole permanent dipole interactions? 


      Only Keesom van der Waals forces. The magnitude of partial positive charge on H atom in H2S is insufficient to donate H bonds. To accept H bonds, not only must the F, O or N atom have at least 1 lone pair available, there must also be a formal (stronger H bonds) or at least partial (weaker H bonds) negative charge on the F, O or N atom (in the molecule's resonance hybrid).

    • Originally posted by ACA-Milkshake:

      h2 A Level TYS 

      2008 Paper 3 Qns 2e(iii): Why does the amount of H from alkane D times 2? And from the ratio between the moles of C and H, the molecular formula i got is C2H5 instead of C4H10 and i am not sure why.

      2008 Paper 3 Qns 2f: How do they know that the R'CO2 in the reactants, is A?

      2009 Paper 3 Qns 3c(iii): In the answer booklet, the answer for the amount of Na2O = 1/2 x [(9.00x10^-3) - (4.00x10^-3)]. My question is where did they get the 4.00x10^-3 from?  

       


      2008 Paper 3 Qns 2e(iii) :

      Because each mol of H2O contains 2 mol of H (from alkane D). The mole ratio gives you empirical formula of C2H5, which gives a wrong (and non-valid non-integer) degree of unsaturation for an alkane, which should be 0 degrees. Hence multiply by 2 (and if it doesn't work, try 3) to obtain the molecular formula of C4H10 (with the correct degree of unsaturation).

      2008 Paper 3 Qns 2f :

      Because from previous parts of the question, you've already found that A is C3H7COOH, C is C5H12, D is C4H10, which means that C must be R-R' (ie. 2 + 3 = 5), hence if we let R be 2 carbons and R' be 3 carbons, then D (4 carbons) would be R-R, then E must be R'-R' which has 6 carbons (eg. hexane or an isomer of), which also means A must be R'COOH (which is deprotonated to be used as R'COO-), which means B must be RCOOH, which must therefore be C2H5COOH.

      2009 Paper 3 Qns 3c(iii) :

      From the acid-base titration, you know the total mol of OH- from both sodium oxide and sodium peroxide. From the redox titration, based on the mol of MnO4- required, you know the mol of hydrogen peroxide oxidized, and thus by stoichiometry, you also know the mol of sodium peroxide present and the mol of OH- from the hydrolysis of the sodium peroxide.

      Hence, you can calculate the mol of OH- generated from sodium oxide alone (ie. total mol of OH- which is 9.00x10^-3, minus away mol of OH- from peroxide which is 4.00x10^-3, which will give you mol of OH- from oxide, which is 5.00x10^-3).

      Divide this value by 2 (since each mol of sodium peroxide generates by hydrolysis 2 mol of NaOH), to obtain the final answer of 2.50x10^-3 mol of sodium oxide.

    • Originally posted by ACA-Milkshake:

      2009 TYS A Level Paper 3 Qns 4d(ii): In the answer, why will the 2 carbons that are bonded to the Br atoms eventually combine and bond to NH after undergoing nucleophilic subs? I thought they would form 2 NH2 groups, one on each of the 2 Carbons? 


      Always draw out (if you're unable to visualize) the mechanism (for either forward and/or backward reaction), that'll lead you to the correct structure and answer. In this particular question, the forward reaction is more thermodynamically feasible than the backward reaction, but for the sake of elucidating J, we'll imagine the backward mechanism here.

      Br- attacks C, NRH gets kicked out as NRH-, gets protonated as NRH2. Another Br- attacks the other C (of the R group), NH2 gets kicked out as NH2-, gets protonated as NH3. What's left behind is J, which includes the 2 Br atoms.

    • Originally posted by Flying grenade:

      i see, thanks ultima!!

       

      i suspect where cs toh thought of an O.S. of 6.

      image

      so, which S2O32- structure is correct? cstoh&google image or the one with two singly bonded mononegative charged Oatom?

       


      They are all correct, because they are resonance contributors of each other (with full bonds and formal charges shown), or the resonance hybrid (with partial bonds and partial charges shown).

      If Cambridge asks you to draw the S2O3 2- ion (either displayed structure or dot & cross structure), Cambridge will accept any of the valid resonance contributors. If you're in doubt as to whether Cambridge wants a resonance contributor or the resonance hybrid, the exam-smart student will give both answers (labeled, of course).

      It is not entirely correct for you to say that since O is more electronegative than S, the resonance contributor in which the negative formal charge is on O is more valid than when it's on S, because other than electronegativity (which favors O), you must also consider charge density (which favors S). So the actual OS of the 2 S atoms in the resonance hybrid, will be somewhere between 0 and -1 for the terminal S atom, and somewhere between +4 and +5 for the central S atom.

      As far as A level purposes are concerned, you only need to be able to work the overall average OS for the purpose of balancing redox equations. Being able to work out OSes of individual atoms in individual contributors will also be useful for some questions. And if you're asked (unlikely for Cambridge A levels) to work out the OS of the different atoms in the resonance hybrid, just draw out the most relevant resonance contributors with the OSes of the different atoms labeled, and accordingly explain in your answer that the OS of each (labeled) atom in the resonance hybrid should therefore be between such and such (as I've done above).


      Originally posted by Flying grenade:

      for the factor of charge density, a charge is more favoured to be in a bigger atom? e.g. S compared to O? 

      because bigger atom , for the same charge, lower charge density, more stable?


      Varying stability of conjugate base (due to varying charge density) is the real reason (together with endothermicity of H-X bond dissociation enthalpy) why the strength of Bronsted-Lowry acidity is HI > HBr > HCl > HF.

      If Cambridge asks you why and you talk about electronegativity, you're shooting yourself in the foot, and will nullify any correct points you may have also written (because as long as you write contradicting points, Cambridge thinks you're trying to cheat your way through by using rainbow ruses and as such you'll get zero marks for the question).

      Edited by UltimaOnline 04 Jul `16, 5:36PM
    • Originally posted by Flying grenade:

      in this pic extracted from updates to cstoh chem books on step-by-step website,

      he cancelled out the slow and fast, for the SN2 mechanism 

      i think i know it is unnecessary(or is it Wrong) to write that for SG-Cambridge A levels,

      but is it wrong, chemically?

      for my sch, the lect notes didn't put, but tutorial soln put the 'slow' and 'fast' ._. . anyway cstoh and C.K.S's books are 1000x better, compact and organised and better info.



      image


      For reaction mechanisms, the "slow" and "fast" steps should only be written if the Cambridge question asks for it, eg. particular if it's a kinetics question.

      Otherwise, you don't have to write it in. But it's ok if you want to do so, and Cambridge will just ignore it (whether you got it right or wrong) if the question didn't ask for it.

      In CS Toh's update, the reason why it's incorrect to put either "slow" or "fast" is because the structure shown is the transition state, not the intermediate (there is no intermediate for SN2). Note that "slow" and "fast" must refer to actual steps converting reactant to intermediate, intermediate to intermediate, or intermediate to product. Transition states are simply human-imagined states in-between (ie. transitioning) actual steps, and thus cannot be either "slow" or "fast".

      Edited by UltimaOnline 03 Jul `16, 11:58PM
    • Originally posted by Flying grenade:

      omg

      page 51 cs toh advanced guide

      i have been learning wrong things throughout my life 

      cs toh say CO2 molecules have intermolecular pd-pd forces of attraction.

      the next statement below "this is possible because in the solid state, CO2 molecules are fixed in position." does this statement mean only CO2(s) has pdpd? what is that statement for?

      i thought only molecules with (net)dipole moment have intermolecular pd-pd. 

      i am now confused already. is CO2 a polar molecule? it's a symmetrical molecule

      i always thought that in order for a molecule to be polar, the molecule must have net dipole moment 

       

      so CO2 has id-id (all molecules have) , does it have pdpd?

       

      is CO2 a polar molecule without a net dipole moment,

      or is CO2 a non-polar molecule without a net dipole moment but has pdpd?

       

      so in order to have pdpd IMFs, just need two(or more) different atoms chemically combined tgt in a molecule, without a need for net dipole moment?


      I told you liao : don't blindly trust Singapore JCs or your school lecture notes. The Distinction A grade student, will not blindly follow or memorize school lecture notes, but intelligently understand the Chemistry concepts in his/her own way. And the intelligent student will also sometimes disagree with his/her school teacher, because school teachers often disagree with each other as well.

      In CO2, the linear molecular geometry ensures that the 2 individual dipoles cancel out, hence it is an overall non-polar molecule. Being non-polar doesn't necessarily always mean only London dispersion van der Waals forces are present between molecules. Intelligently apply your own chemistry understanding case-by-case. For CO2, all 3 types of van der Waals forces are present : Keesom, Debye and London dispersion forces.

      When comparing the intermolecular interactions for 2 different species separately, eg. when asked to explain the difference in melting or boiling point for 2 different species, you only need to state van der Waals forces versus hydrogen bonding or ionic bonding, and specify which is stronger. Only when both molecules have only van der Waals intermolecular forces, then it will be necessary to specify which type of van der Waals interactions are present between molecules of each species.

      But be careful, contrary to what your school (ie. Singapore JCs) may have taught you, instantaneous dipole - induced dipole London dispersion van der Waals forces may not necessarily be weaker than permanent dipole - induced dipole Debye van der Waals forces which may not necessarily be weaker than permanent dipole - permanent dipole Keesom van der Waals forces, it depends on the total no. of electrons present and the molecular size, and hence the polarizability of electron charge clouds, and hence the magnitudes of partial charges and dipoles, and hence the strength of the electrostatic van der Waals attractions, regardless of which type of van der Waals is present.

      You should be able to figure out such counter examples (which are contrary to the oversimplifications taught in Singapore JCs), including extreme counter examples where even instantaneous dipole - induced dipole London dispersion van der Waals forces may be even stronger than intermolecular hydrogen bonds! Shhhh! even if you know such counter examples, don't post them here and spoil the fun for others, let them figure these out for themselves! ;Þ


      Originally posted by MightyBiscuits:

      Hi UltimaOnline, for CO2, you mentioned that  "In CO2, the linear molecular geometry ensures that the 2 individual dipoles cancel out, hence it is an overall non-polar molecule. Being non-polar doesn't necessarily always mean only London dispersion van der Waals forces are present between molecules. Intelligently apply your own chemistry understanding case-by-case. For CO2, all 3 types of van der Waals forces are present : Keesom, Debye and London dispersion forces.", is this only true for solid state? 


      No, not only for solid state, but Keesom van der Waals forces predominate moreso for the solid state > liquid state > gaseous state, for obvious entropy reasons. If the reasons are not obvious to you, go ask your school teacher or private tutor.


      Originally posted by Flying grenade:

      but which one is the predominant (most extensive) IMFs between CO2 molecules?


      I'll leave you with a final parting statement in reply to your question quoted above : it depends on the state/phase of the CO2, and besides, Cambridge won't ask which predominates for the different states/phases, you need only state "van der Waals forces" (ie. all 3 types of van der Waals forces) as the intermolecular interactions between CO2 molecules, regardless of state/phase.

      Edited by UltimaOnline 05 Jul `16, 3:14AM
    • Originally posted by Liying98:

      https://mobile.twitter.com/itsliyingg/status/750022746466234368

       

      Part g and h


      In an octahedral coordination complex, the dz2 and dx2-y2 orbitals have a higher energy level compared to the dxy, dyz and dxz orbitals.

      Since n = 4 (ie. 4 unpaired electrons) is calculated from the formula given, and since Cr2+ is [Ar] 4s0 3d4, it means that all 4 electrons of the Cr2+ ion are unpaired, which in turn means the Cr2+ ion in this particular octahedral hexaaquachromium(II) coordination complex must be high spin (ie. small magnitude of energy gap between d-d* orbitals). This is consistent with H2O being a (relatively) weak field ligand.

      On the other hand, if the 6 ligands were stronger field ligands, eg. cyanide CN- ligands in the octahedral hexacyanochromate(II) coordination complex, then you would expect to calculate n = 2 (ie. 2 unpaired electrons), which means that all 4 electrons of the Cr2+ ion must occupy the 3 lower energy dxy, dyz and dxz orbitals, which in turn means the Cr2+ ion in this particular octahedral hexacyanochromate(II) coordination complex must be low spin (ie. large magnitude of energy gap between d-d* orbitals).

      To show the electron distribution in the transition metal within the coordination complex as required by the MJC question, check out the low spin vs high spin electron distributions for the Fe3+ (ie. [Ar] 4s0 3d5) example on this Wikipedia page, and modify it for Cr2+ (ie. [Ar] 4s0 3d4).

      https://en.wikipedia.org/wiki/Spin_states_(d_electrons)


      Due to inter-electron electrostatic repulsions, it is both kinetically and thermodynamically less favorable to add a negatively charged electron onto a negatively charged anionic coordination complex such as hexacyanochromate(III), and therefore you can expect the standard reduction potential to be significantly less positive, or concordantly more negative ; compared to a when adding a negatively charged electron onto a positively charged cationic coordination complex with the same transition metal ion with the same oxidation state such as hexaaquachromium(III).

      Edited by UltimaOnline 05 Jul `16, 3:02AM
    • Originally posted by Flying grenade:

      a chem tcher from my sch say during lecture , non polar amino acids contains no electronegative atom

      which is untrue right? 

      page 378 cs toh advanced guide, proline contains N atom, tryptophan contains N atom


      There's no such thing as a "non polar amino acid". What your school teacher is referring to, is the R group of the amino acid. And whether an R group is protic polar, aprotic polar, or non-polar, is a matter of common sense that should be instantly obvious to you from a 1-second glance, which you sure as hell shouldn't need to rely on your school teacher (or whatever he/she says) to figure out, ffs.

      I would like to give you the benefit of the doubt, and imagine you're asking the more intelligent questions on why tryptophan's R group cannot accept H bonds despite having an N atom, and is thus hydrophobic ; or why there isn't a 3rd pKa value for tryptophan despite a protonated N atom conjugate acid seemingly possible.

      But then you cock this up by asking the coconaden question about why proline's R group is non-polar?!?

      image

      As to the more intelligent questions (that you didn't explicitly ask) : I'll give you a generous hint : RESONANCE.

      Edited by UltimaOnline 05 Jul `16, 11:56AM
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