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    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
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    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
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    8,018 posts since May '05
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    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
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    8,018 posts since May '05
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    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
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    • Originally posted by Flying grenade:

      oh i got it liao. H bonds.

      water higher bp than ethanol cos can form 2 H bonds per molecule

      sulfur compounds only have weaker(compared to H bonds) intermolecular PdPd


      Correct. But don't anyhow *pom* out a number for H bonds. Each water molecule can form up to 4 H bonds, while each ethanol molecule can form up to 2 H bonds.

      So what you need to say, is that water molecules form a larger number of H bonds between them, compared to the number of H bonds between ethanol molecules.

      But you also need to add that ethanol molecules also have intermolecular van der Waals forces (all 3 types of van der Waals forces, which you also need to state are also the intermolecular interactions between ethoxyethane molecules), which you have to specify are weaker than H bonds.

      So just becoz u got the idea of what's happening liao, but u will still lose marks in the A levels if ur answer doesn't specifically have all the specific points.

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    • Originally posted by Flying grenade:

      Explain why the boiling points of water(H2O), ethanol(CH3CH2OH), ethoxyethane(C2H5OC2H5) are in reverse order of their molecular masses, unlike those of their analogous sulfur compounds : H2S, C2H5SH , C2H5SC2H5


      image

      Also, I told u liao, don't just post qns only (I'm not here to do ur homework for u), post ur own answer, together with ur school's given answer, and ask exactly what u dun understand or agree with, then I'll guide & advise.

      Edited by UltimaOnline 28 Jun `16, 2:00AM
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    • 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
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    • 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, 3:13PM
  • Moderator
    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
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    8,018 posts since May '05
  • Moderator
    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
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    8,018 posts since May '05
  • Moderator
    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
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    8,018 posts since May '05
  • Moderator
    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
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    8,018 posts since May '05
  • Moderator
    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
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    8,018 posts since May '05
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    • Originally posted by Flying grenade:

      ultima i confused...


      Yeah, I knew u would be. U got too many weaknesses in ur understanding of this topic. U need face-to-face help. Go ask ur school teacher or private tutor. With ur great wall of text, u didn't seriously think I would reply ur 1001 qns on the forum, did u?



      Originally posted by Flying grenade:

      it can also undergo WA-SB reaction


      And quit ur nonsensical acronym usage. All I see is WASABI. Be prepared to be heavily penalized by Cambridge if u use acronyms in the A levels.

      image

      Edited by UltimaOnline 27 Jun `16, 1:57AM
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    • Originally posted by Flying grenade:

      Page 158 cstoh advanced guide polybasic acids

      for example 1

      where only NaHSO4 and H2O is left in the soln, the soln is still pH≈2 , strongly acidic

      is the acidity due to

      HSO4- 《》SO42- + H+ 

      or

      NaHSO4 + H2O 《》H2SO4 + H+      ?? idk 

       

      why for only NaHSO4 and H2O in soln, soln is acidic? 



      HSO4- is technically amphiprotic, and if you're given both Ka1 and Ka2 values (unlike for A levels, since Cambridge usually simplifies it to be a strong diprotic acid), go ahead and use the amphiprotic formula to calculate the pH.

      Otherwise, because even the Ka2 value (for HSO4-) is still much larger than the Kb1 value (for SO4 2-), you're allowed to just regard it as acidic rather than amphiprotic : either method of calculation should give the same pH value to 3 sig fig.

      Edited by UltimaOnline 26 Jun `16, 10:49PM
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    • Originally posted by Flying grenade:

      page 223 cs toh advanced guide 

      how do we know element Y exists in two oxidation states, and how do we know it could belong to group IV(group 14) or V(grp 15) ?

      group 16 chlorides also decomposes on heating to liberate Cl2

      from pg 218 , the oxidation number of an element corresponds to the number of e- used for bonding 

      but still can't deduce why element Y exists in two oxidation states

       


      Because as part of the H2 syllabus, you're supposed to be familiar with (ie. have already memorized) the various chlorides that exist for the different Groups. Eg. PbCl2 & PbCl4 for Group 14, and PCl3 & PCl5 for Group 15.

      And the data given in the question tells you that 2 different formula of the chloride of Y are able to exist (coz 1 sample was heated to liberate some Cl2, giving a different chloride of Y), that's why you're expected to deduce element Y exists in two oxidation states in its chloride.

      It's true that Group 16 is also possible (SCl4 is unstable under standard conditions, but indeed 2SCl2 can decompose into S2Cl2 and Cl2), but it's not within the H2 syllabus, so you're not required to include Group 16 in your answer, but go ahead and write it in if you like, as a 3rd possibility.

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    • [Singapore] - The Red Butterfly : Notorious Women Gangsters of the 1960s

      image

      Dressed in black tight-fitting clothes, the woman gangsters often used their belts to whip their female victims into submission. Sometimes, they would brutally attack the victims’ faces using stone-removed ring prongs as jagged-edged weapons. Those who refused to pay them money were facially disfigured and maimed.

      Organized crime, secret societies and gang fights were rampant between the fifties and seventies. Each secret society controlled its own territory tightly, operating illegal businesses like chap ji kee (a lottery game), gambling dens, opium dens and brothels. It also gained other sources of income through protection money, extortion, robberies and kidnapping. Clashes over territories, interests or revenges were so frequent that they were almost like daily affairs.

      The major secret society groups in the sixties were the 108, 24, 32 and 36. Over the years, they expanded so fast and large that they often had branches or small triad groups under them. For example, the Pek Kim Leng (White Golden Dragon) was under 108, and it controlled territories from Chinatown to Bugis.

      Different areas in Singapore were “owned” by different secret societies and gangs. The Tiong Bahru vicinity was, for example, “ruled” by Ang Peh Hor and Hai Lock San, whereas Ang Soon Tong was the active triad around Nee Soon and Sembawang Road. See Tong, affiliated to the infamous drug-dealing Ah Kong gang, made their presence felt at North Bridge Road, Seah Street and Beach Road. Sio Oh Leng, Leng Hor San and Sar Ji each established their respective “strongholds” at River Valley Road, Havelock Road and Boat Quay.

      The Red Butterfly did not vie for territories with other secret societies; they were mainly active at the nightclubs and bars around Cecil Street. But they did often get into fights with other women gangs at Clifford Pier, Geylang, Jalan Besar, Sungei Road and near the Capitol Theatre. Police arrested the Red Butterfly gang members several times but could not bring charges to them as there were either not enough evidences or the victims were too terrified to testify against them.

      https://remembersingapore.org/2016/06/01/sixties-gang-the-red-butterfly/

      Edited by UltimaOnline 26 Jun `16, 5:01PM
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    • 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.

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    • 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.

      Edited by UltimaOnline 26 Jun `16, 2:32PM
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    • 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.

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    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
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    • 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.

  • Moderator
    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
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    8,018 posts since May '05
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    • Originally posted by Flying grenade:

      2013/p1/qn36

      how do we know if group III(now group 13)salts form (highly) colored salts or not?

      we know NaCl, Ag salts, (insoluble) sulfate salts, carbonate salts, transition metal salts, all have color

      dont think we have learn color of group 3 salts in whole jc life before?

      a wikipedia search shows that AlCl3 : "appearance white or yellow solid " . will these count as highly colored (ionic) salts, as posited by an option from the qn?

       

      Thanks Ultima

       


      As far as the A level Chemistry syllabus is concerned, you'll ONLY be asked to explain the origin in color of transition metal coordination complexes, for the flame tests, or why non-transition metal complexes or salts (eg. zinc salts), or certain transition metal complexes or salts with certain OSes (eg. copper(I) salts) do not have color.

      You will be required to memorize colors of all other species relevant to the H2 syllabus, but questions will NOT be asked on how any color present in any other species (whether covalent or ionic, inorganic or organic) arises. Eg. you need to know the color of halogens (on it's own in gaseous/liquid/solid states, or as a solute in aqueous solvent or organic solvent), but you won't be asked on why each halogen has that color.

      Bottomline, as far as color is concerned, don't worry about Group 13 compounds (not examinable in syllabus), just focus on memorizing and explaining the colors of transition metal coordination complexes.

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