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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. -
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)
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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.
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Thanks Ultima
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i understand fully, thanks ultima
pka indicator = pkin
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my school cher say that benzene is not a functional group??
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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.
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THANKS ULTIMAAAAAA
super detailed and witty reply !!!!
and thanks for being so godly psychic telepathy, knowing and answering my follow up questions before i ask them !!
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page 306 cs toh advanced guide
step 1 of the electrophilic addition of Br2 to ethene
should the curly full headed arrow reach the Br with the partial positive charge?
i don't get it why the arrow in the middle of space
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testing
can see this photo?
if cant, then use external link
https://www.dropbox.com/s/t1sjfew4inz8u5t/Screenshot_2016-07-02-00-15-26-1.png?dl=0
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ahh yes, useful photobucket
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Originally posted by Flying grenade:
page 306 cs toh advanced guide
step 1 of the electrophilic addition of Br2 to ethene
should the curly full headed arrow reach the Br with the partial positive charge?
i don't get it why the arrow in the middle of space
You're right, it's actually a better practice (especially for students) to show the curved arrow directly attacking the partial +ve Br atom.And remember to circle your formal charges (which CS Toh neglected to do for the uninegative formal charges on the Br- and OH- ions).
Also, it's actually more correct to show H2O nucleophile, rather than OH- nucleophile, attacking the carbocation to generate the halohydrin product.
Notwithstanding, CS Toh's A Level Study Guide for H2 Chemistry (he has updated it to the new 2017 syllabus) are still strongly recommended for all Singapore JC students, and you'll be better off using CS Toh's book, over your school's lecture notes. Of course, be sure you (ie. all Singapore JC students reading this) buy Chan Kim Seng / Jeanne Tan's and George Chong's books for H2 Chemistry too.
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THANKSSS ULTIMAAA!!!
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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?
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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). -
for thiosulfate molecule, the central S atom has 5 bondpairs(10 e- around it). so is it formal charge on S atom is -4? and doubly bonded 1 an O atom and singly bonded to two S atoms
i got o.s. of S atom in S2O32- to be +2, by memorising that singly bonded O has a mononegative charge. so o.s. of S = (0+4)/2 = +2
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Originally posted by Flying grenade:
for thiosulfate molecule, the central S atom has 5 bondpairs(10 e- around it). so is it formal charge on S atom is -4? and doubly bonded 1 an O atom and singly bonded to two S atoms
i got o.s. of S atom in S2O32- to be +2, by memorising that singly bonded O has a mononegative charge. so o.s. of S = (0+4)/2 = +2
Formal charge on each of the 2 S atoms in the S2O3 2- ion is zero. OSes of the 2 S atoms are 0 and +4, hence giving an average of +2 for S in S2O3 2- ion.Formal charge on each of the 4 S atoms in the S4O6 2- ion is zero. OSes of the 4 S atoms are +5, 0, 0 and +5, hence giving an average of +2.5 for S in S4O6 2- ion.
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for O(or S) to have a uninegative charge and stable octet, 3lp, 1bp
for O to have no formal charge and stable octet, 2lp, 2bp
for O to have a unipositive charge and stable octet, 3bp, 1lp
this technique dont apply for expanded octet is it?
was puzzled why several online pics show diff structures
but ok i think one must be discerning , and trust your own concept and structure to be correct
a resonance contributor of S2O32- is , central S atom singly bonded to S, singly bonded to Two O atom and double bonded to one O atom
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i think i figured out. negative formal charge on an ion must reside in a more electronegative atom. O in the case of S2O32-.
for stable octet, a mononegative charge nor dinegative charge cannot exist in a doubly bonded, or 2bp(&2lp) O atom.
a dinegative charge cannot exist in a singly bonded O.
hence two singly bonded O atom with mononegative charge must exist in S2O32- .
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cs toh advanced guide, page 15 and 16.
error?
book stated sodium thiosulfate(VI) , three times in total for both pages. should it be thiosulfate(II) ?
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Originally posted by Flying grenade:
cs toh advanced guide, page 15 and 16.
error?
book stated sodium thiosulfate(VI) , three times in total for both pages. should it be thiosulfate(II) ?
Because the 2 S atoms each have a different OS (in both various resonance contributors and thus also in the resonance hybrid), the name of this ion should not include the OS. As such, Cambridge will only use thiosulfate ion as the name. -
i see, thanks ultima!!
i suspect where cs toh thought of an O.S. of 6.
so, which S2O32- structure is correct? cstoh&google image or the one with two singly bonded mononegative charged Oatom?
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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.
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Originally posted by Flying grenade:
i see, thanks ultima!!
i suspect where cs toh thought of an O.S. of 6.
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).
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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.
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".
