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Originally posted by Flying grenade:
A classmate of mine asked a qn in class in which the tcher was stumped : why electron shell closer to nucleus, has lower energy level?
For a simplified version (which will suffice for A level purposes), you can think of it as : to maintain your orbit further from the nucleus, you (as an electron in the outer electron shells) naturally require more energy to sustain your further orbit and resist the electrostatic pull of the positively charged nucleus.Going any deeper like the classic scientific enigmas of "Why don't electrons fall into the nucleus?", and "Why don't the protons in the nucleus repel each other and cause the nucleus to fall apart?" (also see Cordus physics) that goes significantly beyond A levels (so we we're *not* going discuss it here, and it's a safe bet that neither your school teacher nor your private tutor will entertain you on this), and at University levels, physicists and chemists have come up with a few different theories or explanations or ways of looking at these enigmas. You may come across some of these ideas in the future if you choose to study Physics or Chemistry in the University. For now, if you're still curious (though you should of course focus on your A levels), you can google out some preliminary reading for your entertainment and satisfaction.
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Thanks so much for your reply !!!
But, an e- distance further away from nucleus experiences weaker electrostatic attraction.
Ok nvm, ur preamble will suffice !
One theory to explain why an e- cannot be found inside a atom's nucleus is because of Heisenberg's uncertainty principle.
The nucleus is extremely small compared to the size of an atom, if a e- is found within the diameter of the nucleus, the velocity of it will exceed speed of light hence impossible, hence impossible to find an e- inside a nucleus.
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One way to explain why electron shell with higher principal quantum has higher energy is :
Electrical potential Energy of a charged particle(EPE) = 1/4πε( Qproton Qe-) /r, As r increase, EPE becomes less -ve(i.e. more +ve) , indicating higher energy level. Since e- occupies that certain discrete energy level of the electron shell, only for that particular e-shell for that particular energy level, one can substitute EPE= energy of Electron shell.
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Hi ultima, hope u could help me with this,
Daniel c harris book
Qn 10-10
I copied down the final answer.
For vol NaOH added =100ml (i.e. at equivalence point,
I tried to use pH = 0.5 [pka(phenylammonium) + pka(phenylamine)]
I found ka of phenylamine by finding kb of phenylamine on the internet.
Then i took kw/kb to find ka of phenylamine.
But i dont get the answer
Hope you can help check, thanks !
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Originally posted by Flying grenade:
Hi ultima, hope u could help me with this,
Daniel c harris book
Qn 10-10
I copied down the final answer.
For vol NaOH added =100ml (i.e. at equivalence point,
I tried to use pH = 0.5 [pka(phenylammonium) + pka(phenylamine)]
I found ka of phenylamine by finding kb of phenylamine on the internet.
Then i took kw/kb to find ka of phenylamine.
But i dont get the answer
Hope you can help check, thanks !
For the sake of all visitors to the forum, post the photo of the question (unless it's a TYS or Prelim paper question in which all visitors are expected to have their own access to). And your working is obviously wrong. There is no (meaningful) Ka or pKa value for phenylamine, since it is basic and not acidic. Ask yourself, what is the species present at equivalence point? Therefore, what is the relevant formula to use? What is the name given to the following formula, and exactly when do you use this formula? pH = 1/2 (pKa1 + pKa2). Your problem is you're blindly memorizing formulae without understanding them, and hence you dunno when to use which formula. -
I don't have the photo of the qn, but i wrote it down, so I'll type it out
Qn : write the reaction for titration of 100ml of 0.100M anilinium bromide (aminobenzene • HBr) with 0.100M NaOH. Sketch the titration curve for points Vb=0, 0.1Ve, 0.5Ve, 0.9Ve, Ve, 1.2Ve
Species present at equivalence pt is phenylamine.
Oh use equation of hydrolysis of phenylamine ?
Why cannot use amphiprotic formula in this case?since phenylamine is the cj base of phenylammonium cation
So not every time at equivalence point use amphiprotic formula, is it?
Thanks for those guidance in thought process that you've written down !!
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Originally posted by Flying grenade:
I don't have the photo of the qn, but i wrote it down, so I'll type it out
Qn : write the reaction for titration of 100ml of 0.100M anilinium bromide (aminobenzene • HBr) with 0.100M NaOH. Sketch the titration curve for points Vb=0, 0.1Ve, 0.5Ve, 0.9Ve, Ve, 1.2Ve
Species present at equivalence pt is phenylamine.
Oh use equation of dissociation of phenylamine ?
Why cannot use amphiprotic formula in this case?since phenylamine is the cj base of phenylammonium cation
So not every time at equivalence point use amphiprotic formula, is it?
Thanks for those guidance in thought process that you've written down !!
Use amphiprotic formula only when the species present is amphiprotic. At the various equivalence points for acid-base titrations, the species present could be either only acidic (eg. H3PO4), only basic (eg. PO4 3-), or amphiprotic (eg. H2PO4- or HPO4 2-).Instead of blindly memorizing formulae without understanding (like the majority of pitiful Singapore JC students), you should intelligently understand what's going on, intelligently identify on a case-by-case basis what's the nature of the species present, and hence intelligently apply the relevant formula.
And btw, phenylamine is a base, bases do not undergo dissociation (as erroneously taught in Singapore JCs), bases undergo hydrolysis.
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Why if a hybrid orbital (e.g. sp2) have more s character then more electron withdrawing by induction?
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Originally posted by Flying grenade:
Why if a hybrid orbital (e.g. sp2) have more a character then more electron withdrawing by induction?
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How to get a fully protonated/deprotonated form of amino acid?
Say, to fully deprotonate amino acid, we add NaOH , to protonate it we add say, HCl? But how separate out the amino acids?
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Originally posted by Flying grenade:
How to get a fully protonated/deprotonated form of amino acid?
Say, to fully deprotonate amino acid, we add NaOH , to protonate it we add say, HCl? But how separate out the amino acids?
Titration need not necessarily begin with fully deprotonated or fully protonated forms, it can also begin with a zwitterionic form (which is the form most easily obtained). And as long as you know the pKa values, with pH data loggers you can easily settle the problem. -
One reason why when a hybrid orbital has more s character ,more electronegative, and consequently more e- withdrawing by induction.
S orbitals hold electrons more tightly to nucleus than P orbitals . This implies that S orbitals are effectively more electronegative . Now, in sp2 carbon the character of each orbital has 33% of s orbital characteristics whereas in each sp3 orbital s character is 25% . So sp2 has more s character and therfore more electronegativity.
So carbons are more electronegative. The larger the s character of the atom the closer the electrons are held to the nucleus. Sp3 carbons are the least electronegative as the p character forces the electrons to exists further from the nucleus, decrease the electronegativity2nd reason : sp^2 hybridized carbon has more lower-binding-energy electrons.
Picture for 2nd reason and source is from : https://www.quora.com/How-is-sp-2-or-sp-hybridized-carbon-more-electronegative-and-stronger-than-sp-3
3rd reason : Bent's rule
Bent’s rule explains deviations in structures predicted from the VSEPR theory The rule states: “atomic s character tends to concentrate in orbitals that are directed toward electropositive groups and atomic p character tends to concentrate in orbitals that are directed toward electronegative groups.. to maximize bonding energy.. s orbitals are closer to the nucleus, allowing for greater stabilization of the lone pair.
Source : https://in.answers.yahoo.com/question/index?qid=20111105080151AAjDAy7 and https://en.m.wikipedia.org/wiki/Bent%27s_rule
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Originally posted by Flying grenade:
One reason why when a hybrid orbital has more s character ,more electronegative, and consequently more e- withdrawing by induction.
S orbitals hold electrons more tightly to nucleus than P orbitals . This implies that S orbitals are effectively more electronegative . Now, in sp2 carbon the character of each orbital has 33% of s orbital characteristics whereas in each sp3 orbital s character is 25% . So sp2 has more s character and therfore more electronegativity.
So carbons are more electronegative. The larger the s character of the atom the closer the electrons are held to the nucleus. Sp3 carbons are the least electronegative as the p character forces the electrons to exists further from the nucleus, decrease the electronegativity2nd reason : sp^2 hybridized carbon has more lower-binding-energy electrons.
Picture for 2nd reason and source is from : https://www.quora.com/How-is-sp-2-or-sp-hybridized-carbon-more-electronegative-and-stronger-than-sp-3
3rd reason : Bent's rule
Bent’s rule explains deviations in structures predicted from the VSEPR theory The rule states: “atomic s character tends to concentrate in orbitals that are directed toward electropositive groups and atomic p character tends to concentrate in orbitals that are directed toward electronegative groups.. to maximize bonding energy.. s orbitals are closer to the nucleus, allowing for greater stabilization of the lone pair.
Source : https://in.answers.yahoo.com/question/index?qid=20111105080151AAjDAy7 and https://en.m.wikipedia.org/wiki/Bent%27s_rule
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Hi ultima , i would like to ask,
I have written down the kb value from the soln book,
I am confused by this part
In part i), the ka value is for phenylammonium cation, correct?
For the part of vol base added = vol equivalence, i used kw/ka(phenylammonium) to get kb(phenylamine)
So, when i use kw/ka is get kb of the cj base ah??
Damnit, all the while i thought its ka and kb of the same molecule !
When i apply kw/ka to find kb, is the kb the kb of same molecule or kb of its cj base!
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Originally posted by Flying grenade:
Hi ultima , i would like to ask,
I have written down the kb value from the soln book,
I am confused by this part
In part i), the ka value is for phenylammonium cation, correct?
For the part of vol base added = vol equivalence, i used kw/ka(phenylammonium) to get kb(phenylamine)
So, when i use kw/ka is get kb of the cj base ah??
Damnit, all the while i thought its ka and kb of the same molecule !
When i apply kw/ka to find kb, is the kb the kb of same molecule or kb of its cj base!
Ka of HA x Kb of A- = KwKa of BH+ x Kb of B = Kw
Ka of conjugate acid x Kb of conjugate base = Kw
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Thank u ultima saviour! !!
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Why is Mno4- a stronger oxidising agent compared to Cr2O72-?
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One of my teacher say that the reason why H- cannot reduce alkene is because alkene C=C bond is neutral(i.e. not polar)
All along i was under this impression.
Today, another different teacher say the reason is because the lone pair on H- repels with the pi e-cloud of C=C bond
:/ i am flustered
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Pang cheng cheong's book and my school cher say reaction of carbonyl cpds with 2,4dnph is at r.t. , but cs toh book say heat, and making sense's reagent and conditions say need warm
How??
These reagents and conditions makes me agitated !
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Help ultima..
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Idk but this is one of my thought
24dnph reaction(condensation/nu sub) can be carried out under r.t.
But perhaps warm, accelerates the reaction , right?
So both are acceptable.
My doubt, would cambridge mark both correct??
If i write heat will they know cos i know the reaction will be quicker? Or will they think i am dumb , the reaction requires heat to overcome Ea of reaction?
So write 24dnph, r.t. more safe is it?
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Sian reagents and conditions damn variable
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Originally posted by Flying grenade:
Why is Mno4- a stronger oxidising agent compared to Cr2O72-?
Because of a complex combination of reasons, which you needn't worry about for A levels. Most relevantly, the oxidation state of the heteroatom in the oxidizing agent (ie. +7 versus +6), but also the electronegativity, the charge density, the electron affinity enthalpy, hydration enthalpy, electron configuration, reaction mechanism, etc.So when you explore these issues at Uni level, you'll realize there are no longer any short, simple answers (those are the O level days of naivety), but a beautifully complex balance of separate (and oftentimes opposing factors). A levels is a transition state or intermediate between naively simplistic O levels, and beautifully complex Uni levels.
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Originally posted by Flying grenade:
One of my teacher say that the reason why H- cannot reduce alkene is because alkene C=C bond is neutral(i.e. not polar)
All along i was under this impression.
Today, another different teacher say the reason is because the lone pair on H- repels with the pi e-cloud of C=C bond
:/ i am flustered
See if you're intelligent enough to see how the following Singaporean link explains the answer to this H2 Chemistry question.http://mothership.sg/2016/05/wear-white-movement-to-be-held-on-the-same-day-as-pink-dot-yet-again/
Whether you get it or not, don't post further on this topic on this forum, don't spoil the fun for others! *evil laugh*
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Originally posted by Flying grenade:
Pang cheng cheong's book and my school cher say reaction of carbonyl cpds with 2,4dnph is at r.t. , but cs toh book say heat, and making sense's reagent and conditions say need warm!
How??? These reagents and conditions makes me damn f**king agitated!
Idk but this is one of my thought 24dnph reaction(condensation/nu sub) can be carried out under r.t. But perhaps warm, accelerates the reaction , right? So both are acceptable. My doubt, would cambridge mark both correct?? If i write heat will they know cos i know the reaction will be quicker? Or will they think i am dumb , the reaction requires heat to overcome Ea of reaction? So write 24dnph, r.t. more safe is it?
Allow BedokFunland JC to enlighten thee :As far as 2,4-DNPH is concerned, Cambridge doesn't give a flying f**k whether you write "room temperature" or "warm" or "heat" or even if you don't specify the temperature at all.
However, don't mistake Cambridge's kind reasonable tolerance for weak bochup nonchalance : between PCl3 and PCl5, you *must* specify "heat" for 1 of these reagents, or Cambridge will penalize you. If Cambridge asks you to explain why this difference, do you know how to answer?
(let's hope for more such out-of-the-box questions for this year's and all future Singapore H2 Chemistry papers, shall we? That will shut up those armchair critics who take perverse pleasure in criticizing the entire Singapore education system to be all about blind memorization without deeper understanding... such criticism applies more for PSLE, N and O levels, but less so for A levels and certainly not Uni levels; and among A level subjects, such criticism is even less valid for Chemistry than other A level subjects, hopefully Cambridge will continue this trend for Singapore's H2 Chemistry).