BedokFunland JC's A Level H2 Chemistry Qns (Part 2)
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Originally posted by hoay:
Amines are basic since the lone pair of nitrogen is available for donation. Amides are not neutral because the lone pair of N is delocalized over the C and O atoms.
What about Nitriles e.g CH3-CN.?? the lone pair is present....Are they basic?
secondly, i am curious to know is basicity has anything to do with hybridization? Because in methylamine the C is sp3 hybridized, in amide it sp2 and in -CN it sp hybridized.
You're correct. There are two distinct factors at work here : resonance, and % s-orbital character.Resonance
If the lone pair is delocalized by resonance (through the sideways overlapping of p orbitals), such as in the case of amides and phenylamine, the lone pair will accordingly be less available to accept a proton.
Percentage s-orbital character
Since s-orbitals are located closer to the nucleus compared to p-orbitals, when a lone pair occupies an orbital with greater % s-orbital character (eg. sp hybridized orbital in nitriles > sp2 hybridized orbital in amides > sp3 hybridized orbital in amines), the lone pair is accordingly more strongly held back electrostatically by the positively charged nucleus, and consequently has a reduced basicity as well as nucleophilicity.
Amides are non-basic for both reasons, while nitriles are non-basic due to the % s-orbital character factor.
Based on the factor of % s-orbital character, nitriles are even less basic compared to amides. However, amides have the additional factor of resonance, and thus both amides and nitriles may be considered approximately equally non-basic overall.
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The lattice energy of a metal is small compared with that of an ionic compound. The lattice energy of Ca is -178 kJmol^-1 and that of Ca(OH)2 is -2506 kJmol^-1
Comment on the difference between their lattice energies.------------------------------------------------------------------------------------
For the same metallic element, metallic bonds (ie. metallic lattice dissociation/formation enthalpy) are generally weaker compared to ionic bonds (ie. ionic lattice dissociation/formation enthalpy), because unlike ionic bonding where, taking for instance Ca(OH)2, we have an actual high charge density dipositive Ca2+ cation present, and thus strong ionic bonds (ie. electrostatic attraction between cation and anion), in contrast for metallic bonding the Ca atom only bears a partial positive charge, because in metallic bonding the 3s valence electrons are not fully dissociated from the Ca atoms, in turn because the high charge density Ca2+ cation electrostatically attracts back the delocalized 3s valence electrons to a partial extent.
Consequently, not only is the actual charge on the Ca atom (in metallic bonding) much lower in magnitude (compared to Ca2+ cation vis-à-vis ionic bonding), but furthermore and simultaneously, due to the same reason (of partial only dissociation of the valence 2s electrons), the availability of the valence 2s electrons to participate in metallic bonding is also reduced.Therefore, for both (intimately related) reasons above, metallic bonding is significantly weaker compared to ionic bonding (for the same metallic element), and hence the metallic lattice dissociation/formation enthalpy is significantly less endothermic/exothermic compared to its corresponding ionic lattice dissociation/formation enthalpy (for the same metallic element).
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Explain how you would expect the numerical magnitude of the LE of sodium chlorate (I) to compare with that of sodium chloride.
The LDE/LFE of NaClO will be less endothermic/exothermic compared to the LDE/LFE of NaCl, because ClO- has lower charge density compared to Cl-.
(i) Explain why the pKa values are increasing down the table.
Successive proton dissociation constants are always smaller, because the increasing magnitude of negative charge on the successive conjugate bases result in decreasing stabilities of the species. Furthermore, negatively charged bases will repel other negatively charged bases.
(ii) Identify the acid/conjugate base pair used to make a phosphate buffer of pH 6.8
At pH = pKa, we have equal molarities of both members of a conjugate acid-base pair, and we have maximum buffer capacity. Hence we will need a slightly higher molarity of H2PO4 - and a slightly lower molarity of HPO4 2- to achieve a buffer with pH of 6.8 (ie. slightly below 7.21).
When aq. iron (III) chloride is shaken with aq sodium carbonate, no ppt of iron (III) carbonate is formed. Instead, effervescence and a red-brown ppt of iron (III) hydroxide are observed.
2 reasons :
The high charge density Fe3+ ion will polarize and distort the anionic charge cloud of CO3 2-, resulting in it's decomposition (becoming less endothermic).
Simultaneously, the high charge density Fe3+ ion is strongly electron withdrawing by induction, withdrawing electron density from its water ligands through the coordinate dative bonds, weakening the O-H bonds in the water ligands, resulting in proton dissociation from the water ligands (becoming less endothermic).
This results in an acidic solution whereby the CO3 2- ions are protonated to generate H2CO3, which exists in equilibrium with, and hence can decompose into, CO2 and H2O.
The water ligands that were deprotonated earlier constitutes the OH- ions that now combine with the Fe3+ ions (when ionic product Qsp exceeds solubility product Ksp) to generate the Fe(OH)3 ppt.
Iron (II) iodide can be crystallised from an aq mixture of iron (II) sulfate and potassium iodide but iron (III) iodide cannot be produced by mixing aq iron (III) sulfate and potassium iodide.
Fe3+ will oxidize I- to I2, and I- will reduce Fe3+ to Fe2+.
Cell potential = Reduction potential @ Cathode + Oxidation potential @ anode.
Use Data Booklet values to show that standard cell potential is positive, and therefore you cannot obtain FeI3(aq) under standard conditions.
Suggest an advantage of using nickel carbonyl complex in the mond process.
CO complexes with nickel readily and reversibly, allowing convenient extraction from nickel oxides. On heating, nickel tetracarbonyl decomposes to give the desired nickel metal, and the carbon monoxide gas, which can hence be easily separated.
0.382 g of the nikel carbonyl complex was completely vaporised in a gas syringe at 120 deg cel and 1 atm. 72.4 cm3 of vapor was produced. calculate the % by mass of CO in the complex.
Note that the final vapour is purely CO, as nickel would present as a solid residue. Using PV=nRT, find the moles of CO. Hence find sample mass of CO. Therefore find % by mass of CO in the complex.
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kickme asked :
1) Is amide soluble in dilute HCl?
Due to the electron withdrawing carbonyl group, the lone pair of electrons on N is not available for coordination to a proton. Hence it is not soluble in HCl. However, what is the true meaning of being soluble? Does it mean forming bonds with the solvent?
2) Will a N-O bond break in alkaline hydrolysis?
In alkaline hydrolysis for amide, the C-N bond breaks.
If i have a compound, C6H5-CONH-OCH2CH3, will the N-O bond break too?
Q1. It means any new bonds or interactions between the would-be solute and would-be solvent, must be comparable to, or more favourable than, the existing solute-solute and solvent-solvent bonds or interactions. Amides are not soluble in dilute HCl because it is not sufficiently basic to be protonated to generate a conjugate acid cation capable of ion - permanent dipole interactions with water (it is insufficiently basic because the lone pair on the N atom is delocalized by resonance to form a pi bond with the carbonyl group, and is thus less available to accept a proton).Q2. Due to the resonance delocalization of the lone pair on the N atom to form a pi bond with the carbonyl group, the resonance hybrid's C-N peptide bond of amide groups have double bond character, and does will not cleave readily (and also does not rotate freely, which has important implications for the secondary and tertiary structure of proteins). If you wish to perform alkaline hydrolysis (more correctly called base-promoted hydrolysis), you will need to heat under reflux with concentrated alkali, in order to cleave the peptide bond of the amide group as part of the hydrolysis process.
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Originally Posted by kickmeThank you, I understood your explanation to question 1.
For question 2, however, I feel that you may have misunderstood the bond which I am talking about. The bond is not part of the peptide bond but in particular, the N-O bond which I have highlighted in red. My question was whether the N-O bond would break under alkaline hydrolysis to break the peptide bond.
C6H5-CONH-OCH2CH3
Ahh yes, my bad.You're referring to the alkoxyamine O-N bond. I misunderstood because the alkoxyamine functional group is not taught by JCs, but with the trend of increasing difficulty of H2 Chem exams, it's understandable for JCs and Cambridge to start bringing in more and more stuff beyond the syllabus.
The answer to your question is : the alkoxyamine O-N bond is fairly stable, and certainly much more stable compared to the peptide bond of the amide functional group. The amide functional group may not be very basic (and even less so acidic), and neither is it very nucleophilic nor electrophilic, but if forced to react, it (specifically the partial positive carbonyl C atom) will behave (primarily, although in chemistry as in real life, be open to other possibilities) as an electrophile, and be attacked by OH- nucleophile or H2O nucleophile during hydrolysis, resulting in the cleavage of the peptide bond of amide groups.
In contrast, the alkoxyamine bond is more stable because the alkoxyamine (in which both the O and N atoms may have either hyrogens or alkyl groups bonded to each them) is neither very basic nor acidic nor nucleophilic nor electrophilic. Why? Because note that both N and O atoms are highly electronegative. When bonded to each other, the O atom being slightly more electronegative, bears a slight partial negative charge, and the N atom, being slightly less electronegative, bears a slight partial positive charge.
Normally, N atoms are significantly more nucleophilic and basic compared to O atoms, for instance, (phenylamines versus phenols, or amines versus alcohols) for two reasons : firstly O is significantly more electronegative (ie. selfish with electrons) compared to N, and thus O is less willing, or inclined, to donate dative bonds (which is what nucleophiles and Bronsted-Lowry bases, both being subclasses of Lewis bases, do) compared to N. Secondly, due to their different valencies (by virtue of the number of valence electrons), N usually has an OS of -3 while O usually only has an OS of -2, which makes N usually more electron-rich (and at the same time because N is less electronegative due to lower magnitude of effective nuclear charge) and hence more willing, or inclined, to donate dative bonds.
However, in alkoxyamines, because the adjacent O atom is more electronegative and thus electron-withdrawing by induction onto the N atom, this results in the N atom becoming significantly less nucleophilic and less basic. And because the N atom is only slightly less electronegative than the O atom, the O atom isn't terribly electron-rich, and thus not very nucleophilic or very basic, either.
Furthermore, the small difference in electronegativities between O and N, mean that the alkoxyamine ON bond is not particularly polarized, and therefore its bond dissociation enthalpy is fairly highly endothermic, and not readily cleaved.
Therefore, to answer your question, hydrolysis (whether acid-catalyzed, or base-promoted) would more readily cleave the peptide bond of amide groups, compared to the hyroxylamine bond.
However, prolonged heating under reflux, particularly under strongly acidic conditions (why acidic rather than basic? the answer lies in whichever mechanisms are more kinetically and thermodynamically feasible), may still cleave the alkoxylamine O-N bond, as follows : protonation of the O atom results in a positive formal charge on the O atom, and an even stronger partial positive charge on the N atom (because you should have realized that an atom with a positive formal charge becomes even more strongly electron-withdrawing by induction, something JCs unfortunately don't teach H2 students), making it rather electrophilic.
Upon (the electrophilic N atom experiencing an) attack by a nucleophile, the alkoxyamine O-N bond cleaves heterolytically, (much to the relief of the O atom, because O being the 2nd most electronegative element, hates the positive formal charge), generating an alcohol and an N-(further)substituted amine.
For 'A' level H2 Chem purposes, Cambridge will specify in the question, whether or not the alkoxyamine functional will react (eg. as amine, as alcohol, as nucleophile, as electrophile, as base, etc), or behave as an inert functional group, for the purpose of the exam question.
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It is often that I encounter a compound with more than 1 electronegative (F, O, N) atom and more than 1 lone pairs of electrons. However, the number of hydrogen they often are said to form in the answer scheme do not reflect the maximum possible number. How do I know exactly how many hydrogen bonds are formed?
Of all the maximum possible of hydrogen bonds possible, at any given time, there exists only less than half of this number. Several reasons for this include : sterics (hinderance), electronics (induction), entropy (balanced with enthalpy).
Hence, multitudes of hydrogen bonds are being broken and formed at every moment. If you could see them as light (for instance, in a cloud of water vapour), you'll be seeing rapid flashes (macham lightning) rather than a constant neon tube.
Furthermore, you mentioned lone pair. Some students have the misunderstanding that, if an F, O or N atom has 3 lone pairs, it means it can form 3 H bonds. Not true. Any atom is capable of participating (ie. either donating or accepting) at most one H bond at any given time.
This is because a H bond is somewhere between an ionic bond, and a dative covalent bond, in terms of it's nature (though weaker than both). Recall that the moment an atom donates a dative covalent bond, it will gain a unipositive formal charge. For instance, in the NH4+ ion and the H3O+ ion, the positive formal charge is on N atom and O atom respectively (even though lesser JC teachers might erroneously insruct you to show the positive formal charge on the H atom), because the N and O atoms donated a dative bond to accept the incoming proton (H+ ion).
When you donate a dative bond, in terms of stable octet, you haven't lost any electrons. But in terms of formal charge, you've lost an electron. Hence the positive formal charge on the dative bond donor.
Similarly, for H bonds, once a lone pair of an F, O or N atom is 'donated' to participate in H bonding, it becomes less electron-rich and therefore cannot use it's remaining lone pairs to donate any more H bonds.
For A level purposes, don't worry about which of the possible H bonds you should illustrate in your answer. Just illustrate (and label!) at least 2 different types of H bonds possible (eg. between NH3 and H2O, one type of H bond is donated by NH3 to H2O, the other type of H bond is donated by H2O to NH3, draw one of each types to secure full marks).
Google/Wikipedia to understand the difference between H bond donor, and H bond acceptor (something most JCs don't usually teach to their students).
Finally, note that, as far as possible (for some molecules, it can admittedly be difficult to reflect this), your partial positive H atom's H bond and covalent bond, both should be 180 deg or linear to each other. (To understand why, recall what I said earlier : H bonds are akin to ionic bonds; what's the shortest distance between two oppositely charged particles electrostatically attracted to each other?) Even though most JCs don't usually teach their students this, but Cambridge Mark Schemes sometimes penalize the violation of this point.
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Originally posted by atomos:
Have a question on predicting entropy signs, in the reaction for combustion of ethane,
2C2H6 (g) + 7O2(g) --> 4CO2 (g) + 6H2O(l)
To deduce the whether entropy is increasing or decreasing, should we say
1) entropy is decreasing as , entropy of gases >> liquids > solids, and we have less molecules of gases (9 --> 4) in the end.
or
2) entropy is increasing as we have more molecules of substances ( 9--> 10) in the end?
Gases have much greater entropy compared to liquids or solids. Hence, for most questions, it'll suffice to compare number of moles of gases on LHS vs RHS. In your qn above, moles of gas has been reduced from 9 (on LHS) to 4 (on RHS), and therefore it can be deduced that entropy change is negative. -
Originally posted by hoay:
How you distinguish between phenol and methoxy benzene?
I think Adding neutral FeCl3 (aq) will distinguish. Only phenol will give a violet-color complex while methoxy benzene will have no reaction.
Correct.Methoxybenzene, unlike phenol, does not have an acidic proton which can be lost, which is a prerequisite for its function as a ligand in ionic coordination complex formation presenting as a visible violet coloration, as only the phenoxide ion (rather than phenol or methoxybenzene) is sufficiently electron-rich to function effectively as the ligand.
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Originally posted by hoay:
(i) SnO2 reacts with concentrated sulfuric acid to form a colourless solution with no
evolution of gasSnO2 + 2H2SO4 Sn(SO4)2 + 2H2O
(ii) PbO2 reacts with concentrated sulfuric acid to give a white solid, B, and oxygen gas.
PbO2 + H2SO4 PbO + PbSO4 + O2 + H2O
B
(iii) (a) PbO2 reacts with cold concentrated hydrochloric acid to give a yellow solution
containing the [PbCl 6]2– ion, with no evolution of gas. (b) Warming this yellow solution causes the evolution of Cl 2 gas, leaving a colourless solution which on cooling in ice precipitates a white solid, C.(a) PbO2 + 6HCl [PbCl6]2- + H2O
(b) [PbCl6]2- Cl2 + [PbCl4]2-
C
Please make any corrections. Thank you.
Interesting inorganic chemistry deductive elucidation questions.Your answers all appear to be fine, but the very last question warrants further discussion.
Heating provides the activation energy for a redox reaction to occur, in which lead(IV) (in the dinegative hexachloroplumbate(IV) ion) is reduced to lead(II) (thus generating the dinegative tetrachloroplumbate(II) ion), and 2 of the chloride ligands are oxidized to chlorine gas.
However, because the question specified that the resulting product is a "colourless solution which on cooling in ice precipitates a white solid, C", it is unlikely to be the soluble dinegative tetrachoroplumbate(II) ion, which as a dinegative ionic species will experience thermodynamically favourable ion-permament dipole bonding with strongly polar protic water.
The white solid C is therefore more likely to be relatively insoluble lead(II) chloride.
As described earlier, heating provides the activation energy for the transfer of electrons from 2 of the chloride ligands to reduce the lead(IV) (in the dinegative hexachloroplumbate(IV) ion) to lead(II) (thus generating the dinegative tetrachloroplumbate(II) ion), and the 2 chloride ligands are oxidized to chlorine gas, and so we get the reaction exactly as you (correctly) described :
[PbCl6]2-(aq) <---> Cl2(g) + [PbCl4]2-(aq)
However, that is probably not the final equation that generates C.
Bearing in mind that "cold concentrated hydrochloric acid" was used to generate the yellow complex ion hexachloroplumbate(IV) (which btw, is yellow due to other electron transitions other than d-d*, since lead is not a d-block transition metal and does not possess partially filled d-orbitals, a possible common exam trick question that many 'A' level students fall for), therefore the coordination compound containing the yellow complex ion, is dihydrogen hexachloroplumbate(IV). So there are plenty of protons to serve as the positive counter ion where necessary.
The tetrachloroplumbate(II) ion itself exists in equilibrium with its constituent species, as illustrated in the equation for the formation constant (Kf) a.k.a. stability constant (Kstab) as follows :
Pb2+(aq) + 4Cl-(aq) <---> [PbCl4]2-(aq)
and that all relatively insoluble precipitates nonetheless exist in equilibrium with their constituent species, as illustrated in the equation for the solubility product, Ksp :
PbCl2(s) <---> Pb2+(aq) + 2Cl-(aq)
Because C is rather insoluble at low temperatures (as specified in the question), it's conceivable that, should the forward reaction (for either or both Kf-Kstab and Ksp equations above) be endothermic, lowering the temperature will shift the position of equilibrium to the left, generating PbCl2(s), with the protons (available as mentioned earlier) acting as counter ions for the Cl- ions, in effect generating HCl(g), which in gaseous state leaves the reaction mixture, pulling the position of equilibrium over to the right as predicted by Le Chatelier's principle, with a favourable positive entropy change as a thermodynamic bonus, too.
H2[PbCl4] <---> PbCl2(s) + 2HCl(g)
To summarize in chronological order, the equations would be :
Coordination complex formation :
PbO2(s) + 6HCl(l) <---> [PbCl6]2-(aq) + H2O(l)
Redox reaction :
[PbCl6]2-(aq) <---> Cl2(g) + [PbCl4]2-(aq)
Coordination complex equilibria :
[PbCl4]2-(aq) <---> Pb2+(aq) + 4Cl-(aq)
Ionic solubility equilibria :
Pb2+(aq) + 2Cl-(aq) <---> PbCl2(s)
Overall chemical equation :
PbO2(s) + 6HCl(l) ---> H2[PbCl6](aq) + H2O(l) ---> Cl2(g) + H2[PbCl4](aq) ---> Cl2(g) + Pb2+(aq) + 2Cl-(aq) + 2HCl(g) ---> Cl2(g) + PbCl2(s) + 2HCl(g)
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Originally posted by hoay:
Wow!! Excellent explanation. Thank you very much Sir.
But PbO is yellow so what about my equation of PbO2 with H2SO4.
PbO2 + H2SO4 PbO + PbSO4 + O2 + H2O.
Is it correct?
Oops, yes thanks for pointing that out.Lead(IV) compounds are unstable, and thus often functions as oxidizing agents as they act to reduce themselves to more stable lead(II) compounds.
Here, sulfur already has a maximum OS of +6, and thus cannot be oxidized further.
Therefore, it is oxygen that is oxidized : one (mole) of the oxide anion loses (two moles of) electrons to (one mole of) the eager I-cant-wait-to-be-reduced lead(IV) ion which is reduced to (one mole of) the more stable lead(II), and the oxide anion itself becomes oxidized to (half mole of) molecular dioxygen gas.
The other oxide anion, being a strong base (due to its dinegative high charge density) is protonated by the strong sulfuric(VI) acid, generating water.
When ionic product exceeds solubility product, the lead(II) ion combines with the sulfate(VI) anion, generating the white solid B, ie. PbSO4(s) precipitate.
PbO2(s) + H2SO4(l) --> PbSO4(s) + H2O(l) + 1/2 O2(g)
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Originally posted by hoay:
I did not pay any attention to the comparative rate of hydrolysis of proteins and amides. Indeed proteins would require a longer time than ester and amides. Thank you very much Sir.
The use of acid halides as halogentaing agent is common. But there is confusion regarding their conditions. I am aware of the following reagents and conditions.
PCl3 heat
PCl5 room temperature
SOCl2 room temperature
Or P + Cl2 + heat
In case of Br:
PBr3 heat
P + Br2 + heat
I think you cannot quote PBr5 directly as it does not exist.
I am not aware of use of iodine.
please make any corrections.
Yes, your stated conditions are all correct.
However, for generating phosphorous halide in situ, it is advised that the 'A' level student specify the use of red phosphorous, as opposed to white phosphorous ('A' level students should be able to explain why; see Wikipedia on phosphorous).
There are at least 2 different reasons why PCl5 exists, but not PI5. Cambridge may ask the 'A' level student to "suggest" reasons for this.
The first reason has to do with oxidizing power. Iodine has much weaker oxidizing strength compared to chlorine ('A' level students should be able to explain this, based on redox potentials as well as atomic structure), and thus while Cl has the capacity to oxidize P to an OS of +5 in phosphorous(V) chloride, iodine only has the capacity to oxidize P to an OS of +3 in phosphorous(III) iodide.
The second reason has to do with steric considerations. Iodine has a much larger atomic radius compared to chlorine, and therefore an analogous PI5 would experience too much steric crowding and van der Waals repulsion (specifically, lone pair - lone pair repulsions) between the 5 iodine atoms in the trigonal bipyramidal geometry, destabilizing the molecular structure.
Being an intermediate between chlorine and iodine, bromine is still able to form PBr5, but it is an unstable species and students should just quote the use of PBr3 instead.
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Originally posted by atomos:
Hi all, need help in checking approach for these 2 questions.
Titration Question
Calculate the pH of the solution when 20.0 cm^3 of 0.1 M of NaOH was added to 20.0 cm^3 of 0.1 M of H2SO4.
Assume the first ionisation is complete while the Ka for 2nd stage of ionisation is =0.011M
Approach
1st stage: NaOH + H2SO4 --> NaHSO4 + H2O
Moles of HSO4- in solution = 0.002 mol
[HSO4- ]= 0.002 mol/ 0.04 dm^3 = 0.05M
Assuming little ionisation of HSO4-,
[H+] = sqrt ( Ka x [HSO4-]) = sqrt (0.011 x 0.05M) = 0.002345
pH = -log (0.002345) = 1.63
and
Originally posted by atomos:Ksp Question
i) Calculate the [OH-] of 0.1 M of NH3 solution. (Kb = 1 x 10-5 M)
[OH-] = sqrt ( Kb x [NH3]) = sqrt (0.00001 x 0.1) =0.001 M
ii) Calculate the ionic product of Ca(OH)2 when 0.1 M of CaCl2 is added to equal volumes of NH3 solution
[Ca2+] = 0.1/2 = 0.05 M
[OH-] = ( Kb x [NH3])= sqrt [ 0.00001 x (0.1)/2]= 0.000707 M
as opposed to just dividing [OH-] from part i) by 2.
[Ca2+][OH-]^2 = (0.05) (0.000707)^2 = 2.5 x 10-8
Yes, both your workings are essentially correct. Just a couple of comments.
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The 2nd proton dissociation of H2SO4 is significant. Afterall, your own calculations indicate the pH to be less than 2. Therefore, it's technically inaccurate* to say "Assuming little ionisation of HSO4-", rather, what you meant was "Assuming it's mathematically valid to approximate equilibrium molarity of HSO4- back to its initial molarity". Mathematically, such an approximation would be valid only if the initial molarity was at least 1000x larger than the Kc (eg. Ka) value, which it's not, in this case. Which means you cannot use approximation, and have to solve the quadratic equation (eg. via a graphing calculator). However, Cambridge has specified that for 'A' level purposes, they will ensure that in their questions, the initial molarity is always at least 1000x larger than the Kc value, so that the approximation is mathematically valid.
(* there's a 2nd reason why your statement is technically inaccurate. "Ionization" (unless you specify "2nd ionization energy", etc) implies to generate an ion from a non-ion, hence to say "H2SO4 ionizes to HSO4-" is fine, but to say "HSO4- ionizes to SO4 2-" is technically inaccurate. Hence, a better phrasing for Bronsted-Lowry acids would be "proton dissociation" rather than "ionization", and a better phrasing for Bronsted-Lowry bases would be "hydrolysis" rather than "ionization" or "dissociation").
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Yes, your method for determining molarity of OH- is correct : you have to re-use the Kb formula in the new volume, instead of just dividing the initial molarity by 2. This is because the position of equilibrium shifts (as predicted by Le Chatelier's principle) when more water is added, providing for a greater availability of protons for the NH3 base to abstract, hence generating more OH- ions in the new equilibrium compared to simply dividing the previous molarity of OH- ions by half, ie. 0.0007 > (0.001)/2, but because the new molarity of NH3 is itself halved in the newly doubled volume, hence the new molarity of OH- is still less than the previous molarity, 0.0007 < 0.001.
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I have found it especially confusing in predicting what would a species be oxidised or reduced to. Say Fe3+ is present. Then does it reduce to Fe2+ or Fe, assume that both reactions are feasible.
You have to write step-wise redox reactions, and show that each stage is thermodynamically feasible (ie. cell potential > 0).The classic context to explore this (which can be asked in the Cambridge H2 'A' level exam), is that of Vanadium.
Because Zinc has a more positive oxidation potential compared to Tin which has a more positive oxidation potential compared to iron in the form of Fe2+, hence Zinc can reduce Vanadium(V) all the way down to Vanadium(II) with a colour change of yellow to violet, while Tin can only reduce Vanadium(V) down to Vanadium(III) with a colour change of yellow to green, while Fe2+ (the weakest of the 3 reducing agents) can only reduce Vanadium(V) to Vanadium(IV) with a colour change of yellow to blue (in practice, a green colour is observed, due to mixing of blue Vanadium(IV) and yellow Fe3+).
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Memo asked :
Why is dilution of concentrated acid give off a large amount of heat? Is it because of Delta H of solution is *very* negative?
Tetrahedron replied :
During dilution, a large amount of ion dipole is formed between water and ions of acid. In addition, due to sulfuric acid having 2 H+ ions in 1 molecule of acid, the amount of ion dipole formed is larger than other acids which are monoprotic.
oGarLiCo added :
& to finish that explanation off, bond forming releases energy
UltimaOnlined judiciously advised them all :
Very good, you guys are doing very well, Tetrahedron and oGarLiCo, keep it up!
Proton dissociation is virtually complete for the 1st proton, and significant for the 2nd proton, as the conjugate bases SO4 2- and HSO4- are effectively stabilized by having their negative charges dispersed by induction and delocalized by resonance over 4 electronegative electron-withdrawing O atoms.
Bond forming is exothermic because the products are more stable (due to loss of kinetic energy), and hence (because energy cannot be created or destroyed) the energy is lost to the surroundings as heat energy.
The new bonds or interactions formed, as a result of proton dissociation, are actually a combination of the following 3 :
1) new covalent bonds in the hydroxonium ions generated (from the protonation of water solvent).
2) new and stronger (stronger because an atom with a +ve formal charge is more strongly electron-withdrawing by induction, compared to if it has no formal charge, hence there is a larger magnitude of partial positive charge on the H atoms bonded to the +ve formal charged O atom in the hydroxonium ion, resulting in stronger) hydrogen bonds between conjugate base anions and hydroxonium ions, with (strongly polar, protic) water solvent.
3) new ion - permanent dipole interactions formed between the conjugate base anion and hydroxonium ions, with (strongly polar, protic) water solvent.
Hence, with a large number of multiple types of new bonds and interactions formed, from the diprotic proton dissociation of H2SO4, it's no wonder that dissolving concentrated or liquid H2SO4 in water, needs to be done gradually and cautiously, using dilution. -
BlackToast asked :
Why is the melting point of sodium ethanoate lower than the melting point of sodium chloride?
The melting point of sodium ethanoate is significantly lower than that of sodium chloride, due to weaker ionic bonds or a less endothermic/exothermic lattice dissociation/formation enthalpy, due to a wider separation of charges and a lower anionic charge density, because ethanoate ion is much larger than the Cl- ion, and also because the negative charge on the carboxylate ion is delocalized by resonance, and also because sodium ethanoate has a significantly lower degree of cation-anion symmetry, and thus packs less compactly compared to sodium chloride. The looser the packing, the lower the activation energy required to dissociate the solid lattice, and therefore the lower the melting point.On a related note, here's a fun question :
At 'A' levels, it's often thought that melting point and boiling point of a compound should parallel each other. For instance, water has both a higher melting and boiling point, compared to methane. Yet, as isomeric structural branching of an alkane increases, its melting point increases, but simultaneously its boiling point decreases. Why???
Answer :
http://masterorganicchemistry.com/2010/07/09/chemical-tetris/ -
A mixture containing 0.100 mol of CaCO3 and 0.100 mol of CaO are introduced into an evacuated 0.01m3 vessel. The vessel is heated to a constant temperature of 385 K such that the equilibrium below is established.
CaCO3 (s) <--> CaO(s) + CO2 (g)
The partial pressure of CO2 at equilibrim is then found to be 0.220atm.
What is the total equilibrium after an additional 0.300 atm of CO2 is pumped into the vessel.
0.530g
10.0g
12.5g
19.5g
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Vongole asked :
Q7
Which of the following molecules do not have all the atoms lying on the same plane?
A.PH3 - trigonal pyramidal
B.AlCl3 - trigonal planar
C.ICl3 - t-shaped
D.H2O - bent
Ans is PH3.
What I don't understand is how can you tell whether all the atoms lie on the same plane. Research told me that it's due to the presence of pi-bonds that results in a planar structure. But then in this qns, al-cl, i-cl and h-o-h bonds are all single bonds so no pi-bonds.
Another qns: Are covalent bonds STRICTLY for giant molecular structures like diamond and graphite, and absolutely not for molecules with van der waals' forces of attraction?
Work out the electron geometry, which for PH3 is tetrahedral (do not confuse electron geometry with molecular/ionic geometry, which for PH3 is trigonal pyramidal; molecular/ionic geometry is based on electron geometry). Any electron geometry other than linear or trigonal planar, constitutes non-planarity.
sp hybridization achieves an electron geometry of linear
sp2 hybridization achieves an electron geometry of trigonal planar
sp3 hybridization achieves an electron geometry of tetrahedral
sp3d hybridization achieves an electron geometry of trigonal bipyramidal
sp3d2 hybridization achieves an electron geometry of octahedralFor a look at geometries beyond octahedral (eg. pentagonal bipyramidal, square antiprismatic, etc), see : http://en.wikipedia.org/wiki/VSEPR_theory
Memo asked :
I get confused with T-shaped and trigonal planar. take a look at this... o.o
exemplified by ICl3. With 3 bond pairs and 2 lone pairs, is the molecular/ionic geometry T-shape or trigonal planar?UltimaOnline explained :
For trigonal bipyramidal electron geometry, lone pairs occupy the equatorial positions.For octahedral electron geometry, lone pairs occupy the axial positions.
Unfortunately, Memo still didn't fully get it :Oh I see... but this is a little bit counter intuitive.
If I am the lone pair of electrons, I will try to minimize repulsion as far as possible
i.e to be 180 degree away from each other instead of 120 degree.UltimaOnline patiently enlightened him :
It's precisely because lone pairs wish to minimize repulsion (even moreso than bond pairs), which is why lone pairs wisely occupy the equatorial position (in which they only need to endure significant repulsion with just the two orthogonally adjacent axial electron charge clouds), as opposed to foolishly occupying the axial position (in which they would suffer significant repulsion with the three orthogonally adjacent equatorial electron charge clouds).
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For covalent (either simple molecular or giant covalent) compounds, covalent bonds are the relevant intramolecular bonds, while van der Waals forces are the relevant intermolecular interactions (together with hydrogen bonds, if the molecule is protic; also take note that permanent dipole - permanent dipole Keesom van der Waals forces between polar molecules, permanent dipole - induced dipole Debye van der Waals forces between polar and non-polar molecules, and instantaneous dipole - induced dipole London dispersion van der Waals forces, are all types of van der Waals forces).
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Prototype asked :
The way I have been taught hybridisation is that you see the number of bond pairs about the central atom
4 bp => sp3
3 bp => sp2
2 bp => spWrong.
Eg. CO3 2- has 4 bond pairs around C, but the C atom's orbitals are sp2 hybridized.
Eg. CO2 has 4 bond pairs around C, but the C atom's orbitals are sp hybridized.The purpose of orbital hybridization is to achieve the ideal bond angles as predicted by VSEPR theory, which is to maximize stabilities by minimizing electron pair repulsions.
Hence, orbital hybridization is all about the electron geometry.
Eg. CO3 2- has 3 electron charge clouds around C, hence the electron geometry is trigonal planar, and hence in order to achieve this trigonal planar electron geometry, we cannot use our orthogonal unhybridized p orbitals, but instead must hybridize our orbitals (here, after promoting an electron from the 2s to the 2pz orbital, we mix the 2s orbital, the 2px orbital and the 2py orbital) in order to obtain 3 equivalent sp2 hybridized orbitals with the desired 120 deg bond angles. Note that the 2pz orbital is left unhybridized and available to overlap sideways to form the pi bond with an O atom.
Eg. CO2 has 2 electron charge clouds around C, hence the electron geometry is linear, and hence in order to achieve this linear electron geometry, we cannot use our orthogonal unhybridized p orbitals, but instead must hybridize our orbitals (here, after promoting an electron from the 2s to the 2pz orbital, we mix the 2s orbital and the 2px orbital) in order to obtain 2 equivalent sp hybridized orbitals with the desired 180 deg bond angles. Note that the 2py and 2pz orbitals are left unhybridized and available to overlap sideways to form the pi bonds with both O atoms.
kickme asked :
Does the unhybridised orbital always either form pi bonds or remain as an lone pair of electron rather than forming sigma bonds??
First of all, an electron pair (either lone or bond) occupies an orbital (either hybridized or unhybridized). The statement "the unhybridised orbital remains as an lone pair of electron" is erroneous, because "electron pairs" and "orbitals" are two separate things, just like "student" and "classroom".
Depending on the element and specific molecule, any unhybridized p orbitals may or may not be used to form pi bonds.
Note that pi (starting with the letter p) bonds are always formed from the sideways or side-on overlap of unhybridized p orbitals.
Note that sigma (starting with the letter s) bonds are always formed from the head-on or end-on overlap of either (unhybridized) s orbitals or hybridized orbitals with some s orbital character.