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

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    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
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    13,902 posts since May '05
    • Originally posted by Flying grenade:

      Hmm i try,

      O is more electronegative than N. O withdraws some e- density from N. Magnitude of delta -ve on O decreases. The O-H bond in hydroyxlamine is less polar compared to O-H bond in water. Hence strength of H bond between hydroxylamine molecules is weaker than strength of Hbond between water molecules.

      Hence water has a higher B.P. than hydroxylamine even though hydroxylamine can from more Hbonds per molecule than water.

      Indeed, the O-H bond in hydroxylamine is less polar, compared to in H2O or in alcohols, due to the electronegative N atom. So far so good. But here's where your school cocked up : the boiling point of hydroxylamine should actually be higher, not lower, than H2O, as number of H bonds formed is more important than strength of H bonds. Simply look at their melting points, and you'll see I'm right. However, in practice, hydroxylamine never gets to reach its theoretical boiling point (ie. above 100 deg C), because it readily decomposes (at 58 deg C) into water, ammonia, and either dinitrogen monoxide or nitrogen gas (depending on mechanism pathway and reaction conditions).

    • Originally posted by Flying grenade:

      'Maximum buffer capacity' section

      Is it that for an acidic buffer or alkaline buffer, theoretically, the buffer can have the pH of the solution remains(almost) unchanged, when acid or base of concentration of up to 10 times the concentration of the weak acid / weak base,  added to the buffer solution? 

      Yes, mathematically, a solution can be considered a valid buffer as long as the ratio of molarities of conjugate acid-base pairs lies within 1:10 and 10:1, ie. pH = pKa (+1 or - 1)

      Originally posted by Flying grenade:

      Woah damn cool add 5moldm-3 of acid/base to a buffer solution comprising of 0.5moldm-3 weakacid/base , ph still remains unchanged 

      A buffer solution's pH changes *slightly* when a *small* amount of strong acid or base is added to it. Note that highlighted words. And Cambridge will require you to write equations to explain exactly how a buffer works. The exact equations will of course, depend on exactly what is the conjugate acid-base pair of the buffer solution.

      Edited by UltimaOnline 13 Oct `17, 2:08AM
    • Originally posted by Flying grenade:

      Hi ultima, hope you could lend help for this qn,

      Qn 1 in here : https://www.dropbox.com/s/3o7vr03sfgpyhm6/20160418_193429.jpg?dl=0

      Qn and attempted answer is in the pic. Suggested soln is C, but i cant figure out why. I dun have solution for mcq, only final answer alphabet



      2015 ACJC P1 Q9

      Work out the Kc value for original conditions.

      Kc = (183x183)/90 = 372.1

      Apply the same Kc value for new conditions.

      Concordantly, equilibrium partial pressure of hydrogen gas = Square root of (150x372.1) = 236.25 kPa.

      Originally posted by Flying grenade:

      How would u know the values for CO (g) ?? 

      By avogadro's law? But n= v/vm , vm same for all gaseous products, since same temp and pressure, all gaseous products is 1 mol from this eqn, wouldn't volume be different? 

      So next time i just use the same partial pressure value of a gaseous product, of same mol, for the partial pressure value of a unknown gaseous molecule of same mol ??

      I've already given you detailed solution liao. If you still can't figure out this little bit by yourself (even if it didn't occur to you before reading my solution, after reading my solution you should now be able to figure out why), you'll have to go ask your school teacher or private tutor. Try to refrain from repeatedly asking too much on a single question, especially after I've already given you the answer on the forum.

      Edited by UltimaOnline 18 Apr `16, 9:36PM
    • Originally posted by Flying grenade:

      Hi ultima, hope you could help me here,

      For qn 2



      Suggested answer is D. 

      Qn: how to draw the structure for BrF2+? Is 2 bond pairs, 1.5 lone pairs with respect to(w.r.t.) Br atom correct? 


      2015 DHS P1 Q4

      Since the electrical conductivity of BrF3 decreases (ie. position of equilibrium shifts to LHS) with increasing temperature, hence heat may be regarded as a product. Hence reaction is exothermic.

      The cation is the unipositive BrF2+, for which due to its larger atomic radius (to minimize steric strain caused by van der Waals repulsions), as well as the fact that a positive formal charge is more stable on a less electronegative atom (Br is less electronegative than F), means that the central atom must be the Br atom with a unipositive formal charge, which means that the Br atom must have (in order to have a unipositive formal charge yet have a stable octet configuration) must have 2 lone pairs and 2 bond pairs, which means the electron geometry is tetrahedral, and making lone pairs invisible, means the ionic geometry of BrF2+ is therefore V-shape or bent.

      Originally posted by Flying grenade:

      Thank you for godly explanation.


      But i dun understand, br has 7 valence e-, bond with 2 F atoms, left 5 unpaired e-, lose 2 electrons as indicated by 2+ ionic charge of polyhalogen cation, shouldn't br be left with 3 unpaired e- , thats why im confused

      For every question's discussion, there will come a point where your own further doubts and misconceptions will have to be clarified face-to-face with a school teacher or private tutor, as it will be too inefficient to attempt further elaboration on the forum, especially after I've already described and explained the essential answer for everyone reading this (ie. the essential solution is already given, each person reading it will have unique further thoughts and doubts, for which they will have to figure out the nitty-gritty details for themselves, and/or consult their school teacher or private tutor).

      You've reached that point for this question. Have a nice day.

      Edited by UltimaOnline 18 Apr `16, 9:51PM
    • Originally posted by Flying grenade:

      Cs toh advanced study guide page 238 "group VII elements are typical non metals ( each element has 7 valence electrons) "

      I know by observation that non metals have large number of valence e-, but is large number or valence e- a *characteristic* of non metals?


      From wiki " ...chemically, they tend to have high ionization energy and electronegativity values, and gain or share electrons when they react with other elements or compounds. " 

      By definition, metals are electropositive, non-metals are electronegative. To be electronegative, you need more protons in your nucleus vis-à-vis the other elements in your same period. Condordantly, as a non-metal you need to be on the right side of the periodic table. Ergo, you'll have a large number of valence electrons. So it's not a definition of non-metals, it's just a characteristic. Transition metals also have a large number of valence electrons, because both the (n-1)d and ns electrons (eg. 3d and 4s), are both considered valence electrons.

    • Originally posted by Flying grenade:

      thinning of ozone layer *reduces* global warming, not increases it!


      Ideally ozone layer should be thin ,without holes, is it?

      But if the ozone layer is thin, it will have holes. You can't have your cake and eat it too. You can't lepak for A levels and still score distinction A grade (unless you're me, Mr BedokFunland JC).

    • Originally posted by UltimaOnline:

      Hi I would like to ask something about Senior Chem Olympiad (for JC1) and whether I should take it

      So recently, i somehow got selected by my school for SChO training, altho im like not very good, just a decent chem student and the ppl who got selected are all quite beast in their studies, like half of them are china scholars.

      So i went for my first session and apparently for the olympiad, the whole syllabus for A level chem needs to be covered as well as some other university chem modules, which is pretty scary given that the trainer only has now until end of Nov to finish it. And my school also not very good at chem olympiads, for last years olympiad 3 silver, 3 bronze, 1 merit

      So right now im not very sure if i should continue as i need to dedicate 2h per week for training and perhaps even more to complete the homework/readings that the external trainer gives. And im not very sure if it helps with my A levels Chem or is it too far fetched. Im afraid it would affect my time for studies and other stuff as in JC i find myself always havng v little time.

      So any advice on whether to go for it would be appreciated!

      Don't bother going for formal Olympiad Chem training or H3 Chem lessons in school, it just takes up valuable time you yourself said you can not afford. Olympiad and H3 are only for students who are already scoring all As in all their H2 subjects, are confident they have the time to do so, and wish to apply for Medicine and/or scholarships.

      If you like, read up and self-learn on Olympiad or H3 stuff yourself, for giving yourself the extra edge in H2 Chem. Or if you (ie. anyone reading this) join my BedokFunland JC tuition in the future, I'll teach you only the Olympiad and H3 concepts and strategies that will help you to get A grade in your H2 Chem.

    • Originally posted by Flying grenade:


      Atoms(e.g. Cl) of Halogen molecules(e.g. Cl2) already have complete octet, why are they still reactive? 


      Update : (Noticed bonds between halogens are quite weak, as seen from bond strength from data booklet),  but still, principal question above^

      Because the Cl atom in covalent molecules (ie. bond-Cl), and the Cl- ion in ionic compounds, also have a stable octet, plus an even more stable OS of -1 (instead of 0 in Cl2).

      Originally posted by Flying grenade:

      Why O.S. of -1 more stable? 

      Doesn't mean the more -ve o.s. is, more stable the atom is correct? 

      Ya know, it's annoying when you (ie. Singapore JC students) say things like that, and you'll lose marks in the A level exams for such ambiguous phrasing. They way you phrase it, makes you appear simple minded (ie. oversimplifying matters due to lack of depth of understanding) or dogmatic (ie. blindly memorizing school notes without deeper understanding), because it makes you seem to think "the more -ve o.s. is, more stable the atom is" holds true for all elements.

      Obviously (at least it's obvious to me and should be obvious to anyone who truly understands Chemistry, but apparently this doesn't necessarily include the majority of Singapore JC students or even some teachers and tutors, does it?), metals being electropositive will find positive OSes more stable, and non-metals being electronegative will find negative OSes more stable.

      As for the exact unique OSes preferred by each individual element, you can go figure out for yourself, and/or research online, and/or ask your school teacher and/or private tutor. At A levels, if you really understand your A level Chemistry, then you'll begin to get a glimpse of understanding as to the myriad of chemistry factors (some opposing each other) contributing to why different elements (eg. different transition metals) have different unique preferred OSes.

      Edited by UltimaOnline 22 Apr `16, 1:05AM
    • Originally posted by Flying grenade:

      When do we write equation/reaction thermodynamically/entropically favourable and thermodynamic/entropically feasible and spontaneous ?

      When delta H is -ve,=, or delta S is +ve, then the -ve delta H, or +ve delta S, is called thermodynamically favorable (but reaction may or may not be thermodynamically feasible). When delta G is -ve, then the reaction is thermodynamically feasible.

    • Originally posted by Flying grenade:

      So for maximum buffer capacity (half neutralisation point) always can skip icf and ice table?

      First, if you're not sure it's maximum buffer capacity (eg. because the question guai-lan and give different volumes and different molarities of both acid and base), do the ICF table to confirm it's maximum buffer capacity. With Cambridge questions getting tougher each year, it may be near, but not exactly maximum buffer capacity. So when in doubt, just do ICF table (which is arguably more important than ICE table, especially since Cambridge has stated they will accept the approximation of equilibrium molarity back to initial molarity).

      If it's really exactly maximum buffer capacity, obviously neither ICF nor ICE table is required, and not even the Henderson-Hasselbalch equation is required (since log 1 = 0). For maximum buffer capacity, pH = pKa1 or pKa2 or pKa3 (depending on which maximum buffer capacity it is during the titration of a polyprotic or multiprotic acid or base).

    • Originally posted by Flying grenade:

      T-T my answer wrong means my working epic fail means my understanding epic fail T_T

      Pls help


      Q8) A student dissolved 10.5 g of sodium fluoride in 250 g of water. The following data is obtained:

      Lattice energy of NaF −918 kJ mol-1
      Enthalpy change of hydration of F- −457 kJ mol-1
      Enthalpy change of hydration of Na+ −390 kJ mol-1
      Given that the specific heat capacity of water is 4.2 J g-1 K-1, what would be the final temperature of the solution, if the initial temperature of water is 30 oC?

      BedokFunland JC guidance :

      Calculate mol of NaF.
      Hence calculate total enthalpy change for the following process :
      NaF(s) --> Na+(g) + F-(g) --> Na+(aq) + F-(aq)
      Plug Q, m, c into formula Q = m x c x delta T
      Solve for delta T.
      Hence obtain the final temperature, and answer for this MCQ.

    • Originally posted by Flying grenade:

      Is it right that Exothermic deltaHsoln (delHsoln <0 means  soluble in water? ) means water gets more hot(gain energy hence temp rise) , as ions release energy(which outweighs LEdissociation) on hydration? 

      Solution Enthalpy = (endothermic) Lattice Dissociation Enthalpy + (exothermic) Hydration Enthalpy

      Solution Entropy = (positive) Lattice Dissociation Entropy + (negative) Hydration Entropy

      Gibbs Free Energy of Solution = Solution Enthalpy - Temperature x Solution Entropy

      A species is soluble in water (ie. solution process is thermodynamically feasible, ie. when energetically and entropically considered at a given temperature) only if Gibbs Free Energy change is negative.

    • Originally posted by Flying grenade:

      Page 334 advanced cstoh guide

      What does 'esterification can be used as a test for alcohol - the ester formed floats on water and has a sweet smell' means?

      Isnt this a test of presence of ester instead?as we deduce ester present from sweet smell?

      You can test for the presence of alcohols by adding an acyl halide. If a sweet smell is observed, meaning that an ester is generated, which in turn means the original analyte must have contained an alcohol.

    • In case you're wondering why potassium iodide is often prescribed in cases of radiation exposure, and how it works (only in a very specific and limited way).


    • Originally posted by Flying grenade:

      Hi Ultima,


      for part ii) , why is this thought process wrong : i want find [H+] . i use ka formula. tried, but didnt get the correct answer. 

      (p.s. , i've internalised the 2 methods/approach that was taught to me , but want to know why above concept wrong. )

      is it because, inside the solution, its not just the weak acid inside,  there's the cj base too? Help my concept please T-T

      thanks ultima

      Being a buffer soln, you *can* indeed use Ka, or even Kb. It's just that the Henderson-Hasselbalch equation saves you time as you can work in moles instead of molarities. If you didn't get the correct answer, it means your molarities (ie. moles / new total volume of soln) of both HA and A- at that point (double-equivalence) are wrong. Obviously you need to plug in the molarities of both conjugate acid and conjugate base into the Ka (or Kb) equation, and not all questions will be as kind as this question to let you have maximum buffer capacity, so you better start practicing tougher questions in which the moles or molarities of both conjugate acid and conjugate base are different. Then try using all 3 formulae : Ka, Kb, and Henderson-Hasselbalch equation, to prove *to yourself* that all 3 formulae will give you the same final answer of the correct pH.

    • Originally posted by Flying grenade:

      Is it okay/fine/chemically correct to write


      For Mg(OH)2 to be precipitated, Qsp (Mg(OH)2) >Ksp (Mg(OH)2)


      Looking at 2015 jc prelims solutions, all wrote  I.P. or just Solubility = ____

      None wrote Qsp

      My school uses I.P. , but sometimes i forgot, i wrote Qsp a number of times already ,but now trying to get the habit back of writing I.P. But is it wrong to write 'Qsp' ?


      Btw, SAJC uses ICF table, lol, cool. that makes BJC students, and them ,that know about it 


      Both are acceptable by Cambridge, use whichever you prefer, IP or Qsp, to be compared with Ksp. Personally, I'll recommend Qsp (since it's closer in form to Ksp). For solubility (which is not the same as Ksp. ie. solubility product), always specify "molar solubility" (mol/dm3) or "mass solubility" (g/dm3).

    • Originally posted by Flying grenade:

      DHS/2015/P2/Q3 part iv )


      I understand the solution.

      But i dun understand the qn, Why to determine solubility of CaC2O4 at high pH, use NaOH? What are the alternative method? 

      I was thinking, CaC2O4 》《Ca2+ + C2O42- , let solubility of CaC2O4 be x, hence solubility is just root(Ksp)

      Thanks ultima 

      Your molar solubility based strictly on Ksp ignores the effect of pH, which is the usual case within the H2 syllabus. This DHS question goes beyond the H2 syllabus (which Cambridge often does) to include the effect on pH, which is why your answer is inadequate and thus wrong. If you do not add NaOH, how are you going to make the pH high (ie. alkaline)? That's why you add NaOH, which increases solubility by precipitating out the cation (in contrast, adding acid increases solubility by protonating the anion). There's no alternative method ; Wilbur's equation only covers acidic to neutral pH.

    • Originally posted by Flying grenade:

      Thanks for ur explanation !!!

      I'm not fully understanding the problem yet

      Is it :

      I want to study the solubility of CaC2O4.

      In this case, the solubility of CaC2O4 in high pH(alkaline conditions)

      In a way, we want to study the effect of [H+] in the solubility of CaC2O4. (Can be seen from Wilber's equation)

      In part iii) , the [H+] plays a role, in protonating anion, in turn having an effect in the solubility of CaC2O4.

      However in part iv) , we actually want to study the effect of [H+] in high pH, but however, the NaOH actually causes increased solubility of CaC2O4, as it precipitates out the cation, instead of due to the effect of [H+] , which we want to study. 

      Is it?

       For part iii) , the solution elucidates about the effect of LCP for the explanation. But can i also explain , based on the equation given in part iii, solubility increases, as total [H+] in numerator increases faster than total [H+] in denominator, hence solubility increases 




      Wrong, in high pH, it's OH- that's relevant, not H+. "We" do *not* want to study the effect of [H+] in high pH, "We" only want to study the effect of pH on solubility, so Wilbur's equation (involving H+) is obsolete and can be retrenched.

      Your mathematics-based answer will not be acceptable, because firstly, the H2 Chem syllabus is always testing students about shifting of position of equilibrium (use of acronyms like LCP not allowed in A level exams); and secondly, the actual question (on Wilbur's work, or similar) in the A levels might not give you an mathematical equation or formula at all, so if all you can give are mathematical reasonings (as opposed to chemical reasoning that's required by the mark scheme), you'll be screwed in the A levels.

      BedokFunland JC exam tip : If you're exam-smart, you'll give a variety of reasons, whatever you can think of. As long as they don't contradict each other, you'll be given credit for whichever answer Cambridge wanted.

    • Originally posted by Flying grenade:

      I understand the discussion about OH- , with the cation, can precipitate out an ionic solid, and , in acidic pH, anion is protonated. But i dont see how these is relevant for the explanation in iv) .

      I dont see how "NaOH might precipitate out Ca(OH)2 , which would decrease [Ca2+] , therefore potentially increasing solubility of CaC2O4 at high pH" (given answer solution) addresses the question of iv) , which is to study the effect of (high) pH on solubility.

      We want to study the effect of pH on solubility. 
      For part iv) , understood, Wilber's equation is obsolete for alkaline pH.

      For part iv) , Then, why is adding NaOH a problem?  it should be Wilber's equation that's the problem.

      i dont see how adding NaOH is a problem studying effect of pH on solubility.

      Adding NaOH IS A Problem if we want to study effect of pH on solubility, USING Wilber's equation.

      I dont see a link here. maybe i dun understand/disagree with the phrasing of the qn, and the ans. might be my command/understanding of english is inadequate.

      Ok real sorry, just move on, i dont mean to annoy you, sorry, its ok if u dont reply to this qn
      Just move on


      Just skip the damn question in the A levels if you don't like Cambridge's phrasing. Oh wait, only I can do that and still get an A grade. *You* can't.

      Yeah, go chat with (and annoy) your school teacher or private tutor instead.

      Edited by UltimaOnline 02 May `16, 1:46PM
    • Louistanshiehsit asked :

      Hi, I would like to ask how do I go about preparing for planning questions? I am doing pretty well in the normal papers but the 5% planning really a big handicap for myself. Im a private candidate and have never been to JC before, so im abit lost at the planning question. Hope to get some advice ! thank youuuuuuuuu :D

      Louis, first of all, don't worry too much about the Planning Qn. You'll be better off focusing on polishing your remaining 95%, then the 5% whose preparation is notoriously difficult and not-worth-planning (pun intended) for.

      For those of you who are still worried sick about the Planning Qn for H2 Chemistry, there are only 2 reliable sources for this (you can of course, try to buy online from ex-students, various Singapore JC notes on Planning, but note that I said *reliable* sources), namely Chan Kim Seng & Jean Tan's H2 Chemistry Planning Book "Understanding Experimental Planning for Advanced Level Chemistry: The Learner's Approach" (see my website), and Cambridge Mark Schemes on Planning Qns (I've included links to such websites here).

    • Originally posted by theophilus:

      Several questions!

      Inorganic Chem:

      For transition metal complex right, how do you determine the coordination number. I understand that most of them are 6 but sometimes it can be 4 or 2, and i get stumped when i see the answer key. so how do i predict the coordination number?

      Physical chem:

      Do you mind explaining how do you predict solubility? isit as simple as polar solvents dissolve polar reagents and non polar dissolve non polar? refering to TYS 2009 P2 Q2(b) I dont understand why E values are used D:

      The topic of complex ions is actually more complex (pun intended) and cannot be fully understood at A levels. At A levels, you only need to know the basic principles and concepts, and you've to memorize the coordination numbers, geometries and colors of the commonly encountered coordination complexes.

      For A level purposes, 3 simple concepts are relevant here : size of metal cation (if small, ie. period 2, only 4 ligands are possible, due to lack of vacant, energetically accessible d orbitals to accommodate an expanded octet), size of ligands (eg. water ligands small, so 6 are possible ; chloride ligands large, so only 4 are possible, to avoid excessive steric strain caused by van der Waals repulsions), charge on ligands (eg. water ligands neutral, so 6 are possible ; hydroxide ligands are negatively charged, so only 4 are possible, to avoid excessive steric strain caused by anionic repulsions between ligands).

      Beyond these A level concepts, University level chemistry such as the Jahn-Teller effect, will be required to explain deeper questions on coordination complexes, eg. why [Cu(NH3)4]2+ has a square planar geometry instead of tetrahedral geometry.

      Solubility is also a complex issue spread over several topics in the H2 syllabus, eg. chemical bonding, thermodynamics, solubility equilibria, Group 2, electrochemistry, organic chemistry, etc. The required answer at A levels depends on the context and H2 Chem syllabus topic that the question falls under. So there's no simple way to predict solubility (simple ways are for O levels, A levels are preparation for Uni level, in which everything becomes increasingly complex*). For the particular question you mentioned, falls under electrochem : only the oxidation potentials of Mg, Zn and Al, but not Cu, are sufficiently positive for their oxidation into aqueous and soluble metal cations when reacted with H+ from HCl (aq), to be thermodynamically feasible.

      * In primary school, you thought you already knew everything in the Universe that could be known. In secondary school, you begin to realize you don't everything, but you still feel confident that you know almost everything. In JC and poly, you begin to understand that you don't really know all that much, and that the Universe is more complex than you thought. In University, you're humbled by the understanding that the Universe is so infinitely complex that you, and all sentient beings on Earth, actually know almost nothing relative to the infinitely vast Universe, but we know just enough to begin to appreciate and respect its beauty and complexity.

    • Originally posted by Flying grenade:

      I understand this : 

      By Hess law,

      🔺Hsoln =  🔺Hhyd + 🔺LEdissociation


      Can help me check where i've gone wrong thinking in this way?? : 

      🔺Hsoln = energy released on hydration by hydrated ions Minus Energy absorbed by Ionic solid to dissociate     


      So 🔺Hsoln = (-ve enthalpy change) minus (positive enthalpy change )

      If i put modulus on the RHS terms, then it becomes all +ve

      Help please thanks! 

      Your problem is you assume magnitude of hydration enthalpy always outweighs magnitude of endothermic lattice dissociation enthalpy, which is false. Either could outweigh the other.

      Hence, without worrying about which magnitude outweighs which, you should simply add up the endothermic lattice dissociation enthalpy change (ie. positive value) with the exothermic hydration enthalpy change (ie. negative value) to obtain solution enthalpy change, which could be either endothermic or exothermic (ie. positive or negative value).

    • Chan Kim Seng appears in the video at 3:10 onwards :


    • Originally posted by MightyBiscuits:

      Hi, I need help for this question. Adopted from MJC tutorial. 

      "Graphite is one of the allotropes of carbon. The graphite structure consists of hexagonal layers. Each carbon atom has one electron that can be delocalized parallel to the layers to make graphite a good electrical conductor.  

       Different atoms can be fitted in between the layers of carbon atoms to produce an usual set of “graphite compounds”. When the graphite lattice contains atoms of alkali metal such as caesium, then a bronze-coloured solid is formed which has a greater conductivity than pure graphite.  

      In hexagonal boron nitride , the structure is composed of layers of hexagonal B3N3 ring. The boron atom utilizes all the three valence electrons to form covalent bonds with three neighbouring nitrogen atoms. Each nitrogen atom still has a lone pair of electrons which it utilizes to form a dative bond with an adjacent boron atom. Although graphite and hexagonal boron nitride have similar structures, they differ in electrical conductivity. Unlike graphite, boron nitride is an insulator.   

      When heated under pressure, this form of boron nitride is converted into another form which is an extremely hard solid .  

      (g)(i) Suggest and describe the structure adopted by the new material formed"

      Giant covalent tetrahedral lattice structure similar or analogous to diamond, in which each N atom has a unipositive formal charge, and each B atom has a uninegative formal charge, because all N and B atoms have a stable octet of 4 bond pairs and 0 lone pairs (just like the C atoms in diamond).

    • Originally posted by Flying grenade:

      Hi Ultima, i have some queries 

      Page 97 cs toh advanced guide 

      Btm of the page "C=O bond energy in CO2 is significantly stronger than average bond energies given in data booklet"

      Is the reason that why not all C-H or C=O bonds are the same, because the C or O atom might be attached to another (different) atoms in different compounds?

      Thanks for helping

      Yes, that's right.

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