BedokFunland JC's A Level H2 Chemistry Qns (Part 2)
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ilovee37e asked :
Why MnCl2 can be found in the laboratory but not MnCI3?
I explained this using ideas about the difference in ionization energies( 1st 2nd and 3rd). Is it applicable here?
Yes, it's relevant. Another similar example along this line of reasoning, will be "why doesn't MgCl3 exist?" It's a matter of considering the magnitudes of endothermic ionization energy versus exothermic lattice energy, in whether the formation of MgCl, MgCl2 or MgCl3 is most thermodynamically favoured.
Another separate relevant point you should include in your answer, is that Mn3+ is a strong oxidizing agent, and will oxidize Cl- to Cl2. Thus Mn3+ will not co-exist together with Cl-, because they would react immediately with each other in a redox reaction. Note that in contrast, MnF3 exists, because the oxidation potential of F- to F2 is (-2.87V) significantly less positive than the oxidation potential of Cl- to Cl2 (-1.36V). Another similar example along this line of reasoning, will be "why doesn't FeI3 exist?"
In the 'A' level exam, give all points that you feel may be relevant. You won't be penalized for giving additional points (as long as they are correct), you'll only be penalized for giving contradictory points, chemically illogical or chemically invalid arguments.
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ilovee37e asked :
Why does hydration of ethyne produce ethanal instead of ethenol?
Hydration of ethyne generates ethenol, which undergoes an internal proton transfer reaction to convert into its more stable tautomer (ie. an isomer which differs only in the position of one proton), ethanal.H2 'A' level students are encouraged to attempt drawing the mechanism for both the hydration process and the tautomerization process : ethyne --> ethenol --> ethanal.
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ilovee37e asked :
Why does the reaction of Na2Co3(aq) and AICl3(aq) does not produce Al2(CO3)3?. The reaction produced NaCI,Co2, and Al2O3.
The high charge density on the small, tripositive Al3+ ion, distorts and polarizes the anionic electron charge cloud of the large, polarizable CO3 2- ion, decomposing it into CO2 and the O2- anion. Consequently, aluminium carbonate doesn't exist, and instead aluminium oxide and carbon dioxide are generated. The Na+ and Cl- ions are just spectator ions. -
blacktoast94 asked :
my book says NH3 and PH3 have different bond angles and shapes and i don't udnerstand how that can possibly be so?
Memo posted :
shape is the same right? :x anyways for bond angle explanation, N is more electronegativity than P hence N attract electrons towards itself more than P therefore, larger expulsion between the electrons
thus, bigger bond angle.
blacktoast94 posted :Memo: T.T sian, got such thing? so meaning shorter bond length = more repulsion??? btw, does a double bond repel more than a single bond?
Memo posted :
nonono, need wait Ultima to clarify... so yeah i learning too :x but generally Higher electronegativity of central atom = higher bond angle. btw, bond length got decrease when electrons are more attracted to N meh?
UltimaOnline rescued the two poor lost souls :There are two reasons why the bond angles in NH3 differ from that of PH3. However, both reasons are beyond the H2 syllabus, and although some JCs teach one of the reasons, it is unlikely that Cambridge will test this, as it is strictly speaking, beyond the H2 syllabus.
The first reason (and the reason more commonly taught by JCs), is that of electronegativity, which Memo has already mentioned. In summary, the more electronegative N will have its bond pairs closer to the nucleus, and accordingly, the bond pairs will also be closer to each other and thus repel each other more. Similarly and conversely, the less electronegative P atom will have its bond pairs further from the nucleus, and accordingly, the bond pairs will also be further from each other and thus repel each other less.
The other reason (which is actually more correct, but JCs don't teach this reason because they fear JC students will be confused and can't handle the truth), has to do with the extent/degree of hybridization. (Most JCs teach about hybridization at the end of JC1, under the topic of "Intro to Organic Chem").
Notice that based on VSEPR theory, the electron geometries of both the N and P atoms (in NH3 and PH3) should be tetrahedral (don't confuse this with molecular geometry, which in this case is trigonal pyramidal, based on the tetrahedral electron geometry) and therefore sp3 hybridized (recall the purpose of hybridization : to achieve the ideal bond angles as predicted by VSEPR theory, which in turn, is to maximize stability by minimizing electron pair repulsion). However, because P is one entire electron shell larger than N, the electron pairs (both lone pairs and bond pairs) are much further apart, and therefore there is much less repulsion, and accordingly, much less need for, and therefore much less extent/degree of hybridization. (The PH3 bond angles are closer to the orthogonal 90 degrees angles that would result from the end-on overlap of the unhybridized p orbitals of P with the s orbitals of H).
Both explanations (ie. electronegativity and extent/degree of hybridization) are correct and both contribute to the observable (ie. experimentally verifiable) phenomenon that the ideal bond angles as predicted by VSEPR theory, holds most accurately true for hybridization of period 2 elements. Atoms of elements in periods 3 and above, do not conform well to VSEPR theory, for the two reasons stated previously.
However, do not worry too much about this, because Cambridge (by right, according to the H2 syllabus specifications) will not test you regarding this concept. So (again by right), Cambridge shouldn't ask you to explain why the NH3 bond angles differ from the PH3 bond angles. But some JCs teach it, but when they do, they only teach the electronegativity explanation, because they think it is less confusing for the average JC student.
Lastly, regarding the query about double bonds versus single bonds, it is indeed the case that double bonds (and triple bonds) do experience slightly greater degree of repulsion with neighbouring electron pairs, compared to single bonds (since the sigma + pi electrons of a double bond take up more space, compared to just the sigma electrons of a single bond).
However again, this too, is beyond the H2 syllabus, and not only does Cambridge not expect 'A' level students to know this, even JCs (including the few JCs which teach about how electronegativity affects VSEPR theory) usually do not teach about this double bond versus single bond aspect of VSEPR theory at all.
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blacktoast94 posted
Does electronegativity affect bond length?
Electronegativity does not affect bond length directly. Instead, atomic radius (ie. number of electron shells) is the most important factor. For instance, why is HI a stronger acid compared to HF? First reason : the H-I bond is longer hence weaker than the H-F bond (since the atomic radius of the I atom is much larger than the atomic radius of the F atom), therefore the H-I bond has a significantly less endothermic enthalpy of dissociation. Second reason : the I- conjugate base has a lower charge density and is hence more stable, compared to the F- conjugate base. The more stable the conjugate base, the stronger the acid.
Another factor is the % s character of the bonds. For instance, take the C-C single bond in propane versus the C-C single bond in propyne. In propane the C-C sigma bond is formed from the end-on overlap of the sp3 hybridized orbitals of both carbons (ie. the sigma bond is sp3-sp3). In propyne, the C-C sigma bond (ie. the single bond) is formed from the end-on overlap of the sp3 hybridized orbital of the tetrahedral carbon, with the sp hybridized orbital of the linear carbon (ie. the sigma bond is sp-sp3).
Consequently, the sp-sp3 propyne sigma bond (ie. the single bond) has greater % s character (alternatively, smaller % p character) compared to the sp3-sp3 propane sigma bond. Since the s orbital is located slightly closer to the nucleus compared to p orbital (for the same quantum electron shell, of course), hence covalent bonds with greater % s character are shorter compared to bonds with greater % p character.
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blacktoast94 posted :
1. For NCLO2, should the N or the CL be in the centre of the molecule? i'm not sure cos both seem to look ok.... but answer is N is in the centre. then the question after that was whether NCLO2 can dissolve in water. however, if the answer given was right, it doesn't have lone pairs.... so how can form H-bonds with water?
2. why does CsCl have a different coordination number from Nacl, Kcl and MgO?
Let's look at Q2 first.
The term "coordination number" in chemistry has two slightly different meanings, depending on the context. In complex ions of transition metals, it refers to how many coordinate dative covalent bonds exist between the central metal ion and the donor atoms of ligands.
In this question however, it is referring to the lattice structure of ionic compounds. The coordination number here, is defined as the number of its nearest oppositely charged neighbours in the ionic lattice structure.
For an ionic compound, the coordination number is determined by the stoichiometry of the compound and by the sizes of the ions. Because Cs+ is much larger than Na+ and K+, hence more Cl- neighbour ions can be packed around each Cs+ ion, therefore the coordination number is larger for CsCl compared to NaCl and KCl. As for MgO, not only is Mg2+ much smaller than Cs+, but furthermore the anion size is different (since Cl- ion is much larger than O 2- ion), hence the coordination number for MgO will be different from that of CsCl for two reasons (cationic AND anionic size).
And here's another example of Cambridge asking a question that's somewhat beyond the syllabus. Coordination number was never asked before in previous years, so JCs considered this as beyond the syllabus and didn't teach their students. In 2009, Cambridge asked about this in Paper 2 Qn 1. As I warned, every year the H2 Chemistry paper is getting increasingly tough and beyond-the-syllabus. Get ready for the 2011 and 2012 papers!
Q1. For neutral molecules such as NClO2, the ionic charge is zero. Since ionic charge is the sum of formal charges, you expect any formal charges on the atoms to balance out. And you would suspect a unipositive formal charge on the N atom, because nitrogen being in period 2, does not have vacant, energetically accessible 3d orbitals to use to expand its octet.Hence the N atom in NClO2 has 4 bond pairs, 0 lone pairs, a unipositive formal charge (since N is in Grp V but now it only has 4 valence electrons in terms of formal charge, and 8 valence electrons in terms of stable octet).
The Cl atom, we begin by assuming no formal charge, and since it is in Grp VII, it must have 1 bond pair, 3 lone pairs (ie. 7 valence electrons in terms of formal charge, and 8 valence electrons in terms of stable octet).
That leaves us with 3 bond pairs with 2 O atoms. Hence, we deduce it must be doubly bonded to an O atom (with no formal charge; doubly bonded O atoms have 6 valence electrons in terms of formal charge, and 8 valence electrons in terms of stable octet), and singly bonded to an O atom with a uninegative formal charge (ie. 7 valence electrons in terms of formal charge, and 8 valence electrons in terms of stable octet).
Everytime you see a unipositive formal charge and a uninegative formal charge on adjacent atoms, you suspect a dative covalent bond, and indicate it thusly if you're drawing the displayed structural formula in physical chemistry context (if it is a dot-and-cross diagram, the dative bond must be indicated with a cross-cross or dot-dot, to distinguish it from a normal dot-cross covalent bond).
(Note : If the structure is to be drawn as part of an organic chemistry mechanism, you must NEVER include the dative bond symbol (ie. the bond with the arrowhead), because in organic chemistry, the organic chemist uses curved / curly arrows to indicate the flow of electrons in explicit stages like a flow-chart diagram, whilst physical & inorganic chemists (being lazier, or more efficient, your point-of-view depending on whether you're an organic chemist or physical / inorganic chemist) combine both stages of the mechanism into one diagram, and use a shortcut by way of a dative bond arrowhead symbol instead of curved / curly arrows.)
The oxidation state of N in this molecule, is :
Oxidation State = Formal Charge + Electronegativity consideration
= (+1) + (-1) + (+3) = +3To understand the calculations working above :
Formal charge is unipositive, then it gains one electron from Cl (since N is slightly more electronegative than Cl) and loses 3 electrons to the more electronegative O atoms (since it has 3 bond pairs with O atoms). Don't worry about the dative bond (regard it as a normal bond when it comes to calculating oxidation states), because it's dative nature is already taken into consideration by the formal charges.The oxidation state of Cl in this molecule, is :
Oxidation State = Formal Charge + Electronegativity consideration
= (0) + (+1) = +1The oxidation state of the doubly bonded O atom in this molecule, is :
Oxidation State = Formal Charge + Electronegativity consideration
= (0) + (-2) = -2The oxidation state of the singly bonded O atom in this molecule, is :
Oxidation State = Formal Charge + Electronegativity consideration
= (-1) + (-1) = -2Check :
Ionic charge = sum of formal charges = (+1) + (-1) = 0 which is correct, since this is known to be a neutral molecule.Check :
Ionic charge = sum of oxidation states = (-2) + (-2) + (+3) + (+1) = 0 which is correct, since this is known to be a neutral molecule.Regarding solubility, NClO2 is slightly soluble because it is capable of some hydrogen bonding (notice that it isn't terribly polar, because all atoms within the molecule are highly electronegative, so the dipoles are terribly great). The two O atoms both have a significantly large (approx 0.5- charge) partial negative charge in the resonance hybrid (and in the resonance contributors, one O atom with a negative formal charge, and one atom with a partial negative charge) and have available lone pairs, and therefore can have some hydrogen bonding with the partially positively charged H atoms of solvent water molecules.
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blacktoast94 posted:
1) X+ compared to X2+ , adding an electron to X2+ will release more energy than to X+ ... why?
2) lol i got problem using the aufbau principle. -.- how to write electronic configuration of Cs? i should be able to figure out from there.
3) my sch never tell me .... F subshell can accomodate how many electrons in total?
4) AlP is similar to the shape of diamond, can it conduct electricity?
I was thinking maybe cos P have 1 lone pair, so can conduct. is it?Q1. Stronger electrostatic attraction between X2+ and e- (compared to X+ and e- ), hence more energy released. Another way to look at it, is the magnitude of energy difference and stabilities (more stable = less energy) between the reactants and products are greater for X2+ versus X+. Remember : the central dogma of chemistry is that charge is destabilizing and everything wants to become more stable. X2+ is a lot more unstable and hence possess a lot more energy compared to X+. Consequently, when the more stable (ie. less energy) X+ is formed from X2+ and e-, where did the energy (from the unstable, high energy X2+ ion) go to? Lost to the environment, as heat energy. This is why bond forming (in this case, the electrostatic bond between the electron added and the positively charged nucleus) is exothermic.
Q2. For H2 Chem, you don't have to care about the names Afbau's, Hund's, Pauli's, etc. Just read the electron configuration off the periodic table.
Cs is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s1
Although many JC teachers insist you write the electrons in ascending order of quantum shells (ie. 4d10 before 5s2), but Cambridge will still accept it if you write it the way you read off the periodic table.
Q3. That's because f orbitals are beyond the H2 Chem syllabus. There are 7f orbitals, so a maximum of 14 electrons can occupy the f orbitals of any quantum shell (from n = 4 and upwards).
Q4. Aluminium phosphide is a semi-conducter of electricity. It doesn't conduct electricity as well as graphite does, because of the elecronegativity difference between Al and P causing the bonds to be polar. And it's due to the delocalized pi bond electrons, rather than lone pairs, that enable aluminium phosphide to be a semi-conductor, similarly to graphite (which is also able to conduct electricity due to the delocalized pi bond electrons). -
Topic : Chemical Bonding
Memo asked :
O2 and S8 are both in the same group, but they have different structure. explain. tmr it's my chem test! RAWR!
Same reason why N exists as N2 while P exists as P4, or why CO2 is simple molecular while SiO2 is giant covalent.
The strength of pi bonds formed, depend on the effectiveness of the sideways overlap of p orbitals. If the p orbitals are more diffused (eg. 3p orbitals), the pi bonds formed are weaker. If the p orbitals are less diffused (eg. 2p orbitals), the pi bonds formed are stronger.
In N2, the pi bonds formed are strong, due to the effective sideways overlap of the less diffused 2p orbitals. In P2, the pi bonds formed are weak, due to the ineffective sideways overlap of the more diffused 3p orbitals. Therefore, for nitrogen, the N2 structure is more stabilizing (since the pi bonds formed are stronger) compared to the N4 structure; while for phosphorus, the P4 structure is more stabilizing compared to the P2 structure (since the pi bonds formed are weaker).
In the simple molecule CO2, the sideways overlap of the 2p orbitals of carbon and the 2p orbitals of O, are effective and the pi bonds formed are strong. In the (hypothetical) simple molecule SiO2, the sideways overlap of the 3p orbitals of silicon and the 2p orbitals of O, are ineffective and the pi bonds formed are weak. Therefore, the oxide of carbon exists as the more stable simple molecule CO2, while the oxide of silicon exists as the more stable giant covalent lattice of SiO2.
And similarly for oxygen versus sulfur, why the former exists as O2 while the latter exists as S8. (There are of course, other allotropes of oxygen and sulfur, but O2 and S8 are the most common allotropes for each element respectively).
For this discussion, we're focusing on the strength of pi bonds formed, relative to the strength of sigma bonds (although the strength of sigma bonds is also dependent on the bond length and (exactly which) quantum shells of the atoms involved; but in comparison, the strength of pi bonds vary to a more dramatic extent from 2p to 3p, and is therefore the deal-breaker (or the camel's back-breaker) when it comes to explaining why period 2 elements prefer to form double bonds while period 3 elements prefer to form single bonds).
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ilovee37e posted :
To prepare deionised water, tap water containing ions ,including CI ions, is passed thru a column of resin that helps removes the ions present. If the concentration of CI- ions in the water that leaves the column is high enough to cause the precipitation of AgCI when 1cm3 of water is added to 1 cm3 of 1moldm^-3 silver nitrate solution,the water is no longer considered deionised and hence, the resin in the column has to be replaced. What is the max. concentration of CI ions that can be present in the water that leaves the column? Ksp of AgCI=1.8*10^-10 mol2dm ^-6
2) Solubility of CaCO3 in pure water is 7.0x10^-5 mol dm^-3. What is the amt of CaCO3 that would dissolve in 100cm3 of 0.05 mol dm^-3 CaCI2 solution?
Hints / Clues :
Q1. From Ksp, find molar solubility. Hence find molarity of Cl- in combined 2cm3 soln. Therefore find molarity of Cl- in original 1cm3 water added.Q2. From molar solubility, find Ksp. Let moles of CaCO3 that would dissolve in 100cm3 be x moles. Given molarity of (the common ion) Ca2+ in CaCl2 soln, work out (in terms of algebraic variable x) the equilibrium molarities of Ca2+ and CO3 2- present in the saturated solution, after adding x moles of CaCO3. Since x is defined as the moles of CaCO3 to be added to reach (not exceed) a saturated solution, we are at the equilibrium of a saturated solution and therefore ionic product = solubility product Ksp. Therefore we can solve for x.
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Qn. Why is MgO is suitable for use as a refractory material for lining blast furnances?
Answer :
This is a consequence of MgO's especially high melting point, which is a consequence of the highly endothermic lattice dissociation energy, which is a consequence of the strong ionic bonds within MgO, which is a consequence of the high charge densities of the Mg2+ and O2- ions.
(To truly excel in chemistry, you need to truly understand it (including the causative-consequence relationships and interconnectivity of all aspects of chemistry), instead of merely blindly memorizing school notes which is the disgraceful failing of the Singapore education system.)DrBiology asked :
Very clear! Thank you Ultima! I have another qn.
"Molecules with a higher Mr has a higher b.p because there are more electrons to be polarised, therefore there are stronger dispersion forces" . What does it mean by 'electrons to be polarised?'. From my understanding, polarisation is the distortion of the electron cloud of the anion due to the polarising power (high charge density) of the cation. How is the concept of polarisation applied in this case? Thanks!When electron charge clouds are polarized, it means that one end of the molecule has a partial positive charge, while the other end has a partial negative charge. The greater the molar mass, the greater the number of electrons (resulting in greater inter-electron repulsion) and the larger the size of the molecule (resulting in weaker electrostatic attraction between positively charged nucleus and the valence electrons), both of which has the consequence of the molecule's electron charge clouds being less tightly attracted by the nucleus and therefore having greater polarizability, which in turn has the consequence of having a greater *magnitude* of partial positive and partial negative charges within the molecule, which has the consequence of stronger electrostatic attraction (known as van der Waals forces and dispersion forces) between the molecules.
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blacktoast94 posted :
predict if the CL-P-CL bond angle in PCL3 is larger than the H-P-H bond angle in PH3.
Memo posted :
bond angle for PCl3 > PH3 (from wikipedia)
so answer is wrong, i guess. arrr this one cannot compare electronegative of central atom. :qmarkangry:[/QUOTE]blacktoast94 posted :
the answer says that PCL3 < PH3, reason: bond paairs in PCL3 are further apart than those in PH3, hence lesser repulsion for PCL3 compared to PH3.
i don't get it. the bond pairs in PCL3 are further apart from each other - isn't that due to more repulsion?
and if it is further apart, doesn't that mean bond angle of PCL3 > PH3? :qmarkblackface:huhh. T.T but i have two sources saying that PH3 > PCL3...
nvm... wait for Ultima... hahhaahaha.
UltimaOnline replied :This question is a good example of where exam-smarts come in.
Before we go into who's right and who's wrong, notice that this qn is asking about the geometry about a period 3 element phosphorus, which is strictly speaking, beyond the H2 syllabus, which will only examine you for geometries about Period 2 elements.
If this qn was about an N atom instead of a P atom, (period 2 elements are way more likely asked by Cambridge), then you'd be required to state that the electron and molecular geometries for both NH3 and NCl3 are the same, namely tetrahedral and trigonal pyramidal respectively, and accordingly, both have the same bond angles of approximately 107 deg C. (In point of fact, the bond angle of NCl3 is very slightly smaller than that of NH3, due to inter-electron repulsion of the lone pairs of Cl atoms; this also applies to PH3 versus PCl3, more of this below).
However, for PH3 and PCl3, notice that the bond angles vary more significantly from the theoretical tetrahedral and trigonal pyramidal electron and molecular geometries, of NH3 and NCl3. There are two different valid explanations for this difference (ie. N versus P), one explanation has to do with electronegativity, the other explanation involving the different atomic radii and corresponding extent of hybridization required. (Again, strictly according to the H2 syllabus, Cambridge is unlikely to test you on geometries about period 3 elements.)
Now, to answer the question on which bond angles are larger : PH3 or PCl3?
If this question was indeed asked by Cambridge in the 'A' level exams, the exam-smart thing to do, is to give BOTH answers, *but with explanations for BOTH*.
You see, exactly like in real life, oftentimes in Chemistry (which is a fundamental aspect of Universal reality, including human reality and human psychology), there will be often enough, conflicting motivations, principles and patterns.
There are valid rationales as to why the bond angles in PH3 *may be expected* to be larger than the bond angles in PCl3. There are also valid rationales as to why the bond angles in PCl3 *may be expected* to be larger than the bond angles in PH3. Professional chemists will understand and appreciate both conflicting principles, and based on experimental evidence, conclude which principle outweights which principle, *but only in a specific context*.
Let's look at them briefly :
Because the P-Cl bond length is expected to be longer than the P-H bond length, accordingly we might expect less inter-atomic repulsion between the 3 atoms (ie. Cl atoms in PCl3, and H atoms in PH3) bonded to the central P atom. Accordingly, one might surmise that based on this principle, the bond angles of PCl3 *might be expected* to be smaller than the bond angles of PH3.
Then again, one has to consider the fact that the Cl atoms have 3 lone pairs, while the H atom has no lone pairs. With 3 lone pairs around each Cl atom, one would expect a greater degree (pun intended!) of inter-electron repulsion between the Cl atoms, compared to the lone-pair-less H atoms. Accordingly, one might surmise that based on this principle, the bond angles of PCl3 *might be expected* to be larger than the bond angles of PH3.
Experimental evidence has proven that the bond angles of PCl3 is slightly larger than that of PH3. Then again, it is not uncommon that (due to either human error, experimental limitations, or different variables or contexts involved) some other experiments might appear to prove otherwise. Therefore, it is unsurprising that different sources, textbooks, or teachers might give a different opinion on whether PCl3 or PH3 has a larger bond angle.
As you go higher up the University and professional levels, there is more and more uncertainty, (as opposed to PSLE and 'O' levels' simplified science "this is black, this is white", because 'A' levels is approaching Uni level, there begins to be) more shades of grey, and more possibilities to be explored with a more open mind.
Look back at yourself in your PSLE years, 'O' level years, and now 'A' level years. And into the future, into your Uni years and professional working years. You come to realize that "the more you know, the more you know that you don't know". In PSLE you thought you knew everything. In 'O' levels, ok there seems to be more to this universe. At 'A' levels you realize damn all this is just the tip of the ice-berg. And so on.
Back to the debate about PCl3 versus PH3. Cambridge is (usually) fair. 'A' level students will *not* be expected to have conducted or read up on experiments that investigate whether PCl3 or PH3 have a larger bond angle. But 'A' level students *are* expected to be able to *think* for themselves (a common criticism of Sg education system is that students don't think, they just blindly memorize school notes), based on their current understanding of chemistry, deduce and explain why the bond angles in PCl3 *might be expected* to be larger than in PH3, or vice-versa.
Therefore, the exam-smart student who can say "both are possible" and gives correct explanations for BOTH possibilities, will earn the full credit (ie. score full marks) for the question. If the Cambridge mark scheme is fair (and it usually is, though not always), then full marks may still be obtainable if the student gives EITHER explanation with the correct relevant conclusion. But if you're good enough to work out both explanations, then be exam-smart and give both (with the correct relevant conclusions for both) and your marks will be more safely assured.
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blacktoast94 posted :
A sample of m g of a compound was vapourised in a gas syringe. the vapour occupid v dm^3 at T degrees celcius and p atm. what is the Mr of the compound?
answer : (m x 22.4 x (T+273)) / (p x v x 273)the thing i don't get is why is there a 273 at the bottom?
UltimaOnline replied :The denominator term is the gas constant R, which is usually 8.31 or 0.0821, etc (depending on the units for the variables), but for such exam questions, you have to either take molar volume as 24.0 or 22.4 and make R the subject accordingly.
PV = nRT
1atm x 22.4 = 1 x R x 273
R = 22.4/273
Sub the above into PV = nRT
(p) (v) = (m/Mr) (22.4/273) (T+273)
Mr = (m x 22. 4 x (T+273) ) / (p x v x 273) -
Nish asked :
When a substance dissolves in a solvent, the compound can be either molecular or ionicall dissociated, depending on the nature of the solvent. Predict and explain the form of PCl5's existence when dissolved in CCl4.
In strongly polar solvents, PCl5 dissociates into ions, as the consequent exothermic ion - permanent dipole solvation interactions make the (slightly) endothermic dissociation process worthwhile.
As CCl4 is a non-polar solvent, and because any ion - induced dipole interactions (should PCl5 autoionize) would be weak (and not worth the effort to autoionize), consequently PCl5 remains simple molecular, with only instantaneous dipole - induced dipole van der Waals interactions with the non-polar CCl4 solvent.
Bonus :
The reason why non-polar solutes dissolve in non-polar solvents cannot be explained by enthalpy changes alone, since any enthalpy change involved would be minimal. The reason why solvation of non-polar solutes in non-polar solvents is feasible, spontaneous and can occur, is the positive (hence favourable) entropy change that would result from mixing of both non-polar species. -
therealist asked :
Describe a reaction to distinguish between chloromethylbenzene and 4-chloromethylbenzene.
Reflux with NaOH(aq), followed by addition of acidified (with HNO3) AgNO3(aq). White ppt of AgCl will be observed only for chloromethylbenzene.
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Originally posted by Pinkbow6826:
In a reaction, 25.0 cm3 of hydrogen peroxide solution was reduced to water by 40 cm3 of excess 0.250M iron(II) ions under acidic conditions. 10.0 cm3 of the resultant solution was found to react completely with 13.5 cm3 of 0.020 M potassium managate (VII) solution.
Calculate the concentration of the 25.0 cm3 hydrogen peroxide solution.
In Chem calculations, algebra is your best friend.Let x be the molarity of H2O2 solution.
Write half-equations and balanced redox equations.Fe2+ oxidized to Fe3+, and H2O2 reduced to H2O.
Fe2+ oxidized to Fe3+, and MnO4- reduced to Mn2+.
In terms of x, find the moles of H2O2 present.
Apply stoichiometry to find moles of Fe2+ reacted (oxidized) away by the H2O2, in terms of x.Apply stoichiometry to find moles of Fe2+ reacted (oxidized) away by the MnO4-, in actual numerical value.
Find the total moles of Fe2+ present, ie. moles = molarity x volume.
(moles of Fe2+ reacted by H2O2) + (moles of Fe2+ reacted by MnO4-) = (total moles of Fe2+ present)
Solve for x.
Note that you have to consider the fact that you're only taking 10cm3 out of 25 + 40 = 65 cm3 resulting solution. After you've found the moles of Fe2+ oxidized / reacted away by MnO4-, you've to multiply this no. of moles by 6.5 times. This will give you the moles of Fe2+ in excess (ie. not oxidized / reacted away by H2O2).
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blacktoast94 posted :
how would you expect the volume of a given amount of hydrazine at a presssure of 10x10^5 pa and 285K to quantitavely compare with that of an ideal gas under similar conditions? explain.
asnwers said: the volume of N2H4 is less than that of an ideal gas at high pressure.
notes said: real gases tend to show positive deviation at high pressure. this is because at high pressure, the gass molecules are closer, causing the collective volume of the particles to be significant with respect to the volume that they occupy. the gas thus becomes less compressible and the volume will be larger than expected.??? which one is correct and why?
UltimaOnline replied :
Both are correct, insofar as the principles involved. As in real-life (for which Chemistry is an integral aspect), tis often the case that diametrically opposing trends vie for dominance over particular aspects for consideration.
Observe that the structure of hydrazine (you should be able to visualize the structure; if you're unable to, then draw it out) makes intermolecular hydrogen bonding possible, thus violating one of the basic tenets of the ideal gas behaviour. That hydrazine is capable of inter-molecular hydrogen bonding, is the first fact that the A level candidate is expected to be able to state in the exam answer.
As pressure increases, the gas molecules will be pushed very closely together, which will result in, at least initially, a decreased volume due to more extensive inter-molecular hydrogen bonding, keeping the gas molecules closer together and consequently a negative deviation in gaseous volume as compared to ideal gases.
However, once pressure exceeds a certain point, steric van der Waals repulsion (due to inter-electron repulsion and/or inter-nuclei repulsion) will force the real gas molecules apart when they get too close for comfort, unlike a theoretical ideal gas which (impossibly) occupies zero volume at absolute zero temperature. Consequently, a positive deviation in gaseous volume is observed, as compared to ideal gases.
1 atm = 1.01325 bar = 1.01325 x 10^5 Pa. This is indeed fairly high pressure. But it is beyond the capability of an A level candidate, to be able to discern exactly where such a pressure lies, on the scale of events I have indicated in the preceding paragraphs.
For the exam smart 'A' level candidate then, write both alternative answers (ie. both positive and negative deviations are possible), *but with qualifications and associated explanations*. In other words, explain both why increasing pressure would initially result in the gaseous volume of a real gas to deviate negatively from an ideal gas, and eventually why too high a pressure would eventually result in the gaseous volume of a real gas to deviate positively from an ideal gas.
As pressure increases, the gas molecules will be pushed very closely together, which will result in, at least initially, a decreased volume due to more extensive inter-molecular hydrogen bonding, keeping the gas molecules closer together and consequently a negative deviation in gaseous volume as compared to ideal gases.
However, once pressure exceeds a certain point, steric van der Waals repulsion (due to inter-electron repulsion and/or inter-nuclei repulsion) will force the real gas molecules apart when they get too close for comfort, unlike a theoretical ideal gas which (impossibly) occupies zero volume at absolute zero temperature. Consequently, a positive deviation in gaseous volume is observed, as compared to ideal gases.
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A level chemistry calculations qn.
Challenging O level chemistry calculations qn.A sample of impure HCL contaminated with solid KOH was used to titrate against 25.0cm^3 of NAOH, conc. of 0.25mol/dm^3. it was discovered that it required 10.25cm^3 of the aicd mixture to turn phenolpthalein colourless. the original conc. of the acid mixture was 1.0mol/dm^3. calculate the mass of KOH that was added to 250cm^3 of the acid mixture, assuming that no change in volume occured.
Solution :
Find the moles of HCl reacted with NaOH.
6.25 x 10^-3 mols.
Find the (real) molarity of HCl.
0.60976 mol/dm3
Find the difference in original versus real molarities. This is the molarity of KOH added (since both acid and base are monoprotic).
0.39024 mol/dm3
Let x be the sample mass of KOH added to 250cm3 of HCl.
x grams
Find moles (in terms of x) of KOH added to 250cm3 of HCl.
(x / 56.2) mol
Find molarity of KOH added to 250cm3 of HCl.
(x / 56.2) / (250/1000) = 0.071174x mol/dm3
Solve for x.
0.071174x mol/dm3 = 0.39024 mol/dm3
Hence x = 5.4829 = 5.48g (3 sf) -
Originally posted by LastRide7:
Sulfur dioxide and oxygen in the mole ratio 2:1 were mixed at a constant temperature of 1110K and a constant pressure of 9atm in the presence of a catalyst. At equilibrium, one-third of the sulfur dioxide had been converted into sulfur trioxide. Calculate the equilibrium constant Kp for this reaction under these conditions.
Since moles and partial pressures are directly proportional (when temperature and volume remain constant), hence Initial partial pressures of SO2, O2 and SO3 are (2/3)(9)atm, (1/3)(9)atm and 0 atm respectively.Since at equilibrium, one-third of the sulfur dioxide had been converted into sulfur trioxide, Change will be -(1/3)(2/3)(9)atm for sulfur dioxide and applying stoichiometry you can determine the corresponding partial pressure Changes for oxygen and sulfur trioxide.
Accordingly, you can determine the Equilibrium partial pressures for all 3 species, and consequently the Kp value.
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ElectroChem
Originally posted by donkhead333:How do you choose what reactions to use, when it involves water?
From what I know, equations with H+ are chosen when the reaction occurs in an acidic medium, while equations with OH- are chosen when it occurs in an alkaline medium.
So, in acidic solution, and assuming H2O is oxidised,
H20 -> H2O2 + 2H+ + 2e- and
H2O -> O2 + 4H+ + 4e-
and in alkaline solution, assuming H2O is reduced,
O2 + 2H2O + 4e- -> 4OH- and
2H2O + 2e- -> H2 + 2OH-
What's the basis for selection? Electrode potential such that Ecell will be maximum/minimum? Thanks in advance.
It is the protons from water which are reduced, and it is the hydroxide ions from water which is oxidized.
Do not use reactions that generate H2O2, unless you're specifically told to do so in the question (most usually, H2O2 will be a reactant, not a product, ie. use the right to left equation instead of left to right).
Notice that in your equation O2 + 2H2O + 4e- -> 4OH-, it is not water that is reduced, but molecular dioxygen (ie. O2 gas) which is reduced.
Again, unless the question specifically dictates molecular dioxygen (ie. O2 gas) being reduced at the cathode, do not use that equation. Use the one in which (the protons from) water are reduced to molecular dihydrogen (ie. H2 gas) instead.
For electrolytic cells, you need not worry about the cell potential, because the source of the electricity comes from elsewhere (eg. the wall socket). Therefore, you need not worry about the feasibility of whether you should use water or protons as the reactant to be reduced @ cathode, or whether you should use water or hydroxide ions as the reactants to be oxidized @ anode. Cambridge will accept either choice of reactants, for both.
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Nish posted :
Platinium(IV) chloride combines with NH3 to form compound Z in which the co-ordination number of Pt is 6. When solution Z is treated with excess Ag(NO3)2 (aq), only 1/2 of the total chloride present is precipitated as AgCl.
What is the formula of compound Z?
A) Pt(NH3)6Cl4
B) Pt(NH3)5Cl4
C) Pt(NH3)3Cl4
D) Pt(NH3)4Cl4UltimaOnline replied :
Platinum(IV) chloride is PtCl4, which reacts with NH3 ligands to generate a complex ion in which some Cl- ions function as ligands within the complex ion, and some Cl- ions function as counter ions outside the complex ion. Since the question specifies that exactly 50% of the Cl- ions are counter ions (ie. outside the complex ion and hence can be precipitated out as solid AgCl), hence the formula of the coordination compound (ie. coordination compound = complex ion + counter ion) must be one in which :
Pt has OS of +4
There are two Cl- ligands
There are two Cl- counter ions
There are a total of 6 ligands (note that both NH3 and Cl- are monodentate ligands) within the complex ion.Hence, the complex ion is diaminedichloroplatinum(IV) [Pt(NH3)4(Cl)2]2+ and the counter ions are the two chloride ions 2Cl-.
Therefore the answer is option D (ie. diaminedichloroplatinum(IV) chloride). -
Organic Chemistry
Draw the mechanisms for :
1) Reacting HNO2 with HCl to generate the nitrosonium NO+ electrophile, including the two resonance contributors of the nitrosonium NO+ electrophile.
2) Having phenylamine nucleophile attack the nitrosonium NO+ electrophile to generate benzene diazonium chloride (electrophile).
3) Having the water nucleophile attack the benzene diazonium chloride electrophile generated in step (2) to generate phenol (nucleophile).
4) Having the phenol nucleophile generated in (3) attack the benzene diazonium chloride electrophile generated in (2), to generate the orange azo dye (para-hydroxyphenyl)azobenzene.(Partial) Solution :
Mechanism for step (2)http://en.wikipedia.org/wiki/File:Nitrosonium_ion_with_amine_reaction.png
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