BedokFunland JC's A Level H2 Chemistry Qns (Part 1)

last SG · 2008-09-19T13:00:14+00:00

'O' & 'A' Level Qns. Qn 1. When 1.01g sample of potassium nitrate(V) is heated above its melting point, oxygen gas is evolved. The mass of the sample dec...

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'A' level Qn. (Supplementary thermodynamics question; not strictly required in H2 syllabus).

For the reaction involving a Group I metal and water, briefly describe (qualitatively) the expected difference between energy change and enthalpy change.

Solution :

The reaction is exogernic and exothermic, and produces hydrogen gas. Under standard conditions, there is an increase in no. of moles of gas from LHS (none) to RHS (hydrogen gas).

Therefore, we expect the energy change to be slightly more exogernic as compared to the enthalpy change; this is due to some of the energy (produced from the reaction) being used up (by the hydrogen gas produced) to do work to push back atmospheric air molecules.

Energy change is measured under constant volume (not constant pressure).

Enthalpy change is measured under constant pressure (not constant volume).

For the reaction :

2Na(s) + 2H2O(l) --> 2NaOH(aq) + H2(g)

Enthalpy change has been measured to be approximately -367.5kJ/mol (of H2 gas produced).

Energy change can been calculated to be approximately

Energy change(kJ/mol) = Ethalpy change(kJ/mol) - (Pressure(atm) x Volume(dm3) change)

= -367.5kJ - (1.0atm x 24.5dm3)

= -367.5kJ - (2.45kJ)

= -369.5kJ/mol (of H2 gas produced).

These values support our expectation that the energy change of the reaction is slightly more exogernic (ie. stronger) than the enthalpy change measured, as some energy is used up to do work (to push surrounding air molecules).

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'A' Level Qn.

Explain why point of maximum buffer capacity exists when pH = pKa.

Solution :

When the pH is lower (ie. more acidic) than the pKa value, the majority of the species exists in protonated form; and has the capacity only to donate protons to buffer incoming bases and/or OH- ions.

When the pH is higher (ie. more alkaline) than the pKa value, the majority of the species exists in deprotonated form; and has the capacity only to accept protons to buffer incoming acids and/or H+ ions.

When the pH is at pKa value, half of the species exists in protonated form, and half of the species exists in deprotonated form; and hence has the maximum capacity to buffer both incoming acids as well as bases.

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'A' Level Qn.

Question :

When trichloroethanal is heated with aqueous sodium hydroxide, two products are formed in a single reaction pathway. One of the products is trichloromethane. Draw the mechanism for this pathway and hence identify the other product.

Solution :

1) Hydroxide ion nucleophile attacks delta +ve carbonyl carbon.

2) Electron density shifts up; pi-bond pair becomes a lone pair on oxygen.

3) When electron density shifts back down (lone pair becomes pi-bond pair, reforming the carbonyl group), the trichloromethyl group leaves as an anion.

4) The trichloromethyl anion is capable of being a leaving group (ie. it is stable enough to exist on its own if only briefly), only because of the 3 highly electronegative and hence strongly electron-withdrawing Cl groups.

5) However, carbanions are still more unstable compared to alkanoate ions, hence the next step is

6) proton transfer (ie. acid-base reaction) from methanoic acid to the carbanion (ie. lone pair on carbanion becomes bond pair with proton (H⁺); bond pair between proton (H⁺) and oxygen becomes lone pair on oxygen), forming

7) trichloromethane and sodium methanoate (negative charge of HCOO^- counterbalanced by Na⁺; note that these ions are aqueous meaning dissolved in water, having a hydration shell, ion-dipole interactions with polar water molecules).

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'A' Level Qns.

1) Compare the structures of carbon dioxide and silicon dioxide.

2) Compare the solubilities of calcium fluoride vs calcium chloride (and contrast these, in turn, to the solubilities of silver fluoride vs silver chloride)

3) Compare the structures of beryllium fluoride vs calcium fluoride.

4) Compare the structures of beryllium fluoride vs beryllium chloride.

Solutions :

1) Compare the structures of carbon dioxide and silicon dioxide.

At room temperature and pressure, carbon dioxide is a simple (linear) covalent molecule while silicon dioxide has a giant tetrahedral 3D structure.

The preference for silicon to form 4 single bonds (hence tetrahedral macromolecule), rather than 2 double bonds (as in carbon dioxide) can be traced to the decreased ease of overlap of its 3p orbitals, compared to the 2p orbitals in carbon, to form pi bonds. In other words, due to the greater atomic radius of silicon (compared to carbon) and hence greater inter-atomic distance between silicon and oxygen (compared to carbon and oxygen), it is easier and more stable to form single bonds between silicon and oxygen, while in contrast, it is easier and more stable to form double bonds between carbon and oxygen.

2) Compare the solubilities of calcium fluoride vs calcium chloride.

Calcium fluoride is insoluble while calcium chloride is soluble. Let's consider qualitatively why calcium fluoride is insoluble (as compared to calcium chloride).

In terms of enthalpy, the (endothermic) lattice dissociation enthalpy outweighs the (exothermic) hydration enthalpy, for calcium fluoride; while the (exothermic) hydration enthalpy outweighs the (endothermic) lattice dissociation enthalpy, for calcium chloride.

This is due to the much higher charge density on the fluoride ion, resulting in a much more strongly endothermic lattice dissociation enthalpy for calcium fluoride (compared to its chloride). The exothermic hydration enthalpy also falls from fluoride (higher charge density) to chloride (lower charge density), but the decrease is not as sharp as the fall in lattice dissociation enthalpy.

(Note that the difference in hydration enthalpies is apparent when you consider that silver fluoride is soluble while silver chloride, silver bromide and silver iodide are insoluble. In this case, because the silver cation has low charge density, hydration enthalpy changes more significantly than lattice dissociation enthalpy, between fluoride and chloride.)

Overall, lattice dissociation enthalpy falls more sharply than hydration enthalpy, when we go from calcium fluoride to calcium chloride. Enthalpy argument hence favours the solution/dissolving of the chloride (ie. it's worth the cheap endothermic lattice dissociation investment, we get back even more heat energy from the ion-dipole interactions with water) over the solution of the fluoride (ie. it's not worth it, we can't get back our expensive endothermic lattice dissociation investment).

In terms of entropy, the fluoride anion has a much higher charge density (compared to chloride) and hence far stronger ordering power over surrounding polar water molecules. Entropy (ie. disorderliness) hence favours the solution of the chloride over the fluoride.

Interestingly (and fortunately for stressed students), both enthalpy and entropy arguments favour the solution of the chloride over the fluoride.

Acknowledgement & Reference :

Jim Clark's ChemGuide.co.uk : http://www.chemguide.co.uk/inorganic/group2/problems.html

3) Compare the structures of beryllium fluoride vs calcium fluoride.

Calcium fluoride has an ionic lattice structure (wherein calcium is coordinated to eight fluoride anions and each F⁻ ion is surrounded by four Ca²⁺ ions). Beryllium fluoride has a covalent structure (even though it's more ionic compared to say, beryllium chloride; due to the fact that fluorine, being the #1 king of electronegativity in the entire periodic table, is even more electronegative than both chlorine and the Be²⁺ cation), and exists as a simple covalent (linear molecular geometry) molecule in the gas phase, while in the solid crystalline phase as a silicon dioxide-like tetrahedral 3D structure with four coordinate beryllium (ie. one fluoride atom each from 2 neighbouring molecules give dative bonds to the beryllium atom of a central molecule, allowing beryllium to have a stable octet; consequently and structurally, every Be atom is bonded to 4 F atoms, and every F atom is bonded to 2 Be atom; formal charges of -2 and +1 exist on the beryllium and fluorine atoms respectively).

4) Compare the structures of beryllium fluoride vs beryllium chloride.

In the gas phase, both beryllium fluoride and beryllium chloride exist as a simple linear covalent molecule (see answer to Q3 above).

In the solid phase, while beryllium fluoride has a silicon dioxde-like tetrahedral 3D structure (see answer to Q3 above), beryllium chloride polymerises to make long (virtually) 1-dimensional polymer chains. One chlorine atom each from 2 neighbouring molecules will give a dative bond with a beryllium atom from a central molecule, allowing beryllium atoms to enjoy a stable octet. So a similar process for a similar objective occurs between the beryllium atoms and the halogen atoms, in both beryllium fluoride and beryllium chloride, but the end structure is different (akin to the alpha helix vs beta pleated secondary structure of polypeptides due to hydrogen bonding).

In this case, the end-result difference in beryllium halide structure is naturally related to the difference in properties between fluorine vs chlorine, resulting in an alternative most-stable beryllium halide solid structure (the specifics are beyond the requirements of the 'A' level or H2/H3 syllabuses.)

Wikipedia image of beryllium chloride long chain polymer

http://en.wikipedia.org/wiki/File:Beryllium-chloride-xtal-3D-balls-A.png

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'A' Level Qn.

You have in your possession a Singapore 50 dollars note (bill). There are several ink or dirt smudges on it (of more than 1 colour or appearance). Describe and explain, using easily available (ie. non-controlled) chemical substances and/or tools, to determine if the smudges is covalent or ionic in nature.

Solution :

Using a tissue paper moistened with water, wipe the smudges. If some of the smudges come off, the nature of the composite pigment could either be soluble ionic, or polar covalent (possibly with the capacity to form hydrogen bonds with water). If the smudges do not come off, the pigment could either be insoluble ionic, or non-polar covalent.

To determine which, wipe the remaining smudges with an alcohol (eg. isopropyl alcohol a.k.a. propan-2-ol) swab that can be purchased from any pharmacy (eg. NTUC, Guardian, Watsons, etc). If the smudges come off, the pigment must have been non-polar covalent. If the smudges remain, the pigment must be insoluble ionic.

An ionic compound is soluble in water, if the Gibbs Free Energy for the solution process (ie. lattice dissociation + hydration enthalpies) is feasible. Otherwise, it is insoluble ionic.

A covalent compound that is soluble in water, is either strongly polar, and/or has the capacity to form hydrogen bonds with water. Water would rather hydrogen bond within itself (ie. water molecules with other water molecules), than interact favourable with non-polar covalent compounds.

(Indeed, this is the basis for the so-called hydrophobic interactions of non-polar 'R' groups of amino acids in proteins. They get pushed away by water molecules (which prefer to bond within themselves, macham elitist), until they (the non-polar groups) end up in close proximity, afterwich induced dipole-dipole van der Waals interactions take over to hold them together (hence, "hydrophobic interactions").)

Alcohol is relatively non-polar (to be precise, the longer the non-polar or "fatty" hydrocarbon chain, the more non-polar the alcohol is; this explains why fats are "fatty" - they have long non-polar hydrocarbon chains). Hence, relatively non-polar and/or hydrogen bonding incapable covalent compounds, that are insoluble to water, may be soluble in alcohols (as well as oil/fats), due to favourable induced dipole-dipole interactions between solute and solvent. Which explains why the alcohol swab is able to wipe away non-polar and/or hydrogen bonding incapable covalent pigments that water cannot.

Tip : for you students who write messily and end up with ink-stained hands after taking an exam paper, wipe your hands with alcohol swabs (which can be bought cheaply from any pharmacy, eg. NTUC, Guardian, Watsons, etc). Often, the ink pigments are relatively non-polar and/or hydrogen bonding incapable covalent compounds, that can be wiped away easily with alcohol swabs.

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'A' Level Qn.

Assuming we're talking about atoms of the same element (eg. carbon), why is it that a double bond is less than twice the strength of a single bond?

Solution :

A single bond consists of 1 sigma bond.

A double bond consists of 1 sigma bond and 1 pi bond.

A sigma bond is formed when orbitals from 2 atoms overlap head-to-head (the sigma bond electron density is concentrated between the 2 nuclei and is responsible for holding the atoms together against the mutual repulsion of the positively charged nuclei).

A pi bond is formed when the p orbitals of the 2 atoms overlap sideways, with the pi bond electron density above and below the sigma bond.

For C-C, the sigma bond consists of sp3-sp3 orbitals overlapping end-on-end.

For C=C, the sigma bond consists of sp2-sp2 orbitals overlapping head-to-head; and the pi bond consists of the unhybridized pz-pz orbital overlapping sideways.

The pi bond is weaker than the sigma bond, due to the following reasons :

(i) for the pi bond, there is significantly less electron density overlap between the component p-orbitals due to their parallel directional orientation in 3D space (in contrast to the head-on-head overlap of the sigma bond).

(ii) the pi bond electron density is further away from the positive charge of the nucleus and hence uses more energy to sustain and is therefore less stable (compared to the more stable sigma bond electron density located directly (and "resting lazily") between the two positively charged nuclei).

(iii) the pi bond orbitals are unhybridized while the sigma bond orbitals are hybridized (eg. sp3 or sp2), which means the energy level of the hybridized sigma orbitals (ie. the weighted average of the different orbital types; recall that the s orbital always has a slightly lower energy than the p orbitals at the same energy level or electron shell) is lower (ie. more stable) compared to the higher energy (ie. more unstable) unhybridized p orbital.

Although the pi bond by itself is weaker than a sigma bond, pi bonds are often components of multiple bonds, together with sigma bonds. The combination of pi and sigma bond is stronger than either bond by itself.

For instance, as the following excerpt from the Data Booklet states, the average bond energy (or enthalpy) of C-C single bond is 350kJ/mol, while that of C=C double bond (consisting of 1 sigma bond + 1 pi bond) is 610kJ/mol, which is less than twice the strength of the single bond.

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'A' Level Qn.

by Dr Rod Beavon (http://home.clara.net/rod.beavon)

Inorganic Chemistry Problem1

http://home.clara.net/rod.beavon/inprob1.htm

A is a white compound of lead; on heating it gives yellow B and a colourles gas that turns limewater milky. If B is heated in air for several hours at 470oC it is converted into a scarlet powder C. C contains 90.66% lead and 9.34% oxygen by mass.

C on heating with dilute aqueous nitric acid gives a colourless solution D and a brown solid E. If sodium hydroxide solution is added to solution D, a white gelatinous precipitate is formed, which with excess sodium hydroxide solution gives a colourless solution G.

Compound E reacts with concentrated hydrochloric acid to give a white solid I and a green gas J. I is soluble in hot water but insoluble in cold, and forms soluble complex ions such as K with excess chloride ions. With potassium iodide solution J gives a brown solution of L, which on shaking with hexane gives a deep purple organic layer.

With potassium iodide solution D gives a bright yellow solid M; this is insoluble in cold water, but in hot water gives a colourless solution.

Identify all the substances A - M, and write equations for all the reactions involved.

Solution :

http://home.clara.net/rod.beavon/inans1.htm

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'A' Level Qn.

by Dr Rod Beavon (http://home.clara.net/rod.beavon)

Inorganic Chemistry Problem2

http://home.clara.net/rod.beavon/inprob2.htm

Copper reacts with 50% nitric acid to give a blue solution A and a brown gas B. If the solution A is diluted and sodium hydroxide solution added cautiously, a gelatinous blue precipitate C is obtained, which if warmed forms a black solid G.

Addition of concentrated ammonia solution to C gives a deep blue solution that contains the ion D. Addition of concentrated hydrochloric acid to C gives a green solution of the ion E, which on dilution with water gives a solution of the ion F.

If the brown gas B is passed into water a mixture of acids H and I is formed in a disproportionation reaction.

Identify A - I, giving equations for and explaining each reaction as far as possible.

Solution :

http://home.clara.net/rod.beavon/inans2.htm

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'A' Level Qn.

by Dr Rod Beavon (http://home.clara.net/rod.beavon)

Inorganic Chemistry Problem3

http://home.clara.net/rod.beavon/inprob3.htm

A white powder A on heating turns yellow and evolves a gas C that turns limewater milky, as well as water vapour. The yellow residue B turns white on cooling, but will turn yellow again when heated.

B reacts with dilute sulphuric acid to give a colourless solution D. If dilute sodium hydroxide solution is added to D, a white precipitate E is produced which with excess alkali gives a colourless solution F.

With dilute ammonia solution D gives a white precipitate G; adding excess ammonia causes the mixture to become clear, giving solution H.

Identify all the substances A – H, and account as thoroughly as you can for the observations.

Solution :

http://home.clara.net/rod.beavon/inans3.htm

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'A' Level Qn.

by Dr Rod Beavon (http://home.clara.net/rod.beavon)

Inorganic Chemistry Problem4

http://home.clara.net/rod.beavon/inprob4.htm

A is a metal that reacts moderately quickly with dilute sulphuric acid to give a pale green solution B and a gas C. If solution B is allowed to crystallise, the pale green solid obtained gives on strong heating a solid H, and two gases I and J. These gases are compounds of non-metallic elements, one of which is in a different oxidation state in each compound. If J is passed into a solution of barium chloride there is a vigorous reaction and a white precipitate K is formed.

A reacts with steam in an equilibrium reaction to give a solid M, which contains A in two different oxidation states, and the gas C; this reaction was at one time used industrially to make C. A reacts with gaseous HCl on heating to give K, a white solid, and C. Hydrated K is pale green. With chlorine, A gives a brown covalent solid L which sublimes on heating; aqueous solutions of the hydrate of L react with copper metal, and for that reason are used to etch printed-circuit boards in electronics.

Solution B gives a dirty-green precipitate D if sodium hydroxide is added; this precipitate remains in excess sodium hydroxide. The same reaction is seen with the addition of ammonia. If D is allowed to stand in air it forms a foxy-red compound E, which is also obtained if sodium hydroxide solution is added to solutions of L.

Acidified solutions of B decolourise potassium manganate(VII) solution giving the manganese-containing ion N as one of the products.

Identify all of the substances A – N, and deduce as much as you can about the processes occurring.

Solution :

http://home.clara.net/rod.beavon/inans4.htm

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'A' Level Qn.

by Dr Rod Beavon (http://home.clara.net/rod.beavon)

Organic Chemistry Problem1

http://home.clara.net/rod.beavon/problem1.htm

Compound W, C₃H₈O, reacts with hot concentrated phosphoric acid to give X, C₃H₆, which rapidly decolourises a solution of bromine in tetrachloromethane. On treatment with ethanoyl chloride, W gives Y, C₅H₁₀O₂. Oxidation of W with potassium dichromate in dilute sulphuric acid gives Z, C₃H₆O, which gives an orange precipitate with 2,4-dinitrophenylhydrazine but does not react with an ammoniacal solution of silver nitrate.

Deduce the structures of W, X, Y and Z, explaining carefully what each piece of evidence tells you.

Solution :

http://home.clara.net/rod.beavon/answerA.htm

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'A' Level Qn.

by Dr Rod Beavon (http://home.clara.net/rod.beavon)

Organic Chemistry Problem2

http://home.clara.net/rod.beavon/problem2.htm

An oily liquid A is insoluble in water, but on heating with aqueous sodium hydroxide solution for half an hour it dissolves. From the reaction mixture a liquid B can be distilled, which gives a yellow precipitate with iodine and sodium hydroxide solution. On careful oxidation B gives an aldehyde, C, which also gives a yellow precipitate with iodine and sodium hydroxide solution. If sulphuric acid is added to the solution obtained from heating A with sodium hydroxide solution, a white precipitate D is obtained. D liberates carbon dioxide from sodium hydrogen carbonate solution, one mole of D reacting with sodium hydrogen carbonate to give one mole of carbon dioxide. Heating D with soda-lime* converts it to benzene.

Find the structures of A, B, C and D, giving your reasoning.

* Soda-lime is calcium oxide that has been slaked with sodium hydroxide solution. Carboxylic acids, on heating with soda-lime, are decarboxylated, that is the -COOH group is replaced by -H.

Solution :

http://home.clara.net/rod.beavon/answer2.htm

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'A' Level Qn.

by Dr Rod Beavon (http://home.clara.net/rod.beavon)

Organic Chemistry Problem3

http://home.clara.net/rod.beavon/problem3.htm

Compound A, C₄H₈O₂, gives an orange precipitate with 2,4-dinitrophenylhydrazine reagent, and a silver mirror with ammoniacal silver nitrate. A is converted to B, C₄H₆O, on treatment with acid. B may exist as a cis-or trans- isomer; it reacts with hydrogen chloride to give C, C₄H₇OCl, and with chlorine to give D, C₄H₆OC1₂.

Reduction of C gives E, C₄H₉OCl, which is hydrolysed by dilute alkali to F, C₄H₁₀O₂. Both A and F give a positive iodoform (CHI₃) test, whereas E does not.

Identify compounds A to F.

Solution :

http://home.clara.net/rod.beavon/answer3.htm

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'A' Level Qn.

by Dr Rod Beavon (http://home.clara.net/rod.beavon)

Organic Chemistry Problem4

http://home.clara.net/rod.beavon/problem4.htm

An alcohol has the molecular formula C₄H₇OH. Oxidation of this alcohol gives a ketone. Ozonolysis of this alcohol gives methanal as one of the products.

Giving your reasoning in full, deduce the structural formula for this alcohol and predict whether it will be optically active.

Ozonolysis is a technique for finding the position of C=C double bonds in a molecule. The compound is reacted with ozone-enriched oxygen in a suitable inert solvent, and the product is hydrolysed with ethanoic acid in the presence of zinc. This produces a mixture of carbonyl compounds, which may be separated and identified, for example by means of their 2,4-DNPH derivatives.

Solution :

http://home.clara.net/rod.beavon/answer4.htm

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'A' Level Qn.

by Dr Rod Beavon (http://home.clara.net/rod.beavon)

Organic Chemistry Problem5

http://home.clara.net/rod.beavon/problem5.htm

Iodoethane CH₃CH₂I reacts with aqueous sodium hydroxide solution to give A, with ammonia to give B, and with potassium cyanide in ethanolic solution to give C, C₃H₅N. C reacts with lithium aluminium hydride in dry ether solution to give D. C with aqueous sodium hydroxide solution followed by dilute hydrochloric acid gives E, C₃H₆O₂, which with phosphorus(V) chloride gives F.

Identify compounds A to F, giving your reasons and equations where appropriate.

Solution :

http://home.clara.net/rod.beavon/answer5.htm

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'A' Level Qn.

by Dr Rod Beavon (http://home.clara.net/rod.beavon)

Organic Chemistry Problem6

http://home.clara.net/rod.beavon/problem6.htm

A, C₃H₇Cl, reacts with aqueous sodium hydroxide solution to give C, C₃H₈O. C reacts with phosphorus pentachloride to give steamy fumes, and can be oxidised with acidified potassium dichromate solution to give D, C₃H₆O, and then E, C₃H₆O₂. D reacts with both 2,4-dinitrophenylhydrazine and ammoniacal silver nitrate. E reacts with phosphorus pentachloride and with sodium hydrogen carbonate solution, in the latter case giving carbon dioxide.

A will react with benzene in the presence of anhydrous aluminium chloride to give B, C₉H₁₂. The structure of this compound is not what might have been expected from A. B can be made from benzene and F as the expected product, however, F being an isomer of A. Sodium hydroxide solution converts F to G, an isomer of C. G can be oxidised with acidified potassium dichromate solution to H, an isomer of D, but no further. H gives a positive result with the 2,4-DNP test but not the ammoniacal silver nitrate test; it also gives a positive result in the iodoform reaction.

On heating with alkaline potassium manganate(VII) solution B gives an aqueous solution of I, which on treatment with dilute sulphuric acid gives a white precipitate J.

(a) Identify all of the substances A – J, and explain each of the observations as fully as you can with supporting equations.

(b) Suggest why A and F give the same product with benzene.

Solution :

http://home.clara.net/rod.beavon/answer6.htm

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'A' Level Qn.

by Dr Rod Beavon (http://home.clara.net/rod.beavon)

Organic Chemistry Problem7

http://home.clara.net/rod.beavon/problem7.htm

A, C₂H₄, reacts with hydrogen bromide in the gas phase to give B, C₂H₅Br. This reacts with aqueous sodium hydroxide solution to give C, C₂H₆O, which can be oxidised by acidified potassium dichromate solution to give successively D, C₂H₄O, and E, C₂H₄O₂. D reacts with both 2,4-dinitrophenylhydrazine and ammoniacal silver nitrate.

With magnesium in dry ether B gives F, C₂H₅MgBr; this reagent will react with D to give G, C₄H₁₀O. Oxidation of this compound with acidified potassium dichromate solution gives H, C₄H₈O. This will also react with F, giving I, C₆H₁₄O. I is not oxidised by acidified potassium dichromate solution.

If A is reacted with bromine in the presence of sodium chloride, two compounds are obtained; J, C₂H₄Br₂, and K, C₂H₄BrCl. No C₂H₄Cl₂ is formed.

(a) Identify each of the compounds A – K, and explain all the observations as fully as you can.

(b) Why is no C₂H₄Cl₂ formed in the reaction given in the last paragraph?

Solution :

http://home.clara.net/rod.beavon/answer7.htm

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