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

last SG · 2010-07-26T23:40:35+00:00

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kickme asked :

Distinguish between CH2COOCH3 and HCOOCH2CH3 using a chemical test.

I see that both are esters. But i dont see any test which i can use to distinguish between the 2 esters. One of them is acidic and the other is not. Thats all i can see...

Solution :

Heat under reflux with acidified KMnO4.

Upon acidic hydrolysis (since you're heating under acidic conditions) to generate carboxylic acids and alcohols, followed by oxidation, the ester HCOOCH2CH3 will yield methanoic acid, which would be further oxidized to which will be further oxidized to carbonic(IV) acid, which exists in equilibrium with, and hence can decompose into, carbon dioxide gas (gaseous products will leave the reactant mixture, pulling the position of equilibrium over to the right, as predicted by Le Chatelier's principle) and water.

It's ironic that "kickme" didn't know the answer to this (admittedly fairly challenging for JC students) question, but he knew something else that 99% of most JC students didn't : that one ester is acidic and the other is not.

CH2COOCH3 is acidic because when deprotonation occurs with (ie. the alpha proton is abstracted away by) a strong base (eg. sodium metal Na, or sodium amide NaNH2), the conjugate base -CHCOOCH3 can be stabilized by having its negative charge delocalized by resonance over to the electronegative O atom of the carbonyl ester group. This is not possible for the other ester, HCOOCH2CH3.

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Source :
http://www.titrations.info/back-titration

1.435 g sample of dry CaCO₃ and CaCl₂ mixture was dissolved in 25.00 mL of 0.9892 M HCl solution. What was CaCl₂ percentage in original sample, if 21.48 mL of 0.09312 M NaOH was used to titrate excess HCl?

Solution :

During titration 21.48×0.09312=2.000 mmole HCl was neutralized. Initially there was 25.00×0.9892=24.73 mmole of HCl used, so during CaCO₃ dissolution 24.73-2.000=22.73 mmole of acid reacted. As calcium carbonate reacts with hydrochloric acid 1:2 (2 moles of acid per 1 mole of carbonate), original sample contained 22.73/2=11.37 mmole of CaCO₃, or 1.137 g (assuming molar mass of CaCO₃ is 100.0 g). So original sample contained 1.137/1.435×100%=79.27% CaCO₃ and 100.0-72.27%=20.73% CaCl₂.

Direct complexometric Al³⁺ determination is difficult, as Al³⁺ reacts with EDTA very slowly. To solution containing some unknown amount of Al³⁺ cations 50.00 mL of 0.05000 M EDTA solution was added. After 30 minutes excess EDTA was titrated with 0.04875 M Zn²⁺ solution. What was the amount of Al³⁺ if 17.58 mL of titrant was used?

Solution :

EDTA reacts with both Al³⁺ and Zn²⁺ in 1:1 ratio. There was 50.00×0.05000=2.500 mmole of EDTA used, and 17.58×0.04875=0.8569 mmole was found to be left. Thus there was 2.500-0.8569=1.643 mmole of Al³⁺ in the sample.

UltimaOnline SG

Ah thank you very much. Really enlighten me about this.

I have a last question relating to ions.

Iodine Tetrafluoride, IF4 - is a anion.

Iodine, the central atom, has 4 bond pairs and 2 lone pairs. The Fluorine atom has 1 bond pair, 3 lone pairs.

This gives Iodine a formal charge of -1. Each fluorine atom has a charge of 0.

Why doesn't the compound IF4 exist, since without the extra electron attached to Iodine, Iodine would have a 4 bond pairs, and 3 electrons attached to it. The formal charge of Iodine would therefore be 7-3-(8/2) = 0 Why does it exists as an anion instead of a compound ?

For it to exist as a molecule (ie. ionic charge = zero), one or more atoms (to be precise, the I atom) would have an unpaired electron (to be precise, 1.5 lone pairs, and the ion would be a free radical) and is thus too unstable to exist.

IF4-

The -ve formal charge is on the central I atom (since it has a larger atomic radius and a lower charge density and is thus more stable, despite its lower electronegativity).

In addition, note that F (being in period 2) cannot expand its octet, and thus to have a stable octet, it must have 1 bond pair and 3 lone pairs, and hence no formal charge (being in Grp VII).

Since I is in Grp VII, to have a -ve formal charge, we deduce that it has 8 valence electrons, ie. 4 bond pairs, 2 lone pairs.

Hence the electron geometry is octahedral, and the ionic geometry is square planar.

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25.0 cm3 of a solution of M2O5 of concentration 0.100moldm-3 is reduced by sulphur dioxide to a lower oxidation state. To re-oxidise M to its original oxidation number, this required 50.00cm3 of 0.0200moldm-3 potassium manganate (VII) in acidified medium. To what oxidation number was M reduced to by sulfur dioxide?

I am really confused by the question. How do I use the mol ratio to find the oxidation number. I can't solve similar questions due to this problem too....

Thanks in advance

Write reduction half-equation of MnO4- to Mn2+.

MnO4– + 8H+ + 5e– Mn2+ + 4H2O

Find moles of MnO4- required.

50.00cm3 of 0.0200moldm-3 = 1 x 10^-3 mol

Hence find moles of electrons transferred.

1 x 10^-3 x 5 = 5 x 10^-3 mol

Find moles of M2O5 present.

25.0 cm3 of 0.100moldm-3 = 2.5 x 10^-3 mol

Find moles of M atoms present.

2.5 x 10^-3 x 2 = 5 x 10^-3

Hence find moles of electrons transferred per mole of M atoms.

(5 x 10^-3) / (5 x 10^-3) = 1

Find OS of M in M2O5.

Hence find reduced OS of M.

(+5) + (-1) = +4

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kickme asked :

Suggest what must be done to the orchid before it can be an electrode in the electroplating process and name the substance used for the other electrode.

My answer is to soak the entire orchid in Au3+ and the other electrode should be made of Au.

Am i right?

Yes, the anode should be made of gold. The flower however, should be coated with graphite powder or paint.

http://en.wikipedia.org/wiki/Electrotyping

-----------------------------

kickme persisted :

Oh... Thanks. But wont the powder be washed off?

On another note, there are so many equation involving water in the data booklet. How do we know which is relevant to the reduction or oxidation of water?

Nope, won't fall off.

Graphite powder is in the form of a sticky paste. Graphite paint is naturally sticky as well (like all other paints, otherwise the customer will demand a refund!).

For electrolytic cells, just use H2O itself as the reactant (for both oxidation and reduction equations).

For galvanic-voltaic cells, apply chemistry sense : if the solution is acidic, use reduction of H+ and oxidation of H2O; If the solution is alkaline, use reduction of H2O and oxidation of OH-.

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So we can say that all the atoms in BF3 are co-planar while this is not the case in CH4?

Correct.

Co-planar implies exactly planar, while peri-planar implies near or almost planar (eg. some bonds/atoms may deviate slightly due to van der Waals repulsion).

Both terms may be used interchangeably for 'A' level purposes.

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kickme asked :

Why is it that the boiling point of CCl4 is greater than that of SiCl4?
Shouldn't SiCl4 have the higher boiling point given the greater electronegativity difference and larger molecular size?
2)Do ionic character in a covalent compound strengthen or weaken the bond?
3)Do covalent character in an ionic compound strengthen or weaken the bond?

Q1.
Silicon is a metalloid, and as such is significantly less electronegative compared to carbon. The Cl atoms in SiCl4 consequently have a larger magnitude of partial negative charge (compared to in CCl4), which results in a greater degree of van der Waals repulsion (here : lone-pair lone-pair repulsion). Hence SiCl4 has a lower boiling point compared to CCl4.

Q2 & Q3.
Case by case basis (because depending on multiple factors such as charge densities and effectiveness of orbital overlaps, sometimes ionic bonds are stronger, sometimes covalent bonds are stronger).
Sometimes (eg. sodium versus silver halides), with greater covalent character, the (part-covalent-part-ionic) bond strength increases. Sometimes (eg. carbon versus silicon halide) the opposite is true.
For 'A' level purposes, use whichever explanation helps your case, or helps explain the data given in the question, or helps to answer the question.
However, what is more generally consistent (although there are always exceptions, eg. CCl4 versus SiCl4), is that the greater the covalent character, the lower the boiling point.

UltimaOnline SG

CCl4 vs SiCl4

UltimaOnline posted:

The Cl atoms in SiCl4 consequently have a larger magnitude of partial negative charge (compared to in CCl4), which results in a greater degree of van der Waals repulsion (here : lone-pair lone-pair repulsion). Hence SiCl4 has a lower boiling point compared to CCl4.

Both molecules are symmetrical. Why are you not taking into account the larger vander Waal's forces in case of SiCl4 due to the larger number of elctrons as compared to CCl4?

Are you saying that the difference of just 6 more electrons in SiCl4 are not going to make it having a higher boiling point than CCl4.? Please explain.

Can you mention other similar cases?

In Chemistry just as in real-life, there are multiple factors in every situation, and commonly enough, some of these factors are contradictory.

Yes, the greater number of electrons and greater molecular size is indeed in favour of SiCl4 having stronger van der Waals over CCl4.

At the same time however, the greater magnitude of partial negative charges on the Cl atoms of SiCl4 is also in favour of CCl4 having stonger van der Waals over SiCl4.

So which outweighs which? We humans can argue all we want based on theoretial principles, but in the end, the Universe has already decided (by mathematics) which outweighs which, and we humans have to carry out experiments to find out what the Universe has already decided : in this context, experiments have proven that CCl4 has a higher boiling point over SiCl4, and therefore we humans can deduce that the second factor outweighs the first factor.

Another similar example, would be that of comparing the bond angles of NH3 versus NCl3. There are two contradictory factors in this regard :

Because the N-Cl bond lengths are significantly greater than N-H bond lengths, one might expect the inter-atomic-nuclei repulsions, and thus bond angles, to be smaller for NCl3 compared to for NH3.

But because Cl has a significantly larger atomic radius and has lone pairs, while H has a significantly smaller atomic radius and has no lone pairs, one might expect the inter-lone-pair repulsions, and thus bond angles, to be larger for NCl3 compared to for NH3.

So which factor wins? As it turns out, experimental evidence has shown that the bond angles of NCl3 and NH3 are almost identical, the former 107.1 deg and the latter 107.8 deg. And therefore we humans can deduce that both factors cancel each other out almost equally.

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Explain why the C-C bond dissociation enthalpy in cyclobutane is significantly less endothermic compared to the C-C bond dissociation enthalpy given in the Data Booklet.

Solution :

Ring strain (ie. angle strain and steric strain) weakens the C-C bond. Molecules including C atoms with 4 bond pairs and 0 lone pairs, are most stable when the C atoms are sp3 hybridized with a bond angle of 109.5 deg. Cyclobutane's bond angles vary, depending on conformation, but all bond angles are significantly smaller than 109.5 deg, constituting a significant deviation (from the ideal tetrahedral bond angles) that results in significant angle strain (ie. inter-electron repulsion between bond pairs) and steric strain (ie. van der Waals repulsion between groups of atoms) which weakens the C-C bonds of cyclobutane (and all similarly small cycloalkanes).

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Ok, need some urgent help here.

How do I convert BrCH2CH2Br to HO2C-CO2H?

Thank you!

Heat under reflux with aqueous NaOH, followed by heating under reflux with acidified K2Cr2O7 (not KMnO4!).

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Convert...

CH3CH2CH2Br to CH3CH(CH3)COOH.

There should be 3 steps between the reactant and the product.

DIdn't occur to me that I should have started with elimination of halogenoalkane. And I totally didn't want to give up on such a supposedly simple question.

Thought process :

CH3CH2CH2Br to CH3CH(CH3)COOH.

Since the final product is branched, a simple nucleophilic substitution will not suffice.

Elimination of HX following by addition of HX, is useful for converting terminal alkyl halides to internal alkyl halides.

Notice that the product has one more C atom than the reactant (which means we need a C nucleophile, such as the cyanide nucleophile, or a Grignard reagent, or a Wittig reagent).

Hence, substitute away the halogen (good leaving group) with CN by heating under reflux with NaCN(alc).

Carry out acidic hydrolysis of CN to convert it into (an amide intermediate which is further hydrolyzed to) COOH.

btw, TS mentions 3 steps, utima your method is technically 4 steps

Step 1: Alc KOH, refulx

Step 2: HX, rtp

Step 3: Alc KCN, reflux

Step 4: H+/H2O heat

There is a way to cut out step 2, but H2 Chem JC students aren't familiar with this process.

Step 1 (Elimination of HX) : KOH (alc), heat under reflux

Step 2 (Hydrocyanation) : HCN (alc), Ni/Cu/Pd catalyst, heat under reflux

Step 3 (Acid-catalyzed Hydrolysis) : H+/H2O, heat under reflux

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I request to review a question i set for AS. It relates to the difference in bonding of CO2 and SiO2 and their relation to bond energy, bond polarity.

In part (a) I have given the relavane melting point of CO2 and SiO2 and asked students to draw the structures of CO2 and SiO2.

Then in part (b) using these diagram I asked from the perspective of bond energy and bond polarity the difference in the bonding of these two oxides using the data from the data booklet. I made two headings bond energy and bond polarity. This is what I expect students to state and the relevant problems I am facing with this question.

"Si=O bonds are not stable" is given in the question.

BOND ENERGY: C-O, Si-O, C=O values quoted from the data booklet.

C=O bond is stronger than C-O in CO2 therefore a simple molcular structure is favoured over a giant covalent. while in SiO2, Si-O bonds are stronger because as given Si=O does not exist therefore a giant -covalent structure is likely to occur than a simple molecular structure. [Now should I give the information regarding Si=O or is there any other way of rephrasing the question?]

BOND POLARITY: Si-O bonds are more polar in SiO2 than C=O in CO2 since the difference in the electronegativities is larger in case of Si than C because Si has is less electronegative than C. Hence a three-dimensional network of Si-O bonds in SiO2 is favoured. [ I also like to involve the Si=O bond here is it workable?]

I would also like to involve the parameter of bond length here. Since multiple bonds are stable than single bonds but double bonds are shorter than single bond. Here is what i wish to stae w.r.t this case.

shorter C=O bond is favoured in CO2 as compared to long C-O bonds if CO2 were to be giant covalent because both C and O are small atoms they can easily overlap to form stronger C=O bond. In SiO2 silicon has a lrager atomic radius than C and thus will form weaker Si=O bonds or a lot of longer Si-O yet strong covalent bonds thus a giant covalent stritcure is more favoured over a simple molecular form.

I am looking forward to your suggestions.

UltimaOnline SG

a) By comparing the lengths and strengths of single bonds versus double bonds. state the correlation between the two parameters.

b) By comparing the relative atomic radii of C vs Si, suggest the difference in bond lengths that the C and Si atoms would form with O atoms.

c) Hence, deduce whether single bonds or double bonds, are favoured when C atoms, and Si atoms, bond with O atoms.

Personally, I would prefer the student explain this in terms of atomic orbitals (for which the question can be similarly structured, as in the example above, to guide the student towards this explanation).

Alternatively, if you leave your question as more open-ended essay style rather than structured, your mark scheme could accept a variety of points. For such essay style questions, the Cambridge mark scheme often has "accept any 5 of the following 7 points" that will suffice for the student to gain full marks for the question.

Carbon and silicon are both Group IV elements. Why then, under standard conditions, does carbon dioxide exist as a simple covalent molecule, while silicon dioxide exist as a giant covalent lattice?

To form the pi bond between C and O atoms, involves the more effective sideways or side-on overlap between the (less diffused) unhybridized 2p orbitals of both C and O atoms. To form the pi bond between Si and O atoms, involves the less effective sideways or side-on overlap between the (more diffused) unhybridized 3p orbital of Si and the (less diffused) unhybridized 2p orbital of O.

Consequently, the C=O pi bond is significantly stronger than the (hypothetical) Si=O pi bond, and hence it is more energetically favourable for C atoms to form (relatively strong) double bonds with O, and in contrast for Si atoms to form (a greater number of stronger) single bonds with O.

Conclusion : it is energetically favoured that (under standard conditions), carbon dioxide has a simple covalent molecular structure, while silicon dioxide has a giant covalent lattice structure.

UltimaOnline SG

Thank you for your invaluable suggestions.

I would like this question to be structure-one rather than essay-type since CIE exam papers prefers structured question for AS. However essay-type questions cannot be completely ruled out. They are often asked in A2.

Secondly your "hybridization answer approach" will prove quite challenging for students I have included the comparison of atomic radii of C and Si in the question.

Finally, after this part (a) and part (b) I have also introduced part (c) which relates to the use and property of three compounds.

part (c)

The industrial processes use different compounds carrying out different purposes. Choose one use and property on which that particular use depends of each the following material listed below.

In each case choose a diffrerent property on which the use depends.

Al2O3

Use Abrasive / catalyst in cracking of hydrocarbons /

Property Inert

SiO2

use Ceramics (porcelain, chaina dish) / stationary phase in TLC, HPLC, GLC /

Property high melting point inert/ can be derivatized on Silanol groups

MgO

use Refractory material in furnace lining

Property absorbs acidic impurites and electrical insulator

Kahynickel, an excellent exam question that you've written for your students, as usual.

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