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

last SG · 2010-07-26T23:40:35+00:00

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Hoay posted :

To UltimaOnline

you mentioned the following generalization regrading the spliting pattern:

Strong ligands are not the same as strong field ligands.

Weak ligands are not the same as weak field ligands

Would you please elaborate it some examples?

First of all, this aspect is totally beyond the H2 Chemistry syllabus. And it's unlikely, even when Cambridge on occasion go beyond the syllabus in their questions, for Cambridge to ask directly on this. In other words, H2 Chem students really needn't concern themselves regarding this.

As an example, consider the water molecule versus the hydroxide ion, as ligands. Obviously, the hydroxide ion is a far stronger ligand, because it's a lot more electron-rich (the O atom in OH- has a negative formal charge, while the O atom in H2O only has a partial negative charge), and therefore it's a far stronger ligand (and nucleophile, and base) compared to the water molecule.

(A Lewis base is a dative bond donor. If you're a Lewis base donating dative bond to a metal ion, you're called a ligand. If you're a Lewis base donating dative bond to a proton, you're called a Bronsted-Lowry base. If you're a Lewis base donating dative bond to an electrophile, you're called a nucleophile. Notice that NH3, functions well as all 3 types of Lewis bases.)

Experimental evidence have shown, however, that when complexed with water ligands, the magnitude of the energy difference of the non-degenerate d orbitals (of the metal ion) after d-d* splitting, is greater than when complexed with hydroxide ligands. In other words, we say that the water ligand is a stronger-field ligand compared to the hydroxide ligand.

What determines whether a ligand is a strong or weak ligand? The strength of a ligand parallels it's basicity and nucleophilicity (even though occassionally for some species, basicity and nucleophilicity do not parallel each other, eg. F- is a stronger base compared to I-, but I- is a stronger nucleophile compared to F-. H2 Chem students should be able to figure out why).

What determines whether a ligand is a strong field or weak field ligand? Several factors, including the electronegativity of the coordinate dative bond donor, the charge density of the ligand molecule/ion, etc. To understand this fully requires a thorough understanding of crytal field theory, ligand field theory and molecular orbital theory, much of which are *way* beyond the H2 syllabus.

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

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

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

For those interested (ie. not required for H2 Chem syllabus), the most relevant content has been reproduced below :

The size of the gap Δ between the two or more sets of orbitals depends on several factors, including the ligands and geometry of the complex. Some ligands always produce a small value of Δ, while others always give a large splitting. The reasons behind this can be explained by ligand field theory. The spectrochemical series is an empirically-derived list of ligands ordered by the size of the splitting Δ that they produce (small Δ to large Δ; see also this table):

I⁻ < Br⁻ < S²⁻ < SCN⁻ < Cl⁻ < NO₃⁻ < N₃⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < CH₃CN < py < NH₃ < en < 2,2'-bipyridine < phen < NO₂⁻ < PPh₃ < CN⁻ < CO

It is useful to note that the ligands producing the most splitting are those that can engage in metal to ligand back-bonding.

The oxidation state of the metal also contributes to the size of Δ between the high and low energy levels. As the oxidation state increases for a given metal, the magnitude of Δ increases. A V³⁺ complex will have a larger Δ than a V²⁺ complex for a given set of ligands, as the difference in charge density allows the ligands to be closer to a V³⁺ ion than to a V²⁺ ion. The smaller distance between the ligand and the metal ion results in a larger Δ, because the ligand and metal electrons are closer together and therefore repel more.

UltimaOnline SG

Organic Chem

>>> on the basis of bond energies, what could be the products of the following reaction? <<<

CH3*+CH3CH2Cl ---> ???
A. CH4+CH3CH*Cl
B. CH3CH2CH2*+HCl
C. CH3CH2CH3+Cl*
D. CH3CH2CH2Cl+H*

This is a poor qn, as it doesn't specify if multiple steps or only one step is involved. If multiple steps are involved, then more than one answer is eventually possible, except for option D (as far as the H2 syllabus on free radical substitution is concerned, you *never* get a H radical). But assuming only one step, the best answer here would be C. The C-Cl bond is weaker than the C-H bond. The methyl radical can therefore induce the homolytic cleavage of the C-Cl bond, and combine with either the resulting ethyl radical or the resulting chlorine radical. Since the C-C bond is stronger than the C-Cl bond, enthalpy considerations will favour the methyl radical combining with the ethyl radical, rather than the chlorine radical.

>>> why does the reaction between propane and chlorine not give a high yield of 2 - chloropropane? <<<

There are two factors that determine product distribution where free radical substitution mechanism is involved : 1) No. of H atoms substitutable 2) Stability of alkyl radical intermediate. The first factor is simpler, and compulsory to know for the H2 syllabus. The 2nd factor is more complicated, and is optional to know for the H2 syllabus (but if you're gunning for a distinction, you better find out about it). To get 1-chloropropane, you can substitute any of the 6 hydrogens on the 1st and 3rd carbon. To get 2-chloropropane, you can only susbtitute either of the 2 hydrogens on the 2nd carbon. Therefore, in terms of this factor (No. of H atoms substitutable), you expect 1-chloropropane to be the major product, with 75% yield (2-chloropropane with 25% yield).

(The following is optional for the H2 syllabus)

However, based on the other factor (Stability of alkyl radical intermediate), because 2-chloropropane involves a more stable secondary alkyl radical intermediate (compared to the primary alkyl radical intermediate required to form 1-chloropropane), and recall that the more stable the intermediate, the lower the activation energy required to reach the intermediate, hence the major product (ie. lower Ea = faster reaction = major product) based on this factor, would be 2-chloropropane. In practice, because both factors are often contradictory, the actual experimental yield will be somewhere between the two extreme predictions based on either factor alone, ie. a weighted average. There have been a number of prelim paper qns which require you to consider both factors. Because Cambridge is increasing the difficulty of the H2 Chem papers, students gunning for a distinction in H2 Chem should be prepared to go deeper than the basic requirements of the syllabus, and explore optional areas such as these.

UltimaOnline SG

A pure sample of N2O4 (l) is introduced into an evacuated vessel. The vessel, of constant volume, is heated to a constant temperature such that the equilibrium below is established : 2NO2 (g) <---> 2NO (g) + O2 (g)

The value of the pressure p is then found to be 20% greater than if only NO2 (g) were present. Calculate the mole fraction of oxygen in this equilibrium mixture.

Solution :

2NO2 (g) 2NO (g) + O2 (g)
1 | 0 | 0
-2x | +2x | +x
1-2x | 2x | x

Total pressure = 1+x

Since (1+x)/(1) = 120/100, hence :
1 + x = 1.2
x = 0.2

Therefore mole fraction of oxygen = (0.2)/(1.2) = 0.16666 = 0.167 (to 3 sf)

UltimaOnline SG

Animizer asked :

Anyone knows how to determine the suitability of the substance of salt bridges for half cells? Thanks!!

UltimaOnline replied :

Choose inert ions that are virtually never oxidized or reduced under aqueous conditions, and do not precipitate out with the other ions present. eg. Na+ or K+ and NO3-.

Animizer continued asking :

Ooh thanks! One more ques,

Why is oxygen and hydrogen produced in electrolysis of aq calcium nitrate?
The hydrogen part i understand. But the oxygen...

O2 + 4H+ + 4e- ---> 2H2O E=1.23
NO3- + 2H+ + e- ---> NO2 + H2O E = 0.81

In this case, NO2 should be preferentially oxidized at the anode.. But since oxygen is produced, it shows H2O is preferentially oxidized.. I don't understand.

Oh, and what about Hg(NO3)2? Hg is not inert.. But I see other inert ions being used as salt bridges too...

UltimaOnline continued replying :

Try not to use mercury salts, as they are toxic, and as you say, they're not as inert as Na+ or K+. If you're referring to a particular setup that uses Hg(NO3)2 as a setup, scan and upload the material and post it here, so I can have a look at it and comment further on it.

NO3- ions are inert under aqueous conditions, because being anionic, it will be electrostatically attracted to the anode. But only oxidation occurs at the anode, and it is not possible to oxidize NO3- ions any further (since N is in Grp V, and it's OS here is already +5), therefore it remains unreacted. Similarly for SO4 2- ions. These ions can only be reduced and not oxidized, but being anionic, are attracted to the wrong electrode (the anode, where only oxidation occurs) instead.

seller_john felt obliged to interject with a random, extra remark :

what a difficult forum session .

Animizer concluded the thread with :

Ah. Thanks alot! I didn't realise NO2 wasn't in the solution hahaha oops.

UltimaOnline SG

Nish asked :

Since the questions for Chemistry at A levels are becoming more ''out of the box" type, I would like to know why Aluminium, not being categorised as a Transition Metal(as far as the A level syllabus is at), is able to form [Al(OH)4]- complex.
Thanks in advance!
P/s I searched via google and nothing much came up.

UltimaOnline replied :

Aluminium is NOT a transition metal, because it does NOT form one or more ions with partially filled d orbitals.

Your misconception is that only transition metals can form complex ions. In reality, *any* metal can form complex ions, as long as its charge density is high enough to polarize the electron charge clouds (we visualize this as a lone pair) from the ligand (ie. donor atoms of the ligand molecule or anion) towards itself, forming what we call coordinate dative covalent bonds with the central metal ion (which may or may not be a transition metal).

Transition metals, due to their capacity or feasibility to exist in various oxidation states (with comparable stabilities), commonly have ions with charge densities high enough to form complex ions, which is why we say one property of transition metals is that they commonly form complex ions.

But just as we could say "Asian cultures traditionally have a closer affinity to martial arts, compared to Western cultures", it doesn't mean all Asians know martial arts, or that there's no such thing as a Caucasian man who knows martial arts.

Al3+ is just one of many non-transition metal ions, which are capable of forming complex ions. Another would be Be2+ (which has a similarly high charge density as Al3+).

Truth be told, it's a spectrum, rather than binary black and white, the transition from ion-dipole interaction to dative bond formation in complex ions. We can (more or less) confidently say the charge density of Be2+ is sufficiently high to form complex ions with actual dative bonds from the water ligands, while the charge density of Sr2+ is not high enough to form complex ions, but only high enough to have ion - permanent dipole interactions with water molecules.

But what about Mg2+ and Ca2+? You could say that the true state of these ions in solution, have the charactersitics of both complex ions, as well as ion - permanent dipole interactions. In other words, somewhere in between. It's only in human minds that we go "huh? how can be somewhere in between?", coz the Universe couldn't care how humans think : to the Universe, Mg2+(aq) is Mg2+(aq) regardless of puny human concepts about "complex ions" or "ion - permanent dipole interactions".

Just as you were lied to at O levels that there's only "covalent bond" versus "ionic bond", now at A levels you realize that oh, it's actually "covalent bond with ionic character", and "ionic bond with covalent character". Similarly, there's a lot of lies (or at least, half-truths) that you're still being taught at A levels, a necessary evil because of the oversimplification process of education at various levels.

Notice that the higher level you go, the less black and white that chemistry, and life in general, becomes. At PSLE you thought you knew everything in the Universe. At O levels you thought you knew a lot about the Universe. At A levels you realize there's a lot more to the Universe that you don't know. At University levels you start to realize the Universe is infinitely more beautiful, complex, astounding and wonderous than humans can even begin to comprehend.

UltimaOnline SG

Daniel798 asked :

The oxidation number of phosphorous in PCL5 is +5.
But PCL5 is formed as phosphorous expands it's octet into the 3d subshell and forms 5 covalent bonds with the 5 chlorine atoms.
But it did not lose and electron so why is it's oxidation number here +5? (I think i have a conceptual misunderstanding of what the oxidation number means).

i thought that oxidation state/number showed how many electrons the atom has lost because i didn't know the definition of oxidation state/number so i got confused since PCL5 has number +5 so i though it lost 5 electrons when in actual fact, it did not lose any electrons.

UltimaOnline replied :

At much higher levels of Chemistry, there is indeed a technical difference between Oxidation State (OS) and Oxidation Number (ON), but for A level H2 Chemistry purposes, OS and ON are to be considered identical.

The true meaning and formula for OS (or ON) is Formal Charge + Electronegativity consideration.

OS (or ON... from here on I'll just use the term OS) is a type of charge on the atom, one that considers electronegativity.

Formal charge is another type of charge on the atom, one that doesn't consider electronegativity.

For instance, consider this exam question :
In the case of carbon monoxide, if an electric field is generated, would the C atom or the O atom be attracted to the positive field? and negative field?

The typical JC student would assign partial charges based on electronegativity, ie. delta negative on O, and delta positive on C. Based on such imagined partial charges, the student would deduce that the O atom is polarized towards the +ve electric field, while the C atom is polarized towards the -ve electric field.

Wrong.

The reason for the typical JC student's error, is because he/she failed to consider formal charges.

Because the C and O atom both have 5 valence electrons from 1 lone pair and 3 bond pairs (ie. triple bond between the C and O atoms), and because C is in Group IV and O is in Group VI, the C atom has a -ve formal charge and the O atom has a +ve formal charge.

(In terms of stable octet though, both atoms have a full stable octet. "Stable Octet" and "Formal Charge" are two different ways chemists count valence electrons. A third way to count electrons, is the Oxidation State aka Oxidation Number).

Based on formal charges, you would expect the C atom to be attracted to the +ve electric field, and the O atom to be attracted to the -ve field.

Based on oxidation states (OS) however :
The OS on the C atom is (formal charge + electronegativity consideration) = -1 + +3 = +2
The OS on the O atom is (formal charge + electronegativity consideration) = +1 + -3 = -2

"Electronegativity consideration" refers to the concept that both electrons in the bond pair would be polarized away from the less electronegative atom and towards the more electronegative atom.

Based on OS, you would expect the O atom to be attracted to the +ve electric field, and the C atom to be attracted to the -ve field.

Notice that while the preceding statement appears, ostensibly or superficially, similar to the reasoning of the typical JC student, the critical difference is that, it is reasonable, valid and correct to deduce the polarity (of a polar molecule when exposed to an electric field) based on OS, and based on formal charges, but not (reasonable, valid or correct to base) on imagined partial charges on atoms that actually have formal charges.

Back to the CO molecule : since the two different models/concepts of Formal Charges and Oxidation States lead us to different (to be precise : opposite) conclusions as to the polarity of the CO molecule when exposed to an electric field, what are we to do? Which concept is more correct?

The concepts are both equally correct, because they are two separate and distinct concepts that do not, within their own framework, contradict each other; the more relevant question to ask is which concept outweighs which concept in the case of carbon monoxide being exposed to an electric field?

As it turns out, experimental evidence has shown that (Chemistry is a microcosm of real life, which is not black and white, so similarly for Chemistry : be aware it's often a case-by-case basis, but) in the case of carbon monoxide, the -ve formal charge and +ve oxidation state of the C atom, and the +ve formal charge and -ve oxidation state of the O atom, largely cancel each other out.

But ultimately, experimental evidence does indicate that the effect of the formal charges do indeed (slightly) outweigh the effect of the oxidation states, ie. the C atom in CO is slightly polarized towards the +ve electric field, and the O atom slightly polarized towards the -ve electric field.

Finally, in case any JC student is reading this and feels disturbed by what I've said in this post, do not worry too much : Cambridge will be reasonable. Cambridge will accept and award full marks to the student, regardless of which of the 3 explanations he/she gives : based on formal charges (the most correct), or based on oxidation states, or based on partial charges (the least correct).

As long as your explanation tallies with your prediction (ie. if you say the C atom has a -ve formal charge, but go on to say therefore the C atom would be attracted to the -ve field, you will get zero marks), Cambridge will give you full marks. This is perfectly understandable, because they just want to see if the candidate is able to make the correct conclusion based on the relevant arguement/concept (and since all 3 arguments/concepts are arguably, all somewhat reasonable, all 3 arguements/concepts will be accepted by Cambridge).

Edited to add :

In PCl5, the P atom has 0 lone pairs and 5 bond pairs. According to this structure :

In terms of a "stable octet", we say it has an "expanded octet" of 10 valence electrons. But in terms of "formal charge", we say it has 5 valence electrons.

Oxidation State (OS) = formal charge + electronegativity consideration

Thus the OS of the P atom in PCl5 = (0 + +5) = +5

In terms of stable octet, it hasn't lost any electrons at all (in fact, it's gained another 5 electrons!)

In terms of formal charge, it hasn't lost any electrons at all.

In terms of oxidation state or oxidation number (OS or ON), it has lost 5 electrons to the more electronegative chlorine atoms.

So has the P atom actually lost any electrons or not?!? It depends on whether you're looking at it from the viewpoint of stable octet, formal charge, or oxidation state.

Have you lost or gained when you donate some cash to the poor? In material/monetary/financial terms, you've lost. In compassionate/ethical/spiritual terms, you've gained. Whether it's chemistry or real life, It's always a matter of perspective.

UltimaOnline SG

Topic : Equilibria

Note that Kstab (stability constant) = Kf (formation constant) = the Kc value for the formation / stability of any complex ion. You should be able to write out the Kstab equation for any complex ion.

For more info :
http://en.wikipedia.org/wiki/Stability_constants_of_complexes

1. The DiNitrogen TetrOxide Gas Qn

If 0.50g of dinitrogen tetroxide is placed in a vacant 2dm³ container at 25°C, molecular dinitrogen tetroxide dissociates partially into molecular nitrogen dioxide. Calculate the partial pressure exerted by the dinitrogen tetroxide at equilibrium, given that the K_p for the reaction at 25°C is = 0.114 atm.

Ans : P_N2O4 = 0.0349 atm

2. The Cadmium Sulfide Ppt Qn

At 25°C, the K_c for the following reaction is 1 x 10⁸¹ :

2H⁺(aq) + 2NO₃^-(aq) + 3H₂S(aq) ↔ 2NO(g) + 4H₂O(l) + 3S(s)

Given the dissociation constant of hydrogen sulfide is 1 x 10^-19 and the solubility product of cadmium sulfide is 8 x 10^-27, calculate the K_c for the following reaction :

3CdS(s) + 8H⁺(aq) + 2NO₃^-(aq) ↔ 2NO(g) + 4H₂O(l) + 3S(s) + 3Cd²⁺(aq)

Ans : Kc = 5.12 x 10^⁵⁹

3. The MonoHydroxoChromium(III) Ion Qn

At 25°C, calculate the pH of a 0.10 mol/dm³ solution of Cr³⁺(aq) ion, given that
the K_stab of the dipositive monohydroxochromium(III) ion = 1.0 x 10¹⁰_.

You may assume the counter anion does not undergo significant hydrolysis.

Ans : pH = 2.5

4. The TetraamineCopper(II) Ion Qn

Given that the mass solubility of copper(II) hydroxide in pure water is 3.315 x 10^-5 g, what molarity of ammonia must be maintained at equilibrium in a solution in order to fully dissolve 0.10 mol of copper(II) hydroxide in 1dm³ of solution?

Given that the K_stab of tetraaminecopper(II) ion = 1.0 x 10¹². All temperatures are at 25°C.

Ans : 12.7 mol/dm3 (since the molarity of NH3 must be slightly greater than 12.63 mol/dm3)

UltimaOnline SG

A rant regarding how secondary schools and JCs teach poorly in the topic of electrolysis.

UltimaOnline SG

1. Below shows the electronic configuration of the three d-block elements in the Periodic Table.

Element X: [Ar]3d^7 4s^2

Element Y: [Ar]3d^8 4s^2

Element Z: [Ar]3d^10 4s^1

Which one of the following statement is incorrect?

A The electronic configuration of central metal ion for [Y(CN)6]4- is [Ar]3d^8

B Upon reduction from ZCl2(aq) to [ZCl2]-(aq), the solution turned colourless.

C The E* value of the X3+/X2+ is less positive than that of Z3+/Z2+

D X is likely to exist as K2X2O7.

Answer is D. I know B and C are out, but I chose A. I thought Y would lose 2 electrons to form Y2+, then forming the complex ion, which has the correct electronic config?

2. 2,2-dimethylpropane reacts with chlorine gas in the presence of ultraviolet light to give a mixture of products. Which one of the following statements is correct regarding this reaction?

A. Only one mono-substituted product is formed.

B. Only the propagation steps involves C-Cl bond formation

C. Only the initation step of the mechanism involves homolytic fission.

D. Both propagation and termination steps produce hydrogen chloride.

Answer is A. I chose C, because I thought the one in propagation step only produces 1 radicals; homolytic fission must produce 2 radicals right? Why is it A anyway? I thought they can be a lot of chlorine substitution going on in one molecule.

3. An organic compound X undergoes the following reactions.

i) It decolourises a solution of bromine in tetreachloromethane.

ii) It reacts with phosphorus pentachloride giving copious white fumes of HCl.

iii) It reacts with hot alkali to give a compound with two alcohol functional groups.

Which compound could be X?

A. HOH2CCH=CHCH=CHCH2Cl

B. Cl2CHCH=CHCH2COOH

C. BrCH2CH2CHClCH2COCl

D. ClCH2CH2Ch=CHCH(Cl)CH2OH

Answer is A. I know C and D are out, but gosh, A and B seems to be both correct!

4. (Choice A- statements 1,2,3 are all corect. B - statement 1,2 are correct. C - statement 2,3 are correct. D - Only statement 1 is correct)

GABA is a neurotransmitter released by red algae which encourages shellfish larvae to settle on the ocean bed.

GABA: H2NCH2CH2CH2CO2H

Which of the following statements are correct?

1. It is a 2-aminocarboxylic acid.

2. It is soluble in water due to zwitterion formation

3. It migrates to the anode of an electrolytic cell at pH2.

Answer is C (statement 2,3). I know 3 is correct, but how can it form a zwitterion like that? I thought there must be a so called alpha carbon, where the NH2 and COOH group must belong to the same carbon atom. What's a aminocarboxylic acid too? :/

sorry for the wall of texts, thanks for your replies man. these are from Vjc 2009 prelim paper.

Q1.
The electron configuration of an uncomplexed ion (ie. without ligands) and a complexed ion (ie. with ligands) is different : dative bonding means the electrons of the ligands enter into the vacant orbitals of the metal ion.

Q2.
Initiation and propagation both involve bond homolytic cleavage (or fission, but cleavage sounds sexier). When only monosubstitution occurs (eg. excess alkane, limited halogen), then for 2,2-dimethylpropane, all hydrogens are equivalent (ie. gives the same product).

Q3.
Option B generates a geminal diol, which readily dehydrates to form a ketone (or in this case, aldehyde).

Q4.
Aminocarboxylic acid means both the amine group and the carboxylic acid group are present in the same molecule. In organic chemistry, the alpha carbon is desginated as the C atom directly bonded to the functional group. For amino acids of proteins, the highest priority goes to the carboxylic acid group, and the alpha C atom is the one bonded to the carboxylic acid group, and notice that (for amino acids of proteins) the amine group happens to be directly bonded to the alpha C atom as well, thus amino acids of proteins are called "alpha amino acids".
Any amino acid (whether alpha or beta or gamma, etc) are capable of forming zwitterions, because the amine group is basic and wants to accept a proton, and the carboxylic acid group is acidic and wants to donate a proton. When the amine group is protonated, the N atom gains a +ve formal charge. When the carboxylic acid group is deprotonated, the O atom gains a -ve formal charge. Thus forming a zwitterion.

UltimaOnline SG

forgiveme posted :

Hey , I hope To clarify some Of my chem doubts . Hope u guys can help me me
Most qns are on inorganic chem .

1. Grp 7
How come reactivity of halogen with hydrogen must be explained with bond energy and not by their oxidizing strength ?when to use oxidingsing strength and when to use nod energy ?

2How come complex silver diamine ion is colorless ? I tot complex is colored ? O.o
3.displacement rxn
State obv when Excess cl2 bubbled In kl right
So chlorine become clorine ion , does the ion has colour ?
And how come iodine gas form still need to react with iodine ion to form i3^-when cl2 is in excess . Isn't the iodine ion complete gone ?

Last, sodium iodine and conc h2so4 will form iodine gas right?
Home come my lecture notes wrote violet fumes and black solid ?

Tats about it . Pls help me . T.T

Q1.
2 reasons why bond energy is more directly relevant than redox potentals :
- the bond must first be broken before redox can occur.
- if u wanna quote data booklet redox potentials, the reactants or products must be in the aqueous state (as written in the redox equation), but the reaction of halogen with hydrogen does not involve the aqueous state.

Q2.
Ag+ has completely filled d-orbitals. Not all complex ions have partially filled d orbitals, which is a prerequisite for colours arising from d-d* transitions.

Q3.
Cl- ion is colourless. The I2 generated may be attacked by the I- nucleophile to generate the I3- ion, but unless the exam questions specifies you need to talk about the I3- ion, you can ignore its formation and just write the product as I2.

Q4.
Black solid = solid iodine. Violet fumes = gaseous iodine.

forgiveme posted :

Hey thanks alot
Is it true tat all trans element below cobalt Is fill 3d ?Eg Ag here

And is all halide ion colour colorless ? And for this type of qns we only need to state the colour change for the halide gas changed right ? The salt no need right ?

Q4:bt the equation suggests that the iodine formed
Is a gas . So I jus right violet fumes can alr right ? Or need to state both ?

Thanks again

Q1. No, case-by-case basis. But barring further anomalous patterns for transition metals in period 5 and above (the anomalies for period 5 elements and beyond, akin to Cr and Cu of period 4, are beyond the A level syllabus), you should be able to work out the electron configuration based on the element's position in the periodic table.

Q2. Yes all halide ions are colourless. Depends on the question. Sometimes, in trickier questions, you must consider the colours of both species present, eg. when unipositive VO2+ reacts with Fe2+, you get blue dipositive VO2+ and yellow Fe3+, the mixing of colours would result in a green solution. The exam smart candidate will state all 3 colours, with explanation (ie. specify that one product VO2+ is blue, the other product Fe3+ is yellow, and therefore the mixing of both will result in a green coloured solution), to secure full marks.

Q3. Being kiasu is a good trait for A level exams. Best to state both, but at a minimum, you're right that it's more important to say violet vapour of I2(g) is observed, rather than black solid of I2(s).

forgiveme posted :

Alright ! I understand alr !
Another qns is that
The redox potential for [fe(h2o)]^2+ is +0.77
redox potential for [fe(cn)6]^3- is +o.36v
Suggest a difference .
My tut and state that it's cos cn is a stronger ligand and hence more stable , les easily reduced
But , the ans in tys suggests that it's cause of the negative sign in cn complex,harder to accept electron .
Hmmm , so which one is right ?

2. When comparing ionic radius between anion and cation right ,
Anion has a larger bond length becos is has one more quantum shell right ? Bt how come some ans says it's cos of electron-electron repulsion ?

Haisss .
Sorry to bother u .

Q1. Both are correct, and contribute to the difference in the reduction potentials, and Cambridge expects the 'A' level candidate to know both answers. In the 'A' level exams, you should give both answers, to secure full marks.

Q2. Again, both are correct. But the more direct & important reason is that the anion has one quantum shell more than then cation. The greater shielding effect (ie. inter-electron repulsion) only serves to [I]additionally[/I] increase the anionic radius by a little bit extra. (Eg. if each quantum shell measures an additional 1cm in diameter, then a certain cation would be 2cm, and it's anion would be 3.2cm instead of 3cm; the additional 0.2cm caused by additional shielding effect).

UltimaOnline SG

Guys like Girls, Nucleophiles like Electrophiles.

Guys are nucleophiles, girls are electrophiles. Guys like girls, so nucleophiles like electrophiles. Only nucleophiles got the 'balls' to attack electrophiles (so the mechanism is always a curved arrow from the 'balls' of the nucleophile to the electron-deficient electrophile), the 'balls' (usually) referring a lone pair, though it can sometimes be a pi bond, as in the case with alkene guy (nucleophle) or benzene guy (nucleophile).

Water is more masculine (nucleophilic) than feminine (electrophilic), since the O atom in the middle can withdraw electrons inductively from both H atoms, resulting in the partial -ve charge on the O atom to be twice as great as the partial +ve charge on each H atom.

The Si atom of SiCl4 is in period 3, and thus has vacant d orbitals (remember : as long as you're in period 3, you have energetically accessible s, p and d orbitals, even if your p orbitals are empty). Because Cl is more electronegative than Si, and because there are 4 electron-withdrawing by induction Cl atoms, the Si becomes extremely feminine (electrophilic), and thus is readily attacked by water nucleophiles (guys can't resist such femininity). Isn't this the same scenario with CCl4? Then why doesn't CCl4 undergo hydrolysis readily, like SiCl4?

The difference between the propensity for hydrolysis of SiCl4 versus CCl4 (the former is readily hydrolyzed while the latter is not), is due to 2 different reasons (give both reasons in the A levels if asked), which are :

Reason #1 - Si is able to accept the dative bond from the water nucleophile into its vacant, energetically accessible d orbitals without needing to simultaneously cleave its existing bonds with the Cl atoms (something which the C atom cannot do, as it does not have energetically accessible d orbitals). This means that the hydrolysis of CCl4 has a much higher activation energy (since it needs to first cleave its existing C-Cl covalent bonds in order to accept a new dative bond from the incoming water nucleophile) compared to the hydrolysis of SiCl4 (as the inter-electron repulsion between the new dative bond formed and the existing Si-Cl bonds will lower the activation energy required to cleave the old Si-Cl bonds). Bottomline : the lower the activation energy, the more readily hydrolysis can occur.

Reason #2 - The Si-Cl bond is longer than the C-Cl bond, since Si is in period 3 while C is in period 2. But be careful not to make the mistake of saying "[I]therefore the Si-Cl bond is weaker than the C-Cl bond[/I]" because if you check the Data Booklet, you may be surprised to note that the Si-Cl bond is actually stronger than the C-Cl bond, for reasons beyond the scope of the A level syllabus. So the bond strength isn't the reason for the difference in propensity for hydrolysis, but the bond length still is. How so? Using an analogy, if it's raining heavily in a windy thunderstorm and you don't want your face to get wet, do you put your umbrella near your face or far away from your face, to protect your face from the water nucleophile? Similarly, because Cl is more electronegative than both C and Si, the 4 Cl atoms arranged tetrahedrally about the central C or Si atom, are partial negatively charged, and thus can repel the incoming water nucleophile (remember : guys only like girls, guys don't like guys; the delta -ve Cl atoms repel the delta -ve O atoms of the water nucleophile) so the partially negatively charged Cl atoms function like 4 little umbrellas around your face, shielding your face from water nucleophile during a windy thunderstorm). And because the C-Cl bonds are shorter than Si-Cl bonds, the Cl atoms provide more effective shielding (like an umbrella held more closely to your face) against the incoming water nucleophile in CCl4 compared to in SiCl4.

Guys always like girls. It's Chemistry.

UltimaOnline SG

Hi, i am a A level student and would like to ask this question about arenes.

In arenes, we have learnt about activating and deactivating groups when attached to a benzene ring can have various effects on the rate of electrophilic substitution.

My question is that are groups like -CH2OH, -CH2Cl , -CH2OR ( where R is a R group like an alkyl ) activating or deactivating to the benzene ring?

Can someone please tell me the answer and explain the reasoning behind the answer? please help :D thx

All 3 abovementioned groups are electron withdrawing overall, because they withdraw by induction (since the C atom is partially positively charged, due to the greater electronegativity of the O, Cl and O atoms respectively) and they neither donate nor withdraw by resonance (to donate by resonance the atom directly bonded to the benzene ring must have a lone pair; to withdraw by resonance, the atom directly bonded to the benzene ring must have a pi bond that can shift away when it accepts electrons from the benzene ring by resonance to avoid violating its octet for period 2 elements).

Since these groups withdraw by induction (and neither donate nor withdraw by resonance), they are thus meta-directing.

UltimaOnline SG

Nish asked :

1. When AgNO3 (aq) is mixed with MgCl2(aq), a white ppt forms which is soluble in NH3(aq). When AgNO3(aq) is mixed with FeCl2(aq), a grey ppt forms which does not dissolve in NH3(aq).

Which statements about one or both of these reactions are correct?

1   Ag+ ions act as an oxidising agent.
2   Ammonia forms a soluble complex with silver metal.
3   Fe2+ ions act as a catalyst in decomposing AgCl.

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2. Given weighed samples of the same mixture of MgCO3 and BaCO3, how can the mole fraction of MgCO3 in the mixture be estimated?

1   Add a known volume of [HCl= 0.1], in excess, and back titrate the excess of the acid.
2   Add an excess HCl and measure, at known temperature and pressure, the volume of CO2 liberated.
3   Add an excess of HCl followed by an excess of H2SO4, filter, dry and weigh the ppt.

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3. Two students separately have available equal volumes of 0.1moldm-3 silver nitrate, sodium ethanoate and KBr.

The first student, on mixing the sodium ethanoate and silver nitrate, obtains a white ppt. On adding KBr, the ppt turns cream.

The 2nd student adds AgNO3 to KBr and obtains a cream ppt. On adding CH3COONA, there is no further change.

Which statements about these observations are correct?

1   Silver ethanoate is insoluble.
2   Silver bromide is less soluble than silver ethanoate.
3   Ethanoate can oxidise bromide.

For this question I'm not sure about statement 1 and 2.

Thanks! :qmarkshy:

Q1. Only option 1 is correct. Option 2 is wrong because NH3 ligands complex with Ag+ ion, not Ag metal. Option 3 is nonsense because Fe2+ ions are not regenerated, furthermore AgCl can't be decomposed if it isn't even formed in the first place (since the redox reaction between Ag+ and Fe2+ occurs instantaneously).

Q2. All options 1, 2, 3 are correct. Solve simultaneous equations (algebraic variables x and y) in all 3 options, and the relative amounts of both carbonates can be determined.

Q3. Option 1 is poorly phrased, but can be considered correct if the word "insoluble" is replaced with "sparingly soluble". Option 2 is correct, as evidenced by the 2nd and 3rd paragraphs of the question. Option 3 is nonsense, as no redox reaction occurs during ionic precipitation reactions.

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