Group VII: The Halogens

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The halogens refer to chlorine, bromine and iodine.

1) Colour and volatility
Chlorine is a dense green gas.
Bromine is a brown volatile liquid.
Iodine is a shiny black solid

Down the group, the number of electrons in the halogen molecule increases, hence the strength of van der Waals forces between the halogen molecules increases. This explains why chlorine exists as a gas, bromine exists as a liquid and iodine exists as a solid.

2) Relative reactivity of halogens as oxidizing agents
Chlorine is the strongest oxidizing agent followed by bromine and iodine. Recall that an oxidizing agent causes oxidation and it itself is reduced, i.e. an oxidizing agent has a tendency to gain electrons.


The electrode potential values become less positive down the group, indicating that down the group the halogens has a reduced tendency to gain electrons, hence the oxidizing power of halogen decreases.
3) Reaction of halogens with hydrogen
Chlorine reacts explosively in sunlight with hydrogen. It also reacts slowly in the dark with hydrogen.
Bromine reacts with hydrogen only at high temperatures
Iodine reacts with hydrogen to form an equilibrium mixture of hydrogen, iodine and HI.

4) Relative thermal stability of hydrides (hydrogen halides)
Thermal stability of hydrogen halides decreases down the group.
Down the group, the atomic radii of the halogen increases and the bond dissociation enthalpy of hydrogen halides decreases.
Bond energy of H--Cl : 431 kJ/mol
Bond energy of H--Br: 366 kJ/mol
Bond energy of H--I: 299 kJ/mol

Note: In the exam, students may be asked to quote bond energy values from the Data booklet. You do not have to memorize these bond energies. they are in the data booklet.

5) Reaction of halide ions with aqueous silver ions followed by aqueous ammonia
Chloride ions form a white precipitate with silver ions. White ppt is soluble in aqueous NH3
Bromide ions form a pale yellow ppt with silver ions. Yellow ppt is sparingly soluble in aqueous ammonia.
Iodide ions form a deep yellow ppt with silver ions. Deep yellow ppt is insoluble in aqueous ammonia.

Silver ions is usually supplied as silver nitrate. This is a common test for halides. Aqueous ammonia is added to differentiate between silver chloride and silveer bromide as it is very hard to differentiate between the white and pale yellow ppt.

Aqueous ammonia has to be added slowly as the pale yellow silver bromide is soluble in excess aqueous ammonia. Aqueous ammonia forms a complex with silver ions, shifting the equlibrium to the left, and increases the solubility of AgCl. AgCl is most soluble followed by AgBr and AgI, thus the addition of ammonia can fully solubilise AgCl. AgBr is only soluble in concentrated ammonia while ammonia has no effect on the solubility of AgI.



6) Reaction of halide ions with concentrated sulphuric acid
Concentrated sulphuric acid can act as a proton donor, i.e. it supplies H+ ions which reacts with the halide ion to form hydrides (HCl, HBr and HI). The hydrides appear as steamy white fumes



Concentrated sulphuric acid can also act as an oxidizing agent.
Recall that iodine is the weakest oxidizing agent, hence it is the most easily oxidized. Concentrated sulphuric acid can oxidize iodide ions to iodine, hence you will see purple vapour
Concentrated sulphuric acid can also oxidize bromide ions to bromine gas.
Concentrated sulphuric acid cannot oxidize chloride ions to chlorine gas.

7) Reaction of chlorine with NaOH


Chlorine reacts with cold NaOH to form NaCl and NaClO (sodium hypochlorate)
Note that in this reaction one atom of chlorine is undergoing oxidation and the other atom of chlorine is undergoing reduction.
Chlorine is oxidized. The oxidation number of chlorine increases from 0 in Chlorine gas to +1 in NaClO. Chlorine ia also reduced. The oxidation number of chlorine decreases from 0 in Chlorine gas to -1 in NaCl.



Chlorine reacts with hot NaOH to form NaCl and NaClO3 (sodium chlorate)
In this reaction one atom of chlorine is undergoing oxidation and the other atom of chlorine is undergoing reduction.
Chlorine is oxidized. The oxidation number of chlorine increases from 0 in Chlorine gas to +5 in NaClO3. Chlorine ia also reduced. The oxidation number of chlorine decreases from 0 in Chlorine gas to -1 in NaCl.

End of group VII notes
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Chemistry of transition elements II

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7) Transition elements in redox systems

These transition metals were highlighted in the syllabus. For e.g. Fe3+ can act as a reducing agent while it itself is oxized to Fe2+.

8) Transition metals react with ligands to form complexes
A ligand possess one or more lone pairs of electrons and donates its electrons to the transition metal ion

A complex is formed by the coordination of lone pairs of electrons from a ligand to a cation which has empty orbitals to accomodate them. Transition metal ions have incompletely filled 3d subshells to accomodate the lone pair donated by the ligand.

A coordinate bond is a covalent bond in which the shared paired of electrons is provided by only one of the bonded atoms. In the case of transition ion complexes, the shared paired of electrons is provided by the ligand

9) Ligand exchange
A stronger ligand can displace a weaker ligand from a complex ion.
The order of ligand strength is CN->NH3>Cl->H2O

CO and oxygen are ligands for Fe in haemoglobin
Haemoglobin contains an Fe2+ ion. Oxygen can act as a ligand and form coordinate bonds with Fe2+. This bonding is reversible to allow haemoglobin to release oxygen where it is needed. Carbon monoxide is a stronger ligand than oxygen. It bonds strongly to Fe2+ . This prevents Fe2+ and thus haemoglobin from bonding with oxygen. This prevents the transport of oxygen in the body eventually leading to death.

10) Transition metals can act as catalyst

Transition metals can act as heterogenous catalysts. Reactant molecules are adsorbedon the surface of the catalyst . The 3d electrons of transition metals enable the transition metal to form temporary bonds with reactant molecules, facilitating the breaking of "old" bonds and the formation of "new" bonds leading to the product formation. Examples include iron (catalyst for the Haber process), Ni (catalyst for hydrogenation).

Transition metals can also act as homogenous catalysts due to the variable oxidation states of transition metals.


For example, the oxidation of iodide ions by peroxodisulphate ions is energetically favourable but in reality the reaction takes place very slowly without a catalyst due to the fact that both reactants are negative ions thus they repel each other.

Fe2+ can act as a catalyst for this reaction. First Fe2+ reduces peroxodisulphate ions to sulphate ions and it itself is oxidized to Fe3+.


Subsequently, Fe3+ oxidizes iodide ions to iodine and Fe2+ is regenerated.

11) Transition metal complexes are coloured
In an isolated tansition metal aton, the 3d orbitals are at the same energy level (degenerate) . However in a complex ion , the d orbitals have slightly different energy due to the presence of ligands (d orbital splitting). An electron can be promoted to a higher energy level by absorbing a photon of light. The color of the complex metal ion is the frequencies of light that are not absorbed.



End of chemistry of transition metals
Link to Chemistry of transition elements I
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Chemistry of Transition elements I

Note: My notes based on the order of learning outcomes presented in the syllabus.
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1) Definition of transtion elements
Transition elements are d block elements which form one or more stable ions with incompletely filled subshell of d electrons
Zn for example is not a transition element because the electronic configuration of Zn2+ is [Ar]3d10, i.e. its 3d subshells are completely filled

2) Electronic configuration of transition elements
The syllabus requires that students know the electronic configuration of transition metals from Sc to Zn. Remember that the 4s subshell is filled first before the 3d subshell. There are 5 orbitals in the 3d subshell. All 5 have to be singly filled before the electrons are paired
Special cases:
Cr: [Ar]3d5 4s1
Cu:[Ar]3d10 4s1

In both cases the 4s orbital is not completely filled due to the stability associated with a fully filled d shell and a half filled d shell

3) The atomic radii, ionic radii and first ionization energy of transition elements are similar.

4)Comparison between s block and transition element

Transition elements have higher melting points, higher densities and better conductivity than s block elements. This is due to the stronger metallic bonds present in transition elements. In transition elements, the d electrons and s electrons are available to take part in delocalisation. In contrast only the s electrons are able to delocalize in the s block element.

The atomic radius, ionic radius and first ionisation energy of d block elements are similar to s block element

5) Transition elements tend to have variable oxidation states

The difference in energy between the 3d and 4s electrons is much smaller than the difference in energy between the 3s and 3p electrons, hence the 3d and 4s electrons may be lost in chemical reactions.

6) Prediction of likely oxidation states given a particular electronic configuration

E.g. Fe is likely to have a +2 oxidation state due to the loss of the 2 outer most 4s electrons and a +3 oxidation state due to a loss of the 4s electrons and another 3d electrons giving rise to a stable half filled 3 d configuration

Note that some transition elements have oxidation state which are less obvious to predict. In the case of Ti, it is not easy to see why the main oxidation states of Ti are +3 and +4.

Continued in Chemistry of transition elements II
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N2007/III/5

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5ai)
The requirements needed for 2 molecules to form a hydrogen bond between them are:
one molecule must have a hydrogen atom bonded to either oxygen, nitrogen, or fluorine and
the second molecule must have a lone pair on an oxygen, nitrogen, or fluorine which can then interact with the hydrogen on the first molecule.

For example, hydrogen bonding can occur between 2 water molecules. The hydrogen atom bonded to oxygen in one water molecule can interact with the oxygen atom of the other water molecule.




5aii)

Carboxylic acids consist of a non polar hydrocarbon chain and a polar carboxylate group. Hydrogen bonding occurs between the carboxylate groups of acid molecules. Also Van Der Waals forces exists between the non polar hydrocarbon chains . Water molecules interact with each other through hydrogen bonding.

The first 4 members of the carboxylic acid series has short hydrocarbon chains and the predominant interactions between acid molecules is H bonding which is similar to the H bonding that exists between the water molecules hence they are soluble in water

For members of the the carbocxylic acid series with longer side chains, the predominant interaction between acid molecules is Van Der Waals forces due to the long hydrocarbon chain. This interaction is weaker than the hydrogen bonding that exists between water molecules, hence it is not favourable for the carboxylic acid to dissolve in water.

Side note: this question is surprisingly hard to answer. If anyone has a better answer please post it in the comments. chemguide.co.uk has a reasonably good explanation. in fact i can't find any other explanation on Google. My answer is mainly based on their explanation.

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N2007/III/2

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2ei)
This question maylook intimidating at the first glance. The way to approach this question is to look for clues in the question itself. First you would notice that the ratio of carbon atoms to hydrogen atoms is close to 1:1. Most of the time, this 1:1 ratio suggests that a benzene ring is present. This is supported by the structure shown in C which indeed contains a benzene ring.

A is able to react with thionyl chloride SOCl2. This suggest that an alcohol function group or a carboxylic acid functional group is present. Alcohols react with thionyl chloride to form alkyl chlorides and carboxylic acids react with thionyl chloride to form acyl chlorides

However the alkyl chloride product does not have any oxygen atoms. This suggest that the functional group present is carboxylic acid .


2eii)
Reaction I is an acid base reaction
OH- abstracts a proton from the amide linkage forming water, N- and Na+.

Reaction II is a Nucleophilic substitution reaction.
The N- generated acts as a nucleophile and substitutes bromine.
Recall that the C-Br bond is polar, with carbon carrying a slight positive charge. The slightly positive carbon attracts the N-.

Reaction III is a hydrolysis reaction.
The amide linkage is hydrolyzed to carboxylic acid and amine functional groups

2eiii)
Reagent: HCl
Condition:Heat

Side note: You would notice that for essay questions, the examiners like to mix 2 different topics in one question. In this case the first portion of the question is about inorganic chemistry (which i have not solved) and the second portion is about organic chemistry. This style of questioning is especially detrimental to students who like to "spot" topics. My advice is dont spot topics. Moreover for organic chemistry the topics only make sense if you study all of them together as they are all interlinked.

For those who are interested, the diagram below illustrates how the cyclic amide linkage is formed.

Mark scheme for Nov 2006 Paper 2

I chanced upon this on the cambridge international examinations website.

http://www.cie.org.uk/docs/dynamic/4619.pdf

There are also specimen papers with mark schemes. Do check out their website.

Update:
I think the CIE website is extremely useful. Other than the mark scheme, they also provide examiners report. As a student i never knew that such reports were available online. In addition, due to the recent changes in syllabus the website also provides specimen papers and answers for these specimen papers. I would strongly encourage students to do the questions by themselves first before refering to the mark scheme. I emphasize again, do the questions by yourself first. Then compare your answers to the ones given by the examiners, it would be highly beneficial. If there are any teachers reading this blog, please tell your students about this website.

Based on personal experience, the actual exam is likely to be very similar to the specimen paper in terms of the style of the questions, so familiarize yourself with the specimen paper. Dont just do it once, do it twice or thrice.

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N2007/III/1

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1ci)
Reaction II involves the oxidation of aldehyde to carboxylic acid
Reagent: Potassium dichromate (aq)
Condition: r.t.p

Reaction III involves the reduction of aldehyde to alcohol
Reagents: LiAlH4 in organic sovent
Conditions: Heat

1cii)
butan-1-ol --> 1-bromobutane
this reaction involves the conversion of alcohol to alkyl halides

Reagent: PBr3
Condition: r.t.p


Side note: Alkyl halides are very useful in organic synthesis. For example, you can add carbon atoms to alkyl halides by reaction with CN

1di)
The rate of hydrolysis of 2-bromobutane will be slower than 2 iodobutane. This is because iodine is less electronegative than bromine, thus it forms the weakest bond to carbon and the iodide ion is a better leaving group than bromide ion.

1dii)
Alkyl halides can undergo nucleophilic substitution. The cyanide ion can be used to extend the carbon chain.
LiAlH4 is a reducing agent. It can reduce nitriles to amines

When alkyl halides react with ethanolic NaOH, dehydrohalogenation occurs. As a result an alkene is formed. This is in contrast to aqueous NaOH. Alkyl halides undergo hydrolysis in the presence of NaOH(aq). Subsequently in the presence of cold KMnO4, the alkene is oxidized to a diol



e)
In the presence of UV light , butane undergoes free radical substitution with bromine. There are 6 hydrogen atoms that can be abstracted from the terminal carbon atoms compared to the 4 hydrogen atoms that can be abstracted from the middle 2 carbon atoms. Hence 1 bromobutane should be the more abundant product statistically.

Side note: Experimentally, it has been determined that 2 bromobutane is the more abundant product because of the greater stability of the secondary radical compared to the primary radical.



I am not very sure about the correct answer to this question. This is because i think that the stability of radicals is not in the A level syllabus. Based on the syllabus, students are required to describe the free radical sustitution mechanism. I dont think that includes the stability of radicals.

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How to make full use of the worked examples?

It is important to try to do the questions by yourself first. You will benefit most by thinking through the question first and writing down your answers. By writing down your answers, you will remember them. During revision, i would encourage students to refer to their lecture notes while attempting the questions. As you do more questions, you will realize that you can remember more stuff and you would be less dependent on your notes.

Also if anyone has any questions that you cant solve, please leave a comment, I will try to solve them as soon as possible.

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First worked example: N2007/II/5

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Due to copyright issues, I cannot put up the question online. So i will refer to the question using the standard notation. E.g N2004/II/5a refers to the Nov 2004 Paper II Question 5a

First worked example : N2007/II/5

Look at the different compounds and identify functional groups present in each compound

Compound A contains the alkene functional group and aldehyde functional group

Compound B contains the alkene and ester linkage functional group. Be aware that the ester functional group OCO is different from the carbonyl functional group C=O

Compound C contains the alcohol(phenol) and alkene functional group

Compound D contains the alkene functional group

5a)
the reagent given is a acyl chloride

acyl chloride + alcohol --> ester + HCl

I still haven figure out the best way to draw structural formula and upload online. Once i think of something i will upload the structural formula

Ans: C


dilute HNO3 is able to catalyze the hydrolysis of esters

Esters are hydrolyzed to give carboxylic acids and alcohol

Ans: B

Na is able to react with phenol functional group to form the phenoxide ion

Ans: C

Tollen's reagent is actually an oxidizing agent which can oxidize aldehydes to carboxylic acid. However the reagent is alkaline hence the carboxylate ion is formed. Just be careful when you draw the structural formula.

Ans: A

5bi)
As mentioned earlier compound C has 2 functional groups: the alkene and phenol functional group. Only the alkene functional group will react with Br(aq) . Thus Br and OH will be added across the double bond

5bii)
reflux with hot concentrated acidified manganate will result in the oxidative cleavage of the double bond
As a result 2 carboxylic acids will be formed

5ci)

Pardon my bad drawing. Basically the aldehyde is being reduced to a primary alcohol


5cii)
Stage I
reagent: LiAlH4
Conditions: organic solvent, heat

StageII
reagent: ethanoic acid
Condition: dil HCl, heat

Ok this is the end of the first worked example. Please feel free to leave comments.

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A level Chemistry Syllabus

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One thing that is very important to know is that the examiners will only test stuff that are in the syllabus. The more challenging questions probably require you to connect different pieces of information in the syllabus and come up with a new idea. However all the information can be found in the syllabus. Therefore i encourage all students to at least read through the syllabus which is available in the website below.

http://www.seab.gov.sg/SEAB/aLevel/syllabus/2008_GCE_A_Level_Syllabuses/9251_2008.pdf

The second thing is don't try to know everything. Just understand what is in the syllabus first. Its a fatal error to memorize the whole textbok from front to back. If you really need something to memorize, memorize the notes that your teachers give you.

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