9701/11Paper 1 Multiple Choice May/June 2013start solving chem mcq paper from the last question and go back wards....start from Q:40 and go backwards till Q:1

leave the calculation questions and do them after finishing the rest...it'll save time..Cl-1Cl-

Group 70Cl2

+1ClO-

+3ClO2-

+4ClO2

+5ClO3-

+7ClO4-

1. for ions likeClO4-thecharge on the ion is equal to the sum of the oxidation numbers so Cl is +7, O is -2, so +7 + (4 x -2) = -1 =- for overall charge on ion Note the convention: for oxidation state, the sign comes before the number for charge on an ion, the sign comes after the number======================================================ExtractionofAluminium(continued) -Electrolysis Cell.

Thesteelcontaineriscoatedwithcarbon(graphite) and this is used as thenegative electrode(cathode).Aluminiumoxide(Al2O3) is anioniccompound.When it ismeltedtheAl3+andO2-ionsarefreetomove andconductelectricity.Electrolysisof thealumina/cryolitesolution givesaluminiumat thecathodeandoxygenat theanode.4Al3++ 12e- 4Al(aluminium metalat the (-)cathode)reduction.6O2- - 12e- 3O2 (oxygen gasat the (+)anode)oxidation.Aluminiumismoredensethan thealumina/cryolitesolution and so itfallsto thebottomof thecellwhere it can betapped offaspureliquid metal.Theoverall reactionisaluminiumoxidealuminium+oxygen.2Al2O3(l) 4Al(l) + 3O2(g)Oxygenisgivenoffat thepositivecarbonanode.Carbon dioxideis alsogivenoffat thecarbonanodebecause hotoxygenreactswith hecarbonanodeto formcarbon dioxidegas.carbon+oxygencarbon dioxide.C(s) +O2(g)CO2(g)Thecarbonanodesslowlydisappear becauseeach moleculeofcarbon dioxidewhich isgivenofftakes alittle pieceofcarbonaway with it.Thecarbonanodesneed to bereplacedwhen they becometoo small.A. -Aluminium ions are reduced in this processB. -Aluminium is liberated at the cathode by the reaction Al3+ + 3e Al.C. Cryolite is not a purified aluminium oxide instead a white or colourless mineral consisting of a fluoride of sodium and aluminium. Cryolite(Na3AlF6, sodium hexafluoroaluminate).

-What is cryolite's purpose in the extraction of aluminum?

The alluminium oxide has to be molten for its electrolytic extraction. To melt the electrolyte, heat has to be provided upto temperatures of 2200 decree celsius. The provision of this energy can be very expensive and cost-ineffective.

Cryolite is added as an impurity to the electrolyte. The cryolite, lowers the melting point of alluminium oxide to 980. The rule is as it is for all impurities they lower the melting point and raise the boiling point, so the reduction of the melting point allows the alluminium to melt at lower temperatures, making electrolysis work at lower temperatures, hence requires lesser energy, making the process cheaper, and more cost effective.================================================================== there are more protons in each nucleus so thenuclear chargein each element increases ... therefore theforce of attractionbetween the nucleus and outer electron is increased, and ...

there is a negligible increase inshieldingbecause each successive electron enters the same energy level ... so more energy is needed to remove the outer electron. the first ionization of an atom is always an endothermic processThe law of conservation of energy dictates that the difference

between the energy of the ionizing radiation and the energy of the ejected electron equals the energy required for ionization. Ionization energies are typically reported either in kJ/mol or in eV (1eV= 96.485 kJ/mol). ionization energy generally increases to the right across a period and decreases down a given group.=================================6. hydrolysis reactions of an organic compound like eter require a catalyst - an acid (H+ions) or alkali (OH-ions) as reaction with water is very slow. The dilute acid provides both the acid catalyst and the water.============================================================7. If the pressure is higher, the chances of collision are greater.

==============================================9.As molecules get larger, then dispersion forces will increase, and you may get other intermolecular forces such as dipole-dipole attractions as well. Gases made of molecules such as these will be much less ideal. Like helium, a hydrogen molecule also has two electrons, and so the intermolecular forces are going to be small - but not as small as helium. In the hydrogen molecule, you have two atoms that you can distribute the charges over. A helium molecule consists of a single small atom, and the van der Waals dispersion forces are as low as it is possible for them to be.====================================================== The ideal gas equation is: pV = nRTPressure, p: Pressure is measured in pascals, Pa - sometimes expressed as newtons per square metre, N m-2. These mean exactly the same thing.Be careful if you are given pressures in kPa (kilopascals). For example, 150 kPa is 150,000 Pa. You must make that conversion before you use the ideal gas equation.Should you want to convert from other pressure measurements: 1 atmosphere = 101,325 Pa 1 bar = 100 kPa = 100,000 Pa

Volume, VThis is the most likely place for you to go wrong when you use this equation. That's because the SI unit of volume is the cubic metre, m3-notcm3or dm3.1 m3= 1000 dm3= 1,000,000 cm3

Number of moles, nThis is easy, of course - it is just a number. You already know that you work it out by dividing the mass in grams by the mass of one mole in grams.You will most often use the ideal gas equation by first making the substitution to give:

Therefore Mr = mRT ---------------- pV

=================================================================these quations are given in data booklet so no need to memorize itSn2+ Sn4++ 2e- equation 1(MnO4)+ (8H+)+ 5e (Mn2+)+ 4H2O) equation 2balance the number of electrons so that each equation have same number of electrons multiply equation 1 by 5 and equation 2 by 5 and then add these equations we get(16H+)+(5Sn2+)+2MnO4- yields (5sn4+)+2Mn2+ignore the liquid we are concerned with ions and ions only2moles of mno4- reacts with 5moles of sn2+2/5moles of mn 1 moles sn(9.5/190)*2/5 9.5/190The amount of SnCl2 in the question is 0.05 moles. Because of thestoichiometry of the reaction (worked out using oxidation states) this will produce 0.02 moles of Mn2+.============================================ The hydroxides becomemore solubleas you go down the Group. The sulphates becomeless solubleas you go down the Group The carbonates tend to becomeless solubleas you go down the Group.====================================because iodine is less reactive than bromine so it's 1- ions do not displace the bromine ions in sodium bromide. (It is lower down in the group).================================================================Astatide ion, At-, is a reducing agent. So when it's oxidised, the product is astatine:

2At- ------> At2+ 2e-

HAt is not a redox product, and the others are reduction or neutralisation products of the H2SO4.=================================================17.It can be either Nacl or NaBr uptill the colorless solution but what is happeneing on the addition of excess HNO3.??

HNO3 simply removes the NH3 ligands by protonating them, regenerating the silver halide ppt.The key is the "excess NH3(aq)" used, instead of "excess concentrated NH3(aq)" that isrequired to dissolve AgBr(s). Hence it can be deduced that the ppt is AgCl(s), rather than AgBr(s).======================================================18.Equilibrium constantsaren't changed if you change the concentrations of things present in the equilibrium. The only thing that changes an equilibrium constant is a change of temperature.==========================================19.Ammonium compunds ALWAYS react with alkalis to form salt,water and AMMONIA gas.So the answer would be C since limewater i.e. calcium hydroxide would react with ammonium sulphate to liberate ammonia.===========================================================21.B is correct because the secondary alcohol group in lactic acid will form anester with methanoic acid. C is incorrect because the secondary alcohol group in lactic acid is not acidic enough to liberate CO2 from NaHCO3. It is possible that some candidates did not select B because the given formula of the organic product, CH3CH(O2CH)CO2H, proved difficult to comprehend. In this situationcandidates may make the question easier for themselves by drawing out fully displayed formulae on their question paper.

=======================

22.

Electrophile================================23. trace back the reaction sodium propionateis thesodiumsaltofpropionic acidwhich has thechemical formula Na(C2H5COO).It is produced by the reaction ofpropionic acidandsodium carbonateorsodium hydroxide halogenoalkane reaction with cyanide ions is a useful way of lengthening carbon chains.C2H5Cl +KCN--- C2H5CN +KClFor example, in the equations above, you start with a 3-carbon chain and end up with a 4-carbon chain. There aren't very many simple ways of making new carbon-carbon bonds.C2H5CN

C2H5COONa

======================================================

24.

a curly arrow from a lone pair on the OH ion to the C+ atom of 1-bromopropane=========================================================So consider only

B. 1,5-dichloropentane. No Chiral CarbonsA.1,3-dichloro-3-methylbutaneC.2,3-dichloropent ane

no chiral carbon 2 chiral carbons

A.1,3-dichloro-2,2-dimethylpropaneD.2,4-dichloropentane A. 1,3 dichloropentane . 2 chiral carbons 1 chiral carbon

no chiral carbon

=================================================================== 26.CH 3CH 2 OH +2[ O ]----------- CH 3 COOH +H 2O ) The molar mass of ethanol is trivial to compute: (M( CH3 CH2OH )=6M( H )+2M( C )+M( O )=46.08 g/mol )hence we have (2.30 g /46.08 g/mol approx0.05 mol ) of ethanol. This yields an equal number of moles of acetate under 100% yield. The molar mass of acetate is just (M(CH3COOH )=4M( H )+2M( C )+2M( O ) =60.06 g/mol ) hence we end up with (0.05 mol times60.06 g/mol =3.00 g ) of acetate with ideal yield. To factor in the fact yield is only 70%, compute (60 % times3.00 g =1.80 g ) i.e C is the answer===================================================================27 Amethylglyoxal has twocarbonylgroups, analdehydeand aketone

The presence of that hydrogen atom makes aldehydes very easy to oxidise. Or, put another way, they are strong reducing agents.Because ketones don't have that particular hydrogen atom, they are resistant to oxidation. Only very strong oxidising agents like potassium manganate(VII) solution (potassium permanganate solution) oxidise ketones - and they do it in a destructive way, breaking carbon-carbon bonds.Provided you avoid using these powerful oxidising agents, you can easily tell the difference between an aldehyde and a ketone. Aldehydes are easily oxidised by all sorts of different oxidising agents: ketones aren't.What is formed when aldehydes are oxidised?It depends on whether the reaction is done under acidic or alkaline conditions. Under acidic conditions, the aldehyde is oxidised to a carboxylic acid. Under alkaline conditions, this couldn't form because it would react with the alkali. A salt is formed instead.

In the case of methanal, HCHO, the oxidation goes further. The methanoic acid or methanoate ions formed are easily oxidised to carbon dioxide and water.

Under acidic conditions it is:

. . . and under alkaline conditions: Using acidified potassium dichromate(VI) solution

The orange dichromate(VI) ions have been reduced to green chromium(III) ions by the aldehyde.ketoneNo change in the orange solution.

aldehydeOrange solution turns green.

Using Tollens' reagent (the silver mirror test)

ketoneNo change in the colourless solution.

aldehydeThe colourless solution produces a grey precipitate of silver, or a silver mirror on the test tube.

Using Fehling's solution or Benedict's solution

ketoneNo change in the blue solution.

aldehydeThe blue solution produces a dark red precipitate of copper(I) oxide.

2,4-dinitrophenylhydrazine:Add either a few drops of the aldehyde or ketone, or possibly a solution of the aldehyde or ketone in methanol, to the Brady's reagent. A bright orange or yellow precipitate shows the presence of the carbon-oxygen double bond in an aldehyde or ketone.----------------------------------------------------------------------------------------------------28-D

Generally

Alcohols tend to have higher boiling points than the hydrocarbons because of the strong hy-drogen bond between hydrogen and oxygen atoms.

propyl methanoate is an ester there are no hydrogen bonds among the ester molecules while there is hydrogen bonding involved in the acid therefore ester has lower while acid has higher boiling point.========================================================29-B

=========================================================optically active= Chiral compoundsIf the hydrogen atoms are on the opposite side of the double bonds of the carbon chain, then it is said to be in "trans" configuration.THEREFORE B is the answer.===================================================

for section B [Q: 31 -40]

there are 3 statements 1, 2 and 3. right

for Statement:1 If you are 100% sure that it is wrong => answer is C

for Statement:2 If you are 100% sure that it is wrong => answer is D

Now if Statement:1 is correct, instead of checking for statement:2 , check for statement 3 first.If youre sure its that Statement:3 correct, then your answer will be A.And if youre sure Statement:3 its wrong, then check statement 2.if Statement:2 is correct your answer is B and if Statement:2 is wrong your answer is D.

Note: If you arent sure whether a statement is wrong or correct or not, better go through all of them.

Youll understand this, if you practice a few papersThats how I got all this. Otherwise, I couldnt manage to complete the paper on timeAnd this isnt some special technique. Its just based on how you use your logic

Look how A, B, C and D are classified.

A : 1, 2, 3 correct

B : 1 and 2 correct

C : 2 and 3 correct

D : 1 correct

=======================Diradical : anatomormoleculehavingtwounpairedelectronsMolecular oxygen (O2) is a rare example of a stable biradical (two unpaired electrons having the same spin), with an even number of electrons.Oxygen has two unpaired electrons in separate orbitals in its outer shell. This electronic structure makes oxygen especially susceptible to radical formation.So only oxygen has diradical therefore statement D.================================Ans is bThe melting point of the metals depends upon1. the number of delocalised electrons(greater for group2)2. the size of the atoms or cations3. whether the packing structure is close packed (Mg and Al) or not (Na).The boiling point of the metals depends upon1. the number of delocalised electrons2. the size of the atoms or cations

Having a higher first ionisation energy would make an atom less able to contribute to the sea of delocalized electrons causing a decrease in melting point.==================================================Ans is b1. BH3 is a trigonal planar2. CH3+A trigonal planar3.PH3 is a trigonal pyramidal ==================================================Ans is A1 Elements B, J and R are in Group 0 of the Periodic Table.(highest ionization energy)2 ============================================

Z is therefore Ca(OH)2. When Ca(OH)2 is added to soil it neutralises acid, so the pH of the soil is increased, not decreased, so statement 3 is incorrectStatement 1 correct but statement 3 incorrect therefore ans =D===========================================================Qno 36: b2. the size of the halogen atom: As the halogen atom gets bigger, the bonding pair gets more and more distant from the nucleus. The attraction is less, and the bond gets weaker (statement 1 the strength of the hydrogen-halogen bond)- exactly what is shown by the data. There is nothing complicated happening in this case.This is important in the thermal stability of the hydrogen halides - how easily they are broken up into hydrogen and the halogen on heating.Hydrogen fluoride and hydrogen chloride are very stable to heat. They don't split up into hydrogen and fluorine or chlorine again if heated to any normal lab temperature.Hydrogen bromide splits slightly into hydrogen and bromine on heating, and hydrogen iodide splits to an even greater extent.As the bonds get weaker, they are more easily broken.The results show that the thermal stability of hydrogen halides decreases with increasingrelative molecular mass.The relative stability of hydrogen halides can be explained in terms of bond length and strength. On decesendin gGroup 17, the atomic size of halogens increases, andhence the H-Xbond lengthincreases.The decomposition of hydrogen halides involves breaking the H-X bonds.The longer the bondlength, the weaker the bond and the more easily the bond can be broken.HF>HCl>HBr>HIH-F< H-Cl