OCR Trends and Patterns Past Paper Help

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CalculusMan
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Taken from the June 04

1) Can someone draw me a dot and cross of sulphur dioxide? I think I've drawn it right but I'm not sure.

2) Draw the shape of [Fe(H20)6]3+

Can someone tell me how this is meant to be done? I know to draw an Fe with 6 H20 around it, but how do I know bond angles and what significance is the 3+ charge?

3) Ligand subsitution question (Fe(SCN)x(H20)y]2+: I'm given a table with 6 different test tube results. Each have a different volume of FeCl3 and NH4SCN and an absorbance.

I've plotted these, MS says two lines of best fit needed. Why? I thought it was meant to be a curve shape.

I've worked out holes for max absorbance. How do I deduce x and y?

4) Photochromic glass contains small amounts of AgCl and CuCl. When bright light strikes the glass, AgCl decomposes to make silver atoms and chlorine atoms. This makes the glass darken.

a) Suggest which substance is formed to give the glass its dark colour. Answer is silver. Why?

b) A sample of this glass containing 0.0287g of AgCl is placed in bright sunlight. Calc the max mass of chlorine atoms that can be formed.

Answer is 0.0071g, I can see this is the glass mass divded by 4, but can't see why you do this as its a 1:1 molar ratio.

Thanks.
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winnie the poo
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(Original post by CalculusMan)
Taken from the June 04

1) Can someone draw me a dot and cross of sulphur dioxide? I think I've drawn it right but I'm not sure.

2) Draw the shape of [Fe(H20)6]3+

Can someone tell me how this is meant to be done? I know to draw an Fe with 6 H20 around it, but how do I know bond angles and what significance is the 3+ charge?

3) Ligand subsitution question (Fe(SCN)x(H20)y]2+: I'm given a table with 6 different test tube results. Each have a different volume of FeCl3 and NH4SCN and an absorbance.

I've plotted these, MS says two lines of best fit needed. Why? I thought it was meant to be a curve shape.

I've worked out holes for max absorbance. How do I deduce x and y?

4) Photochromic glass contains small amounts of AgCl and CuCl. When bright light strikes the glass, AgCl decomposes to make silver atoms and chlorine atoms. This makes the glass darken.

a) Suggest which substance is formed to give the glass its dark colour. Answer is silver. Why?

b) A sample of this glass containing 0.0287g of AgCl is placed in bright sunlight. Calc the max mass of chlorine atoms that can be formed.

Answer is 0.0071g, I can see this is the glass mass divded by 4, but can't see why you do this as its a 1:1 molar ratio.

Thanks.
1)Silicon dioxide's dot and cross diagram will look like a central Si atom with one O atom to the left and one to the right. In each section where the atoms overlap you will have a double covalent bond, i.e. 4 electrons ( 2 from Si and 2 from O)

2)The bond angle will be 90 degrees as each electron bond pair repels each other to an equal extent. This is just one of those things you have to remember - the bond pairs of electrons can't get any further away from each other than 90 degrees.

The 3+ charge is simply the charge on the Fe ion as water molecules are neutral and don't affect the overall charge of the complex ion.

3)For these colorimetry graphs you never do a curve, it's always 2 straight lines that intersect. The point of intersection is the useful bit of information.

To go about deducing the formula you need to work out the ratio betwen number of moles of Fe and SCN-. They probably will have mentioned that it is octahedral so x + y = 6.

4a)This is just common sense. Silver is opaque - it let's very little light pass through it.

b)Number of moles of AgCl = 0.0287/143.5 = 0.0002moles.

AgCl will dissociate to give Ag + Cl-

So, the ratio of AgCl : Cl- = 1:1

Therefore, number of moles of Cl- = 0.0002

Mass of Cl- = 0.0002 x 35.5 = 0.0.0071g.

Hope this helped.

EDIT: Beaten to it, damn my slow typing!
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CalculusMan
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Thanks both of you for replies. I'll rep you both for your great answers.

More questions:

Ammonia reacts with water. Aq ammonia is added in drops to aq Cu(ii) ions, a very pale blue precipitate is observed with disappears in excess ammonia to give a deep blue solution.

Write equations to show the formation from aq Cu(ii) ions of the pale blue precipitate, I've put:

Cu2+ + NH3 --> [Cu(NH3)6]2-

This isn't in the mark scheme, why isn't that right?

Equation for the deep blue solution?

2) A transitions metal is yellow, sketch the absorption spectrum you would predict for this solution. I know purple is opposite yellow on a colour wheel, so the max absorbance should be at purple, but how do I know which wavelength this is at? Do I need anything else on this graph?

This compound is K2CrF6. Ligand is Floride. Why?

3) Why wont CsCl react with water? Why is it colourless/go colourless?

4) Mn04- + 8H+ + 5e- -> Mn2+ + 4H2O
C2042- -> 2CO2 + 2e-

Construct a full ionic equation.

I know I need to balance everything, inc electrons, but how do I balance out the electrons?

Thanks again.
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silent ninja
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1) Blue prec is Cu(OH)2(s) forming as ammonia is a source of OH- ions:
NH3 + H2O --> NH4+ + OH-
Write a full equation of the copper complex reacting with hydroxide.

Excess ammonia results in the complex [Cu(NH3)4(H2O)2]2+ which is deep blue
Depending of the wording of the question, you have to either show hydroxide ions from the precipitate being displaced to form the above, or from the initial copper complex in water to the above. I'll let you figure that out :p:

2) Violet-blue is around 400 and red about 700, yellow is in the middle. If its transmitting yellow then the absorbance will be in the blue region, so you want a peak at violet-blue end, and the rest of the curve almost no absorption.

Fluoride is the ligand because K+ has no lone pairs to form a dative covalent bond (definition of a ligand), Cr is a transition element which is the centre of this complex. Fluoride F- like other Group 7 elements such Cl-, can form a dative covalent bond donating its lone pair.

3) Look what group Cs is in. What happens to other chlorides of this group when dissolved?

4) You want same number of electrons in each equation. Multiply first equation by 2, second one by 5 this will give you 10e in both. When you combine the equations after this, as the electrons are on opposite sides of the equation they will cancel out.
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hannah_banana
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(Original post by CalculusMan)

Ammonia reacts with water. Aq ammonia is added in drops to aq Cu(ii) ions, a very pale blue precipitate is observed with disappears in excess ammonia to give a deep blue solution.

Write equations to show the formation from aq Cu(ii) ions of the pale blue precipitate, I've put:

Cu2+ + NH3 --> [Cu(NH3)6]2-

This isn't in the mark scheme, why isn't that right?

Equation for the deep blue solution?
Equation of the deep blue solution is [Cu(NH3)4(H2O)2]2+ because the ammonia is aqueous so there are 4 ammonia ligands and 2 water ligands.

(Original post by CalculusMan)

2) A transition metal is yellow, sketch the absorption spectrum you would predict for this solution. I know purple is opposite yellow on a colour wheel, so the max absorbance should be at purple, but how do I know which wavelength this is at? Do I need anything else on this graph?

This compound is K2CrF6. Ligand is Floride. Why?
Unfortunately you have to learn the wavelengths for the different colours of light. Violet/purple is approx 400nm, and red is 650-700nm, these are the two extremes and the rest of the visible spectrum has wavelengths within this range. Therefore the yellow trasition metal will absorb light at 400nm, but let yellow light (wavelength approx 590nm) pass through so low absorbance there.

The transition metal complex ion is [CrF6]2- Chromium is the transition metal so fluorine must be the ligand. The potassium is 2K+ to balance the 2- charge and form the compound.

(Original post by CalculusMan)

4) Mn04- + 8H+ + 5e- -> Mn2+ + 4H2O
C2042- -> 2CO2 + 2e-

Construct a full ionic equation.

I know I need to balance everything, inc electrons, but how do I balance out the electrons?
To balance the electrons multiply the first half equation by 2 which gives 10e-:
2Mn04- + 16H+ + 10e- -> 2Mn2+ + 8H2O
and multiply the second half equation by 5 to also give 10 electrons so they balance:
5C2042- -> 10CO2 + 10e-
Then you can combine the two half equations and cancel out the electrons to form the ionic equation of:
2Mn04- + 5C2042 + 16H+ -> 2Mn2+ + 8H2O + 10CO2

Hope this makes sense and helps
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