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Complex ions
The following question as been answered, please scroll down (to post #6) to answer another question.
Q. This question is about Reinecke's salt
When ammonium dichromate(VI) is added gradually to molten ammonium thiocyanate, Reinecke’s salt is formed. It has the formula NH4[Cr(SCN)x(NH3)y] and the following composition by mass:
Cr - 15.5 %
S - 38.15 %
N - 29.2 %.
(a) Calculate the values of x and y in the above formula.
(b) Calculate the oxidation number of chromium in the complex.
(c) Suggest a shape for the complex anion.
(d) Draw two possible structures for the anion and state the type of isomerism it exhibits.
Q. This question is about Reinecke's salt
When ammonium dichromate(VI) is added gradually to molten ammonium thiocyanate, Reinecke’s salt is formed. It has the formula NH4[Cr(SCN)x(NH3)y] and the following composition by mass:
Cr - 15.5 %
S - 38.15 %
N - 29.2 %.
(a) Calculate the values of x and y in the above formula.
(b) Calculate the oxidation number of chromium in the complex.
(c) Suggest a shape for the complex anion.
(d) Draw two possible structures for the anion and state the type of isomerism it exhibits.
OK third go at getting this completed.
a) The first ting to do is work out the ratio of composition by mass to actual atomic weight for chromium as we know there is only 1 present in the complex.
So percentage composition divided by atomic weight
Cr: 15.5 / 51.9961 = 0.298
Then divide the percentage compositions of both sulphur and nitrogen by 0.298.
S: 38.15 / 0.298 = 128.02
N: 29.2 / 0.298 = 97.99
Then divide the numbers above by the atomic weights for sulphur and nitrogen to get the number of atoms of both present.
S: 128.02 / 32.065 = 3.99 ~4
N: 97.99 / 14.0067 = 6.99 ~7
This means there are a total of 4 sulphur and 7 nitrogen atoms present in the complex. SCN has a value of 4 and this uses up 4 of the 7 nitrogen atoms and another is used in the NH4 so that leaves 2 present for use in the NH3. So the overall complex’s formula is NH4[Cr(SCN)4(NH3)2]
b) NH4 is positively charged so [Cr(SCN)4(NH3)2] has a single negative charge as the complex is neutral overall. For the oxidation state calculation I am just going to use the [Cr(SCN)4(NH3)2] part so the charge on the complex is -1. NH3 has a formal charge of 0 and SCN has a formal charge of -1.
The oxidation state of the chromium = charge on the complex – the sum of charges on the ligands.
Oxidation state = -1 -(-1x4)
Oxidation state = 3
So the chromium is Cr(III)
c) As the anion has 6 ligands around a central atom then it is going to be octahedral in shape.
d) The two possible structures for the anion are for the two ammonia groups to be opposite one another so have an angle of 180 degrees between them and the other way would be for them to have 90 degrees between them.
a) The first ting to do is work out the ratio of composition by mass to actual atomic weight for chromium as we know there is only 1 present in the complex.
So percentage composition divided by atomic weight
Cr: 15.5 / 51.9961 = 0.298
Then divide the percentage compositions of both sulphur and nitrogen by 0.298.
S: 38.15 / 0.298 = 128.02
N: 29.2 / 0.298 = 97.99
Then divide the numbers above by the atomic weights for sulphur and nitrogen to get the number of atoms of both present.
S: 128.02 / 32.065 = 3.99 ~4
N: 97.99 / 14.0067 = 6.99 ~7
This means there are a total of 4 sulphur and 7 nitrogen atoms present in the complex. SCN has a value of 4 and this uses up 4 of the 7 nitrogen atoms and another is used in the NH4 so that leaves 2 present for use in the NH3. So the overall complex’s formula is NH4[Cr(SCN)4(NH3)2]
b) NH4 is positively charged so [Cr(SCN)4(NH3)2] has a single negative charge as the complex is neutral overall. For the oxidation state calculation I am just going to use the [Cr(SCN)4(NH3)2] part so the charge on the complex is -1. NH3 has a formal charge of 0 and SCN has a formal charge of -1.
The oxidation state of the chromium = charge on the complex – the sum of charges on the ligands.
Oxidation state = -1 -(-1x4)
Oxidation state = 3
So the chromium is Cr(III)
c) As the anion has 6 ligands around a central atom then it is going to be octahedral in shape.
d) The two possible structures for the anion are for the two ammonia groups to be opposite one another so have an angle of 180 degrees between them and the other way would be for them to have 90 degrees between them.
Thank you very much. Could you explain the earlier stages of part a) in a bit more detail. Also, for part d), why only the amino group? Couldn't the SCN arrange itself such that it produces a different possible structure? Or is it because there are 4 of the SCN and hence they arrange themselves in a tetrehedral shape and there's only one way of doing so.
In part (a) I first worked out what the ratio was between actual atomic weight and the percentage mass given in the question. This allows us then to work out the total mass of each of the compounds with the percentage values given in the question. I used chromium as there is only one atom present in the complex and it is not affected by X and Y like both N and S are.
This ratio was then used against the percentage masses of each of the given elements to produce the actual mass of each of these elements present. The actual mass was then divided by the atomic weight to give the number of atoms of each of the two elements present. Then as the sulphur was only present in the X part the value of X was known to be equal to the number of sulphur atoms present and so we arrive at the value 4. As 7 nitrogen atoms were found to be in the complex, 4 can be discounted as being in the X part and one is in the ammonium ion so that left 2 for part Y.
For part (d) I only mentioned the amino groups as they were easier to describe as there are only 2 compared to 4. I have included an image of what the two isomers would be like. The first is where the two NH3's are opposite one another and the second is where the two NH3's are next to one another.
This ratio was then used against the percentage masses of each of the given elements to produce the actual mass of each of these elements present. The actual mass was then divided by the atomic weight to give the number of atoms of each of the two elements present. Then as the sulphur was only present in the X part the value of X was known to be equal to the number of sulphur atoms present and so we arrive at the value 4. As 7 nitrogen atoms were found to be in the complex, 4 can be discounted as being in the X part and one is in the ammonium ion so that left 2 for part Y.
For part (d) I only mentioned the amino groups as they were easier to describe as there are only 2 compared to 4. I have included an image of what the two isomers would be like. The first is where the two NH3's are opposite one another and the second is where the two NH3's are next to one another.
Thank you very much.
Q. This question is about hydroxylamine and its reaction with iron(III) ions
Hydroxylamine, NH2OH, is a base and a reducing agent; it reacts with hydrochloric acid to form the salt hydroxylammonium chloride, NH3OH+Cl–, and with Fe3+ ions to produce Fe2+. 1.00g of hydroxylammonium chloride was dissolved in distilled water and made up to a total volume of 250 cm3. A 25.0 cm3 aliquot of this solution was added to a solution containing an excess of both iron(III) ions and sulfuric acid. The mixture was then boiled and allowed to cool. It was then titrated against a solution of 0.0200 mol dm–3 potassium manganate(VII), KMnO4, which oxidizes the Fe2+ ions back to Fe3+ and is itself reduced to Mn2+ ions; 28.9 cm3 of the potassium manganate(VII) solution was required.
(a) Calculate the ratio of the number of moles of Fe3+ ions to the number of moles of hydroxylammonium chloride which have reacted together.
(b) Suggest which of the following is the nitrogen-containing product formed from the hydroxylammonium chloride: N2, NO, N2O, N2O4, NH3.
(c) Write a balanced equation for the reaction between hydroxylammonium chloride and iron(III) ions
Hydroxylamine, NH2OH, is a base and a reducing agent; it reacts with hydrochloric acid to form the salt hydroxylammonium chloride, NH3OH+Cl–, and with Fe3+ ions to produce Fe2+. 1.00g of hydroxylammonium chloride was dissolved in distilled water and made up to a total volume of 250 cm3. A 25.0 cm3 aliquot of this solution was added to a solution containing an excess of both iron(III) ions and sulfuric acid. The mixture was then boiled and allowed to cool. It was then titrated against a solution of 0.0200 mol dm–3 potassium manganate(VII), KMnO4, which oxidizes the Fe2+ ions back to Fe3+ and is itself reduced to Mn2+ ions; 28.9 cm3 of the potassium manganate(VII) solution was required.
(a) Calculate the ratio of the number of moles of Fe3+ ions to the number of moles of hydroxylammonium chloride which have reacted together.
(b) Suggest which of the following is the nitrogen-containing product formed from the hydroxylammonium chloride: N2, NO, N2O, N2O4, NH3.
(c) Write a balanced equation for the reaction between hydroxylammonium chloride and iron(III) ions
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