CHEMISTRY LAB REPORTPRACTICAL 12

INTRODUCTION:

The atoms of transition metals have electrons of similar energy in both the 3d and 4s levels. This means that one particular element can form a number of different stable ions by losing or sharing different numbers of electrons. Thus, all the transition metals in the first (titanium to copper) can exhibit two or more oxidation states in their compounds.

1) Electronic Structures of V+, V2+ and V3+,V+ - (Ar) 3d3 4s1

V2+ - (Ar) 3d3

V3+ - (Ar) 3d2

2) The most stable electronic structure of vanadium is V3+. This is because of its complete configuration structure in 4s subshell. When the oxidation happened, the electron in the 4s subshell will release first before 3d subshell. In V3+ ions, the electron in 3d can replace the empty space in 4s subshell. So, V3+ ions is the most stable electronic structure of vanadium.

3) Vanadium (111) compounds are much more common than vanadium (11) and vanadium (1) compounds due to the electronic configuration structure.

DATA COLLECTION:

PART I >Oxidation states of vanadium

Solution ObservationNH4VO3 + NaOH

Ammonium vanadate (V) + (sodium hydroxide)

The white powder of NH4VO3 dissolved in NaOH and the colour of the solution change from colourless to pale yellow.

NH4VO3 + NaOH + HClAmmonium vanadate (V) +

sodium hydroxide + hydrochloric acid

The colour of solution change from pale yellow to orange.

NH4VO3 + NaOH + HCl + ZnAmmonium vanadate (V) +

sodium hydroxide + hydrochloric acid

+ granulated zinc

The colour of solution change step by step as the boiling tube is shaken. At first, the orange colour solution changed to green colour solution and then it changed to blue. After a while, the solution change to green coloured solution and as the shaking continue, the solution become violet in colour. After that, the colour of the solution was no longer changes. Bubbles were formed during the reaction.

NH4VO3 + NaOH + HCl + Zn + HNO3

Ammonium vanadate (V) +

The solution colour change step by step as HNO3 was added drop by drop into the solution. At first, the violet solution changes to green solution.

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CHEMISTRY LAB REPORTPRACTICAL 12

sodium hydroxide + hydrochloric acid + granulated zinc + nitric acid

Then, a layer of blue solution becomes blue in colour.

PART II > Oxidation states of Manganese

Solution ObservationKMnO4 + H2SO4

Potassium manganate (VII) + sulphuric acid

The solution remains colourless and after a while, the solution became light purple.

KMnO4 + NaOHPotassium manganate (VII) + sodium

hydroxide

The solution becomes purple in colour

KMnO4 + H2SO4 + MnO2

Potassium manganate (VII) + sulphuric acid + manganese (IV) oxide

The solution become colourless again

KMnO4 + NaOH + MnO2

Potassium manganate (VII) + sodium hydroxide + manganese (IV) oxide

The solution become dark-brown then changes to green in colour.

KMnO4 + H2SO4 + NaOH + MnO2

Potassium manganate (VII) + sulphuric acid + sodium hydroxide + manganese

(IV) oxide

The solution colour changes to almost colourless or cloudy solution.

MnSO4 + H2SO4

Manganese (II) sulphate + sulpkuric acidThe white powder of MnSO4 dissolves in H2SO4. The solution becomes pale pink.

MnSO4 + H2SO4 + KMnO4

Manganese (II) sulphate + sulphuric acid + potassium manganate (VII)

The solution colour change to brown in colour and some precipitate formed.

MnSO4 + H2SO4 + KMnO4 + H2OManganese (II) sulphate + sulpkuric acid

+ potassium manganate (VII) + water

The colour of solution remains brown but become lighter and almost become colourless.

DATA PROCESSING:

PART I > Oxidation States of Vanadium

1. The usual source of vanadium in the +5 oxidation state is ammonium metavanadate, NH4VO3. This isn't very soluble in water and is usually first dissolved in sodium hydroxide solution.

2. The solution can be reduced using zinc and an acid - either hydrochloric acid or sulphuric acid, usually using moderately concentrated acid.

3. The experiment begins with the process of mixing ammonium vanadate (V) with sodium hydroxide and hydrochloric acid. This solution contains dioxovanadium(V) ions, VO2

+, in acid solution. The resulting solution is orange in colour and the oxidation number is +5. The chemical equation is:

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CHEMISTRY LAB REPORTPRACTICAL 12

VO3- (aq) + 2H+

(aq) VO2+

(aq) + H2O (l)

4. Next, are the reduction from +5 to +4 process. Zink amalgam is in this process because Zink is a reducing agent. The colour changes just continue from orange to green and lastly to blue.. It is important to notice that the green colour seen isn't actually another oxidation state. It is just a mixture of the original yellow of the +5 state and the blue of the +4.

5. The chemical equation for the reaction is

6. After a while, the solution colour changes from blue to green. The green solution actually is Vanadium (III) ion. The oxidation number is again reduced from +4 to +3

7. The chemical equation for the reduction process is:

VO2+(aq) + 2H+

(aq) + e- V3+(aq) + H2O (l)

8. Then, the solution again changes it colors from green to violet. The violet solution consists of Vanadium (II) ion. As for the oxidation number, it is reduced from +3 to +2

9. The chemical equation for this reduction is

V3+(aq) + e- V2+

(aq)

10. When the resulting solution is added with nitric acid, the color of the solution will change back from violet to green and lastly to blue. This is because, the nitric acid is actually an oxidizing agent that will cause the vanadium (II) to be oxidize to vanadium (III) and lastly to vanadium (IV).

11. The chemical reactions for the oxidizing process of Vanadium (II) to vanadium (III) are as the following

V3+(aq)

+ e- V2+(aq)

NO3-(aq) + 4H+

(aq) + 3e- NO(g) + 2H2O(l)

12. For the oxidizing of vanadium(III) to vanadium(IV), the chemical equation are:

VO2+(aq) + 2H+

(aq) + e- V3+(aq) + H2O (l)

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CHEMISTRY LAB REPORTPRACTICAL 12

NO3-(aq) + 4H+

(aq) + 3e- NO(g) + 2H20(l)

13. The experiment is summarized as below

Table 5: Oxidation states of vanadiumIon VO2

+ VO2+ V3+ V2+

Name Dioxovanadium (V) ion

Oxovanadium (IV) ion

Vanadium(III) ion

Vanadium(II) ion

Oxidation states

+5 +4 +3 +2

Colour Orange Bluish-green Green Violet

PART II > Oxidation States of Manganese

1) The MnO4- is a very good oxidizing agent

2) A mixture of KMnO4 with sulphuric acid produced the same colour as the colour of KMnO4 which its solution in purple in colour. The colour of KMnO4 does not change

3) A mixture of KMnO4 with sodium hydroxide however produced a green solution. This proved the presence of Mn6+ ions.

4) Therefore, it could conclude that Mn7+ in KMnO4 is reduced to Mn6+ by sodium hydroxide.

Mn7+ + e- → Mn6+

5) When the green solution which contained Mn6+ ions added with sulphuric acid, the solution colour change from green to purple in colour.

6) Therefore, it could conclude that Mn6+ ions had been reduced sulphuric acid to Mn7+

Mn6+ → Mn7+ + e-

7) The mixture of hydrated manganese (II) sulphate with dilute and concentrated sulphric acid produced an aqueous containing Mn3+ ions in brown solution

8) Later, when the brown solution of Mn3+ was added with water, it produced brown precipitate suggesting that Mn3+ did not dissolve in water readily.

Mn2+ → Mn3+ + e-

ConclusionVanadium and manganese have a lot of oxidization states with various colours and chemical properties. Vanadium (III) and Manganese (VII), (IV), (II) are among the most common compounds

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