Shapes and Color of Coordinations Compounds depends upon the Ligands and d-d transition of electrons in the central elements

1. Transition Elements
XII FDC CHEMISTRY
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2. Shapes of Complex Ions
 The coordination number shown by metals in complexes are 2 to 9
 The most common are 2, 4 and 6
 Coordination Number 2:
The complexes having C.N = 2 are linear, since the geometry provides minimum
ligand-ligand repulsion. Examples are:
[Cu(NH3)2]+, [Ag(NH3)2]+, [Ag(CN)2]-, [Au(CN)2]-, [Hg(NH3)2]+2, [Hg(CN)2]

3.  Coordination Number 4:
Complexes with C.N = 4 may be tetrahedral or square planar.
Complexes such as [ZnCl4]-2, [Cu(CN)4]-2, [Hg(CN)4]-2, [Ni(CO)4], [FeCl4]-, [ZnBr4]-2,
[Zn(CN)4]-2 , [Zn(NH3)4]-2 , [CuX4]-2, where X = Cl-
, Br- CNS- are tetrahedral.
Oxyanions such VO4
-, CrO4
-2, FeO4
-2 and MnO4
- are also tetrahedral.
Square planar geometry is found in Complexes of Cu+2, Ni+2, Pt+2, Pd+2, Au+3 etc. ions
For example: [Ni(NH3)4]+2, [Ni(CN)4]+2, [Pt(NH3)4]+2, [PdCl4]-2, [AuCl4]-, [Cu(NH3)4]+2
Shapes of Complex Ions

4. Shapes of Complex Ions
Coordination Number 4:

5.  Coordination Number 6:
Complexes with C.N = 6 are the most common ones.
6 ligands may arranged round the central metal ion M, either at the
corners of hexagonal plane or at the apices of trigonal prism or at the
apices of regular octahedron.
An extensive study of the geometrical and optical isomers of complexes
with C.N= 6 has however shown that arrangement of 6 ligands in a
complex is always octahedral.
Shapes of Complex Ions

6. Shapes of Complex Ions
Coordination Number 6:

7. Color of Complexes
 Color is produced in the substance due to absorption and emission of the
photon in visible region of the spectrum.
 When transition elements make the compounds, then the electrons of d
orbital are involved in bonding.
 d-orbitals split into two levels.
 A higher energy pair of two d-orbitals
 A lower energy pair of three d-orbitals

8.  The electrons present in the lower d-orbitals absorbs the photons of
energy and are promoted to higher level.
 The photon which is involved for the jumping of electron, is responsible
for showing definite color.
 This transition is called d-d transition.
 All simple transition metal ions which have unpaired electrons are
colored.
Color of Complexes

9.  The color of the ions is related to the number of unpaired electrons.
Color of Complexes
No. of unpaired electrons Color of simple ions
0 Zn+2, Ti+4 Colorless
1 Ti+3 (Purple) Cu+2 (Blue)
2 V+3 (Green) Ni+2 (Green)
3 Cr+3 (Deep Green) Co+2 (Pink)
4 Cr+2 (Blue) Fe+2 (Green)
5 Fe+3 (Yellow)

10. CHEMISTRY OF SOME IMPORTANT
TRANSITION ELEMENTS
Vanadium, Chromium, Manganese, Iron, Copper,

11. Oxidation States of Vanadium
 Common oxidation states: -1, 0, +1, +2, +4 and +5
 Most stable oxidation states are +4 and +5
 Examples:
V(-1) compounds: [V(CN)5(NO+)]-5 , [V(CO)6]
V(0) compounds: V(CO)6
V(+1) compounds: V2O
V(+2) compounds: VO or V2O2, VX2
V(+3) compounds: V2O3, V2(SO4)3
V(+4) compounds: VO2 or V2O4, VX4
V(+5) compounds: VX5, V2O5, VOX3 etc

12. Reducing Vanadium (V) to Vanadium (II)
 Physical State and dissociation of V2O5:
Yellowish-red poisonous powder, M.P = 670oC
Slightly soluble in H2O resulting in a pale yellow acidic solution
Reversible dissociation takes place at temperatures 700-1125oC
2V2O5  2V2O4 + O2
 Conversion of NH4VO3 to V2O5
2NH4VO3
𝐼𝑔𝑛𝑖𝑡𝑖𝑜𝑛
V2O5 + 2NH3 + H2O

13.  Reduction of V2O5
3V2O5 + 10Al  5Al2O3 + 6V
 The exact Vanadium ion present in the solution is very complicated and
varies with the pH of solution.
 The reaction is done under acidic conditions when the main ion present
VO2
+ (Dioxovanadium ion)
 The ion is usually written as VO2
+ but is more accurately [VO2(H2O)4]+4
Reducing Vanadium (V) to Vanadium (II)

14. Reduction of Vanadium from +5 to +4

15. Reduction of Vanadium from +4 to +2

16. V2O5 as a catalyst in Contact process
 The overall reaction:
SO2 + ½ O2
𝑉2 𝑂5
SO3
 How the reaction works?
SO2 + V2O5  SO3 + V2O4
V2O4 + ½ O2  V2O5

17. Reactions of V2O5
 V2O5 dissolves in HCl to give salts like Vanadium oxydichloride VOCl2
and VCl5
V2O5 + 6HCl  2VOCl2 + 3H2O + Cl2
 V2O5 oxidizes HCl to Cl2 i.e. it can be used as an oxidizing agent.
 It also react with HCl in following manner:
V2O5 + 10HCl  2VCl5 + 2H2O
 The reaction of V2O5 with HNO3 gives meta vanadic acid HVO3
V2O5 + 2HNO3  2HVO3 + N2O5

18. Oxidation of Chromium (III) to Chromium
(IV)
 An excess of NaOH is added to a solution of the Hexaaquochromium(III) ions to
produce a green solution of Hexahydroxochromate(III) ions
 This is then oxidised by warming it with hydrogen peroxide solution. You eventually
get a bright yellow solution containing chromate (VI) ions.

19. The Chromate(VI)-Dichromate(VI)
Equilibrium
 Chromic acid H2CrO4 forms only normal salts M2CrO4 types which are
called chromates.
 Chromates are soluble in water and give brilliant yellow solution.
 When acidified, the yellow solution changes to orange due to conversion
of CrO4
-2 into Cr2O7
-2 ion

20. Chromates and Dichromates
 Examples of important chromates are: K2CrO4, Na2CrO4, PbCrO4, basic lead
chromate [Pb(OH)2.PbCrO4], basic zinc chromate [Zn(OH)2.ZnCrO4.H2O]
 Examples of dichromates are (NH4)2Cr2O7, Na2Cr2O7, K2CrO7
 The salts of Dichromic acid (H2Cr2O7) are called dichromates.
 In presence of dil sulfuric acid, Cr2O7
-2 act as oxidizing agent as it is reduced to
Cr+3 ion.
Cr2O7
2
14H
 6e
2Cr
3
 7H2O

21. Reactions
 Reaction with Zn
 When K2CrO4 is added to the neutral solution of Zinc salt, basic zinc chromates
Zn(OH)2.ZnCrO4.H2O is obtained
K2CrO4 + 2Zn+2 + H2O  Zn(OH)2.ZnCrO4.H2O + 2K+
 Reaction with acid
 Acidified solution of K2CrO4 act as an oxidizing agent.
2K2CrO4 + 5H2SO4  2K2SO4 + Cr2(SO4)3 + 5H2O + 3[O]

22. Reduction of Cr (VI) to Cr(II)
 For the reduction from +6 to +3
 For the reduction from +3 to +2

23. Using K2Cr2O7 as oxidizing agent in
organic chemistry
 It is used to:
 oxidise secondary alcohols to ketones;
 oxidise primary alcohols to aldehydes;
 oxidise primary alcohols to carboxylic acids.
 For example, with ethanol (a primary alcohol), you can get either ethanal (an aldehyde) or
ethanoic acid (a carboxylic acid) depending on the conditions.
 If the alcohol is in excess, and you distil off the aldehyde as soon as it is formed, you get
ethanal as the main product.

24.  If the oxidising agent is in excess, and you do not allow the product to escape – for
example, by heating the mixture under reflux (heating the flask with a condenser placed
vertically in the neck) - you get ethanoic acid.
 In organic chemistry, these equations are often simplified:
CH3CH2OH + [O]  CH3CHO + H2O
CH3CH2OH + 2[O]  CH3COOH + H2O
Using K2Cr2O7 as oxidizing agent in
organic chemistry

25. Using K2Cr2O7 as an oxidizing agent in
titrations
 It is a powerful oxidizing agent. In presence of dil H2SO4 one molecule of K2Cr2O7 produces 3
atoms of Oxygen
K2Cr2O7 + 4 H2SO4  K2SO4 + Cr2(SO4)3 + 4H2O + 3[O]
 Different reactions of K2Cr2O7 in presence of dil H2SO4
 K2Cr2O7 + 7 H2SO4 + 6KI  4K2SO4 + Cr2(SO4)3 + 7H2O + 3I2
Or Cr2O7
-2 + 14H+ + 6I  2Cr+3 + 7H2O + 3I2
 K2Cr2O7 + 7 H2SO4 + 6FeSO4  K2SO4 + Cr2(SO4)3 + 7H2O + 3Fe2(SO4)3
Or Cr2O7
-2 + 14H+ + 6Fe+2  2Cr+3 + 7H2O + 6Fe+3

26.  Acidified solution of K2Cr2O7 also oxidizes primary alcohols to aldehydes then to acids
K2Cr2O7 + 4H2SO4  K2SO4 + Cr2(SO4)3 + 4 H2O + 3[O]
C2H5OH + O  CH3CHO + H2O
CH3CHO + O  CH3COOH
 Acidified solution of K2Cr2O7 oxidizes chloroform to carbonyl chloride and aniline to p-
benzoquinone
2CHCl3 + 3O  2COCl2 + Cl2 + H2O
Using K2Cr2O7 as an oxidizing agent in
titrations

27. Advantages of using K2Cr2O7
 Potassium dichromate(VI) can be used as a primary standard. That means that it can
be made up to give a stable solution of accurately known concentration. That isn't
true of potassium manganate(VII).
 Potassium dichromate(VI) can be used in the presence of chloride ions (as long as
the chloride ions aren't present in very high concentration).
 Potassium manganate(VII) oxidises chloride ions to chlorine; potassium
dichromate(VI) isn't quite a strong enough oxidising agent to do this. That means
that you do not get unwanted side reactions with the potassium dichromate(VI)
solution.

28. Disadvantages of using K2Cr2O7
 The main disadvantage lies in the colour change. KMnO4 titrations are self-indicating. As you
run the KMnO4 solution into the reaction, the solution becomes colourless. As soon as you add
as much as one drop too much, the solution becomes pink - and you know you have reached
the end point.
 Unfortunately K2Cr2O7 solution turns green as you run it into the reaction, and there is no
way you could possibly detect the colour change when you have one drop of excess orange
solution in a strongly coloured green solution.
 With K2Cr2O7 solution you have to use a separate indicator, known as a redox indicator. These
change colour in the presence of an oxidising agent.
 There are several such indicators - such as diphenylamine sulphonate. This gives a violet- blue
colour in the presence of excess K2Cr2O7 solution. However, the colour is made difficult by the
strong green also present. The end point of a K2Cr2O7 titration isn't as easy to see as the end
point of a KMnO4 one.

29. Testing for Chromate (VI) ions in solution
 The bright yellow colour of a solution suggests that it would be worth testing for
chromate(VI) ions.
 Testing by adding an Acid: If you add some dilute sulphuric acid to a solution
containing chromate(VI) ions, the colour changes to the familiar orange of
dichromate(VI) ions.

30.  Testing by adding Barium Chloride (or Nitrate) Solution: Chromate(VI) ions will
give a yellow precipitate of barium chromate(VI).
Testing for Chromate (VI) ions in solution

31.  Testing by adding Lead(II) Nitrate Solution: Chromate(VI) ions will give a
bright yellow precipitate of lead(II) chromate(VI).
Testing for Chromate (VI) ions in solution

32. End of lesson
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