April 07, 2012

Coordination Compounds Five Marks

1. Explain the type of hybridisation, magnetic property and geometry for [Ni(CN)4]2– and [Ni(NH3)4]2+ using VB theory Or [Ni(CN)4]2- is diamagnetic whereas [Ni(NH3)4]2+ is paramagnetic. Explain (using VB theory). Or Apply VB theory to find out the geometry of [Ni(NH3)4]2+ and calculate its magnetic moment.
1) [Ni(NH3)4]2+
Nickel atom
Outer electronic configuration 3d84s2
                                                        3d                              4s                   4p
Ni atom
↑↓
↑↓
↑↓
↑↓

Ni+2 ion
↑↓
↑↓
↑↓

[Ni(NH3)4]2+
↑↓
↑↓
↑↓
x
x
x
x
x
x
x
x
                                                                                                          
                                                                                         NH3      NH3NH3NH3 
                                                                                              sp3 hybridisation
Number of unpaired electrons = 2
                               μs = 2(2+2)
Paramagnetic moment, μs = 2.83BM
The molecule is paramagnetic.
Since the hybridisation is sp3, the geometry of the molecule is tetrahedral.
2) [Ni(CN)4]2-
                                                         3d                             4s                  4p
Ni+2 ion
↑↓
↑↓
↑↓
The ligand CN- is a powerful ligand. Hence it forces the unpaired electrons to pair up in d orbitals.
Hence this complex ion does not contain unpaired electrons.
Paramagnetic moment, μs = 0
The molecule is diamagnetic.
[Ni(CN)4]2-
↑↓
↑↓
↑↓
↑↓
x
x
x
x
x
x
x
x
                                                                                                         
                                                                          CN-        CN-       CN- CN-                                                                              dsp2 hybridisation
Since the hybridisation is dsp2 hybridization, the geometry of the molecule is square planar.
2. Apply V.B. theory for magnetic properties of [Fe(CN)6]4-and [FeF6]4- and explain the shape. Or In what way does [FeF6]4- differ from [Fe(CN)6]4-? Or Mention the type of hybridisation, magnetic property and geometry of the following complexes using VB theory. Or Write the application of VB theory on the following complexes: i) [FeF6]4- ii) [Fe(CN)6]4-
[Fe(CN)6]4-
1) Nickel atom
Outer electronic configuration 3d64s2
                                                        3d                            4s                    4p
Fe atom
↑↓
↑↓

Fe+2 ion
↑↓

[Fe(CN)6]4-
↑↓
↑↓
↑↓
x
x
x
x
x
x
x
x
x
x
x
x
                                                                                                    
                                                                  CN-CN-        CN-       CN- CN- CN-                                                                      d2sp3 hybridisation
The ligand CN- is a powerful ligand. Hence it forces the unpaired electrons to pair up in d orbitals.
Hence this complex ion does not contain unpaired electrons.
Paramagnetic moment, μs = 0
The molecule is diamagnetic.
Since the hybridisation is d2sp3, the geometry of the molecule is Octahedral.
[FeF6]4- 
Fe atom
Outer electronic configuration 3d64s2
                                                3d                                4s                      4p
Fe atom
↑↓
↑↓

Fe +2 ion
↑↓
                                         3d                                     4s                     4p                           4d
[Fe F6]4-
↑↓
x
x
x
x
x
x
x
x
x
x
x
x



                                                                                                                     
                                                                                     F-             F-    F-    F-           F-     F-                                                                                  sp3d2 hybridisation
Number of unpaired electrons = 4
                               μs = 4(4+2) = 24
Paramagnetic moment, μs = 4.90BM
The molecule is paramagnetic.
Since the hybridisation is sp3d2, the geometry of the molecule is Octahedral.
Or
[FeF6]4– differ from [Fe(CN)6]4–
S.No.1. Diagrammatic electronic configuration of the complexes
                                                3d                          4s                 4p                              4d
[FeF6]4
↑↓
x
x
x
x
x
x
x
x
x
x
x
x



                                                                                                           
                                                                               F–          F   F  F           F    F
[Fe(CN)6]4
↑↓
↑↓
↑↓
x
x
x
x
x
x
x
x
x
x
x
x
                                                                                                    
                                                                 CN CN       CN     CN CNCN
S.No.
[FeF6]4 –
[Fe(CN)6]4 –
2
sp3d2 hybridisation
d2sp3 hybridisation
3
Number of unpaired electrons, n = 4
No unpaired electrons, n = 0
4
Paramagnetic
Diamagnetic
5
Paramagnetic moment, μs = 4.90 BM
Paramagnetic moment, μs = 0
6
F  is a weak ligand
CN is a powerful ligand
3. Explain / Give / Write the postulates of Werner's theory. Or State the postulates of Werner's theory on co-ordination compound.
1) Every metal atom has two types of valencies
i) Primary valency or ionisable valency
ii) Secondary valency or non ionisable valency
2) The primary valency corresponds to the oxidation state of the metal ion.
The primary valency of the metal ion is always satisfied by negative ions.
3) Secondary valency corresponds to the coordination number of the metal ion or atom. The secondary valencies may be satisfied by either negative ions or neutral molecules.
4) The molecules or ion that satisfy secondary valencies are called ligands.
5) The ligands which satisfy secondary valencies must project in definite directions in space. So the secondary valencies are directional in nature whereas the primary valencies are non-directional in nature.
6) The ligands have unshared pair of electrons. These unshared pair of electrons are donated to central metal ion or atom in a compound. Such compounds are called coordination compounds.
4. Explain Co-ordination isomerism and ionisation isomerism with suitable examples.
Coordination isomerism: In a bimetallic complex, both complex cation and complex anion may be present. In such a case the distribution of ligands between the two coordination spheres can vary, giving rise to isomers called the coordination isomers. This phenomenon is called coordination isomerism.
Example:
[CoIII(NH3)6] [CrIII(CN)6] and [CrIII(NH3)6] [CoIII(CN)6]
Ionisation isomerism: Coordination compounds having the same molecular formula but forming different ions in solution are called ionisation isomers. This property is known as ionisation isomerism.
Example:
[Co(NH3)5Br]SO4 [Co(NH3)5SO4]Br
The red-violet isomer yields sulphate ion and the red isomer furnishes bromide ion in solution.
5. For the complexes K4[Fe(CN)6], [Cu(NH3)]SO4 mention  a) IUPAC names b) Central metal Ion c) Ligand d) Co-ordination number e) Charge on the complex ion f) Geometry of the complex g) Nature of complex                                                  
                                             K4[Fe(CN)6]                            [Cu(NH3)4]SO4
a) IUPAC names   - Potassium hexacyanoferrate (II)        Tetraamminecopper (II) sulphate
b) Central metal Ion                      - Fe(II) / Fe2+/ ferrous ion    Cu(II)
c) Ligand                                       – CN-, cyano                      NH3, ammine
d) Co-ordination number              – 6                                         4
e) Charge on the complex ion        - –4                                       +2
f) Geometry of the complex          – Octahedral                           Square planar
g) Nature of complex                     Anionic complex                 Cationic complex
6. How is chlorophyll important in environmental chemistry? Mention its function.
1. Chlorophyll is a magnesium - porphyrin complex. The magnesium is at the centre of the modified porphyrin ring septeon (corrin). The oxidation state of magnesium is +2 (Mg2+). The modified porphyrin acts as the ligand.
2. There are several kinds of chlorophyll that vary slightly in their molecular structure.
3. In plants, chlorophyll ‘a’ is the pigment directly responsible for the transformation of light energy to chemical energy.
4. Hence in plants, the green pigment chlorophyll helps photosynthesis.
5. The conversion of atmospheric carbondioxide and atmospheric moisture into carbohydrate and molecular oxygen in the presence of sunlight, by the plant is called as photosynthesis.
6. Chlorophyll acts as a light sensitiser in this important process.
                           Chlorophyll
7. xCO2 + yH2O      Cx(H2O)y + O2
                              Sunlight
7. In the coordination complex [Co(NH3)6]Cl3 mention the following: a) IUPAC name of the complex b) Ligand c) Central metal ion d) Co-ordination number e) Nature of complex
8. Photosynthesis requires, in addition to chlorophyll, the help of four other metal complexes, a manganese complex, two iron complexes and a copper complex.
9. All oxygenated animals take molecular oxygen through haemoglobin and release CO2. But chlorophyll helps in the conversion of atmosphere CO2 into molecular oxygen which acts as a fuel for human cell.
      a) IUPAC name of the complex -  Hexaamminecobalt (III) chloride
b) Ligand -                                  NH3, ammine
c) Central metal ion -                    Cobalt(III)
d) Co-ordination number -            6
e) Nature of complex -                  Cationic complex
8. Mention the function of haemoglobin in natural process.
1. Haemoglobin in the red blood cells carries oxygen from the lungs to the tissues.
2. It delivers the oxygen molecule to myoglobin in the tissues.
3. When the oxygen has been released for cell respiration, haemoglobin loses its bright red colour and becomes purple.
4. It then combines with the waste carbon dioxide produced by the cells.
5. Deposits in the lungs so that the gas can be exhaled.
9. Explain hydrate and linkage isomerisms with suitable examples.
Hydrate isomerism or Solvate isomerism
The best known examples of this type of isomerism occurs for chromium chloride CrCl3.6H2O” which may contain 4, 5, (or) 6 coordinated water molecules.
1. [Cr(H2O)4Cl2]Cl.2H2O - Bright green
2. [Cr(H2O)5Cl]Cl2.H2O - grey-green
3. [Cr(H2O)6]Cl3 - Violet
These isomers have very different chemical properties and on reaction with AgNO3 to test for Cl- ions, would find 1,2, and 3 Cl- ions in solution respectively.
Linkage isomerism
Linkage isomerism occurs with ambidentate ligands. These ligands are capable of coordinating in more than one way. The best known cases involve the monodentate ligands SCN-/NCS- and NO2-/ONO-
Example:
[Co(NH3)5ONO]Cl2 the nitrito isomer - O attached - red colour
 [Co(NH3)5 NO2]Cl2 the nitro isomer - N attached - yellow colour
10. Explain the following terms: i) Neutral ligand  ii) chelates iii) coordination sphere.
i) Neutral ligand: The neutral ligands are named as such without any special name. But water
is written as ‘aqua : Ammonia is written as ammine. Note that two m’s to distinguish from organic amine
CO-Carbonyl, NO-Nitrosyl, NH2 - CH2 - CH2 - NH2-ethylenediamine (en), Pyridine C5H5N.
ii) chelates: If a ligand is capable of forming more than one bond with the central metal atom (or) ion then the ring structures are produced which are known as metal chelates. Hence the ring forming group is described as chelating agents (or) polydentate ligands.
 
iii) coordination sphere: In a complex compound, it usually, central metal ion and the ligands are enclosed with in square bracket is called as coordination sphere. This represents a single constituent unit. The ionisable species are placed outside the square bracket.
       [M(L)n](n-) (or) (n+)
                   [Fe(CN)6]4- , [Cu(NH3)4]2+
These ions do not ionise to give the test for constituent ions
11. What are the postulates of valence bond theory?
 5. The following table gives the coordination number, orbital hybridisation and spatial geometry of the more important geometries.
1. The central metal atom / ion makes available a number of vacant orbitals equal to its coordination number.
2. These vacant orbitals form covalent bonds with the ligand orbitals.
3. A covalent bond is formed by the overlap of a vacant metal orbital and filled ligand orbitals. This complete overlap leads to the formation of a metal ligand,σ (sigma) bond.
4. A strong covalent bond is formed only when the orbitals overlap to the maximum extent. This maximum overlapping is possible only when the metal vacant orbitals undergo a process called ‘hybridisation’. A hybridised orbital has a better directional characteristics than an unhybridised one.
Coordination number
Types of hybridisation
Geometry
2
sp
linear
4
sp3 
tetrahedral
4
dsp2 
square planar
6
d2sp3 
octahedral
6
sp3d2
octahedral
12. For the complex K3[Cr(C2O4)3].3H2O mention    a) Name   b) Central metal ion  c) Ligan
      d) Co–ordination number  e) Geometry
a) Name                                   Potassiumtrioxalatochromate(III). trihydrate
b) Central metal ion                Chromium
c) Ligand                                 C2O42 - / Oxalato
d) Co–ordination number       6 (3 bidentate ligands present)
e) Geometry                            Octahedral
13. Write : a) IUPAC Name   b) Central metal ion    c) Ligand        d) Coordination number  e) Geometry of the complex [Co(en)3]Cl3
a) IUPAC Name                                     Tris(ethylenediamine)cobalt(III) chloride
b) Central metal ion                                 Co+ 3 / CoIII / Cobalt(III)
c) Ligand                                                 en / ethylenediamine / NH2 - CH2 - CH2 - NH2
d) Coordination number                           6 (3 bidentate ligands present)
 e) Geometry of the complex                     Octahedral 
14. For the complex [Co(NH3)3(NO2)3] Write the following a) IUPAC Name b) Central metal ion     c) Ligand     d) Coordination number e) Geometry
a) IUPAC Name                                 Triamminetrinitrocobalt(III)
b) Central metal ion                            Co+ 3 / CoIII / Cobalt(III)
c) Ligand                                             NH3 / ammine & NO2- / nitro
d) Coordination number                     6 (3 + 3 unidentate ligands present)
e) Geometry                                        Octahedral
15. Mention the the following for the complex [Co(NH3)4Cl2]NO2 a) IUPAC Name b) Central metal ion    c) Ligand          d) Charge on the Co-ordination sphere      e) Co-ordination number
a) IUPAC Name                                             Tetraamminedichlorocobalt(III) nitrite
b) Central metal ion                                        Co+ 3 / CoIII / Cobalt(III)
c) Ligand                                                         NH3 / ammine & Cl- / Chloro
d) Charge on the Co-ordination sphere          + 3 / Cationic complex / Positive Charge
e) Co-ordination number                                 6 (4 + 2 unidentate ligands present)







 ONE MARKS

No comments:

Post a Comment