CHEM BRAINS: 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+

↑↓

↑

↑ ↑ ↑ ↑

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-

↑↓

↑ ↑ ↑ ↑

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-

↑↓

↑ ↑ ↑ ↑ ↑ ↑

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-

↑↓

↑

↑ ↑ ↑ ↑ ↑ ↑

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.

[FeF6]4– differ from [Fe(CN)6]4–

S.No.1. Diagrammatic electronic configuration of the complexes

3d 4s 4p 4d

[FeF6]4 –

↑↓

↑

↑ ↑ ↑ ↑ ↑ ↑

F– F– F– F– F– F–

[Fe(CN)6]4–

↑↓

↑ ↑ ↑ ↑ ↑ ↑

CN– CN– CN– CN– CN– CN–

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) Ligand

d) Co–ordination number e) Geometry

a) Name Potassiumtrioxalatochromate(III). trihydrate

b) Central metal ion Chromium(III) / Cr+ 3 / CrIII

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)

17. Write the application of valence bond theory on the following complexes :

i) [Ni(CN)4]2 – ii) [FeF6]4 –

18. Explain Ionisation and Hydrate isomerisms with suitable examples.

19. For the given complex [Ni(PPh3)2Cl2] Mention (i) IUPAC Name (ii) Central metal ion (iii) Ligands (iv) Coordination number (v) Nature of the complex

20. For the complex [Co(NH3)5Br]SO4, write the i) IUPAC name ii) Central metal ion iii) Ligands