Chapter 5 Coordination Compounds NCERT Solutions Class 12 Chemistry- PDF Download
Exercises
5.1. Explain the bonding in coordination compounds in terms of Werner’s postulates.
Solution
Werner’s postulates explain the bonding in coordination compounds as follows:
- A metal exhibits two types of valencies namely, primary and secondary valencies. Primary valencies are satisfied by negative ions while secondary valencies are satisfied by both negative and neutral ions.
(In modern terminology, the primary valency corresponds to the oxidation number of the metal ion, whereas the secondary valency refers to the coordination number of the metal ion. - A metal ion has a definite number of secondary valencies around the central atom. Also, these valencies project in a specific direction in the space assigned to the definite geometry of the coordination compound.
- Primary valencies are usually ionizable, while secondary valencies are non-ionizable.
Solution
(NH4)2SO4 + FeSO4 + 6H2O → FeSO4.(NH4)2SO4.6H2O
(FeSO4.(NH4)2SO4.6H2O - Mohr's salt)
CuSO4 + 4NH3 + 5H2O → [Cu(NH3)4]SO4.5H2O
([Cu(NH3)4]SO4.5H2O - tetraamminocopper(ii) sulphate)
Both the compounds i.e., FeSO4.(NH4)2SO4.6H2O and CuSO4 + 4NH3 + 5H2O fall under the category of addition compounds with only one major difference i.e., the former is an example of a double salt, while the latter is a coordination compound.
A double salt is an addition compound that is stable in the solid state but that which breaks up into its constituent ions in the dissolved state. These compounds exhibit individual properties of their constituents.
For e.g. FeSO4.(NH4)2SO4.6H2O breaks into Fe2+, NH4+, and SO42− ions. Hence, it gives a positive test for Fe2+ ions.
A coordination compound is an addition compound which retains its identity in the solid as well as in the dissolved state. However, the individual properties of the constituents are lost. This happens because [Cu(NH3)4]SO4.5H2O] does not show the test for Cu2+. The ions present in the solution of [Cu(NH3)4]SO4.5H2O are [Cu(NH3)4)]2+ and SO42-.
5.3. Explain with two examples each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.
Solution(i) Coordination entity: A coordination entity is an electrically charged radical or species carrying a positive or negative charge. In a coordination entity, the central atom or ion is surrounded by a suitable number of neutral molecules or negative ions (called ligands). For example:
[Ni(NH3)6]2+, [Fe(CN)6]4+ = catopnic complex
[PtCl4]2+, [Ag(CN)2]- = anionic complex
[Ni(CO)4], [Co(NH3)4Cl2] = neutral complex
(ii) Ligands: The neutral molecules or negatively charged ions that surround the metal atom in a coordination entity or a coordinal complex are known as ligands.
For example, 
, Cl−, OH−. Ligands are usually polar in nature and possess at least one unshared pair of valence electrons.
(iii) Coordination number: The total number of ligands (either neutral molecules or negative ions) that get attached to the central metal atom in the coordination sphere is called the coordination number of the central metal atom. It is also referred to as its ligancy.
For example:
- In the complex, K2[PtCl6], there as six chloride ions attached to Pt in the coordinate sphere. Therefore, the coordination number of Pt is 6.
- Similarly, in the complex [Ni(NH3)4]Cl2, the coordination number of the central atom (Ni) is 4.
(iv) Coordination polyhedron: Coordination polyhedrons about the central atom can be defined as the spatial arrangement of the ligands that are directly attached to the central metal ion in the coordination sphere.
For example:
(v) Homoleptic complexes:
These are those complexes in which the metal ion is bound to only one kind of a donor group.
For eg: [Co(NH3)6]3+, [PtCl4]2- etc
(vi) Heteroleptic complexes:
Heteroleptic complexes are those complexes where the central metal ion is bound to more than one type of a donor group.
For eg: [Co(NH3)4Cl2]+, [CO(NH3)5Cl]2+
Or,
- Coordination entity: It constitutes of a central atom/ion bonded to fixed number of ions or molecules by coordinate bonds. e.g. [COCl3(NH3)3], [Ni (CO)4] etc.
- Ligand: The ions/molecules bound to central atom/ion in coordination entity are called ligands. Ligands in above examples are Cl, NH3, CO.
- Coordination number: This is the number of bond formed by central atom/ion with ligands.
- Coordination polyhedron: Spatial arrangement of ligands defining the shape of complex. In above cases Co and Ni polyhedron are octahedral and tetrahedral in [CoCl3(NH3)3] and [Ni(CO)4] respectively.
- Homoleptic: Metal is bound to only one kind of ligands.
Example: Ni in[Ni(CO)4] - Heteroletric: Metal is bound to more than one kind of ligands.
Example: Co in [CoCl3(NH3)3]
5.4. What is meant by unidentate, didentate and ambidentate ligands? Give two examples for each.
Solution
A ligand may contain one or more unshared pairs of electrons which are called the donor sites of ligands. Now, depending on the number of these donor sites, ligands can be classified as follows:
Unidentate ligands: Ligands with only one donor sites are called unidentate ligands. For example: 
, Cl- etc.
Didentate ligands: Ligands that have two donor sites are called didentate ligands. For example:
(1) Ethane-1,2-diamine
(2) Oxalate ion
Ambidentate ligands: Ligands that can attach themselves to the central metal atom through two different atoms are called ambidentate ligands. For example:
(The donor atom is N)
(The donor atom is oxygen)
(The donor atom is S)
(The donor atom is N)
5.5. Specify the oxidation numbers of the metals in the following coordination entities:
(i) [Co(H2O)(CN)(en)2]2+
(ii) [CoBr2(en)2]+
(iii) [Pt(Cl)4]2-
(iv) K3[Fe(CN)6]
(v) [Cr(NH3)3Cl3]
Solution
(i) [Co(H2O)(CN)(en)2]2+
Let the oxidation number of Co be x.
The charge on the complex is +2.
(ii) [CoBr2(en)2]+
Let the oxidation number of Co be x.
The charge on the complex is +1.
(iii) [Pt(Cl)4]2-
Let the oxidation number of Pt be x.
The charge on the complex is -2.
(iv) K3[Fe(CN)6]
i.e. [Fe(CN)6]3-
Let the oxidation number of Fe be x.
The charge on the complex is -3.
(v) [Cr(NH3)3Cl3]
Let the oxidation number of Cr be x.
The charge on the complex is 0.
(i) Tetrahydroxidozincate(II)
(ii) Potassium tetrachloridopalladate(II)
(iii) Diamminedichloridoplatinum(II)
(iv) Potassium tetracyanidonickelate(II)
(v) Pentaamminenitrito-O-cobalt(III)
(vi) Hexaamminecobalt(III) sulphate
(vii) Potassium tri(oxalato)chromate(III)
(viii) Hexaammineplatinum(IV)
(ix) Tetrabromidocuprate(II)
(x) Pentaamminenitrito-N-cobalt(III)
Solution
(i) [Zn(OH)4]2−
(ii) K2[PdCl4]
(iii) [Pt(NH3)2Cl2]
(iv) K2[Ni(CN)4]
(v) [Co(NH3)5(ONO)]2+
(vi) [Co(NH3)6]2 (SO4)3
(vii) K3[Cr(C2O4)3]
(viii) [Pt(NH3)6]4+
(ix) [Cu(Br)4]2−
(x) [Co(NO2)(NH3)5]2+
(i) [Co(NH3)6]Cl3
(ii) [Pt(NH3)2Cl(NH2CH3)]Cl
(iii) [Ti(H2O)6]3+
(iv) [Co(NH3)4Cl(NO2)]Cl
(v) [Mn(H2O)6]2+
(vi) [NiCl4]2−
(vii) [Ni(NH3)6]Cl2
(viii) [Co(en)3]3+
(ix) [Ni(CO)4]
Solution
(i) Hexaamminecobalt(III) chloride
(ii) Diamminechlorido(methylamine) platinum(II) chloride
(iii) Hexaquatitanium(III) ion
(iv) Tetraamminechloridonitrito-N-Cobalt(III) chloride
(v) Hexaquamanganese(II) ion
(vi) Tetrachloridonickelate(II) ion
(vii) Hexaamminenickel(II) chloride
(viii) Tris(ethane-1, 2-diammine) cobalt(III) ion
(ix) Tetracarbonylnickel(0)
5.8. List various types of isomerism possible for coordination compounds, giving an example of each.
Solution
Coordination compounds exhibit stereo isomerism and structural isomerism.
Two types of stereoisomerism and their examples are as follows:
- Geometrical isomerism
- Optical isomerism
Four types of structural isomerism are as follows:
- Linkage isomerism: This type of isomerism is found in complexes that contain ambidentate ligands. For example: [Co(NH3)5 (NO2)]Cl2 and [Co(NH3)5 (ONO)Cl2
- Coordination isomerism: This type of isomerism arises when the ligands are interchanged between cationic and anionic entities of different metal ions present in the complex.
For example: [Co(NH3)6] [Cr(CN)6] and [Cr(NH3)6] [Co(CN)6] - Ionization isomerism: This type of isomerism arises when a counter ion replaces a ligand within the coordination sphere. Thus, complexes that have the same composition, but furnish different ions when dissolved in water are called ionization isomers.
For example: Co(NH3)5SO4)Br and Co(NH3)5Br]SO4. - Solvate isomerism: Solvate isomers differ by whether or not the solvent molecule is directly bonded to the metal ion or merely present as a free solvent molecule in the crystal lattice.
For example: [Cr[H2O)6]Cl3 [Cr(H2O)5Cl]Cl2⋅H2O and [Cr(H2O)5Cl2]Cl⋅2H2O
(i) [Cr(C2O4)3]3−
(ii) [Co(NH3)3Cl3]
Solution
(i) For [Cr(C2O4)3]3−, no geometric isomer is possible as it is a bidentate ligand.
(ii) For [Co(NH3)3Cl3], two geometrical isomers are possible.
5.10. Draw the structures of optical isomers of:
(i) [Cr(C2O4)3]3−
(ii) [PtCI2(en)2]2+
(iii) [Cr(NH3)2CI2(en)]+
Solution
(i) [Cr(C2O4)3]3− → [Cr(ox)3]3−
(ii) (cis-isomer only) Cis-[PtCI2(en)2]2+
(iii) (cis-isomer only) Cis-[Cr(NH3)2CI2(en)]+
5.11. Draw all the isomers (geometrical and optical) of:
(i) [CoCl2(en)2]+
(ii) [Co(NH3)Cl(en)2]2+
(iii) [Co(NH3)2Cl2(en)]+
Solution
(i) [CoCl2(en)2]+
Geometrical isomerism
Optical isomerism
Since, only cis isomer is optically active, it shows optical isomerism.
In total, three isomers are possible.
(ii) [Co(NH3)Cl(en)2]2+
Geometrical isomerism
Optical isomerism
Since, only cis isomer is optically active, it shows optical isomerism.
(iii) [Co(NH3)2Cl2(en)]+
Geometrical isomerism
Optical isomerism
Since, only cis isomer is optically active, it shows optical isomerism.
(i) For [CoCl2(en)2]+- Geometrical: 2 (1 cis, 1 trans)
- Optical: 1 (from cis)
- Geometrical: 2 (1 cis, 1 trans)
- Optical: 1 (from cis)
- Geometrical: 2 (1 cis, 1 trans)
- Optical: 1 (from cis)
5.12. Write all the geometrical isomers of [Pt(NH3)(Br)(Cl) (py)] and how many of these will exhibit optical isomerism?
Solution
Three geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] are possible.
These are obtained by keeping the position of one of the ligand, say NH3 fixed and rotating the positions of others. This type of isomers do not show any optical isomerism. Optical isomerism only rarely occurs rarely in square planar or tetrahedral complexes and that too when they contain unsymmetrical chelating ligand.
5.13. Aqueous copper sulphate solution (blue in colour) gives:
(i) A green precipitate with aqueous potassium fluoride and
(ii) A bright green solution with aqueous potassium chloride. Explain these experimental results.
Solution
Aqueous CuSO4 solution exists as [Cu(H2O)4]SO4 which has blue colour due to [Cu(H2O)4]2+ ions.
(i) When KF is added, F– ligands replaces the weak H2O ligands forming [CuF4]2– ions which is a green precipitate.
[Cu(H2O)4]2+ + 4F– → [Cu(F)4]2- + 4H2O
[Cu(F)4]2- : Tetrafluoridocuprate(II) - Green ppt.
(ii) When KCl is added, Cl– ligands replace the weak H2O ligands forming [CuCl4]2- ion which has bright green colour.
[Cu(H2O)4]2+ + 4Cl– → [CuCl4]2- + 4H2O
[CuCl4]2- : Tetrachoridocuprate(II) - Green ppt.
In both these cases, the weak field ligand water is replaced by the F− and Cl− ions
5.14. What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H2S(g) is passed through this solution?
Solution
CuSO4(aq) + 4KCN(aq) → K2[Cu(CN)4](aq) + K2SO4(aq)
i.e. [Cu(H2O)4]2+ + 4CN- → [Cu(CN)4]2- + 4H2O
Thus, the coordination entity formed in the process is K2[Cu(CN)4]. K2[Cu(CN)4] is a very stable complex, which does not ionize to give Cu2+ ions when added to water. Hence, Cu2+ ions are not precipitated when H2S(g) is passed through the solution.
5.15. Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:
(i) [Fe(CN)6]4–
(ii) [FeF6]3–
(iii) [Co(C2O4)3]3–
(iv) [CoF6]3–
Solution
(i) [Fe(CN)6]4−
In the above coordination complex, iron exists in the +II oxidation state.
Fe2+: Electronic configuration is 3d6
Orbitals of Fe2+ ion:
As CN− is a strong field ligand, it causes the pairing of the unpaired 3d electrons.
Since there are six ligands around the central metal ion, the most feasible hybridization is d2sp3. d2sp3 hybridized orbitals of Fe2+ are:
6 electron pairs from CN− ions occupy the six hybrid d2sp3orbitals.
Then,
Hence, the geometry of the complex is octahedral and the complex is diamagnetic (as there are no unpaired electrons).
(ii) [FeF6]3−
In this complex, the oxidation state of Fe is +3.
Orbitals of Fe+3 ion:
There are 6 F− ions. Thus, it will undergo d2sp3 or sp3d2 hybridization. As F− is a weak field ligand, it does not cause the pairing of the electrons in the 3d orbital. Hence, the most feasible hybridization is sp3d2.
sp3d2 hybridized orbitals of Fe are:
Hence, the geometry of the complex is found to be octahedral.
(iii) [Co(C2O4)3]3−
Cobalt exists in the +3 oxidation state in the given complex.
Orbitals of Co3+ ion:
Oxalate is a weak field ligand. Therefore, it cannot cause the pairing of the 3d orbital electrons. As there are 6 ligands, hybridization has to be either sp3d2 or d2sp3 hybridization.
sp3d2 hybridization of Co3+:
The 6 electron pairs from the 3 oxalate ions (oxalate anion is a bidentate ligand) occupy these sp3d2orbitals.
Hence, the geometry of the complex is found to be octahedral.
(iv) [CoF6]3−
Cobalt exists in the +3 oxidation state.
Orbitals of Co3+ ion:
Again, fluoride ion is a weak field ligand. It cannot cause the pairing of the 3d electrons. As a result, the Co3+ ion will undergo sp3d2 hybridization.
sp3d2 hybridized orbitals of Co3+ ion are:
Hence, the geometry of the complex is octahedral and paramagnetic.
5.16. Draw figure to show the splitting of d orbitals in an octahedral crystal field.
Solution
The splitting of the d orbitals in an octahedral field takes place in such a way that dx2-y2, dz2 experience a rise in energy and form the eg’ level, while dxy, dyz and dzx experience a fall in energy and form the t2g level.
5.17. What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
Solution
Spectrochemical Series: The arrangement of ligands in order of their increasing field strengths i.e. increasing magnitude of crystal field splitting energy values is called spectrochemical series,
I– < Br– < SCN– < Cl– < F– < OH– < OX < H2O < EDTA4– < NH3 < en < CN– < Co
The ligand present on the R.H.S. of the series are strong field ligand while L.H.S. are weak field ligand. Also, strong field ligand cause higher splitting in the d-orbitals than weak field ligand.
|
Weak field ligand |
Strong field ligand |
|
The are formed when the crystal field stabilisation energy (∆0) in octahedral complexes is less than the energy required for an electron pairing in a single orbital (p). |
They are formed when the crystal field stabilisation energy (∆0) is greater then the p. |
|
They are also called high spin complex. |
They are called low spin complexes. |
|
They are mostly paramagnetic in nature complex. |
They are mostly diamagnetic or less paramagnetic than weak field. |
5.18. What is crystal field splitting energy? How does the magnitude of Δ0 decide the actual configuration of d-orbitals in a coordination entity?
Solution
When the ligands approach a transition metal ion, the d-orbitals split into two sets, one with lower energy and the other with higher energy. The difference of energy between the two sets of orbitals is called as crystal field splitting energy (Δ0 for octahedral field). If Δ0 < P (pairing energy), the fourth electron enters one of the eg, orbitals giving the configuration t2g3e1g, thus forming high spin complexes. Such ligands for which Δ0 < P are called weak field ligands. If Δ0 > P, the fourth electron pairs up in one of the t2g orbitals giving the configuration t2g4e1g thereby forming low spin complexes. Such ligands for which Δ0 > P are called strong field ligands.
5.19. [Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2− is diamagnetic. Explain why?
Solution
Cr is in the +3 oxidation state i.e., d3 configuration. Also, NH3 is a weak field ligand that does not cause the pairing of the electrons in the 3d orbital.
Cr3+:
Therefore, it undergoes d2sp3 hybridization and the electrons in the 3d orbitals remain unpaired. Hence, it is paramagnetic in nature.
In [Ni(CN)4]2−, Ni exists in the +2 oxidation state i.e., d8 configuration.
Ni2+:
CN− is a strong field ligand. It causes the pairing of the 3d orbital electrons. Then, Ni2+ undergoes dsp2 hybridization.
As there are no unpaired electrons, it is diamagnetic.
5.20. A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2− is colourless. Explain.
Solution
In [Ni(H2O)6]2+, Ni is in + 2 oxidation state and having 3d8 electronic configuration, in which there are two unpaired electrons which do not pair in the presence of the weak H2O ligand. Hence, it is coloured. The d-d transition absorbs red light and the complementary light emitted is green.
In [Ni(CN)4]2– Ni is also in +2 oxidation state and having 3d8 electronic configuration. But in presence of strong ligand CN– the two unpaired electrons in the 3d orbitals pair up. Thus, there is no unpaired electron present. Hence, it is colourless.
5.21. [Fe(CN)6]4− and [Fe(H2O)6]2+ are of different colours in dilute solutions. Why?
Solution
In both the complexes, Fe is in +2 oxidation state with configuration 3d6 i.e., it has four unpaired electrons. In the presence of weak H2O ligands, the unpaired electrons do not pair up but in the presence of strong ligand CN–, they get paired up. Due to this difference in the number of unpaired electron, complex ions have different colours.