NCERT Solutions for Class 12 Chemistry Chapter 9 - Coordination Compounds
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1. Explain the bonding in coordination compounds in terms of Werner’s postulates.
Ans. The main postulates of Werner’s theory of coordination compounds are as follows:
(i) In coordinate compounds metals possess two types of valencies called
(a) primary valency which are ionizable; (b) secondary valency which are non-ionisable
(ii) Primary valency is satisfied by the negative ions and it is that which a metal exhibits in the formation of its simple salts.
(iii) Secondary valencies are satisfied by neutral ligand or negative ligand and are those which metal exercises in the formation of its complex ions.
The secondary valency is equal to the coordination number and is fixed for a metal.
(iv) The ions/group bound by the secondary linkages to the metal have characteristics spatial arrangements corresponding to different coordination number.
Example : Structure of [CoCl2(NH3)4]+ Cl– is as follows: Co has 3 as primary valency and 4 as secondary valency.
2. FeSO4 solution mixed with (NH4)2SO4 solution in 1 : 1 molar ratio gives the test of Fe2+ ion but CuSO4 solution mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu2+ ion. Explain why?
Ans. When FeSO4 and (NH4)2 SO4 solutions are mixed in 1 : 1 molar ratio, a double salt known as Mohr’s salt is formed. It has the formula FeSO4.(NH4)2SO4.6H2O. In aqueous solution, the salt dissociates as :
$$\text{FeSO}_{4}+\text{(NH}_{4})_{2}\text{SO}_{4}\xrightarrow{}\text{FeSO}_{4}.(\text{NH}_{4})_{2}\text{SO}_{4}\xrightarrow{}\text{FeSO}_{4}.(\text{NH}_{4})_{2}\text{SO}_{4}6\text{H}_{2}\text{O}\space(\text{Mohr’s salt})\\\text{FeSO}_{4}.(\text{NH}_{4})_{2}\text{SO}_{4}.6\text{H}_{2}\text{O (aq)}\xrightarrow{}\text{Fe}^{2+}\text{(aq)}+2\text{NH}_{4}^{\normalsize+}\text{(aq)}+2\text{SO}_{4}^{2-}\text{(aq)}+6\text{H}_{2}\text{O}$$
The solution gives the tests for all the ions including Fe2+ ions. On the other hand, when CuSO4 and NH3 are mixed in the molar ratio of 1 : 4 in solution, a complex [Cu(NH3)4]SO4 is formed. Since the Cu2+ ions are a part of the complex entity (enclosed in square bracket), it will not give their characteristic tests as are given by Fe2+ ions.
$$\text{CuSO}_{4}+\text{NH}_{3}\xrightarrow{}[\text{Cu (NH}_{3})_{4}]\text{SO}_{4}$$
3. Explain with two examples each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.
Ans. Coordination entity: 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 e.g., Ni in[Ni(CO)4]
Heteroletric : Metal is bound to more than one kind of ligands eg Coin [CoCl3(NH3)3]
4. What is meant by unidentate, didentate and ambidentate ligands? Give two examples for each.
Ans. Unidentate ligand : The negative ion or neutral molecule having only one donor atom called unidentate ligand.
Example :
Bidentate/didentate ligand : The ions or molecules having two donor atoms are called bidentate ligand
Example :
Ambidentate ligand : Ligand which can ligate through two different atoms is called ambidentate ligand.
Example : NO2–, SCN– ions.
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) [PtCl4]2–
(iv) K3[Fe(CN)6]
(v) [Cr(NH3)3Cl3]
Ans. (i) [Co(H2O)(CN)(en)2]2+ x + 0 + (–1) + (2 × 0) = + 2, x = +3
(ii) [CoBr2(en)2]+ x + 2 × (–1) + (2 × 0) = + 1, x = +3
(iii) [PtCl4]2– x + (–1) × 4 = –2, x = +2
(iv) K3[Fe(CN)6] 1 × 3 + x + 6 × (–1) = 0, x = +3
(v) [Cr(NH3)3Cl3] x + 3 × 0 + 3 × (–1) = 0, x = +3
6. Using IUPAC norms, write the formulae for the following :
(i) Tetrahydroxozincate(II)
(ii) Potassium tetrachloridopalladate(II)
(iii) Diamminedichloridoplatinum(II)
(iv) Potassium tetracyanonicklate (II)
(v) Pentaamminenitrito-O-cobalt(III)
(vi) Hexaaminecobalt(III) sulphate
(vii) Potassium trioxalatochromate(III)
(viii) Hexaammineplatinum (IV)
(ix) Tetrabromidocuprate (II)
(x) Pentaamminenitrito-N-cobalt (III).
Ans. (i) [Zn(OH)4]2–
(ii) K2[PdCl4]
(iii) [Pt(NH3)2Cl2]
(iv) K2[Ni(CN)4]
(v) [Co(NH3)5ONO]2+
(vi) [Co(NH3)6]2 (SO4)3
(vii) K3 [Cr(OX)3]
(viii) [Pt(NH3)6]4+
(ix) [Cu(Br)4]2–
(x) [Co(NH3)5NO2]2+.
7. Using IUPAC norms write the systematic names of the following:
(i) [Co(NH3)6]CI3,
(ii) [Pt(NH3)2Cl (NH2CH3)] Cl
(iii) [Ti(H2O)6]3+
(iv) [Co(NH3)4Cl(NO2)]CI
(v) Mn(H2O)6]2+
(vi) [NiCl4]2–
(vii) [Ni(NH3)6]CI2
(viii) [Co(en)3]3+
(ix) [Ni(CO)4]
Ans. (i) Hexaammine cobalt (III) chloride.
(ii) Diammine chlorido
(methylamine) platinum (II)chloride.
(iii) Hexaaquatitanium (III) ion.
(iv) Tetraammine chlorido nitrito-N-cobalt (IV) chloride.
(v) Hexaaquamanganese (II) ion.
(vi) Tetrachloridonickelate (II) ion.
(vii) Hexaammine nickel (II) chloride.
(viii) Tris (ethane -1,2-diamine) cobalt (III) ion.
(ix) Tetra carbonyl nickel (0).
8. List various types of isomerism possible for coordination compounds, giving an example of each.
Ans. Isomerism in coordination compound is of two types :
(1) Structural isomerism, (2) Stereo isomerism
(1) Structural isomerism: It is of 4 types:
(i) Linkage isomerism [Co(NH3)5NO2] Cl2 and [Co (NH3)5 ONO] Cl2
(ii) Coordination isomerism [Co (NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6]
(iii) Ionisation isomerism [Co(NH3)5SO4]Br and [Co(NH3)5Br]SO4
(iv) Solvate or hydrate isomerism [Cr(H2O)6] Cl2 and [Cr(H2O)5 Cl] Cl.H2O
(2) Stereoisomerism: It is of 2 types
(i) Geometrical isomerism
(ii) Optical isomerism
9. How many geometrical isomers are possible in the following coordination entities?
(i) [Cr(C2O4)3]3–
(ii) [Co(NH3)3CI3]
Ans. (i) [Cr(C2O4)3]3– → No geometrical isomers are possible in this coordination entity.
(ii) [Co(NH3)3 Cl3] → Two geometrical isomers are possible (fac and mer) in this coordination entity.
10. Draw the structures of optical isomers of
(i) [Cr(C2O4)3]3-
(ii) [PtCI2(en)2]2+
(iii) [Cr(NH3)2Cl2(en)]+
Ans. (i) [Cr(C2O4)3]3– ⇒ [Cr(ox)3]3–
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)]+
Ans. (i) [CoCl2(en)2]+ has 2 geometrical isomers. Further cis-form shows optical isomerism.
12. Write all the geometrical isomers of [Pt(NH3)(Br)(Cl) (py)] and how many of these will exhibit optical isomerism?
Ans. 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.
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.
Ans. 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, the weak H2O ligands are replaced by F– ligands forming [CuF4]2– ions which is a green precipitate.
$$[\text{Cu(H}_2\text{O})_4]^{2+} + 4\text{F}^{\normalsize–}\xrightarrow{}\underset{\text{Tetrafluoridocuprate (II)(Green ppt})}{[\text{CuF}_{4}]^{2-}} + 4\text{H}_{2}\text{O}$$
(ii) When KCl is added, Cl– ligands replace the weak H2O ligands forming [CuCl4]2- ion which has bright green colour.
$$\text{[Cu(H}_2\text{O})_4]^{2+}+4\text{Cl}^{\normalsize-}\xrightarrow{}\underset{\text{Tetrafluoridocuprate (II)(Bright Green solution)}}{[\text{CuCl}_{4}]^{2-}}+4\text{H}_{2}\text{O}$$
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?
Ans. When excess of KCN(aq) is mixed with CuSO4(aq), a complex named Potassiumtetra cyanocuprate (II) is formed. Since CN– ions are strong ligands, the complex is quite stable. It is evident from the stability constant value (K = 2.0 × 1027)
$$\text{4KCN(aq) + CuSO}_{4}\text{(aq)}\xrightarrow{}\text{K}_{2}[\text{Cu(CN)}_{4}](\text{aq})+\text{K}_{2}\text{SO}_{4}\text{(aq)}\underset{\text{No cleavage and no production of Cu}^{+2} \text{ions}}{\xrightarrow{+\text{H}_{2}\text{S}}}$$
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–
Ans. (i) [Fe(CN)6]4 –: In this complex Fe is present as Fe2+
Since, all the electrons are paired, the complex is diamegnetic morover (n-1) d–orbitals are involved in bonding, so, it is an inner orbital or low spin complex.
(ii) [FeF6]3– : In this complex Fe is present as Fe3+
Because of the presence of s unpaired electrons, the complex is paramagnetic. Morover, nd–orbitals are involved in bonding, so it is an outer orbital or high spin complex.
(iii) [Co(C2O4)3]3– : In this complex, Co is present as Co3+
Because of the presence of 4 unpaired electron, complex +3 paramagnetic. Since, nd orbitals take part in bonding. It is an outer orbital complex or high spin complex.
16. Draw figure to show the splitting of d-orbitals in an octahedral crystal field.
Ans.
17. What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
Ans. 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 |
|
They are formed when the crystal field stabilisation energy (∆0) is greater then the p. |
|
They are called low spin complexes. |
|
They are mostly diamagnetic or less paramagnetic than weak field. |
18. What is crystal field splitting energy? How does the magnitude of Δ0 decide the actual configuration of d-orbitals in a coordination entity?
Ans. 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 t32ge1g, 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 t42ge1g thereby forming low spin complexes. Such ligands for which Δ0 > P are called strong field ligands.
19. [Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2– is diamagnetic. Explain why?
Ans. In [Cr(NH3)6]3+, the Cr present as Cr3+
Pairing of electrons occurs due to strong CN– ligand
In [Cr(NH3)6]3+ unpaired electrons are present, so it is paramegnetic in nature whereas in [Ni(CN)4]2– pairing of electrons occur, so it is diamagnetic in nature.
20. A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2– is colourless. Explain.
Ans. 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.
21. [Fe(CN)6]4– and [Fe(H2O)6]2+ are of different colours in dilute solutions. Why?
Ans. 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.
22. . Discuss the nature of bonding in metal carbonyls.
Ans. In metal carbonyl, the metal carbon bond (M – C) possess both the σ and π -bond character. The bond are formed by overlap of atomic orbital of metal with that of C-atom of carbon monoxide in following sequence:
(a) σ-bond is first formed between metal and carbon when a vacant d-orbital of metal atom overlaps with an orbital containing lone pair of electrons on C-atom of carbon monoxide (:C = O:)
(b) In addition to σ-bond in metal carbonyl, the electrons from filled d-orbitals of a transition metal atom/ ion are back donated into anti bonding π-orbitals of carbon monoxide. This stabilises the metal ligand bonding. The above two concepts are shown in following figure:
σ-overlap: Donation of lone pair of electrons on carbon into a vacant orbital on the metals.
π-overlap: Donation of electrons from a filled metal d-orbital into a vacant antibonding p-orbital of CO.
23. Give the oxidation state, d-orbital occupation and coordination number of the central metal ion in the following complexes:
(i) K3[CO(C2O4)3]
(ii) cis [CrCl2(en)2]Cl
(iii) (NH4)2[COF4]
(iv) [Mn(H2O)6]SO4
Ans. (i) K3[Co(C2O4)3]
The central atom is Co. The oxidation state
⇒ 1 × 3 + x + 3(–2) = 0
x = +3
d orbital occupation ⇒ 3d6 : t2g6 eg0
Coordination number ⇒ 3 × denticity of C2O4–
= 3 × 2 = 6
(ii) Cis[CrCl2 (en)2]Cl
The central atom is Cr
oxidation number ⇒ x + 2 × 0 + 2 (–1) = +1
x = +3
d-orbital occupation ⇒ 3d3 : t32g
Coordination number = 3 × denticity of en
= 3 × 2 = 6
(iii) (NH4)2 CoF4
The central atom is Co
oxidation number ⇒ 1 × 2 + x (–1) × 4 = 0
x = +2
d orbital occupation ⇒ 3d7 : t52g eg2
Coordination number = 2 × denticity
= 2 × 2 = 4
(iv) [Mn(H2O)6]SO4
The central atom is Mn
oxidation number ⇒ x + 6 × 0 + (– 2) = 0
d orbital occupation ⇒ 3d5 : t32g e2g
Coordination number = 3 × 2 = 6
24. Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex:
(i) K[Cr(H2O)2(C2O4)2]–3 H2O
(ii) [Co(NH3)5Cl]Cl2
(iii) [CrCI3(Py)3]
(iv) Cs[FeCl4]
(v) K4[Mn(CN)6]
Ans. (i) K[Cr(H2O)2(C2O4)2|–3 H2O
IUPAC name is Potassiumdiaquadioxalatochromate (III) hydrate.
Oxidation state of Cr : 1 × 1 + x + 2(0) + 2(–2) + 3(0) = 0
x = + 3
Shape is octahedral
Electronic configuration of Cr3+ = 3d3 = t32ge°g .
Coordination number = 6
Stereochemistry : cis and trans isomers
Megnetic moment
$$\mu=\sqrt{n(n+2)}=\sqrt{3(3+2)}=3.87\space\text{BM}$$
(ii) [Co(NH3)5Cl]Cl2
IUPAC name is Pentaamminechloridocobalt (III) chloride
Coordination number of Co = 6
Shape is octahedral
Oxidation state of Co, x + 5(0) + (–1) + 2 (–1) = 0
x = +3
Electronic configuration of Co3+ = 3d6 = t62ge°g n = 0, μ = 0 .
(iii) CrCI3(Py)3. IUPAC name is trichloridotripyridine chromium (III).
Oxidation state of Cr, x + (–1) × 3 + 0 × 3 = 0
x = +3
Electronic configuration of Cr3+ = 3d6 = t32g e0g
Coordination number of Cr = 6
Stereochemistry = face and mer isomers
Magnetic moment
$$\mu=\sqrt{n(n+2)}=\sqrt{3(3+2)}=3.87\space\text{BM}$$
(iv) Cs[FeCl4]
IUPAC name is Cesium tetrachloridoferrate (III)
Oxidation number of Fe, 1 + x + (–1) × 4 = 0
x = +3
Electronic configuration of Fe3+ = t32ge2g
Coordination number of Fe is 4
Stereochemistry = No stereoisomers
Magnetic moment
$$\mu=\sqrt{n(n+2)}=\sqrt{5(5+2)}=5.92\space\text{BM}$$
(v) K4[Mn(CN)6]
IUPAC name is Potassium hexacyanomanganate (II)
Oxidation state of Mn is (+1) × 4 + x (–1) × 6 = 0
x = +2
Electronic configuration of Mn2+ = 3d5 = t52g e0g
Coordination number of Mn = 6
Stereochemistry = No stereoisomers
Magnetic moment
$$\mu=\sqrt{n(n+2)}=\sqrt{1(1+2)} =\sqrt{3}=1.73\space\text{BM}$$
25. Explain the violet colour of the complex [Ti (H2O)6]3+ on the basis of crystal field theory.
Ans. In ground state, Ti has 23 electrons with electronic configuration 3d3 4s2. The oxidation state of Ti in the given compound is +3.
Hence, it will now have the configuration 3d2. Since, it has 2 unpaired electrons and has to ability to undergo d-d transition, the given complex gives violet colour.
26. What is meant by the chelate effect? Give an example.
Ans. When a didentate or a polydentate ligand contains donor atoms positioned in such a way that when they coordinate with the central metal ion, a five or a six membered ring is formed, the effect is called chelate effect. For example,
27. Discuss briefly giving an example in each case the role of coordination compounds in :
(i) Biological systems,
(ii) Analytical chemistry,
(iii) Medicinal chemistry, and
(iv) Extraction/Metallurgy of metals.
Ans. (i) Role of coordination compounds in biological systems:
(a) Haemoglobin, the oxygen carrier in blood is a coordination coumpound of iron.
(b) The chlorophyll in plants, responsible for photosynthesis is a coordinate compound of magnesium.
(c) Vitamin B12, the anti pernicious anaemia factor is a coordinate compound of cobalt.
(ii) Role of coordination compounds in medical chemistry:
(a) cis platin [cis–[Pt (NH3)2 Cl2] complex of platinum is used in the treatment of cancer.
(b) EDTA complex of calcium is used in the treatment of lead poisoning.
(iii) Role of coordination compounds in analytical chemistry:
(a) Qualitative analysis:
Detection of Cu2+ is based on the formation of a blue tetraamine copper (II) ion.
$$\text{Cu}^{+} +4\text{NH}_3\xrightarrow{}\underset{\text{Deep Blue}}{[\text{Cu(NH}_{3})_{4}]^{2+}}$$
(b) Quantitative analysis:
Gravimetric estimation of Ni2+ is carried out by precipitating Ni2+ as red nickel dimethyl glyoxine complex in the presence of aminoma.
$$\text{Ni}^{2+}+ 2\text{DMG}\xrightarrow{\text{NH}_{3}}\underset{\text{Red ppt.}}{[\text{Ni(DMG)}]}$$
(iv) Role of coordination compounds in extraction/metallurgy of metals:
Extraction of various metals from their ore involves complex formation.
Example: Silver and gold are extracted from their ore by formining cyanide complex.
$$4\text{Au + 3KCN + 2H}_{2}\text{O + O}_{2}\underset{\text{Potassium dicyanoaurate (l)}}{\xrightarrow{}4\text{K}[\text{ Au(CN)}_{2}]+4\text{KOH}}\\2\text{K[Au(CN)}]_{2}+\text{Zn}\xrightarrow{}\text{K}_{2}[\text{Zn(CN)}_{4}]+2\text{Au}\darr$$
28. How many ions are produced from the complex Co(NH3)6Cl2 in solution?
(i) 6
(ii) 4
(iii) 3
(iv) 2
Ans. (i) Coordination number of cobalt = 6
Hence, the complex is [Co (NH3)6] Cl2. It ionizes in the solution as follows:
$$[\text{Co(NH}_{3})_{6}]\text{Cl}_{2}\xrightarrow{+\text{aq}}[\text{Co(NH}_{3})_{6}]^{2+}+2\text{Cl}^{\normalsize-}$$
Thus, three ions are produced. Hence, the correct option is (iii)
29. Amongst the following ions? Which one has the highest magnetic moment value:
(i) [Cr(H2O)6]3+
(ii) [Fe(H2O)6]2+
(iii) [Zn(H2O)6]2+
$$\textbf{Ans.}\space\text{Megnetic moment }\mu=\sqrt{n(n+2)}\\\text{Electronic configuration of Cr}^{3+}=3d^{3},\text{unpaired electrons = 3}\Rarr\mu=\sqrt{3(3+2)}=2.83\space\text{BM}\\\text{Electronic configuration of Fe}^{2+}=3d^{6},\text{unpaired electrons = 4}\Rarr\mu=\sqrt{4(4+2)}=3.16\space\text{BM}\\\text{Electronic configuration of Zn}^{2+}=3d^{10},\text{unpaired electrons = 0}\Rarr\mu=\sqrt{0(0+2)}=0\space\text{BM}$$
The complex [Fe(H2O)6]2+ with maximum number of unpaired electrons has the highest magnetic moment. Thus, correct answer is (ii).
30. The oxidation number of cobalt in K[Co(CO)4] is:
(i) +1
(ii) +3
(iii) –1
(iv) –3
Ans. K[Co(CO)4]
∴ Oxidation number ⇒ 1 + x + 4 × 0 = 0
x = –1
Hence, (iii) is the correct option.
31. Amongst the following, the most stable complex is:
(i) [Fe(H2O)6]3+
(ii) [Fe(NH3)6]3+
(iii) [Fe(C2O4)3]3–
(iv) [FeCl6]3–
Ans. In each of the given complex, Fe is in +3 oxidation state. As C2O42– is didentate chelating ligand, it forms chelate rings and hence (iii) out of complexes given above is the most stable complex.
32. What will be the correct order for the wavelengths of absorption in the visible region for the following:[Ni(NO2)6]4–, [Ni(NH3)6]2+, [Ni(H2O)6]2+ ?
Ans. As metal ion is fixed, the increasing field strengths, i.e., the CFSE values of the ligands from the spectro-chemical series are in the order: H2O < NH3 < NO2–;
Thus, the energies absorbed for excitation will be in the order:
[Ni(H2O)6]2+ < [Ni(NH3)6]2+ < [Ni(NO2)6]4–
$$\text{As}\space \text{E}=\frac{hc}{\lambda}\space\text{the wavelengths absorbed will be in the opposite order.}$$
[Ni(H2O)6]2+ > [Ni (NH3)6]2+ > [Ni(NO2)6]4–
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