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NCERT Class 12 Chemistry Chapter 9 Coordination Compounds
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Coordination Compounds
Chapter: 9
| Part – I |
1. Explain the bonding in coordination compounds in terms of Werner’s postulates.
Ans: Bonding in coordination compounds in terms of Werner’s Postulates are:
(i) In coordination compounds, metals exhibit two types of linkages (valencies): primary and secondary.
(ii) The primary valencies are normally ionisable and are satisfied by negative ions.
(ii) The secondary valencies are non-ionizable and are satisfied by neutral molecules or anions. The secondary valence corresponds to the coordination number and is consistent for a given metal in a neutral compound.
(iv) The ions/groups bound by the secondary linkages to the metal have characteristic spatial arrangements corresponding to different numbers.
In contemporary formulations, these spatial arrangements are referred to as coordination polyhedra. The entities within square brackets are known as coordination complexes, while the ions outside the brackets are termed counter ions.
He further postulated that octahedral, tetrahedral and square planar geometrical shapes are more common in coordination compounds of transition metals. Thus [Co(NH₂)6]3+, [CoCl(NH3)5]²+ and [CoCl2(NH3)4]+ l are octahedral entities, while [Ni(Co)4] and [PtCl4]2- are tetrahedral and square planar respectively.
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: Ferrous sulphate and ammonium sulphate form Mohr’s salt (a double salt), which provides free ferrous ions in solution.
Copper forms complex with ammonia. Free cupric ions are not present in solution.
Formation of Mohr’s Salt:
FeSO4 + (NH4)2SO4 + 6H2O → FeSO4⋅(NH4)2SO4⋅6H2O (Mohr’s Salt)
Formation of the Copper-Ammonia Complex:
CuSO4 + 4NH3 + 5H2O → [Cu(NH3)4]SO4⋅ 5H2O
3. Explain with two examples each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.
Ans: Coordination Entity: A coordination entity constitutes a central atom/ion, usually of a metal, to which are attached a fixed number of other atoms or groups each of which is called a ligand. It may be neutral or charged.
Examples: [Co(NH₃)₆]³⁺, [Fe(CN)₆]⁴⁻, [NiCl₄]²⁻, [PtCI4]2-.
Ligand: An atom or molecule or ion capable of donating a pair of electrons to the central metal or ion and forms a coordinate bond with it. Examples include chloride ions, water and ammonia.
Coordination Number: TThe coordination number of a metal ion is the total number of ligand atoms directly bonded to it. It is also equal to the total number of coordinate bonds in the complex. For example, in [Ni(CO)4], the coordination number of Ni is 4, In [CoCl3(NH3)3], the coordination number of [Co3+] is 6.
Coordination polyhedron: The spatial arrangement of the ligand atoms directly bonded to the central atom/ion. Common polyhedra include octahedral, square planar, and tetrahedral. For instance, [Co(NH₃)₆]³⁺ is octahedral, [Ni(CO)₄] is tetrahedral, and [PtCl₄]²⁻ is square planar.
Heteroleptic: Complexes in which a metal is bound to more than one kind of donor groups, e.g., [Co(NH), Cll are known as heteroleptic.
Homoleptic: In a homoleptic complex, the metal is bound exclusively to one type of donor group. E.g.,include: [Ni(NH3)6]2+, [Co(NH3)6]3+
4. What is meant by unidentate, didentate and ambidentate ligands? Give two examples for each.
Ans: Unidentate Ligand: A ligand that forms only one coordinate bond with the central metal atom or ion.
Example: Ammonia (NH₃)
Didentate Ligand: When the ligand can donate the pair of electrons through two atoms of the ligand, it is called didentate ligand. Example: Ethylenediamine (en).
Ambidentate: It is that unidentate ligand which can ligate through two different atoms present in it to central atom/ion giving two different coordination entities.
Examples are NO2– and SCN– ions. NO2– can ligate through either or O atoms and SCN can ligate through S or N atoms to the central atom/ion of coordination entity. This results in formation of linkage isomers.
5. Specify the oxidation numbers of the metals in the following coordination entities:
(i) [Co(H2O)(CN)(en)2] 2+
Ans: x + 0 − 1 + 0 = +2
∴ x = 3
Oxidation no. of Co = 3
(ii) [CoBr2(en)2]+
Ans: x + 4(-1) = -2
Or x = 2
Oxidation no. of Pt = 2
(iii) [PtCl4]2–
Ans: x + 0 + 3(-1) = 0
x = 3
Oxidation no. of Cr = 3
(iv) K3[Fe(CN)6]
Ans: x + 2(-1) + 0 = +1
Or x = 3
Oxidation no. of Co = 3
(v) [Cr(NH3)3Cl3]
Ans: 3(+1) + x + 6(-1) = 0
Or x = 3
Oxidation no. of fe = 3
6. Using IUPAC norms write the formulas for the following:
(i) Tetrahydroxozincate(II)
Ans: [Zn(OH)4]2−
(ii) Potassium tetrachloropalladate(II)
Ans: K2[PdCl4]
(iii) Diamminedichloroplatinum(II)
Ans: [Pt(NH3)2Cl2]
(iv) Potassium tetracyanonickelate(II)
Ans: K2[Ni(CN)4]
(v) Pentaammine Nitrito-O-cobalt(III)
Ans: [Co(NH3)5(ONO)]2+
(vi) Hexaamminecobalt(III) sulphate
Ans: [Co(NH3)6]2(SO4)3
(vii) Potassium tri(oxalato)chromate(III)
ANs: K3[Cr(C2O4)3]
(viii) Hexammine Platinum(IV)
Ans: [Pt(NH3)6]4+
(ix) Tetrabromocuprate(II)
Ans: [CuBr4]2−
(x) Pentaammine Nitrito-N-cobalt(III)
Ans: [Co(NH3)5(NO2)]2+
7. Using IUPAC norms write the systematic names of the following:
(i) [Co(NH3)6]Cl3
Ans: Hexa amminecobalt(III) chloride.
(ii) [Pt(NH3)2Cl(NH2CH3)]Cl
Ans: Diammine Dichloride(methylamine)platinum(II) chloride.
(iii) [Ti(H2O)6]3+
Ans: Hexaaquatitanium(III) ion.
(iv) [Co(NH3)4Cl(NO2)]Cl
Ans: Tetraamminechloridonitrito-N-cobalt(III) chloride.
(v) [Mn(H2O)6]2+
Ans: Hexaaquamanganese(II) ion.
(vi) [NiCl4] 2–
Ans: Tetrachloronickelate(II).
(vii) [Ni(NH3)6]Cl2
Ans: Hexaamminenickel(II) chloride.
(viii) [Co(en)3] 3+
Ans: Tris(ethylenediamine)cobalt(III) ion.
(ix) [Ni(CO)4]
Ans: Tetra carbonyl nickel(0).
8. List various types of isomerism possible for coordination compounds, giving an example of each.
Ans: Various types of isomerism possible for coordination compounds, along with an example of each are as shown.
(i) Ionisation Isomerism: Occurs when a counter ion swaps places with a ligand in the coordination sphere, changing the ions that are released in solution.
For example, [Co(NH₃)₅SO₄]2-Br and [Co(NH₃)₅Br]SO₄ will produce different ions when dissolved in water.
(ii) Linkage Isomerism: The isomerism in which a ligand can form linkage with metal through different atoms, e.g., nitro group can link to metal either through nitrogen (-NO₂) or through oxygen atom, e.g.,[Co(NH3)5 (NO2)]Cl2 and [Co(NH3)5 (ONO)Cl2
(iii) Coordination Isomerism: Coordination isomerism arises when both the cation and anion are coordination complexes, and they differ in the arrangement of ligands around the metal ions. For instance, [Co(NH₃)₆][Cr(C₂O₄)₃] and [Cr(NH₃)₆][Co(C₂O₄)₃].
(iv) Geometrical Isomerism: coordinated square planar complexes, cis- (when same In tetra are on opposite sides) isomers are possible depending of different trans-diamine dichloro platinum(II).
(v) Optical Isomerism: Optical isomers are those which are not superimposable on their mirror images. Complexes with coordination number six, having bidentate ligands provide examples of optical
9. How many geometrical isomers are possible in the following coordination entities?
(i) [Cr(C2O4)3] 3–
Ans: The two entities are represented as
(ii) [Co(NH3)3Cl3]
Ans: The two entities are represented as
10. Draw the structures of optical isomers of:
(i) [Cr(C2O4)3] 3–
Ans:
(ii) [PtCl2(en)2] 2+
Ans:
(iii) [Cr(NH3)2Cl2(en)]+
Ans:
11. Draw all the isomers (geometrical and optical) of: (i)
(i) [CoCl2(en)2]+
Ans:
(ii) [Co(NH3)Cl(en)2]2+
Ans:
(iii) [Co(NH3)2Cl2(en)]+
Ans:
Geometric isomers, also known as cis-trans isomers, differ in the spatial arrangement of groups around a rigid structure, like a double bond or a ring. When these isomers are unsymmetrical, they can exhibit optical isomerism, meaning they can form non-superimposable mirror images, called enantiomers, labelled as d (dextrorotatory) and l (levorotatory) forms.
12. Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will exhibit optical isomers?
Ans: There geometrical isomers of [Pt(NH3)(Br)(Cl)(py)]
Optical isomerism is not exhibited by this compound with a coordination number of 4 and square planar geometry, due to the presence of a horizontal plane of symmetry.
13. Aqueous copper sulphate solution (blue in colour) gives:
(i) a green precipitate with aqueous potassium fluoride and
Ans: Aqueous copper sulphate (blue) exists as [Cu(H₂O)₄]²⁺ in solution:
(ii) a bright green solution with aqueous potassium chloride. Explain these experimental results.
Ans: The complex [Cu(H₂O)₄]²⁺ is labile, meaning the water ligands (H₂O) are easily replaced by fluoride ions (F⁻) from potassium fluoride (KF) or chloride ions (Cl⁻) from potassium chloride (KCl).
(i) [Cu(H₂O)₄]2+(aq) + 4F−(aq) → [CuF₄]2− + 4H₂O
(Green ppt.)
(ii) [Cu(H₂O)₄]2+(aq)+4Cl−(aq) → [CuCl₄]2−(aq) + 4H₂O
(Bright green solution)
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 KCN(aq) is added to CuSO₄(aq), a complex called potassium tetracyanocuprate(II) is formed. As CN⁻ ions are strong ligands, the complex is highly stable. This stability is evident from its high stability constant (K = 2 × 10²⁷).
15. Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:
(i) [Fe(CN)6] 4–
Ans: The outer electronic configuration of iron (Z = 26) in its ground state is 3d⁶ 4s². In the +2 oxidation state, iron loses two 4s electrons, resulting in a 3d⁶ configuration.
The complex is diamagnetic with no unpaired electrons. In the +2 oxidation state, iron’s 3d⁶ electrons pair up, leaving two vacant 3d orbitals. Hybridization of these with one 4s and three 4p orbitals forms six equivalent dsp³ hybrid orbitals. The complex is octahedral and diamagnetic due to paired electrons.
(ii) [FeF6] 3–
Ans: In [FeF₆]³⁻, iron is in the +3 oxidation state with a coordination number of 6 and an octahedral geometry. The complex is paramagnetic due to five unpaired 3d electrons. Bonding involves the overlap of sp³d hybrid orbitals on Fe³⁺ with the lone pair orbitals of six fluoride ligands.
The Entity is strongly paramagnetic due to five unpaired electrons and is an outer orbital complex.
(iii) [Co(C2O4)3] 3–
Ans:
As didentate oxalate ions approach pairing of electrons in3d orbitals takes place.
Because all electrons are are paired. Hence diamagnetic.
(iv) [CoF6] 3–
Ans:
As there are four unpaired electrons, the complex exhibits paramagnetism.
16. Draw figure to show the splitting of d orbitals in an octahedral crystal field.
Ans: (i) Let us assume that the six ligands are positioned symmetrically along the cartesian axes, with metal atom at the origin. As the ligands approach first there is an increase in energy of d-orbitals relative to that of the free ion just as would be the case in a spherical field.
(ii) In an octahedral crystal field, the d orbitals along the axes (dₓ²-y² and dᵧ²) experience stronger repulsion compared to the d orbitals between the axes (dₓᵧ, dᵧz, dₓz).
17. What is a spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
Ans: A spectrochemical series is a list of ligands (attachments to a metal ion) and metal ions, arranged by the strength of the ligands and the oxidation state of the metal ions. The series is utilised to determine whether a coordination complex is high-spin or low-spin. The order is
I–<Br–< S <2- < CI– < F < -OH- < C2O42- < H2O < NH3 < CN– < CO
| Weak field coordination entity | Strong field ligand |
| They are formed when the crystal field stabilisation energy (Δo) in octahedral complexes is less than the energy required for an electron pairing in a single orbital (p). | They are formed when f the crystal field stabilisation energy (Δo) is greater than the p. |
| These complexes are also referred to as high-spin complexes. | These complexes are referred to as low-spin complexes. |
| These complexes are generally of paramagnetic nature. | They are mostly diamagnetic or less paramagnetic than weak field. |
| Never formed by CH- ligands. | Formed by ligands similar to CN. |
18. What is crystal field splitting energy? How does the magnitude of Do decide the actual configuration of d orbitals in a coordination entity?
Ans: Δ₀ is the energy-difference between the 11 and set of d-orbitals and is called crystal field eg stabilisation energy octahedral complex.
In a d¹ coordination entity, the single electron occupies t₂g orbitals. For d² and d³, electrons fill t₂g orbitals singly due to Hund’s rule. For d⁴ ions, if Δ₀ < P (pairing energy), a weak field, high spin situation occurs. The fourth electron enters an eg orbital, resulting in t₂g³ eg¹ configuration.
19. [Cr(NH3)6] 3+ is paramagnetic while [Ni(CN)4] 2– is diamagnetic. Explain why?
Ans: In The Complex [Cr(NH3)6]3+ chromium (Z = 24) has a ground-state configuration of 3d54s1. In the +3 oxidation state, it loses one 4s electron and two 3d electrons, resulting in a 3d3 configuration. The two vacant 3d orbitals, one 4s orbital, and three 4p orbitals hybridise to form six equivalent d2sp3 orbitals. Each NH3 molecule donates a pair of electrons to these orbitals. The complex adopts an octahedral geometry and has three unpaired electrons, making it paramagnetic.
[Ni(CN42-]: Outer electronic configuration of nickel (Z = 28) in ground state is 3d84s2 Nickel in this complex is in the +2 oxidation states. It achieves +2 oxidation state by the loss of the two 4s- electrons. The resulting N i2+ ion has an outer electronic configuration of 3d8. The two unpaired 3d electrons are forced to pair up.
The vacant 3d, 4s, and two 4p orbitals hybridise to form four equivalent dsp² orbitals. Four cyanide ions each donate an electron pair, resulting in a diamagnetic complex with no unpaired electrons.
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)2+, H₂O molecules are weak field ligands they do not cause electron pairing. As a result, the complex has two unpaired electrons.
Thus, d-d transitions occur in [Ni(H2O)6]2+ due to the absorption of red light, resulting in the emission of the complementary green colour, making the solution appear green. In [Ni(CN)4]2−, cyanide ions are strong field ligands, causing the pairing of all 3d electrons. As there are no unpaired electrons, no d-d transitions occur, making the solution colourless.
21. [Fe(CN)6] 4– and [Fe(H2O)6]2+ are of different colours in dilute solutions. Why?
Ans. In both the complex compounds, Fe is in +2 oxidation state with configuration 3d, 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. Then no unpaired electron is left. Due to this difference in the number of unpaired electrons, both complex ions have different colours.
22. Discuss the nature of bonding in metal carbonyls.
Ans: (i) Formation of an M ← C sigma bond occurs through the donation of the lone pair of electrons from the carbon atom (in CO) into a vacant orbital of the metal atom. This is a type of coordinate bonding.
(ii) Formation of an M → C π bond occurs through the transfer of electron density from the filled d-orbitals of the metal into the empty antibonding π* molecular orbitals of the CO ligand. This interaction is known as π-backbonding.
It is back donation or back bonding. It creates a synergic effect and strengthens the bond between CO and the metal.
The metal-carbon bond in metal carbonyls have both sigma (σ) and pi (π) characters. The metal- carbon o-bond is formed by the donation of lone pairs of electrons of the carbonyl carbon to a vacant orbital of the metal. The metal-carbon w-bond is formed by the donation of a pair of electrons from a filled d-orbital of metal into the vacant antibonding – molecular orbital (m²) of carbon monoxide. The metal to ligand bonding creates a synergic effect which strengthens the bond between CO and the metal.
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]
Ans: (+1) 3 + x + (-2) х 3 = 0
+3 + x – 6 =0
x = +3
Thus, Co is present as Co3+.
Co = [Ar]3d44s2
Co3+ = [Ar]3d6
Oxidation State: +3
Coordination Number: 6
d Orbital Occupation: 3d6(configuration: t62geg0)
(ii) cis-[CrCl2(en)2]Cl
Ans:(+1) х 2 + x + (-1) х 4 = 0
2 + x – 4 = 0
x = +2
Thus, Co is present as Co2+
Cr2+ = [Ar]3d7
Oxidation State: +3
Coordination Number: 6
d Orbital Occupation: 3d3(configuration: t52geg2)
(iii) (NH4)2[CoF4]
Ans: x + (0) 2 + (-1) x 2 + (-1)
x + 0 – 2 – 1 = 0
x = +3
Cr3+ = [Ar]3d3
Oxidation State: +2
Coordination Number: 4
d Orbital Occupation: 3d7(configuration: t32geg0)
(iv) [Mn(H2O)6]SO4
Ans: x +(o) 6 + (-2) = 0
X = + 2
Thus,
Mn2+ = [Ar]3d5
Oxidation State: +2
Coordination Number: 6
d Orbital Occupation: 3d5 (configuration: t32geg2)
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]3H2O
(ii) [Co(NH3)5Cl- ]Cl2
(iii) [CrCl3(py)3]
(iv) Cs[FeCl4]
(v) K4[Mn(CN)6]
Ans:
| Complex | IUPAC Name | Oxidation State | Electronic Configuration | Coordination Number | Stereochemistry | Magnetic Moment |
| K[Cr(H2O)2(C2O4)2]3H2O | Potassium diaquabis(oxalato)chromate(III) trihydrate | +3 | [Ar] 3d³ | 6 | Cis and trans isomers | 3.87 μ_B |
| [Co(NH3)5Cl- ]Cl2 | Pentaamminechlorocobalt(III) chloride | +3 | [Ar] 3d⁶ | 6 | No stereoisomerism | 4.90 μ_B |
| [CrCl3(py)3] | Trichlor Tri(pyridine)chromium(III)chromium(III) | +3 | [Ar] 3d³ | 6 | Fac and mer isomers | 3.87 μ_B |
| Cs[FeCl4] | Caesium tetrachloroferrate(III) | +3 | [Ar] 3d⁵ | 4 | No stereoisomerism | 5.92 μ_B |
| K4[Mn(CN)6] | Potassium hexacyanomanganate(III) | +3 | [Ar] 3d⁴ | 6 | Octahedral | Diamagnetic |
25. Explain the violet colour of the complex [Ti(H2O)6] 3+ on the basis of crystal field theory.
Ans: In [Ti(H2O)6]3+, titanium in the +3 oxidation state has a 3d1 configuration. In the octahedral field, 3d orbitals split into 2g and eg levels. This unpaired electron is excited from t2g level to eg level by absorbing yellow light and hence appears violet coloured.
26. What is meant by the chelate effect? Give an example.
Ans: When a didentate or a poly dentate ligan contains donor atoms positioned in such a way that when they coordinate with the central metal ion, a or 6-membered ring is formed, the effect is known a cheiate effect. Example
27. Discuss briefly giving an example in each case the role of coordination compounds in:
(i) biological systems.
Ans: (a) Haemoglobin the red blood cell which acts as oxygen carrier to different parts of the body is a complex of iron(II).
(b) A zinc(II) complex known as the enzyme CPA aids in the digestion of food.
(ii) medicinal chemistry. and
Ans: (a) The multidentate ligand EDTA (ethylene diamine tetraacetic acid) forms highly stable complexes with metal ions such as Ca²+ and Mg²+. This property is utilised to determine water hardness through a straightforward titration method using
(b) A confirmatory test for nickel consists in adding a solution of dimethyl glyoxime, when a scarlet-red coloured precipitate is formed. due to the formation of a chelate complex.
(iii) analytical chemistry
Ans: (a) Vitamin B_{12} used to prevent anaemia is a complex compound of cobalt.
(b) A confirmatory test for nickel involves adding a solution of dimethylglyoxime, which results in the formation of a scarlet-red precipitate. This occurs due to the creation of a stable chelate complex between nickel and dimethylglyoxime.
(iv) extraction/metallurgy of metals.
Ans: (a) Coordination compounds of silver and gold are used as the constituents of electroplating baths for the controlled delivery of Ag and Au ions, during electro-refining of these metals.
(b) Metals such as gold and silver are extracted by complex formation technique. Cyanide process is used for extraction of silver and gold from its ore.
Ag2S + 4NaCN → 2Na[Ag(CN)2] + Na2S
2Na[Ag(CN)2] + Zn → Na2[Zn(CN)4] + 2Ag
28. How many ions are produced from the complex Co(NH3)6Cl2 in solution?
(i) 6
(ii) 4
(iii) 3
(iv) 2
Ans: The current answer in (iii) 3.
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+
Ans: The oxidation states of metals in the complexes along with the electronic configuration are given
(a) Cr3+ : 3d6 configuration,
No. of unpaired electrons (n) = 3
(b) Fe2+ : 3d 8 configuration, n = 0
(c) Zn 2+ : 3d 10 configuration, n = 0
The ion with the highest magnetic moment is [Fe(H₂O)₆]²⁺. This is because Fe²⁺ has 4 unpaired electrons (3d⁶ configuration), resulting in a higher magnetic moment compared to Cr³⁺ and Zn²⁺, which have fewer unpaired electrons.
Thus, the correct answer is (b).
30. 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: (iii) [Fe(C2O4)3 ]3-
The most stable complex is [Fe(C₂O₄)₃]³⁻ because oxalate is a bidentate ligand and forms stable chelate rings, which greatly enhances the stability of the complex due to the chelate effect.
31. 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: In all these cases the given complexes, the metal ion is N i^ 2+ The increasing field strengths of the ligands present as per electro chemical series are in the order
H2O < NH3 < NO2−
Thus, the amount of crystal-field splitting observed will be in the following order:
[Ni(H2O)6]2+ < [Ni(NH3)6]2+ < [Ni(NO2)6]4−
Hence, the wavelengths of absorption in the visible region will be in the order:
[Ni(H2O)6]2+ > [Ni(NH3)6]2+ > [Ni(NO2)6]4−
