RBSE Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds

Rajasthan Board RBSE Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds Textbook Exercise Questions and Answers.

RBSE Class 12 Chemistry Solutions Chapter 9 Coordination Compounds

RBSE Class 12 Chemistry The Coordination Compounds InText Questions and Answers

Question 9.1. 
Write the formulae for the following coordination compounds:
(i) tetraammine diaquacobalt (III) chloride 
(ii) potassium tetracyanonickelate (II)
(iii) tris (enthane-1, 2-diamine) chromium (III) chloride
(iv) amminebormidochloridonitrito-N-platinate (II)
(v) dichloridobis (ethane-1, 2-diamine) platinum (IV) nitrate
(vi) iron (III) hexacyanoferrate (II)
Answer:
(i) [CO(NH₃)₄(H₂O)₂]Cl₃ 
(ii) K₂[Ni(CN)₄] 
(iii) [(Cr(en)₃]Cl₂ 
(iv) (Pt(NH₃)BrCl(NO₂)] 
(v) [PtCl(en)₂] (NO₃)₂ 
(vi) Fe [Fe(CN)₆]₃

Question 9. 2. 
Write the IUPAC names of the following coordination compounds:
(i) [(CO(NH₃)₆] 
(ii) [(CO(NH₃)₅Cl]Cl₂
(iii) K₃[Fe(CN)₆] 
(iv) K₃[Fe(C₂O₄)₃] 
(v) K₂[PdCl₄] 
(vi) [Pt(NH₃)₂Cl (NH₂CH₃)]Cl 
Answer:
(i) hexaamminecobalt (III) chloride 
(ii) pentaamminechloridocobalt (III) chloride 
(iii) potassium hexacyanoferrate (III) 
(iv) potassium trioxalatoferrate (III)
(v) potassium tetrachloridopalladate (II)
(vi) diamminechlorido (methylamine) platinum (II) chloride

 

Question 9.3. 
Indicate the types of isomerism exhibited by the following complexes and draw the structures of these isomers:
(i) K[Cr(H₂O)₂ (C₂O₄)₂] 
(ii) [CO(en)₃]Cl₃ 
(iii) (CO(NH₃)₅ (NO₂)] (NO₃)₂ 
(iv) [Pt(NH₃)(H₂O)Cl₂]
Answer:
(i) The complex K[Cr(H₂O)₂ (C₂O₄)₂] shows geometrical isomerism structure of its cis and trans isomers are as follows:

(ii) The complex [CO(en)₃]Cl₃ shows optical isomerism and it has two optical isomers.


(iii) The complex [CO(NH₃)₅ (NO₃)] (NO₃)₂ shows ionisation isomerism and linkage isomerism.
(a) Ionisation isomerism: [CO(NH₃)₅ (NO₂)] (NO₃)₂ and CO(NH₃)₅ (NO₃)] (NO₂) (NO₃)
(b) Linkage isomerism: [CO(NH₃)₅ (NO₂) (NO₃)₂ and [CO(NH₃)₅ (ONO) (NO₃)₂

(iv) The complex [Pt(NH₃)(H₂O)Cl₂] shows geometrical isomerism.

Question 9.4. 
Give evidence that [CO(NH₃),Cl]SO₄ and [CO(NH₃)₅SO₄]Cl are ionisation isomers.
Answer:
Ionisation isomers give different ions in aqueous solution. Therefore, they react differentially with different reagents.
[CO(NH₃)₅Cl]SO₄ gives sulphate ions in aqueous solution which gives the precipitate of BaSO₄ on adding of BaCl₂ 


 [CO(NH₃)₅SO₄]Cl gives chloride ions in aqueous solution which gives the precipate of Agcl on addition of AgNO₃.

But it does not react with BaCl₂ solution.


Question 9.5. 
Explain on the basis of valence bond theory that [Ni(CN)₄]^2- ion with square planar structure is diamagnetic and the [NiCl₄]^2- ion with tetrahedral geometry is paramagnetic.
Answer:
In both the complexes, Ni is in +2 oxidation state and has 3d⁸ configuration in the presence of CN^- which is a strong field ligand, the 3d electrons of Ni²⁺ ion pair up leaving no unpaired electron. Therefore it involves dsp² hybridisation and is diamagnetic in nature.
In the presence of Cl^- which is a weak field ligand, the 3d electrons do not pair up. Hence the complex [NiCl₄]^2- contains two unpaired electrons and is paramagnetic.
It can again explained as follow:
Magnetic behaviour of [Ni(CN)₄]^2-: It has CN^- a strong field ligand.
The complex [Ni(CN)₄]^2- has no unpaired e^-, so, it is diamagnetic in nature.
Magnetic behaviour of [NiC₄]^2- It has Cl^-, a weak field ligand.
However, it does not react with AgNO₃.
Ni (28) ground state

Ni (II) ground state

Ni (II) excited state (when ligands approach central metal ion to form coordinate bonds)

dsp³ hybrid orbitals of Nⁱ²⁺ ion

[CO(NH₃)₅SO₄]Cl gives chloride ions in aqueous solution which gives the precipitate of AgCl on addition of AgNO₃.

The complex [Ni(CN)₄]^2- has no unpaired e-, so, it is diamagnetic in nature.

Question 9.6. 
[NiCl₄]^2- is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral. Explain.
Answer:
In the complex [NiCl₄]^2- Ni is in +2 oxidation state and has 3d configuration. In the presence of Cl^- (a weak field ligand), 3d electrons do not get paired up. Hence the complex Ni contains two unpaired electrons and is paramgnetic.
In the complex [Ni(CO)₄] Ni is in zero oxidation state. In the presence of CO, (a strong field ligand), the unpaired d-electrons of Ni gets paired up leaving no unpaired electron Hence the complex [Ni(CO)₄] is diamagnetic.

Question .9.7. 
[Fe(H₂O)₆]³⁺  is strongly paramagnetic whereas [Fe(CN)₆]^3- is weakly paramagnetic. Explain?
Answer:
In both the complexes, Fe is in + 3 oxidation state and has 3d⁵ configuration. In the presence of H₂O (a week field ligand), 3d electrons do not pair up. The hybridisation is sp³d² forming an outer orbital complex containing five unpaired electrons. Hence it is strongly paramagnetic In the presence of CN, (a strong field ligands) the 3delectrons pair up leaving only one unpaired electron. The hybridisation in d²sp³ forming inner orbital complex. Hence the complex [Fe(CN)₆]^3- is weakly paramagnetic.

Question 9.8. 
Explain [CO(NH₃)₆]³⁺ is an inner orbital complex whereas [Ni(NH₃)₆]²⁺  is an outer orbital complex.
Answer:
In the complex (CO(NH₃)₆]³⁺, Co is in + 3 oxidation state and has d6 configuration. In the presence of NH₃ (a strong field ligand), 3d-electrons pair up leaving two 3d-orbitals vacant. Hence it involves d²sp³ -hybridisation and forms inner orbital complex. In the complex [Ni(NH₃)₆]²⁺, Ni is in + 2 oxidation state and has 3d⁸ configuration. In case of Ni²⁺ ion, two empty 3d orbitals are not available even after pairing. Hence Nilion undergoes spd hybridisation using its 4d-orbitals forming outer orbital complex.

Question 9.9. 
Predict the number of unpaired electrons in the square planar [Pt(CN)₄]^2- ion.
Answer:
The atomic number of platinum is 78 and its electronic configuration is 5d⁹ 6s¹. In this complex, the oxidation state of Pt is +2 and has 5d⁸ configuration. For square planer geometry, the hybridisation is dsp². Hence the unpaired electrons in 5d-orbitals pair up to make one d-orbital empty for dsp²-hybridisation. Thus there is no unpaired electron.
Pt (78) ground state

pt (II) ground state

Pt (II) excited state (when ligands approach central metal ion to form covrdinate bonds)

[Pt (CN)₄]^2-

Hence, number of unpaired electrons in [Pt (CN)₄]^2- is zero.

Question 9.10. 
The hexaaquomanganese (II) ion contains five unpaired electron, while the hexacyano ion contains only one unpaired electron. Explain using crystal field
theory.
Answer:
Mn²⁺ ion has 3d⁵-configuration. In the complex [Mn(H₂O)₆]²⁺, water in a weak field ligand. Therefore ∆₀ < P. Hence the configuration of Mn²⁺ will be (t_2g)³  (eg)² i.e., it has five unpaired electrons on the other hand in the complex [Mn(CN)₆]^4-, CN^- is a strong field ligand. Therefore ∆₀ > P. Hence the configuration of Mn²⁺ will be (t_2g)⁵ (eg)⁰. It indicates that it has one unpaired electron.

Question 9.11. 
Calculate the overall complex dissociation equilibrium constant for the [Cu(NH₃)₄]²⁺ ion given that B₄ for this complex is 2.1 x 10¹³.
Answer:
The overall dissociation constant is the reciprocal of the overall stability constant. Thus, overall
dissociation constant  \(\frac{1}{\beta_4}=\frac{1}{2 \cdot 1 \times 10^{13}}\) = 4.7  x 10^-14.
 

RBSE Class 12 Chemistry Coordination Compounds Textbook Questions and Answers

Question 9.1. 
Explain the bonding in coordination compounds in terms of Werner's postulates.
Answer:

1. In coordination compounds metals show two types of linkages (valencies) - primary and secondary.
2. The primary valencies are normally ionisable and are satisfied with negative ions.
3. The secondary valencies are non-ionisable. These are satisfied by neutral molecules or negative ions. The secondary valency is equal to the coordination number and is fixed for a metal.
4. The ions/groups bound by the secondary linkages to the metal have characteristic spatial arrangements corresponding to different coordination numbers.

e.g., Structure of  [COCl₂(NH₃)₄]⁺ Cl^- is as follows: CO has 3 as Primary valency and 4 as Secondary valency.

Question 9.2 
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 rati does not give the test of Cu²⁺ ion. Explain why.
Answer:
FeSO₄ solution mixed with (NH₄)₂SO₄ in 1 : 1 molar ratio gives a double salt. This double salt is FeSO₄ (NH₄)₂SO₄. 6H₂O (Mohr's salt) which dissociates in aqueous solution to give Fetion. Therefore, it gives test for Fe²⁺ ion.
FeSO₄ (NH₄)₂SO₄ → Fe²⁺ + 2NH₄⁺ + 2SO₄^2-
CuSO₄, solution mixed with aqueous ammonia in 1 : 4 molar ratio gives a coordination compound of formula [Cu(NH₃)₄]SO₄. The complex ion [Cu(NH₃)₄]²⁺ does not dissociate in aqueous solution to give Cu²⁺ ion. Hence it does not give test for Cu²⁺ ion.

Question 9.3. 
Explain with two examples each of the following:
Coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.
Answer:
(i) Coordination entity: A coordination entity constitutes a central metal atom or ion bonded to a fixed number of ions or molecules. Example: [CO(NH₃)₃Cl₃] is a coordination entity in which cobalt ion is surrounded by three ammonia molecules and three chloride ions. Other examples are:
(i) [Pt(NH₃)₂Cl₂]
(ii) [Fe(CN)₆]^4-
(iii) [CO(NH₃)₆]³⁺ etc.

(ii) Ligand: The ions or molecules bound to the central metal atom/ion in the coordination entity are called ligands. These may be simple ions as Cl^- and may be large molecule as NH₂CH₂CH₂NH₂ or even macromolecules like proteins.

(iii) Coordination number: It is the number of ligand donor atoms to which the metal is directly bonded. For example, [PtCl₆]^2- the coordination number of Pt is 6 while in [Ni(NH₃)₄]²⁺, the C.N. of Ni is 4.

(iv) Coordination polyhedron: The spatial arrangement of the ligand atoms which are directly attached to the central atom/ion defines a coordination polyhedron about the central atom. e.g., octahedral, square planar and tetrahedral polyhedron.

(v) Homoleptic complexes: Complexes in which a metal is bounded to only one kind of donor groups e.g, [CO(NH₃)₆]³⁺ [Cr(H₂O)₆]³⁺ etc.

(vi) Heteroleptic complexes : Complexes in which a metal is bound to more than one kind of donor groups.e.g. [CO(NH₃)₄Cl₂]⁺ [Cr(H₂O)₄Br₂]⁺ etc.

Question 9.4. 
What is meant by unidentate, bidentate and ambidentate ligands? Give two examples for each.
Answer:
(a) Monodentate ligands: The ligands which can coordinate to central metal ion through single donor atom are called monodentate ligands.
Examples: Cl^-, H₂O, NH₃, CO, NO₃^- NO₂^- etc.

(b) Bidentate ligands : The ligands which can coordinate to central metal ion through two donor atoms are called bidentate ligands.
Examples: NH₂CH₂CH₂NH₂ (ethane-1, 2-diamine) C₂O₄^2- (oxalate ion) etc.

(c) Ambidentate ligands: The unidentate or monodentate ligands which can coordinate to central metal ion through two different atoms are called ambidentate ligands.

Question 9.5. 
Specify the oxidation number of the metals in the following coordination entities:
(i) [CO(H₂O)(CN)(en)₂]²⁺
(ii) [PtCl₄]^2-
(iii) [Cr(NH₃)₃Cl₃]
(iv) [COBr₂(en)₂]⁺
(v) K₃[Fe(CN)₆]
Answer:
(i) [CO(H₂O)(CN)(en)₂]²⁺
x + 0 + (-1) + 2 x 0 = + 2
x - 1 = +2
x = +3

(ii) [PtCl₄]^2-
x + 4 x (-1) = -2
x - 4 = -2
x = +2

(iii) [Cr(NH₃)₃Cl₃]
x + 0 + 3 x (-1) = 0
x + 0 - 3 = 0
x = +3

(iv) [COBr₂(en)₂]⁺
x + 2 x (-1) + 2 x 0 = +1
x - 2 + 0 = +1
x = +3

(v) K₃[Fe(CN)₆]
3 x (+1) + x + 6 x (-1) = 0
3 + x - 6 = 0
x = +3

Question 9.6. 
Using IUPAC norms write the formulae for the following:
(i) tetrahydroxozincate (II) 
(ii) potassium tetrachloridopalladate(IT) 
(iii) diamminedichloridoplatinum (II) 
(iv) potassium tetracyanonickilate (IT) 
(v) pentaamminenitrito-O-cobalt(III) 
(vi) hexammine cobalt(III) sulphate 
(vii) potassium tri (oxalato) chromate (IIT) 
(viii) hexaammine platinum (IV) 
(ix) tetrabromidocuprate (II) 
(x) pentaammine nitrito-N-cobalt(IIT) 
Answer:
(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)]
(viii) [Pt(NH₃)₆]⁴⁺
(ix) [CUBr₄]^2-
(x) [CO(NH₃)₅(NO₂)]²⁺

Question 9. 7. 
Using IUPAC norms write the systematic names of the following:
(i) [CO(NH₃)₆]Cl₃ 
(ii) [Pt(NH₃)₂ Cl (NH₂CH₃)]Cl 
(iii) [Ti(H₂O)₆]³⁺ 
(iv) [CO(NH₃)₄ Cl(NO₂)]Cl 
(v) [Mn(H₂O)₆]²⁺ 
(vi) [NiCl₄]^2-
(vii) [Ni(NH₃)₆]Cl₂ 
(viii) [CO(en)₃]³⁺ 
(ix) [Ni(CO)₄]
Answer:
(i) hexaammine cobalt(III) chloride
(ii) diammine chlorido (methyl amine) platinum (II) chloride
(iii) hexaaquatitanium (III)ion 
(iv) tetraamminechlorido nitrito-N-cobalt(III) chlordie 
(v) hexaquamanganese (II) ion 
(vi) hexaamminenickel (II) chloride 
(vii) trans (ethane-1, 2-diamine) cobalt(III) ion 
(ix) tetracarbonylnickel (0)

Question 9.8. 
List various types of isomerism possible for coordination compounds, giving an example of each.
Answer:
Coordination compounds show two types of isomerism:

(1) Structural isomerism: It can be further classified as follows:
(i) Ionisation isomerism 
Examples:

1. [CO(NH₃)₅CI]Br 
2. [CO(NH₃)NO₂]NO₃ and [CO(NH₃)₅NO₃]NO₂

(II) Linkage isomerism Examples: 

1. [CO(NH₃)₅(NO₂)]Cl₂ and [CO(NH₃)₅ (ONO)]Cl₂ 
2. [CO(NH₃)₅ (CN)Cl₂ and [CO(NH₃)(NC)]Cl₂ 

(iii) Coordination isomerism 
Examples: 

1. [CO(NH₃)₆] [Cr(CN)₆] and [Cr(NH₃)₆] [CO(CN)₆] 
2. [Pt(NH₃)₄] [PtCl₄] and [Pt(NH₃)₃Cl] [PtCl₃(NH₃)] 

(iv) Solvate isomerism 
Example: 

1. [Cr(H₂O)₆]Cl and [Cr(H₂O)₆Cl] Cl₂.H₂O 
2. [CO(H₂O)₆]Br₃ and [CO(H₂O)₅Br] Br₂.H₂O

(2) Stereo isomerism: 
(i) Geometrical isomerism 
Example: 

1. [Pt(NH₃)₂ (Cl)₂] 
2. [CO(en)₂Cl₂] 

(ii) Optical isomerism 
Examples: cis [CO(en)₂Cl₂]

Question 9.9. 
How many geometrical isomers are possible in the following coodination entities:
(a) [Cr(C₂O₄)₃]₃ 
(b) [CO(NH₃)₃Cl₃]
Answer:
(a) [Cr(C₂O₄)³]₃ does not show geometrical inomerism.
(b) [CO(NH₃)₃Cl₃]

Question 9.10. 
Draw the structures of optical isomers of: 
(i) [Cr(C₂O₄)₃]^3-
(ii) pt[Cl(en)₂]²⁺
(iii) [Cr(NH₃)₂Cl₂(en)]⁺
Answer:
(i) [Cr(C₂O₄)₃]^3-

Question 9. 11. 
Draw all the isomers (geometrical and optical) of: 
(i) [COCl₂(en)₂]⁺
(ii) [CO(NH₃)Cl(en)₂]²⁺ 
(iii) [Co(NH₃)₂Cl(en)]⁺
Answer:
(i) [COCl₂(en)₂]⁺

(ii) [CO(NH₃)Cl(en)₂]²⁺ 

(iii) [Co(NH₃)₂Cl(en)]⁺

Question 9.12. 
Write all the geometrical isomers of [Pt(NH₃)(Br)(Cl) (Py)] and how many of these will exhibit optical isomers?
Answer:
This complex has three possible geometrical isomers:

These types of isomers do not show optical isomerism because square planar complexes rarely show optical isomerism

Question 9.13. 
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.
Answer:
Aqueous copper sulphate solution exists as [Cu(H₂O)₄]²⁺ H₂O. This solution has blue colour due to the presence of [Cu(H₂O)₄]²⁺ ions.
(i) When potassium fluoride is added, then F- ligands replaces weak H₂O ligands and form [CuF₄]^2- ions which is a green coloured complex. 

(ii) When potassium chloride is added, then Cl^- ligands replace weak H₂O ligands and form (Cuculions which have bright green colour. 

Question 9.14. 
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?
Answer:
When aqueous solution of KCN is added to an aqueous solution of copper sulphate then first of all cupric cyanide is formed which dissociates to give cuprous cyanide and cyanogen. Cuprous cyanide disselves in excess of KCN to give K₃[Cu(CN)₄] 
CUSO₄ + 2KCN → CU(CN)₂ + K₂SO₄ x 2
2CU(CN)₂ → CU₂(CN)₂ + (CN)₂
CU₂(CN)₂ + 6KCN → 2K₃[CU(CN)₄] + 2K₂SO₄ + (CN)₂↑
In this way, the coordination entitity formed is [Cu(CN)₄]^3- Since CN^- is a strong field ligand, it dissociates to give Cu²⁺ ion. Hence no precipitate is formed on passing H₂S.

Question 9.15. 
Discuss the nature of bonding in the following coordination entites on the basis of valence bond theory: 
(a) [Fe(CN)₆]^4-
(b) [FeF₆]^3- 
(c) [CO(C₂O₂)₃]^3-
(d) [COF₆]^3-
Answer:
(a) [Fe(CN)₆]^4-:
The atomic number of Fe is 26 and its electronic configuraton is 3d⁶4s². The oxidation state of Fe in this complex is +2. 

Since CN^- is a strong field ligand, hence 3d electrons are paired up to make twod-orbitals vacant for ds²sp³ hybridisation d²sp³ hybrid orbitals of Fe (II)

Geometry → Octahedral
Colour → colourless because there is no unpaired electrons
Magnetic Character → Diamagnetic because there is no unpaired electrons
Nature of the complex → Inner orbital complex

(b) [FeF₆]^3-: The atomic number of Fe is 26 and it has 3d⁶4s² configuration. The oxidation state of Fe in this complex is +3. Here, F- is a weak field ligand. Therefore, 3d electrons do not pair up. Fe(26) 3d 4s.


Hybridisation → sp³ d²
Geometry → Octahedral
Colour → Colourless because of unpaired electrons
Magnetic character → Paramagnetic due to the presence of unpaired electrons

\(\mu=\sqrt{n(n+2)} \text { B.M. } \mu=\sqrt{5(5+2)} \text { B.M. } \mu=\sqrt{35} \text { B.M. }\)

(c) [CO(C₂O₄)₃]^3-: The atomic number of cobalt is 27 and has electronic configuration 3d⁷4s². The oxidation state of CO in this complex is +3. Here, C₂O₄^2- is a strong field ligand. Therefore, 3d electrons are paired up and inner orbital (low spin) complex is formed.


Hybridisation: d2sp3
Geometry: Octrahedral
Colour: Colourless because there is no unpaired electrons
Magnetic character: Dimagnetic because there is no unpaired electrons
Nature of the complex: Inner orbital (low spin) complex

(d) [COF₆]^3-: The atomic number of cobalt is 27 and has configuration 3d⁷4s². The oxidation state of cobalt in this complex is +3. 


Hybridisation: sp³d² 
Geometry: Octahedral
Colour: Coloured due to the presence of unpaired electrons 
Magnetic character: Paramagnetic due to the presence of unpaired electrons
μ = μ = μ Nature of the complex: Outer orbital (high spin) complex 

Question 9.16. 
Draw figure to show the splitting of d-orbitals in an octahedral crystal field.
Answer:
When six ligands approach to central metal ion, they give degenerate d-orbitals of central metal ion split into two sets of degenrate orbitals having degenerate energies.
These two sets are tag and e orbitals. The difference in energies between these two sets are known as crystal field splitting energy (CSFE, ∆0)

Question 9.17. 
What is spectochemical series? Explain the differences bn a weak field ligand and a strong field ligand.
Answer:
The ligands can be arranged in a series in the order of increasing field strength. This series is called spectochemical series and is given below:
I^- < Br^- < SCN^- < CI^- < S^2- < F < OH^- < C₂O₄^2- < H₂O < NCS^- < EDTA^4- < NH₃ < en < CN^- < CO.

Question 9.18. 
What is crystal field splitting energy? How does the magnitude of A, decide the actual configuration of d-orbitals in a coordination entity?
Answer:
In an octahedral field, when ligands approach to the central metal ion to form coordination bonds, the five degenerated d-obitals of central metal ion splits in two sets of orbitals having different energies. This splitting of degenerated d-orbitals is called central field splitting and the energy difference between two sets of orbitals is called crystal field splitting energy. 
The configuration of d-orbitals in a coordination complex depends on the value of ∆_o < p, the fourth electron enters one of the orbitals hence configuration becomes (t_2g)z (eg)1 and high spin complex is formed. The ligands if ∆_o > p, fourth electron get paired up in one of the 12 orbitals hence configuration becomes (t_2g)4 (eg)⁰  and low spin complex is formed. The ligands for which pare called strong field ligands.

Question 9.19. 
[Cr(NH₃)₆]³⁺ is paramagnetic while [Ni(CN)₄]^2- is diamagnetic. Explain why?
Answer:
[Cr(NH₃)₆]³⁺: 
Atomic number of cobalt is 24 and electronic configuration is 3d⁵ns¹ and the oxidation state is +3.

NH₃ is a strong field ligand but two d-orbitals are already vacant hence pairing of electron is not required. Cr(III) Excited state 


Hybridisation → d²sp³ 
Geometry → Octahedral
Colour → Coloured due to the presence of unpaired electrons. 
Magnetic character → Paramagnetic due to presence of unparied electrons. 
Magnetic Moment 
= \(\sqrt{n(n+2)} B M\)

= \(\sqrt{3(3+2)} B M\)
= \(\sqrt{15}\)

(b) [Ni(CN)₄]^2-:
Atomic number of Ni is 28, electronic configuration is 3d⁸4s² and oxidation state is +2.
Ni (28): 3d⁸ 4s²


Hybridisation - dsp² 
Geometry - Square planar 
Colour - Colourless because there is no unpaired electrons. 
Magnetic character - Diamagnetic because there is no unpaired electron. 

Question 9.20. 
A solution of [Ni(H₂O)₄]²⁺ is green but a solution of [Ni(CN)₄]^2- is colourless, explain. 
Answer:
[Ni(H₂O)₄]²⁺: Atomic number of nickel is 28, electronic configurationis 3d⁸4s² and oxidation state is +2. 

H₂O is a weak field ligand. Hence it cannot paired up the electrons in 3d orbitals so 4d-orbitals take part in hybridisation So it is green.

Since there are two unpaired electrons, hence d-d transitions would occur so colour of solution become green. 
[Ni(CN)₄]^2-: Atomic number of Ni is 28, electronic configuration in 3d⁸4s² and oxidation state is +2.

Since there is no unpaired electron, no d-d transitions would occur and hence the complex is colourless.

Question 9.21. 
[Fe(CN)₆]^4- and [Fe(H₂O)₆]²⁺ are of different colours in dilute solutions, Why? 
Answer:
In both the complexes, the oxidations state of Fe is +2 and electronic configuration of Fe²⁺ is 3d⁶4s^o. In the presence of CN^- (a strong field ligand), 3d electrons get paired up leaving one unpaired electron in the complex [Fe(CN)₄]^4- but in the presence of H₂O (a weak field ligand) 3d electrons do not get paired up and therefore there are five unpaired electrons in the complex [Fe(H₂O)]²⁺ Since both the complexes have different number of unpaired electrons so different region of visible light will be absorbed and therefore transmitted colour would be different i.e., they will give different colours in their solutions.

Question 9.22. 
Discuss the nature of bonding in metal carbonyls.
Answer:
The metal carbon bond in metal carbonyl possesses both s-and p-character. The M-C σ - bond is formed by the donation of lone pair of electrons on the carbonyl carbon into a empty orbital of the metal. The M-C- π - bond is formed by the donation of a pair of electrons from a filled d-orbitals of metal into vacent antibonding π* orbital of carbon mono-oxide. The metal to ligand bonding creates a synergic effect which strengthens the bond between CO and the metal.

Question 9.23. 
Give the oxidation state, d-orbital configuration and coordination number of the central metal ion in the following complexes: 
(i) K₃[CO(C₂O₄)₃]
(ii) Cis-[Cr(Cn)₂Cl₂]Cl
(iii) (NH₄)₂ [COF₄]
(iv) [Mn(H₂O)₆]SO₄ 
Answer:
(i) K₃[CO(C₂O₄)₃]
= x + 3 (-2) = -3
x = +3

(ii) Cis-[Cr(Cn)₂Cl₂]Cl
= x + 0 + 2(-1)
= +1
x = +3

(iii) (NH₄)₂ [COF₄]
= x + 4(-1) = -2
x = +2

(iv) [Mn(H₂O)₆]SO₄ 
x + 0 = +2
x = +2



Question 9.24. 
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) [CrCl₃(Py)₃] 
(iii) [CO(NH₃)₅Cl]Cl₂ 
(iv) Cs(FeCl₄) 
(v) K₄(Mn(CN)₆] 
Answer:
(i) K[Cr(H₂O)₂ (C₂O₄)₂].3H₂O
IUPAC Name - Potassium diaquadioxalatochromate (III) trihydrate 
Oxidation State - 
K[Cr(H₂O)₂ (C₂O₄)₂]
+ 1 + x + 2 x 0 + 2 x (-2)' = 0 
1 + x - 4 = 0
x = + 3 
Coordination Number - 6
Stereochemistry - Since the coordination number is 6, geometry is octahedral.
Magnetic moment and electronic configuration Cr(24); 4s¹ 3d⁵
Cr³⁺ : 4s⁰ 3d⁹; [(t_2g)³ (eg)^o] 
Unpaired e^- = 3
Three unpaired electrons hence paramagnetic. Magnetic moment
\(\begin{aligned} (\mu) &=\sqrt{n(n+2)} B M \\ &=\sqrt{3(3+2)} B M \\ &=\sqrt{3 \times 5} B M \\ &=\sqrt{15} \end{aligned}\)

= 3.18 BM

(ii) [CrCl₃(Py)₃] 
IUPAC Name:Trichloridotripyridinechromium (III) 
Oxidation number: 
[CrCl₃(Py)₃] 
x + 3 x (-1) + 3 x 0 = 0
X - 3 = 0
Y = +3 
Coordination Number - 6
Stereochemistry - Since the coordination number is 6, geometry is octahedral 
Magnetic moment and electronic configuration:
Cr(24) : 4s¹ 3d⁵
Cr³⁺ (21) : 4s° 3d³ [(t2g)³ (eg)⁰]
d²sp³ hybridization 
Three unpaired electrons hence paramagnetic 
Magnetic moment 
= \(\sqrt{n(n+2)} B M\)

\(\begin{aligned} &=\sqrt{3(3+2)} B M \\ &=\sqrt{3 \times 5} B M \\ &=\sqrt{15} B M \end{aligned}\)
= 3.87 BM

(iii) [CO(NH₃)₅Cl]Cl₂
IUPAC Name - Pentaamminechloridocobalt(III) chloride
Oxidation State - [CO(NH₃)₅Cl]Cl₂
x + 5 x (0) + (-1) + 2 x (-1) = 0
x - 1 - 2 = 0
x = +3
Coodination number - 6
Stereochemistry - Since the cordination number is 6, geometry of the complex is octahedral. 
Magnetic moment and electronic configuration
CO(27): 4s² 3d⁷
CO³⁺ (24): 4S⁰ 3d⁶ [(t2g)⁶ (eg)⁰]
d²sp³ hybridization 
No unpaired electrons hence diamagnetic
Magnetic moment (μ) = 0

(iv) Cs[FeCl₄]
IUPAC name-Caesium tetrachloridoferrate (III) 
Oxidanitn state: Cs[FeCl₄] 
+ 1 + x + 4 x (-1) = 0
X - 3 = 0
x = +3 
Cordiantion number - 4 
Stereochemistry - Tetrahedral 
Magnetic moment and electronic configuration
Fe (26): 4s² 3d⁶
Fe³⁺ (23): 4s^o 3d [(eg)² (t2g)³]
sp³ hybridisation 
Five unpaired electrons hence paramagnetic 
Magnetic moment (u) = \(\sqrt{n(n+2)}\)
\(\begin{aligned} &=\sqrt{5(5+2)} \\ &=\sqrt{5 \times 7} \\ &=\sqrt{35} \mathrm{BM} \end{aligned}\)
= 5.92 B M

(v) K₄[Mn(CN)₆]:
IUPAC Name-Potassium hexacyanomanganate (II) 
Oxidation State: K₄[Mn(CN)₆] 
4 x (+1) + x + 6 x (-1) = 0
4 = N -6 = 0
N = +2
Coordination Number - 6 
Stereochemistry - Octahedral 
Magnetic moment and electronic configuration
Mn (25): 3d⁵ 4s² 
Mn²⁺ (23): 3d⁶ 4s⁰  [(t2g)⁵ (eg)⁰]
d²sp³ hybridisation 
One unpaired electron hence paramagnetic. 
Magnetic moment  (u) = \(\sqrt{n(n+2)}\)
\(\begin{aligned} &=\sqrt{5(5+2)} \\ &=\sqrt{5 \times 7} \\ &=\sqrt{35} \mathrm{BM} \end{aligned}\)

= 1.732 B M

Question 9.25 
What is meant by stability of a coordination compound in solution? State the factors which govern stability of complexes.
Answer:
Mostly the complex ions are highly stable, the reaction between a metal ions and ligands is considered as a Lewis acid-base reaction in which the ligands act as Lewis bases and the central metal ion acts as a Lewis acid. If the interaction between central atom as ion and ligands is strong, a stable complexion is formed. The stability of the complex in solution refers to the degree of association between metal ion and ligands in Volved in the state of equibrum Ine magnitude of the stability as formation) equilibrium constant for the association quantitatively express the stability. If we consider the fol lowing reaction
\(\mathrm{M}+4 \mathrm{~L} \rightleftharpoons \mathrm{ML}_4\)
Then stability constant is given by
\(k=\frac{\left[\mathrm{ML}_4\right]}{[\mathrm{M}][\mathrm{L}]^4}\)
Higher the value of stability contant, greater the amount of ML4 and lesser the amount of free metal ions in solution hence complex will be stable.
Factors affecting the stability of complexes:
(1) In Analytical Chemistry: Coordiantion compounds are used in many qualitative and quantitative chemical analysis. Some metal ions coordinate with a number of ligands (especially chelating ligand), to form columned complexes which form the basis for their detection and estimation by classicial and instrumental methods of analysis. For example,
(i) The presence of CU²⁺ ions are tested by adding ammonia to the solution salt when a blue coloured complex is formed.

(ii) The presence of Fe³⁺ ions are tested by adding potassium ferrocyanide solution to the salt solution when ferriferrocyanide or prussian blue complex is formed.

(iii) Fe³⁺ ions are also detected by adding potassium thioyanate solution to the salt solution when a red coloured complex is formed.

Question 9.26. 
What is mornt by chelate effect? Give an example.
Answer:
When a di - or polydentate ligand coordinates to the central metal atom or ion using its to as more donor atoms, then a ring like structure is formed which increase the stability of the complex. This enhanced stability of the complexes containing bidentate of polydentate ligands (chelating ligands) in called chelate effect :
Example: [Cu(CN)₂]²⁺

Question 9.27. 
Discuss briefly giving an example in each can the role of coordinatioin compounds in (i) biological systems (ii) analytical chemistry (iii) medicinal chemistry (iv) extraction/metallurgy of metals.
Answer:

Factors affecting the stability of complexes:
(1) In Analytical Chemistry: Coordiantion compounds are used in many qualitative and quantitative chemical analysis. Some metal ions coordinate with a number of ligands (especially chelating ligand), to form columned complexes which form the basis for their detection and estimation by classicial and instrumental methods of analysis. For example,
(i) The presence of CU²⁺ ions are tested by adding ammonia to the solution salt when a blue coloured complex is formed.

(ii) The presence of Fe³⁺ ions are tested by adding potassium ferrocyanide solution to the salt solution when ferriferrocyanide or prussian blue complex is formed.

(iii) Fe³⁺ ions are also detected by adding potassium thioyanate solution to the salt solution when a red coloured complex is formed.

Question 9.28. 
How many ions are produced from the complex CO(NH₃)₆Cl₂ in solution
(i) 6 
(ii) 4
(iii) 3
(iv) 2 
Answer:
Oxidation number of cobalt = 6 
The complex [CO(NH₃)₆] Cl₂ ionises as
\(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_2 \stackrel{e q}{\longrightarrow}\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{2+}+2 \mathrm{Cl}^{-}\)
Hence three ions are produced. 
Therefore option (ii) is correct.

Question 9.29. 
Amongst the following ions which one has highest magnetic moment?
(i) [Cr(H₂O)₆]³⁺
(ii) [Fe(H₂O)₆]²⁺
(iii) [Zn(H₂O)₆]²⁺
Here oxidaiton states are - Cr(III), Fe(II), Zn(II)
Electronic configuration of Cr³⁺ = 3d [(t_2g)³ (eg)⁰] 
Number of unpaired electrons = 3 
Electronic configuration of Fe
Electronic configuration of Zn²⁺ = 3d¹⁰ 
Number of unpaired electron = 0
Magnetic moment μ = \(\sqrt{n(n+2)}\)
Since n = 4 is highest, the magnetic moment of option (iii) will be highest.

Question 9.30. 
The oxidation number of cobalt in K[CO(CO)₄] 
(i) +1 
(ii) +3 
(iii) -1 
(iv) -3 
Answer:
K[CO(CO)₄] 
+ 1 + x + 4  x  (0) = 0
1 + x = 0
x = -1 
Hence option (iii) is correct.

Question 9.31. 
Among the following the most stable complex is:
(i) [Fe(H₂O)₆]²⁺
(ii) [Fe(NH₃)₆]³⁺ 
(iii) [Fe(C₂O₄)₃]^3- 
(iv) [FeCl₆]^3-
Answer:
In each of the above complexes, the oxidation state of Fe is +3, Since C₂O₄^2- is a chelating ligand and it forms chelate ring, therefore option (iii) is most stable complex.

Question 9.32. 
What will be the correct order of the wavelengths of absorption in the visible region for the following:
[Ni(NO₂)₆]^4-, [Ni(NH₃)₆]²⁺, [Ni(H₂O)₆]²⁺ 
Answer:
Since the metal ion is fixed, the increasing order of field strength of the ligands from the spectrochemical series is H₂O < NH₃ < NO. Thus, the order of energies absorded for excitation is as follows:
[Ni(NO₂)₆]^4-, [Ni(NH₃)₆]²⁺, [Ni(H₂O)₆]²⁺ 
As E  = \(\frac{h c}{\lambda}\) the wavelengths absorbed will be in the opposite order.