Class 12 Chemistry Chapter 5 Coordination Compounds ncert solutions

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Coordination Compounds Class 12 ncert solutions: Chemistry Class 12 Chapter 5 exercise solutions

TextbookNCERT
ClassClass 12
SubjectChemistry
ChapterChapter 5
Chapter NameCoordination Compounds ncert solutions
CategoryNcert Solutions
MediumEnglish

Are you looking for Class 12 Chemistry Chapter 5 Coordination Compounds ncert solutions? Now you can download Ncert Chemistry Class 12 Chapter 5 exercise solutions pdf from here.

Question 5.1: Explain the bonding in coordination compounds in terms of Werner’s postulates.

Solution 5.1: Werner’s postulates explain the bonding in coordination compounds as follows:

( a ) A metal shows two kinds of valencies viz primary valency and secondary valency. Negative ions satisfy primary valencies, and secondary valencies are filled by both neutral ions and negative ions.

( b ) A metal ion has a fixed amount of secondary valencies about the central atom. These valencies also orient themselves in a particular direction in the space provided to the definite geometry of the coordination compound.

( c ) Primary valencies are usually ionizable, while secondary valencies are non-ionizable.

Question 5.2: 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.

Solution 5.2: When FeSO₄ is mixed with (NH₄)₂SO₄ in a 1:1 molar ratio, the solution still gives a test for Fe²⁺ ions because no strong complex is formed. The Fe²⁺ ions remain free in solution.

In contrast, when CuSO₄ is mixed with ammonia (NH₃) in a 1:4 molar ratio, the Cu²⁺ ions form a stable complex, \([Cu(NH₃)_4]^{2+}\). This coordination complex prevents the free Cu²⁺ ions from being detected, which is why the solution does not give a test for Cu²⁺ ions.

Question 5.3: Explain with two examples each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.

Solution 5.3:

1. Coordination Entity: A coordination entity consists of a central metal atom/ion bonded to ligands.

  • Example 1: \([Co(NH₃)_6]^{3+}\)
  • Example 2: \([Fe(CN)_6]^{4-}\)

2. Ligand: A ligand is an ion or molecule that donates a pair of electrons to the central metal atom/ion.

  • Example 1: \(NH_3\) (Ammonia)
  • Example 2: \(Cl^-\) (Chloride)

3. Coordination Number: The coordination number is the number of ligand donor atoms attached to the central metal atom/ion.

  • Example 1: 6 in \([Co(NH₃)_6]^{3+}\)
  • Example 2: 4 in \([PtCl_4]^{2-}\)

4. Coordination Polyhedron: The spatial arrangement of ligands around the central metal atom/ion forms a specific shape, known as the coordination polyhedron.

  • Example 1: Octahedral in \([Co(NH₃)_6]^{3+}\)
  • Example 2: Square planar in \([PtCl_4]^{2-}\)

5. Homoleptic: A complex where the metal is bonded to only one type of ligand.

  • Example 1: \([Co(NH₃)_6]^{3+}\)  (All ligands are ammonia)
  • Example 2: \([Fe(CN)_6]^{4-}\) (All ligands are carbonyl)

6. Heteroleptic: A complex where the metal is bonded to more than one type of ligand.

  • Example 1: \([Co(NH₃)_4Cl_2]^+\) (Ammonia and chloride ligands)
  • Example 2: \([Pt(NH₃)_2Cl_2]\) (Ammonia and chloride ligands)

Question 5.4: What is meant by unidentate, didentate and ambidentate ligands? Give two examples for each.

Solution 5.4:

1. Unidentate Ligands: A unidentate ligand has only one donor atom that can bind to a central metal ion. This means the ligand forms a single coordinate bond with the metal.

Examples:

  • Ammonia (NH₃): The nitrogen atom has a lone pair of electrons and acts as a donor atom.
  • Chloride ion (Cl⁻): The chlorine atom donates a lone pair of electrons to the metal.

2. Didentate Ligands: A didentate ligand has two donor atoms that can simultaneously bind to a central metal ion. The ligand forms two coordinate bonds with the metal, often resulting in the formation of a ring structure called a chelate.

Examples:

  • Ethylenediamine (en): This molecule has two nitrogen atoms, each of which has a lone pair, allowing it to form two bonds with the metal.
  • Oxalate ion (C₂O₄²⁻): The oxalate ion has two oxygen atoms, each of which can donate electrons to the metal.

3. Ambidentate Ligands: An ambidentate ligand can coordinate to a metal ion through two different atoms, but only one at a time. These ligands have more than one possible donor atom, but they can bind through only one donor atom at a time.

Examples:

  • Nitrite ion (NO₂⁻): The nitrite ion can bind through either the nitrogen atom (as in nitro complexes) or an oxygen atom (as in nitrito complexes).
  • Thiocyanate ion (SCN⁻): This ion can coordinate through the sulfur atom (S-bonded) or through the nitrogen atom (N-bonded).

Question 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) [PtCl4]2–, (iv) K3[Fe(CN)6], (v) [Cr(NH3)3Cl3 

Solution 5.5:

  • (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

Question 5.6: Using IUPAC norms, write the formulas for the following:
(i) Tetrahydroxidozincate(II)
(ii) Potassium tetrachloridopalladate(II)
(iii) Diamminedichloridoplatinum(II)
(iv) Potassium tetracyanonickelate(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 5.6:

  • ( i ) [Zn(OH)4]2 –
  • ( ii ) K2[ Pd Cl4]
  • ( iii ) [ Pt ( NH3)2Cl2]
  • ( iv ) K2[ Ni(CN )4]
  • ( v ) [ Co (NO2) ( NH3)52+
  • ( vi) [ Co( NH3)6]2 (SO4)3
  • ( vii ) K3 [ Cr ( C2O4)3]
  • ( viii ) [ Pt (NH3)64+
  • ( ix ) [ Cu (Br)42−
  • ( x ) [Co ( ONO )( NH3)52+

Question 5.7: Using IUPAC norms, write the systematic names of the following:
(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 5.7:

( i ) Hexaamminecobalt(III) chloride

( ii ) Diamminechlorido(methylamine) platinum(II) chloride

( iii ) Hexaquatitanium(III) ion

( iv ) Tetraamminichloridonitrito-N-Cobalt(III) chloride

( v ) Hexaquamanganese(II) ion

( vi ) Tetrachloridonickelate(II) ion

( vii ) Hexaamminenickel(II) chloride

( viii ) Tris(ethane-1, 2-diamine) cobalt(III) ion

( ix ) Tetracarbonylnickel(0)

Question 5.8: List various types of isomerism possible for coordination compounds, giving an example of each.

Solution 5.8: 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: 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

ii. 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]

iii. 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.

iv .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

(2) Stereoisomerism: It is of 2 types:

( i ) Geometrical isomerism:

( ii ) Optical isomerism:

Question 5.9: How many geometrical isomers are possible in the following coordination entities?
(i) [Cr(C2O4)3]3–, (ii) [Co(NH3)3Cl3]

Solution 5.9:

(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.

Question 5.10: Draw the structures of optical isomers of:
(i) [Cr(C2O4)3]3–, (ii) [PtCl2(en)2]2+, (iii) [Cr(NH3)2Cl2(en)]+

Solution 5.10: (i) [Cr(C2O4)3]3−

NCERT Solutions for Class 12 Chemistry Chapter 9

(ii) [PtCl2(en)2]2+

NCERT Solutions for Class 12 Chemistry Chapter 9

(iii) [Cr(NH3)2Cl2(en)]+

NCERT Solutions for Class 12 Chemistry Chapter 9

Question 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 5.11: (i) [CoCl2(en)2]+ Geometrical isomerism

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Optical isomerism Since only cis isomer is optically active, it shows optical isomerism.

NCERT Solutions for Class 12 Chemistry Chapter 9

In total, three isomers are possible.

(ii) [Co(NH3)Cl(en)2]2+

Geometrical isomerism

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Optical isomerism Since only cis isomer is optically active, it shows optical isomerism.

NCERT Solutions for Class 12 Chemistry Chapter 9

(iii) [Co(NH3)2Cl2(en)]+

NCERT Solutions for Class 12 Chemistry Chapter 9

Question 5.12: Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)], and how many of these will exhibit optical isomers?

Solution 5.12: 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.

Question 5.13: Q13. 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 5.13: 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

Question 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 5.14: When excess aqueous potassium cyanide (KCN) is added to an aqueous solution of copper sulfate (CuSO₄), a complex coordination entity is formed. The cyanide ions (CN⁻) act as ligands and coordinate with copper ions (Cu²⁺) to form the stable complex ion \([Cu(CN)_4]^{2-}\). This complex is known as tetracyanocuprate(II) and is soluble in water. The reaction can be represented as follows:

\(Cu^{2+} (aq) + 4CN^{-} (aq) \rightarrow\)\( [Cu(CN)_4]^{2-} (aq)\)

Formation of Coordination Entity

The formation of the tetracyanocuprate(II) complex occurs due to the strong field strength of the CN⁻ ligand, which effectively stabilizes the copper ion in solution. The coordination number of copper in this complex is 4, resulting in a tetrahedral geometry.

Absence of Precipitate with H₂S

When hydrogen sulfide (H₂S) gas is passed through the solution containing \([Cu(CN)_4]^{2-}\), no precipitate of copper(II) sulfide (CuS) is formed. This can be attributed to the following reasons:

  1. Complex Stability: The cyanide complex \([Cu(CN)_4]^{2-}\) is highly stable and prevents the copper ions from being released back into the solution as free Cu²⁺ ions, which are necessary for the formation of copper sulfide.
  2. Low Solubility Product: Copper sulfide has a low solubility product (Ksp), which means it would typically precipitate when Cu²⁺ ions are present in solution. However, because the concentration of free Cu²⁺ ions is effectively very low due to the formation of the cyanide complex, the conditions necessary for the precipitation of CuS are not met.
  3. Ligand Competition: CN⁻ is a stronger ligand than sulfide ions (S²⁻), and it preferentially binds to Cu²⁺ over the sulfide ions. Therefore, the presence of CN⁻ ions inhibits the reaction that would lead to the formation of CuS.

Question 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 5.15: (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:

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As CN− is a strong field ligand, it causes the pairing of the unpaired 3d electrons.

NCERT Solutions for Class 12 Chemistry Chapter 9

Since there are six ligands around the central metal ion, the most feasible hybridization is d2sp3.

d2sp3 hybridized orbitals of Fe2+ are:

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6 electron pairs from CN− ions occupy the six hybrid d2sp3orbitals.

Then,

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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:

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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:

NCERT Solutions for Class 12 Chemistry Chapter 9

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:

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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+:

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The 6 electron pairs from the 3 oxalate ions (oxalate anion is a bidentate ligand) occupy these sp3d2 orbitals.

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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:

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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:

NCERT Solutions for Class 12 Chemistry Chapter 9

Hence, the geometry of the complex is octahedral and paramagnetic.

Question 5.16: Draw figure to show the splitting of d orbitals in an octahedral crystal field

Solution 5.16:

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The splitting of the d orbitals in an octahedral field takes place in such a way that dx2y2, 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.

Question 5.17: What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.

Solution 5.17: A spectrochemical series is the arrangement of common ligands in the increasing order of their crystal-field splitting energy (CFSE) values. The ligands present on the R.H.S of the series are strong field ligands while that on the L.H.S are weak field ligands. Also, strong field ligands cause higher splitting in the d orbitals than weak field ligands.

I− < Br− < S2− < SCN− < Cl−< N3 < F− < OH− < C2O42− ∼ H2O < NCS− ∼ H− < CN− < NH3 < en ∼ SO32− < NO2− < phen < CO

Question 5.18: What is crystal field splitting energy? How does the magnitude of ∆o decide the actual configuration of d orbitals in a coordination entity?

Solution 5.18: Crystal-field splitting energy is the difference in the energy between the two levels ( i.e., t2g and eg ) that have split from a degenerated d orbital because of the presence of a ligand. It is symbolised as  ∆o.
Once the orbitals split up, electrons start filling the vacant spaces. An electron each goes into the three t2g orbitals, the fourth electron, however, can enter either of the two orbitals:

( 1 ) It can go to the eg orbital ( producing t2g3eg 1 like electronic configuration), or
( 2 ) it can go to the t2g orbitals ( producing t2g4eg 0 like electronic configuration).

This filling of the fourth electron is based on the energy level of ∆o. If a ligand has an ∆o value smaller than the pairing energy, then the fourth electron enters the eg orbital. However, if the value of ∆o is greater than the value of pairing energy, the electron enters t2gorbital.

Question 5.19: [Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2– is diamagnetic. Explain why.

Solution 5.19: 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+

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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+:

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CN− is a strong field ligand. It causes the pairing of the 3d orbital electrons. Then, Ni2+ undergoes dsp2 hybridization.

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As there are no unpaired electrons, it is diamagnetic.

Question 5.20: A solution of [Ni(H2O)6]2+ is green, but a solution of [Ni(CN)4]2– is colourless. Explain.

Solution 5.20: [ Ni ( H2O)2+ consists of Ni+2 ion with 3d8 electronic configuration. In this configuration, there are two unpaired electrons which cannot pair up because H2O is a weak ligand. Thus, the d–d transition absorbs the incoming light and emits a green light, thereby giving a  green colour to the solution.[ Ni ( CN)42- consists of Ni+2 ion with 3d8 electronic configuration. But CN  is present here as a strong ligand, and in its presence, the unpaired electrons pair up. Thus, there is no d–d  transition, so there is no colour.

Question 5.21: [Fe(CN)6]4– and [Fe(H2O)6]2+ are of different colours in dilute solutions. Why?

Solution 5.21: The colour of a particular coordination compound depends on the magnitude of the crystal-field splitting energy, Δ. This CFSE in turn depends on the nature of the ligand. In case of [Fe(CN)6]4− and [Fe(H2O)6]2+, the colour differs because there is a difference in the CFSE. Now, CN− is a strong field ligand having a higher CFSE value as compared to the CFSE value of water. This means that the absorption of energy for the intra d-d transition also differs. Hence, the transmitted colour also differs.

Question 5.22: Discuss the nature of bonding in metal carbonyls.

Solution 5.22: The metal-carbon bonds in metal carbonyls have both σ and π characters. A σ bond is formed when the carbonyl carbon donates a lone pair of electrons to the vacant orbital of the metal. A π bond is formed by the donation of a pair of electrons from the filled metal d orbital into the vacant anti-bonding π* orbital (also known as back bonding of the carbonyl group). The σ bond strengthens the π bond and vice-versa. Thus, a synergic effect is created due to this metal-ligand bonding. This synergic effect strengthens the bond between CO and the metal.

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Question 5.23: 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

Solution 5.23: (i) K3[Co(C2O4)3]

The central metal ion is Co.

Its coordination number is 6.

The oxidation state can be given as:

x − 6 = −3

x = + 3

The d orbital occupation for Co3+ is t2g6eg 0.

( ii ) cis – [ Cr( en)2 Cl2 ]Cl

The central metal ion is Cr.

The coordination number is 6.

The oxidation state can be given as:

x + 2(0) + 2(−1) = +1

x − 2 = +1

x = +3

The d orbital occupation for Cr3+ is t2g3.

(iii) ( NH4)2[ CoF4 ]

The central metal ion is Co.

The coordination number is 4.

The oxidation state can be given as:

x − 4 = −2

x = + 2

The d orbital occupation for Co2+ is eg4t2g3.

( iv ) [ Mn(H2O)]SO4

The central metal ion is Mn.

The coordination number is 6.

The oxidation state can be given as:

x + 0 = +2

x = +2

The d orbital occupation for Mn is t2g3 eg2.

Question 5.24: Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also, give the 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]

Solution 5.24: (i) Potassium diaquadioxalatochromate (III) trihydrate.

Oxidation state of chromium = 3

Electronic configuration: 3d3: t2g3

Coordination number = 6

Shape: octahedral

Stereochemistry:

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(ii) [Co(NH3)5Cl]Cl2

IUPAC name: Pentaamminechloridocobalt(III) chloride

Oxidation state of Co = +3

Coordination number = 6

Shape: octahedral.

Electronic configuration: d6: t2g6.

Stereochemistry:

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Magnetic Moment = 0

(iii) CrCl3(py)3

IUPAC name: Trichloridotripyridinechromium (III)

Oxidation state of chromium = +3

Electronic configuration for d3 = t2g3

Coordination number = 6

Shape: octahedral.

Stereochemistry:

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Both isomers are optically active. Therefore, a total of 4 isomers exist.

Magnetic moment, μ = √n(n + 2)

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(iv) Cs[FeCl4]

IUPAC name: Caesium tetrachloroferrate (III)

Oxidation state of Fe = +3

Electronic configuration of d6 = eg2t2g3

Coordination number = 4

Shape: tetrahedral

Stereochemistry: optically inactive

Magnetic moment:

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(v) K4[Mn(CN)6]

Potassium hexacyanomanganate(II)

Oxidation state of manganese = +2

Electronic configuration: d5+: t2g5

Coordination number = 6

Shape: octahedral.

Streochemistry: optically inactive

Magnetic moment, μ = √n(n + 2)

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Question 5.25: Explain the violet colour of the complex [Ti(H2O)6]3+ on the basis of crystal field theory

Solution 5.25: The stability of a coordination compound in a solution is the degree/level of association among the species involved in a state of equilibrium.
Stability can also be written quantitatively in terms of formation constant or stability constant.
M + 3L   ⇔        ML3
Stability constant , β =\(\frac{[ML_{3}]}{[M][L]^{3}}\) The greater the value of β, the stronger the metal-ligand bond is.
Factors responsible for the stability of a complex:
( 1 ) Charge on the central metal ion – the bigger the charge, the more stable the complex is.
( 2 ) Nature of ligand – chelating ligand produces a more stable complex.
( 3 ) The basic strength of ligand – the more basic a ligand, the more stable its complex.

Question 5.26: What is meant by the chelate effect? Give an example.

Solution 5.26: When a ligand attaches to the metal ion in a manner that forms a ring, then the metal- ligand association is found to be more stable. In other words, we can say that complexes containing chelate rings are more stable than complexes without rings. This is known as the chelate effect.

For example:

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Question 5.27: Discuss briefly giving an example in each case of the role of coordination compounds in:
(i) biological systems, (ii) medicinal chemistry, (iii) analytical chemistry and (iv) extraction/metallurgy of metals.

Solution 5.27: (i) Role of coordination compounds in biological systems: 

We know that photosynthesis is made possible by the presence of the chlorophyll pigment. This pigment is a coordination compound of magnesium. In the human biological system, several coordination compounds play important roles. For example, the oxygen-carrier of blood, i.e., haemoglobin, is a coordination compound of iron.

(ii) Role of coordination compounds in medicinal chemistry:

Certain coordination compounds of platinum (for example, cis-platin) are used for inhibiting the growth of tumours.

(iii) Role of coordination compounds in analytical chemistry:

During salt analysis, a number of basic radicals are detected with the help of the colour changes they exhibit with different reagents. These colour changes are a result of the coordination compounds or complexes that the basic radicals form with different ligands.

(iv) Role of coordination compounds in extraction or metallurgy of metals:

The process of extraction of some of the metals from their ores involves the formation of complexes. For example, in aqueous solution, gold combines with cyanide ions to form [Au(CN)2]. From this solution, gold is later extracted by the addition of zinc metal.

Question 5.28: How many ions are produced from the complex Co(NH3)6Cl2 in the solution?
(i) 6 (ii) 4 (iii) 3 (iv) 2

Solution 5.28: (iii) 3

The given complex  [ Co( NH3)6 ]Cl2 ionises to give three ions, viz one [ Co( NH3)6] + and two Cl  ions.

Question 5.29: 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+

Solution 5.29:

( i ) [ Cr( H2O)] 3+
number of unpaired electrons, n  = 3
Magnetic moment, µ = \(\sqrt{3(3 +2)}\)

= \(\sqrt{3(3 +2)}\)

= \(\sqrt{15}\)

≈ 4BM

( ii ) [Fe ( H2O)62+
number of unpaired electrons, n  = 4
Magnetic moment, µ = \(\sqrt{4(4 +2)}\)

= \(\sqrt{4(4 +2)}\)

= \(\sqrt{24}\)

≈ 5 BM
( iii ) [ Zn ( H2O)62
n = 0
Thus,[Fe(H2O)62+ has the highest magnetic moment value.

Question 5.30: 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–

Solution 5.30: Stability of the Complexes:

  1. [Fe(H₂O)₆]³⁺: Less stable due to weak field ligands and lower field strength.
  2. [Fe(NH₃)₆]³⁺: More stable than [Fe(H₂O)₆]³⁺ because NH₃ is a stronger field ligand that can stabilize low-spin configurations.
  3. [Fe(C₂O₄)₃]³⁻: Likely to be the most stable due to the chelation effect of oxalate, which forms a more stable and rigid complex.
  4. [FeCl₆]³⁻: Less stable due to the weak field strength of Cl⁻ and the likelihood of forming high-spin configurations, which are generally less stable.

The most stable complex among the given options is [Fe(C₂O₄)₃]³⁻. The chelation effect and the overall stability provided by the bidentate oxalate ligands contribute significantly to the stability of this complex.

Question 5.31: 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+?

Solution 5.31: To determine the order of wavelengths of absorption in the visible region for the complexes \([Ni(NO_2)_6]^{4-}\), \([Ni(NH_3)_6]^{2+}\), and \([Ni(H_2O)_6]^{2+}\), we must consider the ligand field strength of the ligands attached to the nickel ion in each complex.

Ligands and Their Field Strengths:

  • Nitrite \((NO_2^-)\): A relatively strong field ligand.
  • Ammonia \((NH_3)\): A moderate field ligand.
  • Water \((H_2O)\): A weak field ligand.

Crystal Field Splitting Energy (Δ):

  • The stronger the ligand, the greater the crystal field splitting energy (Δ), which results in absorption of light of shorter wavelengths (higher energy).

Wavelength Order:

  • Complexes with weaker ligands absorb light at longer wavelengths (lower energy), while those with stronger ligands absorb at shorter wavelengths (higher energy).

Explanation:

  • \([Ni(NO_2)_6]^{4-}\) absorbs at the shortest wavelength due to the strong field strength of the nitrite ligands.
  • \([Ni(H_2O)_6]^{2+}\) absorbs the longest wavelength due to the weak field strength of water.
  • \([Ni(NH_3)_6]^{2+}\) absorbs at a shorter wavelength because ammonia is a stronger field ligand than water.

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