NCERT Solutions Class 12 Chemistry Chapter 5: Coordination Compounds (In-Depth) | LearnCBSEHub.in

NCERT Solutions

Class 12 • Chemistry • Coordination Compounds • In-Depth
Question 5.1
Explain the bonding in coordination compounds in terms of Werner’s postulates.

Werner’s Coordination Theory (1893): Alfred Werner proposed the first successful explanation for the formation of coordination compounds.

  1. Two Types of Valency: Metals in complexes exhibit two types of valencies:
    • Primary Valency: This corresponds to the oxidation state of the metal. It is ionisable and is satisfied by anions. Example: In \([Co(NH_3)_6]Cl_3\), the primary valency is +3, satisfied by three \(Cl^-\) ions.
    • Secondary Valency: This corresponds to the coordination number of the metal. It is non-ionisable and is satisfied by neutral molecules or anions (ligands). Example: In \([Co(NH_3)_6]Cl_3\), the secondary valency is 6, satisfied by six \(NH_3\) molecules.
  2. Fixed Orientation: The secondary valencies are directed towards fixed positions in space, giving the coordination polyhedron a definite geometry (e.g., Octahedral for CN=6, Tetrahedral/Square Planar for CN=4). This explains the existence of isomers.
Question 5.2
\(FeSO_4\) solution mixed with \((NH_4)_2SO_4\) solution in 1:1 molar ratio gives the test of \(Fe^{2+}\) ion but \(CuSO_4\) solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of \(Cu^{2+}\) ion. Explain why?
Case 1: Mohr’s Salt

When \(FeSO_4\) and \((NH_4)_2SO_4\) are mixed in a 1:1 ratio, they crystallize to form a Double Salt called Mohr’s Salt: \(FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O\).

Property: Double salts exist only in the solid state. In aqueous solution, they completely dissociate into their constituent simple ions (\(Fe^{2+}, NH_4^+, SO_4^{2-}\)). Hence, the solution gives a positive test for \(Fe^{2+}\).

Case 2: Complex Salt

When \(CuSO_4\) is mixed with excess ammonia, a Coordination Complex is formed: \([Cu(NH_3)_4]SO_4\).

Property: Coordination compounds retain their identity in solution. The complex ion \([Cu(NH_3)_4]^{2+}\) does not dissociate to give free \(Cu^{2+}\) ions. The copper is bound strongly to ammonia ligands. Hence, it does not give the qualitative test for \(Cu^{2+}\).

Question 5.3
Explain with two examples each: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.
  • Coordination Entity: A central metal atom/ion bonded to a fixed number of ions/molecules.
    Ex: \([CoCl_3(NH_3)_3]\), \([Ni(CO)_4]\).
  • Ligand: An ion or molecule capable of donating a pair of electrons to the central metal atom.
    Ex: \(H_2O\), \(Cl^-\).
  • Coordination Number (CN): The total number of ligand donor atoms to which the metal is directly bonded.
    Ex: In \([PtCl_6]^{2-}\), CN is 6. In \([Ni(NH_3)_4]^{2+}\), CN is 4.
  • Coordination Polyhedron: The spatial arrangement of the ligand atoms directly attached to the central atom.
    Ex: Octahedral, Tetrahedral.
  • Homoleptic Complexes: Complexes where the metal is bound to only one kind of donor group.
    Ex: \([Co(NH_3)_6]^{3+}\), \([Fe(CN)_6]^{4-}\).
  • Heteroleptic Complexes: Complexes where the metal is bound to more than one kind of donor group.
    Ex: \([Co(NH_3)_4Cl_2]^+\), \([Pt(NH_3)_2Cl_2]\).
Question 5.4
What is meant by unidentate, didentate and ambidentate ligands? Give two examples for each.
  • Unidentate Ligand: A ligand that binds to the metal ion through a single donor atom.
    Ex: \(H_2O\) (donor O), \(NH_3\) (donor N).
  • Didentate Ligand: A ligand that binds through two donor atoms simultaneously.
    Ex: Ethane-1,2-diamine (\(en\), \(NH_2CH_2CH_2NH_2\)), Oxalate ion (\(ox\), \(C_2O_4^{2-}\)).
  • Ambidentate Ligand: A unidentate ligand that contains two different potential donor atoms but binds through only one at a time.
    Ex: \(NO_2^-\) (can bind via N-nitrito or O-nitrito), \(SCN^-\) (can bind via S-thiocyanato or N-isothiocyanato).
Question 5.5
Specify the oxidation numbers of the metals in the following coordination entities: (i) \([Co(H_2O)(CN)(en)_2]^{2+}\) (ii) \([CoBr_2(en)_2]^+\) (iii) \([PtCl_4]^{2-}\) (iv) \(K_3[Fe(CN)_6]\) (v) \([Cr(NH_3)_3Cl_3]\).
Calculations
  • (i) \([Co(H_2O)(CN)(en)_2]^{2+}\):
    Let oxidation state of Co be \(x\).
    \(x + 0 (H_2O) + (-1) (CN) + 2 \times 0 (en) = +2\)
    \(x – 1 = +2 \implies x = +3\).
  • (ii) \([CoBr_2(en)_2]^+\):
    \(x + 2(-1) + 2(0) = +1\)
    \(x – 2 = +1 \implies x = +3\).
  • (iii) \([PtCl_4]^{2-}\):
    \(x + 4(-1) = -2\)
    \(x – 4 = -2 \implies x = +2\).
  • (iv) \(K_3[Fe(CN)_6]\):
    \(K\) is +1. The complex ion is \([Fe(CN)_6]^{3-}\).
    \(x + 6(-1) = -3\)
    \(x – 6 = -3 \implies x = +3\).
  • (v) \([Cr(NH_3)_3Cl_3]\):
    \(x + 3(0) + 3(-1) = 0\)
    \(x – 3 = 0 \implies x = +3\).
Question 5.6
Using IUPAC norms write the formulas for the following:
  1. Tetrahydroxidozincate(II): \([Zn(OH)_4]^{2-}\)
  2. Potassium tetrachloridopalladate(II): \(K_2[PdCl_4]\)
  3. Diamminedichloridoplatinum(II): \([Pt(NH_3)_2Cl_2]\)
  4. Potassium tetracyanidonickelate(II): \(K_2[Ni(CN)_4]\)
  5. Pentaamminenitrito-O-cobalt(III): \([Co(NH_3)_5(ONO)]^{2+}\)
  6. Hexaamminecobalt(III) sulphate: \([Co(NH_3)_6]_2(SO_4)_3\) (Charge balance: Complex is +3, Sulphate is -2).
  7. Potassium tri(oxalato)chromate(III): \(K_3[Cr(C_2O_4)_3]\)
  8. Hexaammineplatinum(IV): \([Pt(NH_3)_6]^{4+}\)
  9. Tetrabromidocuprate(II): \([CuBr_4]^{2-}\)
  10. Pentaamminenitrito-N-cobalt(III): \([Co(NH_3)_5(NO_2)]^{2+}\)
Question 5.7
Using IUPAC norms write the systematic names of the following:
  1. \([Co(NH_3)_6]Cl_3\): Hexaamminecobalt(III) chloride
  2. \([Pt(NH_3)_2Cl(NH_2CH_3)]Cl\): Diamminechlorido(methanamine)platinum(II) chloride
  3. \([Ti(H_2O)_6]^{3+}\): Hexaaquatitanium(III) ion
  4. \([Co(NH_3)_4Cl(NO_2)]Cl\): Tetraamminechloridonitrito-N-cobalt(III) chloride
  5. \([Mn(H_2O)_6]^{2+}\): Hexaaquamanganese(II) ion
  6. \([NiCl_4]^{2-}\): Tetrachloridonickelate(II) ion
  7. \([Ni(NH_3)_6]Cl_2\): Hexaamminenickel(II) chloride
  8. \([Co(en)_3]^{3+}\): Tris(ethane-1,2-diamine)cobalt(III) ion
  9. \([Ni(CO)_4]\): Tetracarbonylnickel(0)
Question 5.8
List various types of isomerism possible for coordination compounds, giving an example of each.
Structural Isomerism
  • Linkage Isomerism: Occurs with ambidentate ligands. Example: \([Co(NH_3)_5(ONO)]Cl_2\) (Nitrito-O) and \([Co(NH_3)_5(NO_2)]Cl_2\) (Nitrito-N).
  • Coordination Isomerism: Interchange of ligands between cationic and anionic complex ions. Example: \([Co(NH_3)_6][Cr(CN)_6]\) and \([Cr(NH_3)_6][Co(CN)_6]\).
  • Ionisation Isomerism: Counter ion replaces a ligand. Example: \([Co(NH_3)_5SO_4]Br\) (gives \(Br^-\)) and \([Co(NH_3)_5Br]SO_4\) (gives \(SO_4^{2-}\)).
  • Solvate Isomerism: Solvent molecules act as ligands or crystallisation water. Example: \([Cr(H_2O)_6]Cl_3\) (Violet) and \([Cr(H_2O)_5Cl]Cl_2 \cdot H_2O\) (Grey-green).
Stereoisomerism
  • Geometrical Isomerism: Differ in spatial arrangement of ligands. Example: \([Pt(NH_3)_2Cl_2]\) (Cis-platin and Trans-platin).
  • Optical Isomerism: Non-superimposable mirror images (Chirality). Example: \([Co(en)_3]^{3+}\) exists as dextro (\(d\)) and laevo (\(l\)) forms.
Question 5.9
How many geometrical isomers are possible in the following coordination entities? (i) \([Cr(C_2O_4)_3]^{3-}\) (ii) \([Co(NH_3)_3Cl_3]\)
  • (i) \([Cr(C_2O_4)_3]^{3-}\): This is an octahedral complex of the type \([M(AA)_3]\) (Tris-bidentate). Since all ligands are identical bidentate ligands, no geometrical isomers are possible. (Note: It does show Optical Isomerism, but the question asks for Geometrical). Answer: 0.
  • (ii) \([Co(NH_3)_3Cl_3]\): This is an octahedral complex of the type \([Ma_3b_3]\). Two geometrical isomers are possible:
    • Facial (fac): Identical ligands occupy the corners of a face of the octahedron (adjacent positions).
    • Meridional (mer): Identical ligands occupy the meridian of the octahedron (includes trans positions).
    Answer: 2.
Question 5.10
Draw the structures of optical isomers of: (i) \([Cr(C_2O_4)_3]^{3-}\), (ii) \([PtCl_2(en)_2]^{2+}\), (iii) \([Cr(NH_3)_2Cl_2(en)]^+\).
  • (i) \([Cr(ox)_3]^{3-}\): Exists as \(d\) (dextro) and \(l\) (laevo) enantiomers.
  • (ii) \([PtCl_2(en)_2]^{2+}\): The cis isomer is optically active and exists as \(d\) and \(l\) forms. The trans isomer has a plane of symmetry and is optically inactive.
  • (iii) \([Cr(NH_3)_2Cl_2(en)]^+\): The cis form (where identical ligands are adjacent) can show optical activity. The trans forms generally have symmetry planes.
Question 5.11
Draw all the isomers (geometrical and optical) of: (i) \([CoCl_2(en)_2]^+\) (ii) \([Co(NH_3)Cl(en)_2]^{2+}\) (iii) \([Co(NH_3)_2Cl_2(en)]^+\).
  • (i) \([CoCl_2(en)_2]^+\): Type \([M(AA)_2a_2]\).
    – Geometrical: 2 (Cis and Trans).
    – Optical: Cis form is optically active (d, l). Trans is inactive.
    Total Structures: 3 (Cis-d, Cis-l, Trans).
  • (ii) \([Co(NH_3)Cl(en)_2]^{2+}\): Type \([M(AA)_2ab]\).
    – Geometrical: 2 (Cis and Trans).
    – Optical: Cis form is optically active (d, l). Trans is inactive.
    Total Structures: 3.
  • (iii) \([Co(NH_3)_2Cl_2(en)]^+\): Type \([M(AA)a_2b_2]\).
    – This complex has more geometrical isomers (Cis-Cis, Cis-Trans relations).
    – The form where both \(NH_3\) are cis and both \(Cl\) are cis is optically active.
Question 5.12
Write all the geometrical isomers of \([Pt(NH_3)(Br)(Cl)(py)]\) and how many of these will exhibit optical isomers?

Geometrical Isomers: This is a square planar complex of type \([Mabcd]\). The number of geometrical isomers is 3.

Method: Fix one ligand (say \(NH_3\)) and vary the ligand trans to it.

  • Isomer 1: \(NH_3\) trans to \(Cl\).
  • Isomer 2: \(NH_3\) trans to \(Br\).
  • Isomer 3: \(NH_3\) trans to \(py\).

Optical Isomers: Square planar complexes possess a plane of symmetry (the molecular plane containing the metal and ligands). Hence, none of them will exhibit optical isomerism.

Question 5.13
Aqueous copper sulphate solution (blue in colour) gives: (i) a green precipitate with aqueous potassium fluoride and (ii) a bright green solution with aqueous potassium chloride. Explain these experimental results.

Aqueous \(CuSO_4\) contains the blue ion \([Cu(H_2O)_4]^{2+}\).

  • (i) With KF: Fluoride ions (\(F^-\)) are weak field ligands. However, they cause the precipitation of Copper(II) fluoride (\(CuF_2\)) or a basic fluoride, which is green in colour.
    \([Cu(H_2O)_4]^{2+} + 2F^- \to CuF_2(s) \text{ (Green ppt)}\).
  • (ii) With KCl: Chloride ions (\(Cl^-\)) act as ligands. They replace water molecules to form the tetrachloridocuprate(II) complex ion, which is bright green.
    \([Cu(H_2O)_4]^{2+} + 4Cl^- \to [CuCl_4]^{2-} \text{ (Green solution)} + 4H_2O\).
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 \(H_2S(g)\) is passed through this solution?
Reaction

When excess KCN is added, \(Cu^{2+}\) is first reduced to \(Cu^+\) by \(CN^-\), releasing cyanogen gas \((CN)_2\). The \(Cu^+\) then forms a very stable complex with excess cyanide.

$$2Cu^{2+} + 10CN^- \to 2[Cu(CN)_4]^{3-} + (CN)_2$$

The coordination entity is Tetracyanidocuprate(I) ion \([Cu(CN)_4]^{3-}\).

Reason for no ppt with H2S

The stability constant of the complex \([Cu(CN)_4]^{3-}\) is extremely high. The dissociation of this complex to produce free \(Cu^+\) ions is negligible.

$$[Cu(CN)_4]^{3-} \rightleftharpoons Cu^+ + 4CN^-$$

The concentration of free \(Cu^+\) ions in the solution is so low that the ionic product \([Cu^+]^2[S^{2-}]\) does not exceed the solubility product (\(K_{sp}\)) of \(Cu_2S\). Hence, no precipitate is formed.

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) \([FeF_6]^{3-}\) (iii) \([Co(C_2O_4)_3]^{3-}\) (iv) \([CoF_6]^{3-}\).
(i) \([Fe(CN)_6]^{4-}\)
  • Oxidation State: \(Fe^{2+} (3d^6)\).
  • Ligand: \(CN^-\) is a Strong Field Ligand. It forces pairing of 3d electrons.
  • Hybridisation: The 6 electrons pair up in three 3d orbitals, leaving two 3d orbitals empty. Hybridisation involves \(2(3d) + 1(4s) + 3(4p) = d^2sp^3\) (Inner orbital).
  • Magnetic Nature: All electrons paired \(\to\) Diamagnetic.
(ii) \([FeF_6]^{3-}\)
  • Oxidation State: \(Fe^{3+} (3d^5)\).
  • Ligand: \(F^-\) is a Weak Field Ligand. No pairing occurs.
  • Hybridisation: 4s, 4p, and outer 4d orbitals are used. \(sp^3d^2\) (Outer orbital).
  • Magnetic Nature: 5 unpaired electrons \(\to\) Paramagnetic.
(iii) \([Co(C_2O_4)_3]^{3-}\)
  • Oxidation State: \(Co^{3+} (3d^6)\).
  • Ligand: Oxalate is a chelating strong ligand. Pairing occurs.
  • Hybridisation: \(d^2sp^3\) (Inner orbital).
  • Magnetic Nature: All paired \(\to\) Diamagnetic.
(iv) \([CoF_6]^{3-}\)
  • Oxidation State: \(Co^{3+} (3d^6)\).
  • Ligand: \(F^-\) is a Weak Field Ligand. No pairing.
  • Hybridisation: \(sp^3d^2\) (Outer orbital).
  • Magnetic Nature: 4 unpaired electrons \(\to\) Paramagnetic.
Question 5.16
Draw figure to show the splitting of d orbitals in an octahedral crystal field.

Explanation: In an octahedral field, six ligands approach the metal ion along the x, y, and z axes. The d-orbitals lying along the axes (\(d_{x^2-y^2}\) and \(d_{z^2}\)) experience greater repulsion from the ligand electrons and are raised in energy. These form the \(e_g\) set. The orbitals lying between the axes (\(d_{xy}, d_{yz}, d_{zx}\)) experience less repulsion and are relatively lower in energy. These form the \(t_{2g}\) set. The energy separation is denoted by \(\Delta_o\).

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

Spectrochemical Series: It is the arrangement of ligands in the increasing order of their field strength (ability to split d-orbitals).
Order: \(I^- < Br^- < SCN^- < Cl^- < F^- < OH^- < C_2O_4^{2-} < H_2O < NCS^- < edta^{4-} < NH_3 < en < CN^- < CO\).

Difference:

  • Strong Field Ligand: Causes a large crystal field splitting energy (\(\Delta_o\)). If \(\Delta_o > P\) (Pairing energy), electrons pair up in \(t_{2g}\) orbitals before entering \(e_g\). Forms low spin complexes. Examples: \(CN^-, CO\).
  • Weak Field Ligand: Causes a small splitting energy. If \(\Delta_o < P\), electrons occupy \(e_g\) orbitals before pairing in \(t_{2g}\). Forms high spin complexes. Examples: \(Cl^-, H_2O\).
Question 5.18
What is crystal field splitting energy? How does the magnitude of \(\Delta_o\) decide the actual configuration of d orbitals in a coordination entity?

Crystal Field Splitting Energy (CFSE): The energy difference between the lower energy set of orbitals (\(t_{2g}\)) and higher energy set (\(e_g\)) in a complex is called CFSE, denoted by \(\Delta_o\) for octahedral.

Decision of Configuration (e.g., for \(d^4\)):

  • If \(\Delta_o < P\) (Weak Field): The 4th electron enters the \(e_g\) orbital. Configuration: \(t_{2g}^3 e_g^1\). (High Spin).
  • If \(\Delta_o > P\) (Strong Field): The 4th electron pairs up in the \(t_{2g}\) orbital. Configuration: \(t_{2g}^4 e_g^0\). (Low Spin).
Question 5.19
\([Cr(NH_3)_6]^{3+}\) is paramagnetic while \([Ni(CN)_4]^{2-}\) is diamagnetic. Explain why?
  • \([Cr(NH_3)_6]^{3+}\): Chromium is in +3 state (\(3d^3\)). In an octahedral field, the 3 electrons occupy the three \(t_{2g}\) orbitals singly (\(t_{2g}^3\)) regardless of ligand strength. Since there are 3 unpaired electrons, the complex is paramagnetic.
  • \([Ni(CN)_4]^{2-}\): Nickel is in +2 state (\(3d^8\)). \(CN^-\) is a strong field ligand. It causes the pairing of electrons. The 8 electrons pair up in four 3d orbitals (\(d_{xy}, d_{yz}, d_{zx}, d_{z^2}\)), leaving one d-orbital (\(d_{x^2-y^2}\)) empty. This allows \(dsp^2\) hybridisation (Square Planar). Since all electrons are paired, it is diamagnetic.
Question 5.20
A solution of \([Ni(H_2O)_6]^{2+}\) is green but a solution of \([Ni(CN)_4]^{2-}\) is colourless. Explain.
  • \([Ni(H_2O)_6]^{2+}\): This is an outer orbital octahedral complex containing \(Ni^{2+} (3d^8)\). Since \(H_2O\) is a weak field ligand, the electrons are not paired (\(t_{2g}^6 e_g^2\)). There are 2 unpaired electrons. The complex undergoes d-d transitions by absorbing red light from the visible spectrum, thus appearing complementary green.
  • \([Ni(CN)_4]^{2-}\): This is a square planar complex with strong field \(CN^-\) ligands. All 3d electrons are paired. There are no unpaired electrons available for low-energy d-d transitions in the visible region. Hence, it does not absorb visible light and appears colourless.
Question 5.21
\([Fe(CN)_6]^{4-}\) and \([Fe(H_2O)_6]^{2+}\) are of different colours in dilute solutions. Why?

Both complexes contain Iron in the +2 oxidation state (\(3d^6\)). The colour of a coordination compound depends on the magnitude of the crystal field splitting energy (\(\Delta_o\)), which determines the wavelength of light absorbed for d-d transition.

\(CN^-\) is a strong field ligand causing a large \(\Delta_o\), while \(H_2O\) is a weak field ligand causing a small \(\Delta_o\). Because the energy gaps are different, they absorb different wavelengths of light from the visible region. Consequently, the transmitted (complementary) colours are different.

Question 5.22
Discuss the nature of bonding in metal carbonyls.
Synergic Bonding

The metal-carbon bond in metal carbonyls possesses both \(\sigma\) and \(\pi\) character.

  1. \(\sigma\)-bond Formation: A lone pair of electrons from the carbon atom of the CO molecule is donated into a vacant orbital of the metal atom. (Ligand to Metal).
  2. \(\pi\)-bond Formation (Back Bonding): A pair of electrons from a filled d-orbital of the metal is donated into the vacant antibonding \(\pi^*\)-orbital of carbon. (Metal to Ligand).

This synergic effect strengthens the bond between CO and the metal, making metal carbonyls very stable.

Question 5.23
Give the oxidation state, d orbital occupation and coordination number of the central metal ion in the following complexes: (i) \(K_3[Co(C_2O_4)_3]\) (ii) \(cis-[CrCl_2(en)_2]Cl\) (iii) \((NH_4)_2[CoF_4]\) (iv) \([Mn(H_2O)_6]SO_4\).
  • (i) \(K_3[Co(C_2O_4)_3]\):
    Metal: Co. Ox State: +3. CN: 6 (3 bidentate).
    Config: \(3d^6\). Strong ligand (ox) causes pairing. \(t_{2g}^6 e_g^0\).
  • (ii) \(cis-[CrCl_2(en)_2]Cl\):
    Metal: Cr. Ox State: +3. CN: 6 (2 mono + 2 bidentate).
    Config: \(3d^3\). Always \(t_{2g}^3 e_g^0\).
  • (iii) \((NH_4)_2[CoF_4]\):
    Metal: Co. Ox State: +2. CN: 4.
    Config: \(3d^7\). Weak ligand (F). Tetrahedral. \(e^4 t_2^3\).
  • (iv) \([Mn(H_2O)_6]SO_4\):
    Metal: Mn. Ox State: +2. CN: 6.
    Config: \(3d^5\). Weak ligand (H2O). High spin. \(t_{2g}^3 e_g^2\).
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 stereochemistry and magnetic moment of the complex.
(i) \(K[Cr(H_2O)_2(C_2O_4)_2] \cdot 3H_2O\)

Name: Potassium diaquadioxalatochromate(III) trihydrate.
Ox State: +3. CN: 6. Config: \(t_{2g}^3\).
Stereo: Octahedral. (Shows cis/trans isomers. Cis is optically active).
Moment: 3 unpaired e-. \(\mu = \sqrt{15} \approx 3.87\) BM.

(ii) \([Co(NH_3)_5Cl]Cl_2\)

Name: Pentaamminechloridocobalt(III) chloride.
Ox State: +3. CN: 6. Config: \(t_{2g}^6\) (Low spin).
Stereo: Octahedral. No geometric isomers.
Moment: 0 (Diamagnetic).

(iii) \([CrCl_3(py)_3]\)

Name: Trichloridotripyridinechromium(III).
Ox State: +3. CN: 6. Config: \(t_{2g}^3\).
Stereo: Octahedral. (Shows fac and mer isomers).
Moment: 3.87 BM.

(iv) \(Cs[FeCl_4]\)

Name: Caesium tetrachloridoferrate(III).
Ox State: +3. CN: 4. Config: \(e^2 t_2^3\) (Tetrahedral).
Stereo: Tetrahedral.
Moment: 5 unpaired e-. \(\mu = \sqrt{35} \approx 5.92\) BM.

(v) \(K_4[Mn(CN)_6]\)

Name: Potassium hexacyanidomanganate(II).
Ox State: +2. CN: 6. Config: \(t_{2g}^5\) (Low spin, strong CN).
Stereo: Octahedral.
Moment: 1 unpaired e-. \(\mu = \sqrt{3} \approx 1.73\) BM.

Question 5.25
Explain the violet colour of the complex \([Ti(H_2O)_6]^{3+}\) on the basis of crystal field theory.

\([Ti(H_2O)_6]^{3+}\) contains \(Ti^{3+}\) which has a \(3d^1\) configuration. In the octahedral crystal field of six water molecules, the d-orbitals split into \(t_{2g}\) and \(e_g\) sets.

The ground state configuration is \(t_{2g}^1 e_g^0\). When white light falls on the complex, the single electron absorbs energy corresponding to the yellow-green region and jumps to the higher energy \(e_g\) orbital (d-d transition). The complementary colour observed is Violet.

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

Chelate Effect: When a di- or polydentate ligand binds to a metal ion using two or more donor atoms, it forms a ring structure called a chelate ring. Complexes with chelate rings are significantly more stable than similar complexes with unidentate ligands. This enhanced stability is called the chelate effect (primarily due to favorable entropy change).

Example: \([Ni(en)_3]^{2+}\) (forms 5-membered rings) is much more stable than \([Ni(NH_3)_6]^{2+}\).

Question 5.27
Discuss briefly giving an example in each case the role of coordination compounds in: (i) biological systems (ii) medicinal chemistry (iii) analytical chemistry (iv) extraction/metallurgy of metals.
  • (i) Biological Systems:
    • Chlorophyll: A coordination complex of Magnesium (Mg), essential for photosynthesis.
    • Haemoglobin: A coordination complex of Iron (Fe), transports oxygen in blood.
  • (ii) Medicinal Chemistry:
    • Cis-platin \([PtCl_2(NH_3)_2]\): Used effectively in the treatment of cancer.
  • (iii) Analytical Chemistry:
    • Detection of Ions: \(Ni^{2+}\) is detected using Dimethylglyoxime (DMG), forming a red chelate complex.
  • (iv) Extraction/Metallurgy:
    • Gold/Silver Extraction: Gold is extracted by forming a soluble cyanide complex \([Au(CN)_2]^-\) (MacArthur-Forrest Cyanide Process).
    • Purification of Nickel: Impure nickel is converted to volatile \([Ni(CO)_4]\) and then decomposed to pure metal (Mond Process).
Question 5.28
How many ions are produced from the complex \(Co(NH_3)_6Cl_2\) in solution? (i) 6 (ii) 4 (iii) 3 (iv) 2

The formula is \([Co(NH_3)_6]Cl_2\).

Dissociation: \([Co(NH_3)_6]Cl_2 \to [Co(NH_3)_6]^{2+} + 2Cl^-\).

Total ions = 1 Complex ion + 2 Chloride ions = 3 ions.

Correct Answer: (iii) 3
Question 5.29
Amongst the following ions which one has the highest magnetic moment value? (i) \([Cr(H_2O)_6]^{3+}\) (ii) \([Fe(H_2O)_6]^{2+}\) (iii) \([Zn(H_2O)_6]^{2+}\)

\(H_2O\) is a weak field ligand, so assume high spin.

  • (i) \(Cr^{3+} (3d^3)\): 3 unpaired electrons. \(\mu \approx 3.87\) BM.
  • (ii) \(Fe^{2+} (3d^6)\): High spin configuration is \(t_{2g}^4 e_g^2\). This gives 4 unpaired electrons. \(\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.9\) BM.
  • (iii) \(Zn^{2+} (3d^{10})\): All paired. \(\mu = 0\).
Highest Magnetic Moment: (ii) \([Fe(H_2O)_6]^{2+}\)
Question 5.30
Amongst the following, the most stable complex is: (i) \([Fe(H_2O)_6]^{3+}\) (ii) \([Fe(NH_3)_6]^{3+}\) (iii) \([Fe(C_2O_4)_3]^{3-}\) (iv) \([FeCl_6]^{3-}\)

The stability of a complex increases with the strength of the ligand and, most importantly, with the formation of chelate rings.

  • \(H_2O\) and \(Cl^-\) are monodentate weak ligands.
  • \(NH_3\) is a monodentate strong ligand.
  • \(C_2O_4^{2-}\) (Oxalate) is a bidentate chelating ligand.

Due to the Chelate Effect, the oxalate complex is the most stable.

Correct Answer: (iii) \([Fe(C_2O_4)_3]^{3-}\)
Question 5.31
What will be the correct order for the wavelengths of absorption in the visible region for the following: \([Ni(NO_2)_6]^{4-}\), \([Ni(NH_3)_6]^{2+}\), \([Ni(H_2O)_6]^{2+}\)?

The energy of light absorbed (\(E\)) corresponds to the crystal field splitting energy (\(\Delta_o\)).

From the Spectrochemical Series, the field strength of ligands is: \(H_2O < NH_3 < NO_2^-\).

Therefore, the splitting energy order is: \(\Delta_o(H_2O) < \Delta_o(NH_3) < \Delta_o(NO_2^-)\).

Since wavelength (\(\lambda\)) is inversely proportional to energy (\(E = hc/\lambda\)), the order of wavelengths absorbed will be the reverse of the energy order.

Order of \(\lambda\): \([Ni(H_2O)_6]^{2+} > [Ni(NH_3)_6]^{2+} > [Ni(NO_2)_6]^{4-}\)
learncbsehub.in