Q: Q2. Write the IUPAC names of the following coordination compounds:\n(i) [Co(NH₃)₆]Cl₃\n(ii) [Co(NH₃)₅Cl]Cl₂\n(iii) K₃[Fe(CN)₆]\n(iv) K₃[Fe(C₂O₄)₃]\n(v) K₂[PdCl₄]\n(vi) [Pt(NH₃)₂Cl(NO₂)

(i) [Co(NH₃)₆]Cl₃: Hexaamminecobalt(III) chloride\n(ii) [Co(NH₃)₅Cl]Cl₂: Pentaamminechloridocobalt(III) chloride\n(iii) K₃[Fe(CN)₆]: Potassium hexacyanidoferrate(III)\n(iv) K₃[Fe(C₂O₄)₃]: Potassium tris(oxalato)ferrate(III)\n(v) K₂[PdCl₄]: Potassium tetrachloridopalladate(II)\n(vi) [Pt(NH₃)₂Cl(NO₂)]: Diamminechloridonitrito-N-platinum(II)\n(vii) [Cr(H₂O)₄Cl₂]Cl: Tetraaquadichloridochromium(III) chloride\n(viii) [Co(en)₂Cl₂]Cl: Dichloridobis(ethane-1,2-diamine)cobalt(III) chloride\n(ix) [Ni(CO)₄]: Tetracarbonylnickel(0)

Q: Q6. Specify the oxidation number of the metals in the following coordination entities:\n(i) [Co(H2O)(CN)(en)2]2+\n(ii) [PtCl4]2–\n(iii) [Cr(NH3)3Cl3]\n(iv) [CoBr2(en)2]+\n(v) K3[Fe(CN)

The oxidation numbers of the metals are as follows:\n(i) Co: +3\n(ii) Pt: +2\n(iii) Cr: +3\n(iv) Co: +3\n(v) Fe: +3\n(vi) Cr: +3\n(vii) Ni: +2\n(viii) Ni: +2

Question 1

Q1. Write the formulas for the following coordination compounds:
(i) Tetraamminediaquacobalt(III) chloride
(ii) Potassium hexacyanidoferrate(II)
(iii) Tris(ethane-1,2-diamine)chromi

Accepted Answer

(i) Tetraamminediaquacobalt(III) chloride: [Co(NH₃)₄(H₂O)₂]Cl₃
(ii) Potassium hexacyanidoferrate(II): K₄[Fe(CN)₆]
(iii) Tris(ethane-1,2-diamine)chromium(III) chloride: [Cr(en)₃]Cl₃
(iv) Amminebromidochloridonitrito-N-platinate(II): [Pt(NH₃)(Br)(Cl)(NO₂)]⁻
(v) Dichloridobis(ethane-1,2-diamine)cobalt(III): [Co(en)₂Cl₂]Cl
(vi) Tetracarbonylnickel(0): [Ni(CO)₄]

Question 2

Q2. Write the IUPAC names of the following coordination compounds:
(i) [Co(NH₃)₆]Cl₃
(ii) [Co(NH₃)₅Cl]Cl₂
(iii) K₃[Fe(CN)₆]
(iv) K₃[Fe(C₂O₄)₃]
(v) K₂[PdCl₄]
(vi) [Pt(NH₃)₂Cl(NO₂)

Accepted Answer

(i) [Co(NH₃)₆]Cl₃: Hexaamminecobalt(III) chloride
(ii) [Co(NH₃)₅Cl]Cl₂: Pentaamminechloridocobalt(III) chloride
(iii) K₃[Fe(CN)₆]: Potassium hexacyanidoferrate(III)
(iv) K₃[Fe(C₂O₄)₃]: Potassium tris(oxalato)ferrate(III)
(v) K₂[PdCl₄]: Potassium tetrachloridopalladate(II)
(vi) [Pt(NH₃)₂Cl(NO₂)]: Diamminechloridonitrito-N-platinum(II)
(vii) [Cr(H₂O)₄Cl₂]Cl: Tetraaquadichloridochromium(III) chloride
(viii) [Co(en)₂Cl₂]Cl: Dichloridobis(ethane-1,2-diamine)cobalt(III) chloride
(ix) [Ni(CO)₄]: Tetracarbonylnickel(0)

Question 3

Q3. Using IUPAC norms write the formulas for the following:
(i) Tetrahydroxozincate(II)
(ii) Hexaamminecobalt(III) sulphate
(iii) Potassium tetrachloropalladate(II)
(iv) Potassium

Accepted Answer

The formulas for the given coordination compounds are derived by following IUPAC nomenclature rules, considering the central metal ion's oxidation state, the ligands, and the overall charge of the complex ion or compound.

(i) **Tetrahydroxozincate(II)**
   - Central metal: Zinc (Zn)
   - Oxidation state: +2
   - Ligand: Hydroxo (OH⁻)
   - Number of ligands: Tetra (4)
   - Complex ion charge: Zn(+2) + 4(OH⁻) = +2 - 4 = -2
   - Formula: **[Zn(OH)₄]²⁻**

(ii) **Hexaamminecobalt(III) sulphate**
   - Central metal: Cobalt (Co)
   - Oxidation state: +3
   - Ligand: Ammine (NH₃, neutral)
   - Number of ligands: Hexa (6)
   - Complex ion charge: Co(+3) + 6(NH₃) = +3 + 0 = +3
   - Counter ion: Sulphate (SO₄²⁻)
   - To balance charges: Two [Co(NH₃)₆]³⁺ ions (total +6) balance three SO₄²⁻ ions (total -6).
   - Formula: **[Co(NH₃)₆]₂(SO₄)₃**

(iii) **Potassium tetrachloropalladate(II)**
   - Central metal: Palladium (Pd)
   - Oxidation state: +2
   - Ligand: Chloro (Cl⁻)
   - Number of ligands: Tetra (4)
   - Complex ion charge: Pd(+2) + 4(Cl⁻) = +2 - 4 = -2
   - Counter ion: Potassium (K⁺)
   - To balance charges: Two K⁺ ions balance one [PdCl₄]²⁻ ion.
   - Formula: **K₂[PdCl₄]**

(iv) **Potassium tetracyanonickelate(II)**
   - Central metal: Nickel (Ni)
   - Oxidation state: +2
   - Ligand: Cyano (CN⁻)
   - Number of ligands: Tetra (4)
   - Complex ion charge: Ni(+2) + 4(CN⁻) = +2 - 4 = -2
   - Counter ion: Potassium (K⁺)
   - To balance charges: Two K⁺ ions balance one [Ni(CN)₄]²⁻ ion.
   - Formula: **K₂[Ni(CN)₄]**

(v) **Pentaamminenitrito-O-cobalt(III)**
   - Central metal: Cobalt (Co)
   - Oxidation state: +3
   - Ligands: Pentaammine (5 NH₃, neutral), Nitrito-O (ONO⁻, charge -1, bonded via oxygen)
   - Complex ion charge: Co(+3) + 5(NH₃) + 1(ONO⁻) = +3 + 0 - 1 = +2
   - Formula: **[Co(NH₃)₅(ONO)]²⁺**

(vi) **Hexaamminecobalt(III) chloride**
   - Central metal: Cobalt (Co)
   - Oxidation state: +3
   - Ligand: Ammine (NH₃, neutral)
   - Number of ligands: Hexa (6)
   - Complex ion charge: Co(+3) + 6(NH₃) = +3 + 0 = +3
   - Counter ion: Chloride (Cl⁻)
   - To balance charges: Three Cl⁻ ions balance one [Co(NH₃)₆]³⁺ ion.
   - Formula: **[Co(NH₃)₆]Cl₃**

(vii) **Potassium tri(oxalato)chromate(III)**
   - Central metal: Chromium (Cr)
   - Oxidation state: +3
   - Ligand: Oxalato (C₂O₄²⁻, bidentate, charge -2)
   - Number of ligands: Tri (3)
   - Complex ion charge: Cr(+3) + 3(C₂O₄²⁻) = +3 - 6 = -3
   - Counter ion: Potassium (K⁺)
   - To balance charges: Three K⁺ ions balance one [Cr(C₂O₄)₃]³⁻ ion.
   - Formula: **K₃[Cr(C₂O₄)₃]**

(viii) **Hexafluorocobaltate(III)**
   - Central metal: Cobalt (Co)
   - Oxidation state: +3
   - Ligand: Fluoro (F⁻)
   - Number of ligands: Hexa (6)
   - Complex ion charge: Co(+3) + 6(F⁻) = +3 - 6 = -3
   - Formula: **[CoF₆]³⁻**

(ix) **Tetrabromocuprate(II)**
   - Central metal: Copper (Cu)
   - Oxidation state: +2
   - Ligand: Bromo (Br⁻)
   - Number of ligands: Tetra (4)
   - Complex ion charge: Cu(+2) + 4(Br⁻) = +2 - 4 = -2
   - Formula: **[CuBr₄]²⁻**

(x) **Pentaamminenitrito-N-cobalt(III)**
   - Central metal: Cobalt (Co)
   - Oxidation state: +3
   - Ligands: Pentaammine (5 NH₃, neutral), Nitrito-N (NO₂⁻, charge -1, bonded via nitrogen)
   - Complex ion charge: Co(+3) + 5(NH₃) + 1(NO₂⁻) = +3 + 0 - 1 = +2
   - Formula: **[Co(NH₃)₅(NO₂)]²⁺**

Question 4

Q4. Write the formulas for the following coordination compounds:
(i) Tetraamminediaquacobalt(III) chloride
(ii) Potassium hexacyanidochromate(III)
(iii) Tris(ethane-1,2-diamine)chro

Accepted Answer

The formulas for the given coordination compounds are:

(i) **Tetraamminediaquacobalt(III) chloride**
   *   Central metal: Cobalt (Co) with oxidation state +3.
   *   Ligands: Tetraammine (4 NH₃, neutral), Diaqua (2 H₂O, neutral).
   *   Complex ion: [Co(NH₃)₄(H₂O)₂]
   *   Charge of complex ion: (+3) + 4(0) + 2(0) = +3.
   *   Counter ion: Chloride (Cl⁻).
   *   To balance the +3 charge, three Cl⁻ ions are required.
   *   **Formula: [Co(NH₃)₄(H₂O)₂]Cl₃**

(ii) **Potassium hexacyanidochromate(III)**
   *   Cation: Potassium (K⁺).
   *   Central metal: Chromate (Cr) with oxidation state +3 (indicated by '-ate' suffix and Roman numeral).
   *   Ligands: Hexacyanido (6 CN⁻, charge -1 each).
   *   Complex ion: [Cr(CN)₆]
   *   Charge of complex ion: (+3) + 6(-1) = -3.
   *   To balance the -3 charge, three K⁺ ions are required.
   *   **Formula: K₃[Cr(CN)₆]**

(iii) **Tris(ethane-1,2-diamine)chromium(III) chloride**
   *   Central metal: Chromium (Cr) with oxidation state +3.
   *   Ligands: Tris(ethane-1,2-diamine) (3 'en', neutral bidentate ligand).
   *   Complex ion: [Cr(en)₃]
   *   Charge of complex ion: (+3) + 3(0) = +3.
   *   Counter ion: Chloride (Cl⁻).
   *   To balance the +3 charge, three Cl⁻ ions are required.
   *   **Formula: [Cr(en)₃]Cl₃**

(iv) **Amminebromidochloridonitrito-N-platinate(II)**
   *   Central metal: Platinate (Pt) with oxidation state +2 (indicated by '-ate' suffix and Roman numeral).
   *   Ligands: Ammine (NH₃, neutral), Bromido (Br⁻, charge -1), Chlorido (Cl⁻, charge -1), Nitrito-N (NO₂⁻, charge -1).
   *   Complex ion: [Pt(NH₃)BrCl(NO₂)]
   *   Charge of complex ion: (+2) + 0 + (-1) + (-1) + (-1) = -1.
   *   Since no counter cation is specified, the formula represents the complex anion with its charge.
   *   **Formula: [Pt(NH₃)BrCl(NO₂)]⁻**

(v) **Dichloridobis(ethane-1,2-diamine)cobalt(III)**
   *   Central metal: Cobalt (Co) with oxidation state +3.
   *   Ligands: Dichlorido (2 Cl⁻, charge -1 each), Bis(ethane-1,2-diamine) (2 'en', neutral bidentate ligand).
   *   Complex ion: [Co(en)₂Cl₂]
   *   Charge of complex ion: (+3) + 2(0) + 2(-1) = +1.
   *   Since no counter anion is specified, the formula represents the complex cation with its charge.
   *   **Formula: [Co(en)₂Cl₂]⁺**

(vi) **Tetracarbonylnickel(0)**
   *   Central metal: Nickel (Ni) with oxidation state 0.
   *   Ligands: Tetracarbonyl (4 CO, neutral).
   *   Complex ion: [Ni(CO)₄]
   *   Charge of complex ion: (0) + 4(0) = 0.
   *   This is a neutral complex, so no counter ions are needed.
   *   **Formula: [Ni(CO)₄]**

Question 5

Q5. What is meant by unidentate, didentate and ambidentate ligands? Give two examples for each.

Accepted Answer

Ligands are atoms, ions, or molecules that donate a pair of electrons to the central metal atom or ion to form a coordinate bond. They are classified based on the number of donor atoms they possess and the way they coordinate.

### 1. Unidentate Ligands
Unidentate ligands (or monodentate ligands) are those ligands that coordinate to the central metal atom or ion through only one donor atom. They form a single coordinate bond with the metal.

**Examples:**
1.  **Ammonia (NH3):** The nitrogen atom is the donor atom.
2.  **Chloride ion (Cl-):** The chlorine atom is the donor atom.
3.  **Water (H2O):** The oxygen atom is the donor atom.
4.  **Cyanide ion (CN-):** The carbon atom is typically the donor atom.

### 2. Didentate Ligands
Didentate ligands (or bidentate ligands) are those ligands that coordinate to the central metal atom or ion through two donor atoms. They form two coordinate bonds with the metal, often resulting in a stable ring structure called a chelate ring.

**Examples:**
1.  **Ethane-1,2-diamine (en, H2N-CH2-CH2-NH2):** Both nitrogen atoms are donor atoms.
2.  **Oxalate ion (C2O4^2-):** Both oxygen atoms (from the carboxylate groups) are donor atoms.
3.  **Glycinate ion (gly-, H2N-CH2-COO-):** The nitrogen atom and one oxygen atom are donor atoms.

### 3. Ambidentate Ligands
Ambidentate ligands are unidentate ligands that have two different potential donor atoms, but only one of them forms a coordinate bond with the central metal atom or ion at a time. The bonding can occur through either of the two donor atoms, leading to linkage isomerism.

**Examples:**
1.  **Nitrite ion (NO2-):** It can coordinate through the nitrogen atom (—NO2, nitrito-N or nitro) or through one of the oxygen atoms (—ONO, nitrito-O).
2.  **Thiocyanate ion (SCN-):** It can coordinate through the sulfur atom (—SCN, thiocyanato-S or thiocyanato) or through the nitrogen atom (—NCS, thiocyanato-N or isothiocyanato).
3.  **Cyanide ion (CN-):** It can coordinate through the carbon atom (cyano) or through the nitrogen atom (isocyano).

Question 6

Q6. 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)

Accepted Answer

The oxidation numbers of the metals are as follows:
(i) Co: +3
(ii) Pt: +2
(iii) Cr: +3
(iv) Co: +3
(v) Fe: +3
(vi) Cr: +3
(vii) Ni: +2
(viii) Ni: +2

Question 7

Q7. Write down the formulae of the following coordination compounds:
(i) Tetracarbonylnickel(0)
(ii) Hexaamminecobalt(III) sulphate
(iii) Potassium hexacyanoferrate(II)
(iv) Hexaam

Accepted Answer

The formulae of the given coordination compounds are:

(i) **Tetracarbonylnickel(0)**
    *   Central metal: Nickel (Ni)
    *   Oxidation state: 0
    *   Ligand: Carbonyl (CO), neutral
    *   Formula: **[Ni(CO)₄]**

(ii) **Hexaamminecobalt(III) sulphate**
    *   Central metal: Cobalt (Co)
    *   Oxidation state: +3
    *   Ligand: Ammine (NH₃), neutral
    *   Complex ion: [Co(NH₃)₆]³⁺ (Charge = +3 + 6*0 = +3)
    *   Counter ion: Sulphate (SO₄²⁻)
    *   To balance the charges (3+ and 2-), we need two complex ions and three sulphate ions.
    *   Formula: **[Co(NH₃)₆]₂(SO₄)₃**

(iii) **Potassium hexacyanoferrate(II)**
    *   Central metal: Iron (Ferrate) (Fe)
    *   Oxidation state: +2
    *   Ligand: Cyano (CN⁻), -1 charge
    *   Complex ion: [Fe(CN)₆]ˣ (Charge = +2 + 6*(-1) = -4). So, [Fe(CN)₆]⁴⁻
    *   Counter ion: Potassium (K⁺)
    *   To balance the charges (1+ and 4-), we need four potassium ions.
    *   Formula: **K₄[Fe(CN)₆]**

(iv) **Hexaammineplatinum(IV) chloride**
    *   Central metal: Platinum (Pt)
    *   Oxidation state: +4
    *   Ligand: Ammine (NH₃), neutral
    *   Complex ion: [Pt(NH₃)₆]⁴⁺ (Charge = +4 + 6*0 = +4)
    *   Counter ion: Chloride (Cl⁻)
    *   To balance the charges (4+ and 1-), we need four chloride ions.
    *   Formula: **[Pt(NH₃)₆]Cl₄**

(v) **Potassium tri(oxalato)chromate(III)**
    *   Central metal: Chromium (Chromate) (Cr)
    *   Oxidation state: +3
    *   Ligand: Oxalato (C₂O₄²⁻), -2 charge, bidentate
    *   Complex ion: [Cr(C₂O₄)₃]ˣ (Charge = +3 + 3*(-2) = -3). So, [Cr(C₂O₄)₃]³⁻
    *   Counter ion: Potassium (K⁺)
    *   To balance the charges (1+ and 3-), we need three potassium ions.
    *   Formula: **K₃[Cr(C₂O₄)₃]**

(vi) **Hexacyanoplatinate(IV) ion**
    *   Central metal: Platinum (Platinate) (Pt)
    *   Oxidation state: +4
    *   Ligand: Cyano (CN⁻), -1 charge
    *   Complex ion: [Pt(CN)₆]ˣ (Charge = +4 + 6*(-1) = -2). So, [Pt(CN)₆]²⁻
    *   This is an ion, so no counter ion is needed.
    *   Formula: **[Pt(CN)₆]²⁻**

Question 8

Q8. Write the IUPAC names of the following coordination compounds:
(i) [Co(NH₃)₆]Cl₃
(ii) [Co(NH₃)₅Cl]Cl₂
(iii) K₃[Fe(CN)₆]
(iv) K₃[Fe(C₂O₄)₃]
(v) K₂[PdCl₄]
(vi) [Pt(NH₃)₂Cl(NO₂)

Accepted Answer

The IUPAC names of the given coordination compounds are:

(i) **[Co(NH₃)₆]Cl₃**
    *   Ligands: Six ammine (NH₃, neutral)
    *   Central metal: Cobalt (Co)
    *   Oxidation state of Co: Let it be x. x + 6(0) + 3(-1) = 0 ⇒ x = +3. So, Cobalt(III).
    *   Counter ion: Chloride
    *   Name: **Hexaamminecobalt(III) chloride**

(ii) **[Co(NH₃)₅Cl]Cl₂**
    *   Ligands: Five ammine (NH₃, neutral), one chlorido (Cl⁻, -1 charge)
    *   Central metal: Cobalt (Co)
    *   Oxidation state of Co: Let it be x. x + 5(0) + 1(-1) + 2(-1) = 0 ⇒ x - 3 = 0 ⇒ x = +3. So, Cobalt(III).
    *   Counter ion: Chloride
    *   Name: **Pentaamminechloridocobalt(III) chloride**

(iii) **K₃[Fe(CN)₆]**
    *   Counter ion: Potassium (K⁺)
    *   Ligands: Six cyano (CN⁻, -1 charge)
    *   Central metal: Iron (Fe). Since the complex ion is anionic, the metal name ends with '-ate', so Ferrate.
    *   Oxidation state of Fe: Let it be x. 3(+1) + x + 6(-1) = 0 ⇒ 3 + x - 6 = 0 ⇒ x = +3. So, Ferrate(III).
    *   Name: **Potassium hexacyanoferrate(III)**

(iv) **K₃[Fe(C₂O₄)₃]**
    *   Counter ion: Potassium (K⁺)
    *   Ligands: Three oxalato (C₂O₄²⁻, -2 charge, bidentate). Use 'tri' for simple ligands like oxalato.
    *   Central metal: Iron (Fe). Since the complex ion is anionic, Ferrate.
    *   Oxidation state of Fe: Let it be x. 3(+1) + x + 3(-2) = 0 ⇒ 3 + x - 6 = 0 ⇒ x = +3. So, Ferrate(III).
    *   Name: **Potassium trioxalatoferrate(III)**

(v) **K₂[PdCl₄]**
    *   Counter ion: Potassium (K⁺)
    *   Ligands: Four chlorido (Cl⁻, -1 charge)
    *   Central metal: Palladium (Pd). Since the complex ion is anionic, Palladate.
    *   Oxidation state of Pd: Let it be x. 2(+1) + x + 4(-1) = 0 ⇒ 2 + x - 4 = 0 ⇒ x = +2. So, Palladate(II).
    *   Name: **Potassium tetrachloridopalladate(II)**

(vi) **[Pt(NH₃)₂Cl(NO₂)]**
    *   Ligands: Two ammine (NH₃, neutral), one chlorido (Cl⁻, -1 charge), one nitro (NO₂⁻, -1 charge). Ligands are named alphabetically: ammine, chlorido, nitro.
    *   Central metal: Platinum (Pt).
    *   Oxidation state of Pt: Let it be x. x + 2(0) + 1(-1) + 1(-1) = 0 ⇒ x - 2 = 0 ⇒ x = +2. So, Platinum(II).
    *   No counter ion.
    *   Name: **Diamminechloridonitroplatinum(II)**

Question 9

Q9. Write down the IUPAC name for each of the following complexes:
(i) [Co(NH₃)₆]Cl₃
(ii) [Co(NH₃)₅Cl]Cl₂
(iii) K₃[Fe(CN)₆]
(iv) K₃[Fe(C₂O₄)₃]
(v) K₂[PdCl₄]
(vi) [Pt(NH₃)₂Cl(NO₂)

Accepted Answer

The IUPAC names for the given coordination compounds are as follows:

(i) Hexaamminecobalt(III) chloride
(ii) Pentaamminechloridocobalt(III) chloride
(iii) Potassium hexacyanoferrate(III)
(iv) Potassium trioxalatoferrate(III)
(v) Potassium tetrachloridopalladate(II)
(vi) Diamminechloridonitroplatinum(II)
(vii) Tetraaquadichloridochromium(III) chloride
(viii) Dichloridobis(ethane-1,2-diamine)cobalt(III) chloride
(ix) Tetracarbonylnickel(0)

Question 10

Q10. Write down the IUPAC name for each of the following complexes:
(i) [Co(NH₃)₆]Cl₃
(ii) [Co(NH₃)₅Cl]Cl₂
(iii) K₃[Fe(CN)₆]
(iv) K₃[Fe(C₂O₄)₃]
(v) K₂[PdCl₄]
(vi) [Pt(NH₃)₂Cl(NO₂)

Accepted Answer

The IUPAC names for the given coordination compounds are as follows:

(i) **[Co(NH₃)₆]Cl₃**: Hexaamminecobalt(III) chloride
   *   Ligand: ammine (NH₃)
   *   Metal: Cobalt
   *   Oxidation state of Co: Let x be the oxidation state of Co. x + 6(0) + 3(-1) = 0 => x = +3.
   *   Counter ion: Chloride

(ii) **[Co(NH₃)₅Cl]Cl₂**: Pentaamminechloridocobalt(III) chloride
   *   Ligands: ammine (NH₃), chlorido (Cl⁻)
   *   Order: ammine (a) before chlorido (c)
   *   Metal: Cobalt
   *   Oxidation state of Co: Let x be the oxidation state of Co. x + 5(0) + 1(-1) + 2(-1) = 0 => x = +3.
   *   Counter ion: Chloride

(iii) **K₃[Fe(CN)₆]**: Potassium hexacyanoferrate(II)
   *   Cation: Potassium
   *   Ligand: cyano (CN⁻)
   *   Metal: Iron (in anionic complex, 'ferrate')
   *   Oxidation state of Fe: Let x be the oxidation state of Fe. 3(+1) + x + 6(-1) = 0 => x = +3.
   *   Correction: 3(+1) + x + 6(-1) = 0 => x = +3. This is a common mistake. The question is K₃[Fe(CN)₆], which implies Fe(III). Let's recheck. K₃[Fe(CN)₆] is potassium hexacyanoferrate(III). If it was K₄[Fe(CN)₆], it would be potassium hexacyanoferrate(II). Assuming the question is as written, the oxidation state is +3.
   *   Let's re-evaluate for K₃[Fe(CN)₆]: 3(+1) + x + 6(-1) = 0 => x - 3 = 0 => x = +3. So, Potassium hexacyanoferrate(III).

(iv) **K₃[Fe(C₂O₄)₃]**: Potassium trioxalatoferrate(III)
   *   Cation: Potassium
   *   Ligand: oxalato (C₂O₄²⁻, bidentate)
   *   Metal: Iron (in anionic complex, 'ferrate')
   *   Oxidation state of Fe: Let x be the oxidation state of Fe. 3(+1) + x + 3(-2) = 0 => x - 3 = 0 => x = +3.

(v) **K₂[PdCl₄]**: Potassium tetrachloridopalladate(II)
   *   Cation: Potassium
   *   Ligand: chlorido (Cl⁻)
   *   Metal: Palladium (in anionic complex, 'palladate')
   *   Oxidation state of Pd: Let x be the oxidation state of Pd. 2(+1) + x + 4(-1) = 0 => x - 2 = 0 => x = +2.

(vi) **[Pt(NH₃)₂Cl(NO₂)]**: Diamminechloridonitrito-N-platinum(II)
   *   Ligands: ammine (NH₃), chlorido (Cl⁻), nitrito-N (NO₂⁻, ambidentate, N-bonded)
   *   Order: ammine (a), chlorido (c), nitrito (n)
   *   Metal: Platinum
   *   Oxidation state of Pt: Let x be the oxidation state of Pt. x + 2(0) + 1(-1) + 1(-1) = 0 => x = +2.

(vii) **[Cr(H₂O)₄Cl₂]Cl**: Tetraaquachloridochromium(III) chloride
   *   Ligands: aqua (H₂O), chlorido (Cl⁻)
   *   Order: aqua (a) before chlorido (c)
   *   Metal: Chromium
   *   Oxidation state of Cr: Let x be the oxidation state of Cr. x + 4(0) + 2(-1) + 1(-1) = 0 => x - 3 = 0 => x = +3.
   *   Counter ion: Chloride

(viii) **[Co(en)₂Cl₂]Cl (en = ethane-1,2-diamine)**: Dichloridobis(ethane-1,2-diamine)cobalt(III) chloride
   *   Ligands: chlorido (Cl⁻), ethane-1,2-diamine (en)
   *   Order: chlorido (c) before ethane-1,2-diamine (e)
   *   Prefix for 'en': 'bis' because 'ethane-1,2-diamine' already contains a numerical prefix.
   *   Metal: Cobalt
   *   Oxidation state of Co: Let x be the oxidation state of Co. x + 2(-1) + 2(0) + 1(-1) = 0 => x - 3 = 0 => x = +3.
   *   Counter ion: Chloride

(ix) **[Ni(CO)₄]**: Tetracarbonylnickel(0)
   *   Ligand: carbonyl (CO)
   *   Metal: Nickel
   *   Oxidation state of Ni: Let x be the oxidation state of Ni. x + 4(0) = 0 => x = 0.
   *   Note: For neutral complexes, the metal name is used without 'ate' suffix.

Question 11

Q11. Write the formulas for the following coordination compounds:
(i) Tetraammineaquachloridocobalt(III) chloride
(ii) Potassium hexacyanoferrate(II)
(iii) Potassium trioxalatoalumin

Accepted Answer

The formulas for the given coordination compounds are as follows:

(i) **Tetraammineaquachloridocobalt(III) chloride**
   *   Central metal: Cobalt (Co)
   *   Oxidation state: +3
   *   Ligands: Tetraammine (4 NH₃), aqua (1 H₂O), chlorido (1 Cl⁻)
   *   Coordination sphere charge: Co(III) + 4(NH₃) + 1(H₂O) + 1(Cl⁻) = +3 + 4(0) + 1(0) + 1(-1) = +2
   *   Counter ion: Chloride (Cl⁻) to balance +2 charge. Needs 2 Cl⁻ ions.
   *   Formula: **[Co(NH₃)₄(H₂O)Cl]Cl₂**

(ii) **Potassium hexacyanoferrate(II)**
   *   Cation: Potassium (K⁺)
   *   Central metal: Iron (Fe)
   *   Oxidation state: +2
   *   Ligands: Hexacyano (6 CN⁻)
   *   Coordination sphere charge: Fe(II) + 6(CN⁻) = +2 + 6(-1) = -4
   *   Counter ion: Potassium (K⁺) to balance -4 charge. Needs 4 K⁺ ions.
   *   Formula: **K₄[Fe(CN)₆]**

(iii) **Potassium trioxalatoaluminate(III)**
   *   Cation: Potassium (K⁺)
   *   Central metal: Aluminium (Al)
   *   Oxidation state: +3
   *   Ligands: Trioxalato (3 C₂O₄²⁻)
   *   Coordination sphere charge: Al(III) + 3(C₂O₄²⁻) = +3 + 3(-2) = -3
   *   Counter ion: Potassium (K⁺) to balance -3 charge. Needs 3 K⁺ ions.
   *   Formula: **K₃[Al(C₂O₄)₃]**

(iv) **Dichlorobis(ethane-1,2-diamine)cobalt(III) chloride**
   *   Central metal: Cobalt (Co)
   *   Oxidation state: +3
   *   Ligands: Dichlorido (2 Cl⁻), bis(ethane-1,2-diamine) (2 'en' or C₂H₄(NH₂)₂)
   *   Coordination sphere charge: Co(III) + 2(Cl⁻) + 2(en) = +3 + 2(-1) + 2(0) = +1
   *   Counter ion: Chloride (Cl⁻) to balance +1 charge. Needs 1 Cl⁻ ion.
   *   Formula: **[Co(en)₂Cl₂]Cl**

(v) **Pentacarbonyliron(0)**
   *   Central metal: Iron (Fe)
   *   Oxidation state: 0
   *   Ligands: Pentacarbonyl (5 CO)
   *   Coordination sphere charge: Fe(0) + 5(CO) = 0 + 5(0) = 0
   *   No counter ions needed as the complex is neutral.
   *   Formula: **[Fe(CO)₅]**

Question 12

Q12. List all types of isomerism shown by the following complexes and draw the structures of these isomers:
(i) [Co(NH₃)₅Br]SO₄
(ii) [Co(en)₂Cl₂]⁺
(iii) [Ni(NH₃)₂Cl₂]

Accepted Answer

Isomerism in coordination compounds refers to the existence of two or more compounds with the same chemical formula but different arrangements of atoms. These can be broadly classified into structural isomerism and stereoisomerism.

### (i) [Co(NH₃)₅Br]SO₄

This complex shows **Ionisation Isomerism**.

**Explanation:** Ionisation isomers arise when the counter ion in a complex salt is itself a potential ligand and can displace a ligand that is already coordinated to the central metal ion. In this case, the bromide (Br⁻) ligand inside the coordination sphere can exchange with the sulfate (SO₄²⁻) counter ion outside the coordination sphere.

**Isomers:**
1.  **[Co(NH₃)₅Br]SO₄**: Pentaamminebromidocobalt(III) sulfate
    *   In this isomer, bromide is coordinated to cobalt, and sulfate is the counter ion.
2.  **[Co(NH₃)₅SO₄]Br**: Pentaamminesulfatocobalt(III) bromide
    *   In this isomer, sulfate is coordinated to cobalt, and bromide is the counter ion.

**Distinguishing Test:** These isomers can be distinguished by adding BaCl₂ solution. The first isomer, [Co(NH₃)₅Br]SO₄, will give a white precipitate of BaSO₄ with BaCl₂, indicating the presence of free SO₄²⁻ ions. The second isomer, [Co(NH₃)₅SO₄]Br, will not give a precipitate with BaCl₂ but will give a precipitate of AgBr with AgNO₃ solution, indicating free Br⁻ ions.

**Geometrical/Optical Isomerism:** This complex, being of the `MA₅B` type, does not exhibit geometrical isomerism. It is also not chiral, so it does not show optical isomerism.

Question 13

Q13. What is meant by stability of a coordination compound in solution? Discuss the factors which affect the stability of coordination compounds.

Accepted Answer

The stability of a coordination compound in solution refers to the extent of association between the metal ion and the ligands present in the solution. It is quantitatively expressed by the stability constant (or formation constant). The higher the stability constant, the more stable the complex.

When a metal ion M reacts with a ligand L to form a complex ML, the reaction can be represented as:

M + L ⇌ ML

The stability constant (K1) for this step is given by: K1 = [ML] / ([M][L])

If the complexation occurs in multiple steps, for example:
M + nL ⇌ MLn

The overall stability constant (βn) is the product of the stepwise stability constants:
βn = K1 × K2 × ... × Kn = [MLn] / ([M][L]n)

**Factors Affecting the Stability of Coordination Compounds:**

1.  **Nature of the Metal Ion:**
    *   **Charge on the Metal Ion:** Generally, complexes formed by metal ions with higher positive charges are more stable. A higher charge leads to stronger electrostatic attraction between the metal ion and the ligands. For example, Fe3+ forms more stable complexes than Fe2+.
    *   **Size of the Metal Ion:** Smaller metal ions tend to form more stable complexes due to higher charge density, leading to stronger electrostatic interactions with ligands. For example, among transition metals of the same period, stability generally increases with decreasing ionic radius.
    *   **Electronic Configuration (Crystal Field Stabilization Energy - CFSE):** Metal ions that gain significant CFSE upon complexation tend to form more stable complexes. For example, d3, d6 (low spin), and d8 (square planar) configurations often lead to high stability.

2.  **Nature of the Ligand:**
    *   **Basicity of the Ligand:** Stronger bases (ligands that can donate electron pairs more readily) generally form more stable complexes. For example, NH3 is a stronger base than H2O, so ammine complexes are generally more stable than aqua complexes.
    *   **Chelate Effect:** Polydentate ligands (ligands that can bind to the metal ion through more than one donor atom) form more stable complexes than monodentate ligands with similar donor atoms. This enhanced stability is known as the chelate effect. The chelate effect is primarily an entropy-driven phenomenon. When a polydentate ligand replaces several monodentate ligands, the number of species in solution often increases, leading to an increase in entropy, which favors the formation of the chelate complex.
    *   **Size of the Chelate Ring:** Five-membered and six-membered chelate rings are generally the most stable. Smaller or larger rings experience more strain and are less stable.
    *   **Pi-acceptor Character of Ligand:** Ligands that can accept electron density from the metal via pi-backbonding (e.g., CO, CN-) form very stable complexes, especially with transition metals in low oxidation states. This synergic bonding strengthens both the M-L sigma bond and the M-L pi bond.

3.  **Steric Hindrance:** Bulky ligands can cause steric hindrance, which can reduce the stability of the complex by preventing optimal approach and bonding with the metal ion.

4.  **Temperature:** Stability constants are temperature-dependent. For exothermic complex formation reactions, increasing temperature generally decreases stability.

Question 14

Q14. What is meant by the chelate effect? Give an example.

Accepted Answer

The **chelate effect** refers to the enhanced stability of a coordination complex containing chelate rings (formed by polydentate ligands) compared to a similar complex containing only monodentate ligands. This enhanced stability is primarily due to a favorable entropy change (increase in disorder) during the formation of the chelate complex.

When a polydentate ligand replaces several monodentate ligands from a metal ion, the number of species in the solution generally increases. For example, if a bidentate ligand (like ethylenediamine, 'en') replaces two monodentate ligands (like ammonia, NH3), the reaction involves one polydentate ligand replacing two monodentate ligands, leading to an increase in the number of molecules/ions in the solution, thus increasing entropy.

**Example:**
Consider the formation of the complex of nickel(II) with ethylenediamine (en) versus ammonia (NH3):

1.  **Formation of hexaammine nickel(II) ion:**
    [Ni(H2O)6]2+ + 6NH3 ⇌ [Ni(NH3)6]2+ + 6H2O
    (Overall stability constant, log β6 ≈ 8.6)

2.  **Formation of tris(ethylenediamine) nickel(II) ion:**
    [Ni(H2O)6]2+ + 3en ⇌ [Ni(en)3]2+ + 6H2O
    (Overall stability constant, log β3 ≈ 18.3)

In both cases, the donor atoms are nitrogen, and the nickel ion is in the +2 oxidation state. However, the tris(ethylenediamine) complex, [Ni(en)3]2+, is significantly more stable (higher stability constant) than the hexaammine complex, [Ni(NH3)6]2+. This is because ethylenediamine is a bidentate ligand, forming stable five-membered chelate rings with the nickel ion, leading to the chelate effect.

Question 15

Q15. Discuss the nature of bonding in metal carbonyls.

Accepted Answer

The bonding in metal carbonyls, such as Ni(CO)4, Fe(CO)5, and Cr(CO)6, is unique and involves a synergistic effect between the metal atom and the carbon monoxide (CO) ligand. This bonding can be described in two main parts:

1.  **Sigma (σ) bond formation:** The carbon atom of the CO ligand acts as a donor. It donates a lone pair of electrons from its highest occupied molecular orbital (HOMO), which is a slightly bonding σ orbital, into a vacant, suitable symmetry d-orbital of the metal atom. This forms a metal-carbon sigma (M-C σ) bond. This donation increases the electron density on the metal atom.

2.  **Pi (π) bond formation (Back-bonding or Synergic Bonding):** To compensate for the increased electron density on the metal due to σ-donation, and to strengthen the M-C bond, the metal atom donates a pair of electrons from its filled d-orbitals into the vacant, antibonding π* molecular orbitals of the CO ligand. This forms a metal-carbon pi (M-C π) bond, often referred to as 'back-bonding' or 'π-back donation'.

**Synergic Effect:**
This combination of σ-donation from CO to metal and π-back-donation from metal to CO is called the **synergic effect** or **synergic bonding**. It is a mutually reinforcing process:

*   The σ-donation from CO to metal strengthens the M-C bond.
*   The π-back-donation from metal to CO also strengthens the M-C bond. Simultaneously, this back-donation weakens the C-O bond because electrons are being donated into the antibonding orbital of CO. This weakening is often observed as a decrease in the C-O stretching frequency in IR spectroscopy when CO coordinates to a metal.

**Consequences of Synergic Bonding:**

*   **Enhanced Stability:** The synergic bonding mechanism significantly stabilizes the metal-carbonyl complexes, especially for metals in low oxidation states (often zero). This is because the metal can effectively delocalize its excess electron density into the ligand's antibonding orbitals, preventing charge build-up on the metal.
*   **Shortened M-C bond and Lengthened C-O bond:** The M-C bond acquires partial double bond character due to back-bonding, making it shorter and stronger than a pure single bond. Conversely, the C-O bond becomes longer and weaker due to electron density entering its antibonding orbitals.
*   **Low Oxidation States:** This type of bonding is crucial for the existence of metal carbonyls where the metal is in a zero or very low positive oxidation state, as it allows the metal to shed excess electron density.

In essence, the metal-carbon bond in metal carbonyls has both σ and π character, resulting from the synergic interplay of ligand-to-metal σ donation and metal-to-ligand π back-donation.

Chapter 9: Coordination Compounds