Question:

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:

(i) When aqueous copper sulphate solution is mixed with aqueous potassium fluoride, a green precipitate is formed. This is because the copper ions in the copper sulphate solution react with the fluoride ions in the potassium fluoride solution to form copper fluoride, which is insoluble in water and forms a green precipitate.

(ii) When aqueous copper sulphate solution is mixed with aqueous potassium chloride, a bright green solution is formed. This is because the copper ions in the copper sulphate solution react with the chloride ions in the potassium chloride solution to form copper chloride, which is soluble in water and forms a bright green solution.

Question:

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:

Answer: When excess of aqueous KCN is added to an aqueous solution of copper sulphate, a coordination entity of copper(I) cyanide is formed. This is because the cyanide ion (CN^-) is a strong ligand and can form a strong coordination bond with the copper(I) ion.

The reason why no precipitate of copper sulphide is obtained when H2S(g) is passed through this solution is because the H2S(g) reacts with the CN^- ligands first, forming hydrogen cyanide (HCN) and sulfur, instead of reacting with the copper(I) ion to form copper sulphide.

Question:

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

Answer:

(i) Zn(OH)4 (ii) K2PdCl4 (iii) Pt(NH₃)2Cl₂ (iv) K4[Ni(CN)4] (v) [Co(NO2)5(NH₃)5]O (vi) [Co(NH₃)₆]SO₄ (vii) K3[Cr(C₂O₄)3] (viii) [Pt(NH₃)₆]Cl4 (ix) CuBr4 (x) [Co(NO2)5(NH₃)5]N

Question:

Explain the bonding in coordination compounds in terms of Werners postulates.

Answer:

1. First, explain what Werners postulates are. Werners postulates are a set of rules proposed by Alfred Werner in 1893 that explain the geometry of coordination compounds. The postulates state that the coordination number of the central atom is equal to the number of ligands attached to it, and that the ligands are arranged in an octahedral, tetrahedral, or square planar geometry.

2. Next, explain how Werners postulates can be used to explain the bonding in coordination compounds. Coordination compounds are formed when a central metal atom is surrounded by ligands. The ligands are typically organic molecules or anions that form coordinate covalent bonds with the central atom. The number of ligands attached to the central atom determines the coordination number, and Werners postulates can be used to explain the geometry of the coordination compound. The postulates state that the coordination number of the central atom is equal to the number of ligands attached to it, and that the ligands are arranged in an octahedral, tetrahedral, or square planar geometry. This explains why some coordination compounds have a particular shape or geometry.

Question:

How many geometrical isomers are possible in the following coordination entities? (i) [Cr(C₂​O₄​)₃]^3- (ii) [Co(NH₃)₃Cl₃]

Answer:

(i) [Cr(C₂​O₄​)₃]^3-

There are no geometrical isomers possible in [Cr(C₂​O₄​)₃]^3- since the ligands (C₂​O₄​) are all the same and bonded to the central atom (Cr) in the same way.

(ii) [Co(NH₃)₃Cl₃]

There are 6 geometrical isomers possible in [Co(NH₃)₃Cl₃]. This is because the ligands (NH₃ and Cl₃) are different and can bond to the central atom (Co) in different ways.

Question:

Amongst the following, the most stable complex is: (i) [Fe(H₂O)₆]³⁺ (ii) [Fe(NH₃)₆]³⁺ (iii) [Fe(C₂​O₄​)₃]^3- (iv) [FeCl6​]^3-

Answer:

Step 1: Understand the question.

The question is asking which of the given complexes is the most stable.

Step 2: Analyze the complexes.

(i) [Fe(H₂O)₆]³⁺: This complex has six water molecules surrounding the Fe3+ ion.

(ii) [Fe(NH₃)₆]³⁺: This complex has six ammonia molecules surrounding the Fe3+ ion.

(iii) [Fe(C₂​O₄​)₃]^3-: This complex has three oxalate molecules surrounding the Fe3+ ion.

(iv) [FeCl6​]^3-: This complex has six chloride ions surrounding the Fe3+ ion.

Step 3: Determine the most stable complex.

The most stable complex is [Fe(NH₃)₆]³⁺ because the ammonia molecules are strong field ligands, which means they can stabilize the Fe3+ ion more than the other complexes.

Question:

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

Answer:

1. The geometrical isomers of [Pt(NH₃)(Br)(Cl)(py)] are:

A. cis-[Pt(NH₃)(Br)(Cl)(py)] B. trans-[Pt(NH₃)(Br)(Cl)(py)]

2. Both of these geometrical isomers can exhibit optical isomers, as the py ligand can exist as a pair of enantiomers. Therefore, there are a total of four optical isomers of [Pt(NH₃)(Br)(Cl)(py)].

Question:

A solution of [Ni(H₂O)₆]²⁺ is green but a solution of [Ni(CN)₄]^2- is colourless, Explain.

Answer:

1. Nickel is a transition metal which can exist in different oxidation states, ranging from +2 to +4.

2. [Ni(H₂O)₆]²⁺ is a complex ion with nickel in its +2 oxidation state. This complex ion is green in colour.

3. [Ni(CN)₄]^2- is a complex ion with nickel in its +4 oxidation state. This complex ion is colourless.

4. The colour of the complex ion is due to the d-d transition of electrons in the metal’s outermost shell. In the +2 oxidation state, there is an energy gap between the d-orbitals and electrons can absorb energy to move from one orbital to another, resulting in the absorption of visible light and the complex ion appears green.

5. In the +4 oxidation state, the energy gap between the d-orbitals is too small for electrons to absorb energy and move from one orbital to another, resulting in no visible light being absorbed and the complex ion appearing colourless.

Question:

[Cr(NH₃)₆]³⁺ is paramagnetic while [Ni(CN)₄]^2- is diamagnetic. Explain why?

Answer:

1. Paramagnetism and diamagnetism are both types of magnetism. Paramagnetic materials are attracted to a magnetic field, while diamagnetic materials are repelled by a magnetic field.

2. The [Cr(NH₃)₆]³⁺ ion contains six NH₃ molecules, each of which has an unpaired electron. These unpaired electrons are attracted to an external magnetic field, which is why [Cr(NH₃₃)₆]³⁺ is paramagnetic.

3. The [Ni(CN)₄]^2- ion does not contain any unpaired electrons, so it is not attracted to an external magnetic field. This is why [Ni(CN)₄]^2- is diamagnetic.

Question:

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⁺?

Answer:

1. [Ni(NO₂)₆]^4-

2. [Ni(NH₃)₆]²⁺

3. [Ni(H₂O)₆]²⁺

4. Red

5. Orange

6. Yellow

7. Green

8. Blue

9. Indigo

10. Violet

Question:

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

Answer:

1. The chelate effect is a type of bonding that occurs when a metal ion forms a ring-like structure with a ligand, which is an atom or molecule that binds to the metal ion.

2. An example of the chelate effect is the formation of EDTA, a complex molecule that binds to metal ions such as calcium, magnesium, iron, and zinc. EDTA is often used as a chelating agent in cleaning and industrial processes.

Question:

Give the oxidation state, d orbital occupation and coordination number of the central metal ion in the following complexes: (i) K3​[Co(C₂​O₄​)₃] (ii) cis−[CrCl₂(en)₂]Cl (iii) (NH₄)2​[CoF4​] (iv) [Mn(H₂O)₆]SO₄

Answer:

(i) Oxidation state: +3, d orbital occupation: 5, coordination number: 6

(ii) Oxidation state: +3, d orbital occupation: 6, coordination number: 4

(iii) Oxidation state: +2, d orbital occupation: 6, coordination number: 4

(iv) Oxidation state: +2, d orbital occupation: 5, coordination number: 6

Question:

Draw all the isomers (geometrical and optical) of: (i) [CoCl₂(en)₂]⁺ (ii) [Co(NH₃)Cl(en)₂]²⁺ (iii) [Co(NH₃)2​Cl₂(en)]⁺

Answer:

(i) [CoCl₂(en)₂]⁺

Geometrical isomers:

1. cis-[CoCl₂(en)₂]⁺
2. trans-[CoCl₂(en)₂]⁺

Optical Isomers:

1. [CoCl₂(en)₂]⁺ (R)-form
2. [CoCl₂(en)₂]⁺ (S)-form

(ii) [Co(NH₃)Cl(en)₂]²⁺

Geometrical isomers:

1. cis-[Co(NH₃)Cl(en)₂]²⁺
2. trans-[Co(NH₃)Cl(en)₂]²⁺

Optical Isomers:

1. [Co(NH₃)Cl(en)₂]²⁺ (R)-form
2. [Co(NH₃)Cl(en)₂]²⁺ (S)-form

(iii) [Co(NH₃)2​Cl₂(en)]⁺

Geometrical isomers:

1. cis-[Co(NH₃)2​Cl₂(en)]⁺
2. trans-[Co(NH₃)2​Cl₂(en)]⁺

Optical Isomers:

1. [Co(NH₃)2​Cl₂(en)]⁺ (R)-form
2. [Co(NH₃)2​Cl₂(en)]⁺ (S)-form

Question:

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​(C₂​O₄​)2​].3H₂O (ii) [Co(NH₃₃)5Cl^-]Cl₂ (iii) [CrCl₃(py)₃] (iv) Cs[FeCl₄] (v) K₄​[Mn(CN)₆]

Answer:

(i) Potassium dichromate dihydrate, Oxidation state: +6, Electronic configuration: [Ar] 3d³ 4s0, Coordination number: 6, Stereochemistry: octahedral, Magnetic moment: 0

(ii) Pentaamminechlorocobalt(III) chloride, Oxidation state: +3, Electronic configuration: [Ar] 3d⁷ 4s², Coordination number: 6, Stereochemistry: octahedral, Magnetic moment: 1.73

(iii) Trichloridotris(pyridine)chromium(III), Oxidation state: +3, Electronic configuration: [Ar] 3d³ 4s0, Coordination number: 6, Stereochemistry: octahedral, Magnetic moment: 0

(iv) Cesium tetrachloroferrate(II), Oxidation state: +2, Electronic configuration: [Ar] 3d6 4s², Coordination number: 4, Stereochemistry: tetrahedral, Magnetic moment: 0

(v) Tetrakis(cyanomanganate) potassium, Oxidation state: +4, Electronic configuration: [Ar] 3d5 4s1, Coordination number: 6, Stereochemistry: octahedral, Magnetic moment: 0

Question:

What is crystal field splitting energy? How does the magnitude of Δ0​ decide the actual configuration of d orbitals in a coordination entity?

Answer:

Answer:

1. Crystal field splitting energy (Δ0) is the energy difference between the energy levels of the d orbitals in a coordination entity when it is exposed to an external electric field.

2. The magnitude of Δ0 determines the actual configuration of d orbitals in a coordination entity. A higher magnitude of Δ0 will result in the d orbitals being split into two sets of energy levels, one higher and one lower, while a lower magnitude of Δ0 will result in the d orbitals being split into more sets of energy levels. The actual configuration of the d orbitals will depend on the magnitude of Δ0 and the arrangement of the ligands around the central atom.

Question:

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

Answer:

(i) Role of Coordination Compounds in Biological Systems: Coordination compounds play an important role in biological systems. For example, hemoglobin, the protein responsible for transporting oxygen in the blood, contains a coordination compound called heme, which contains iron. Heme is responsible for binding oxygen molecules, allowing them to be transported around the body.

(ii) Role of Coordination Compounds in Medicinal Chemistry: Coordination compounds are also used in medicinal chemistry. For example, some coordination compounds are used as drugs to treat diseases, such as platinum-based chemotherapy drugs used to treat cancer.

(iii) Role of Coordination Compounds in Analytical Chemistry: Coordination compounds are also used in analytical chemistry. For example, they are used in spectroscopic techniques such as infrared spectroscopy and nuclear magnetic resonance (NMR) spectroscopy to identify and characterize unknown compounds.

(iv) Role of Coordination Compounds in Extraction/Metallurgy of Metals: Coordination compounds are also used in the extraction and metallurgy of metals. For example, coordination compounds are used to extract metals from ores, such as the use of copper sulfate to extract copper from copper ore.

Question:

Draw figure to show the splitting of d orbitals in an octahedral crystal field.

Answer:

1. Draw an octahedral shape with six sides.

2. Inside the octahedral shape, draw three circles to represent the d orbitals.

3. Label each circle with the appropriate letter (dxy, dyz, and dxz).

4. Draw arrows pointing outward from each circle to represent the splitting of the orbitals.

5. Add a legend to the figure to explain the arrows.

Question:

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

Answer:

(i) 6 ions are produced from the complex Co(NH₃)₆Cl₂ in solution. This includes 6 NH₃ ions, 2 Cl^- ions, and 1 Co³⁺ ion.

Question:

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]

Answer:

(i) [Co(H₂O)(CN)(en)₂]²⁺: Co = +2, H = +1, C = -4, N = -3, en = -1

(ii) [CoBr₂(en)₂]+: Co = +3, Br = -1, en = -1

(iii) [PtCl₄]^2-: Pt = +4, Cl = -1

(iv) K₃[Fe(CN)₆]: K = +1, Fe = +3, C = -4, N = -3

(v) [Cr(NH₃)₃Cl₃]: Cr = +3, N = -3, H = +1, Cl = -1

Question:

FeSO₄​ solution mixed with (NH₄)2​SO₄​ 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?

Answer:

1. FeSO₄​ is a salt composed of iron (Fe) and sulfate (SO₄​) ions. (NH₄)2​SO₄​ is a salt composed of ammonium (NH₄) and sulfate (SO₄​) ions. When mixed in a 1:1 molar ratio, the FeSO₄​ solution and (NH₄)2​SO₄​ solution form a complex ion called iron ammonium sulfate. This complex ion is able to give a test for the presence of iron (Fe²⁺).

2. CuSO₄​ is a salt composed of copper (Cu) and sulfate (SO₄​) ions. When mixed with aqueous ammonia in a 1:4 molar ratio, the CuSO₄​ solution does not form a complex ion with the ammonia. Therefore, it is unable to give a test for the presence of copper (Cu²⁺).

Question:

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)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]

Answer:

(i) Hexaamminechlorochromium(III) chloride (ii) Diamminechloroethanaminoplatinum(IV) chloride (iii) Hexaaquatitanium(III) ion (iv) Tetraamminechloronitrito-cobalt(III) chloride (v) Hexaaquamanganese(II) ion (vi) Tetrachloronickel(II) ion (vii) Hexaamminenickel(II) dichloride (viii) Ethylenediaminetri-cobalt(III) ion (ix) Tetracarbonylnickel(0)

Question:

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

Answer:

1. Spectrochemical series is a list of ligands arranged in order of their ability to split the d-d energy levels of a transition metal complex.

2. A weak field ligand is a ligand that does not cause a large splitting of the d-d energy levels. This ligand is usually of low-spin configuration and has a low-spin coupling constant.

3. A strong field ligand is a ligand that causes a large splitting of the d-d energy levels. This ligand is usually of high-spin configuration and has a high-spin coupling constant.

Question:

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

Answer:

Answer: Unidentate ligands are ligands with one binding site, such as ammonia (NH₃) and chloride (Cl^-).

Didentate ligands are ligands with two binding sites, such as ethylene diamine (en) and oxalate (C₂O₄).

Ambidentate ligands are ligands with more than one type of binding site, such as nitrite (NO^2-) and thiocyanate (SCN^-).

Examples of unidentate ligands include ammonia (NH₃) and chloride (Cl^-).

Examples of didentate ligands include ethylene diamine (en) and oxalate (C₂O₄).

Examples of ambidentate ligands include nitrite (NO^2-) and thiocyanate (SCN^-).

Question:

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⁺

Answer:

Step 1: Magnetic moment is a measure of the strength of the magnetic field generated by an object.

Step 2: The magnetic moment of an ion is determined by its electron configuration.

Step 3: Chromium (Cr) has an electron configuration of 3d5 4s1, while iron (Fe) has an electron configuration of 3d⁶ 4s² and zinc (Zn) has an electron configuration of 3d¹⁰ 4s².

Step 4: Chromium has the highest magnetic moment value of the three ions, since it has the largest number of unpaired electrons (5).

Question:

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

Answer:

Coordination Entity: A coordination entity is a species formed when a central atom or ion is surrounded by a group of ligands. Examples include [Fe(CN)₆]^4- (hexacyanoferrate(III)) and [Co(NH₃)₆]³⁺ (hexamminecobalt(III)).

Ligand: A ligand is an atom or molecule that binds to a central atom or ion to form a coordination entity. Examples include CN^- (cyanide) and NH₃ (ammonia).

Coordination Number: The coordination number of a central atom or ion is the number of ligands that are bound to it. Examples include [Fe(CN)₆]^4- (hexacyanoferrate(III)) with a coordination number of 6, and [Co(NH₃)₆]³⁺ (hexamminecobalt(III)) with a coordination number of 6.

Coordination Polyhedron: A coordination polyhedron is a three-dimensional structure formed when a central atom or ion is surrounded by a group of ligands. Examples include octahedron (formed by six ligands) and tetrahedron (formed by four ligands).

Homoleptic: A homoleptic compound is one in which all the ligands bound to a central atom or ion are the same. Examples include [Fe(CN)₆]^4- (hexacyanoferrate(III)) and [Co(NH₃)₆]³⁺ (hexamminecobalt(III)).

Heteroleptic: A heteroleptic compound is one in which the ligands bound to a central atom or ion are different. Examples include [Fe(CO)4]2- (tetracarbonylferrate(II)) and [NiCl₂(NH₃)4] (tetraamminechloronitrosylnickel(II)).

Question:

[Fe(CN)₆]^4- and [Fe(H₂O)₆]²⁺ are of different colours in dilute solutions. Why?

Answer:

1. The two ions, [Fe(CN)₆]^4- and [Fe(H₂O)₆]²⁺, are composed of different elements, iron and water, respectively.

2. The two ions have different structures, with the [Fe(CN)₆]^4- ion having a hexagonal structure and the [Fe(H₂O)₆]²⁺ ion having an octahedral structure.

3. The two ions have different charge densities, with the [Fe(CN)₆]^4- ion having a higher charge density than the [Fe(H₂O)₆]²⁺ ion.

4. The two ions interact differently with light, with the [Fe(CN)₆]^4- ion absorbing light in the visible spectrum and the [Fe(H₂O)₆]²⁺ ion reflecting light in the visible spectrum.

5. As a result, the two ions appear different colors in dilute solutions, with the [Fe(CN)₆]^4- ion appearing purple and the [Fe(H₂O)₆]²⁺ ion appearing green.

Question:

What is meant by stability of a coordination compound in solution? State the factors which govern stability of complexes.

Answer:

Answer: Stability of a coordination compound in solution refers to the ability of a complex to resist changes in its chemical structure or composition when exposed to different environmental conditions. Factors which govern the stability of complexes include the nature of the ligands, the charge on the central metal ion, the coordination number, the oxidation state of the metal ion, and the nature of the solvent.

Question:

Discuss the nature of bonding in metal carbonyls.

Answer:

1. First, define what metal carbonyls are: metal carbonyls are compounds that contain metal-carbon bonds, which are formed by the combination of a metal atom and a carbon monoxide (CO) molecule.

2. Next, explain the nature of the metal-carbon bond in metal carbonyls: the metal-carbon bond is covalent, meaning that the electrons in the bond are shared between the metal atom and the carbon monoxide molecule.

3. Describe the stability of the metal-carbon bond: the metal-carbon bond is relatively stable due to the fact that the metal atom has a higher electronegativity than the carbon monoxide molecule, resulting in a partial positive charge on the metal atom and a partial negative charge on the carbon monoxide molecule.

4. Explain the different types of metal carbonyls: there are two main types of metal carbonyls, namely, neutral metal carbonyls and ionic metal carbonyls. Neutral metal carbonyls are composed of a metal atom and a carbon monoxide molecule, while ionic metal carbonyls are composed of a metal cation and a carbon monoxide anion.

5. Finally, discuss the implications of metal carbonyls: metal carbonyls are important in many industrial processes, such as the production of fuels and chemicals, and they can also be used as catalysts in certain reactions.

Question:

Draw the structures of optical isomers of: (i) [Cr(C₂​O₄​)₃]^3- (ii) [PtCl₂(en)₂]²⁺ (iii) [Cr(NH₃)2​Cl₂(en)]⁺

Answer:

(i) [Cr(C₂O₄)3]^3-

Optical Isomer 1:

Optical Isomer 2:

(ii) [PtCl₂(en)2]²⁺

Optical Isomer 1:

Optical Isomer 2:

(iii) [Cr(NH₃)2Cl₂(en)]⁺

Optical Isomer 1:

Optical Isomer 2:

Question:

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

Answer:

1. Structural Isomerism: Structural isomerism occurs when the same atoms are connected in different arrangements. An example of structural isomerism in coordination compounds is cis-diaquabis(ethylenediamine)cobalt(III) chloride and trans-diaquabis(ethylenediamine)cobalt(III) chloride.

2. Linkage Isomerism: Linkage isomerism occurs when the same atoms are connected in different ways. An example of linkage isomerism in coordination compounds is diaquabis(ethylenediamine)cobalt(III) chloride and diaquabis(ethylenediamine)cobalt(III) bromide.

3. Coordination Isomerism: Coordination isomerism occurs when the same atoms are connected in different ways and the coordination number of the metal ion is different. An example of coordination isomerism in coordination compounds is diaquabis(ethylenediamine)cobalt(III) chloride and diaquadichlorocobalt(III) chloride.

4. Geometric Isomerism: Geometric isomerism occurs when the same atoms are connected in different ways and the geometry of the molecule is different. An example of geometric isomerism in coordination compounds is cis-diaquabis(ethylenediamine)cobalt(III) chloride and trans-diaquabis(ethylenediamine)cobalt(III) chloride.

Question:

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(C₂​O₄​)₃]^3- (iv) [CoF₆]^3-

Answer:

(i) [Fe(CN)₆]^4-: The bonding in [Fe(CN)₆]^4- is a combination of covalent and ionic bonding. The Fe-C and Fe-N bonds are covalent in nature and the Fe-CN bonds are ionic in nature.

(ii) [FeF₆]^3-: The bonding in [FeF₆]^3- is mainly ionic in nature. The Fe-F bonds are ionic due to the difference in electronegativity of Fe and F.

(iii) [Co(C₂​O₄​)₃]^3-: The bonding in [Co(C₂​O₄​)₃]^3- is a combination of covalent and ionic bonding. The Co-C bonds are covalent in nature and the Co-O bonds are ionic in nature.

(iv) [CoF₆]^3-: The bonding in [CoF₆]^3- is mainly ionic in nature. The Co-F bonds are ionic due to the difference in electronegativity of Co and F.