CHEM 471- The Lewis structure and the hybridization of molecules

CHEM 471- The Lewis structure and the hybridization of

molecules

Problems and Solutions

1. Consider the reaction of XeF4 with water:

a) Draw the Lewis structure of XeF4 and XeO2F2. b) Determine the hybridization of

Xe in both molecules.

2. Explain the lanthanide contraction and discuss two significant consequences of this

phenomenon in inorganic chemistry.

Solution:

Lanthanide contraction: - Poor shielding of 4f electrons - Increased effective nuclear

charge across the series - Decrease in atomic and ionic radii across the series

Two significant consequences: 1. Similar sizes of later lanthanides and early

transition metals (e.g., Lu3+ and Y3+) 2. Difficulty in separating lanthanides due to

similar chemical properties

3. For the complex [Fe(CN)6]4−: a) Determine the oxidation state of Fe. b) Predict

whether the complex is high-spin or low-spin. c) Calculate its magnetic moment in

4. Describe the bonding in the linear [AuCl2]− complex using Molecular Orbital theory.

Solution:

- Au+ has a d10 configuration - Linear geometry suggests sp hybridization - Two sp

hybrid orbitals form σ bonds with Cl 3p orbitals - Filled d orbitals participate in π

back-bonding with empty Cl 3p orbitals - This π back-bonding contributes to the

stability of the linear geometry

5. Compare and contrast the acid-base behavior of BF3 and BCl3.

Solution:

Similarities: - Both are Lewis acids due to empty p orbital on boron - Both form

adducts with Lewis bases

[CoF6]3−: - Co3+ is d6 - F− is a weak-field ligand - High-spin configuration: (t2𝑔)4

7. Propose a synthetic route for the preparation of [Ru(bpy)3](PF6)2 from RuCl3 ·

nH2O.

Solution:

1. RuCl3 · nH2O + 3 bpy → [Ru(bpy)3]Cl3 (in ethanol, reflux) 2. [Ru(bpy)3]Cl3 + Zn

dust → [Ru(bpy)3]Cl2 (reduction) 3. [Ru(bpy)3]Cl2 + 2 NH4PF6 → [Ru(bpy)3](PF6)2

+ 2 NH4Cl

8. Describe the structure and bonding in ferrocene, Fe(C5H5)2.

Bonding: - 18-electron rule satisfied: Fe(0) + 2(C5H5−) = 8 + 2(5) = 18 e− - Fe 3d

9. Explain the concept of isolobal analogy and provide an example.

Solution:

Isolobal analogy: - Fragments with similar frontier orbitals in terms of symmetry,

energy, and electron occupation - These fragments can often replace each other in

molecules

Example: CH3+ and Mn(CO)5+ are isolobal - CH3+ has an empty sp3 hybrid orbital -

Mn(CO)5+ has an empty d𝑧2 orbital of similar symmetry and energy - Both can form

similar types of bonds with ligands

10. Describe the fluxional behavior of PF5 and explain how it can be studied using NMR

spectroscopy.

atoms, appearing as a quintet.

11. Consider the reaction of XeF4 with water:

a) Draw the Lewis structure of XeF4 and XeO2F2. b) Determine the hybridization of

Xe in both molecules.

Solution:

a) Lewis structures:

XeF4: F | F–Xe–F | F

XeO2F2: O || F–Xe–F || O

b) Hybridization: XeF4: sp3d2 (square planar geometry) XeO2F2: sp3d (see-saw

geometry)

Lanthanide contraction: - Poor shielding of 4f electrons - Increased effective nuclear

similar chemical properties

13. For the complex [Fe(CN)6]4−: a) Determine the oxidation state of Fe. b) Predict

whether the complex is high-spin or low-spin. c) Calculate its magnetic moment in

Bohr magnetons.

Solution:

a) Oxidation state of Fe: 2+ (CN− is -1, so 6 × (-1) + x = -4; x = +2)

b) Low-spin complex (CN− is a strong-field ligand)

c) Magnetic moment: Fe2+ (d6) in low-spin: (t2𝑔)6 (e𝑔)0 No unpaired electrons, so μ

= 0 B.M.

14. Describe the bonding in the linear [AuCl2]− complex using Molecular Orbital theory.

Similarities: - Both are Lewis acids due to empty p orbital on boron - Both form

adducts with Lewis bases

Differences: - BF3 is a weaker Lewis acid than BCl3 - BF3 has stronger B-F bonds due

to p𝜋-p𝜋 back-bonding - BCl3 has weaker B-Cl bonds, making it more reactive - BF3

is planar, while BCl3 is slightly pyramidal due to larger Cl atoms

16. Explain why [Co(NH3)6]3+ is diamagnetic while [CoF6]3− is paramagnetic.

Solution:

[Co(NH3)6]3+: - Co3+ is d6 - NH3 is a strong-field ligand - Low-spin configuration:

[CoF6]3−: - Co3+ is d6 - F− is a weak-field ligand - High-spin configuration: (t2𝑔)4

17. Propose a synthetic route for the preparation of [Ru(bpy)3](PF6)2 from RuCl3 ·

+ 2 NH4Cl

Structure: - Sandwich compound with Fe between two parallel cyclopentadienyl

rings - Each C5H5− ring is planar and aromatic

Bonding: - 18-electron rule satisfied: Fe(0) + 2(C5H5−) = 8 + 2(5) = 18 e− - Fe 3d

orbitals interact with π orbitals of C5H5− rings - Delocalized bonding between Fe

and both rings - High stability due to aromatic nature of rings and 18-electron

configuration

19. Explain the concept of isolobal analogy and provide an example.

Solution:

Isolobal analogy: - Fragments with similar frontier orbitals in terms of symmetry,

energy, and electron occupation - These fragments can often replace each other in

Solution:

Fluxional behavior of PF5: - Trigonal bipyramidal structure - Rapid Berry

pseudorotation exchanges axial and equatorial F atoms

NMR study: - At low temperatures: two 19F NMR signals (axial and equatorial F) - At

high temperatures: single averaged 19F NMR signal - Variable temperature NMR can

determine the energy barrier for this process - 31P NMR shows coupling to all F

atoms, appearing as a quintet

21. Consider the reaction of XeF4 with water:

a) Draw the Lewis structure of XeF4 and XeO2F2. b) Determine the hybridization of

Xe in both molecules.

Differences: - BF3 is a weaker Lewis acid than BCl3 - BF3 has stronger B-F bonds due

Isolobal analogy: - Fragments with similar frontier orbitals in terms of symmetry,

stability of the linear geometry

37. Propose a synthetic route for the preparation of [Ru(bpy)3](PF6)2 from RuCl3 ·

nH2O.

Solution:

1. RuCl3 · nH2O + 3 bpy → [Ru(bpy)3]Cl3 (in ethanol, reflux) 2. [Ru(bpy)3]Cl3 + Zn

dust → [Ru(bpy)3]Cl2 (reduction) 3. [Ru(bpy)3]Cl2 + 2 NH4PF6 → [Ru(bpy)3](PF6)2

+ 2 NH4Cl

38. Describe the structure and bonding in ferrocene, Fe(C5H5)2.

Solution:

Structure: - Sandwich compound with Fe between two parallel cyclopentadienyl

stability of the linear geometry

67. Propose a synthetic route for the preparation of [Ru(bpy)3](PF6)2 from RuCl3 ·

nH2O.

Solution:

1. RuCl3 · nH2O + 3 bpy → [Ru(bpy)3]Cl3 (in ethanol, reflux) 2. [Ru(bpy)3]Cl3 + Zn

dust → [Ru(bpy)3]Cl2 (reduction) 3. [Ru(bpy)3]Cl2 + 2 NH4PF6 → [Ru(bpy)3](PF6)2

+ 2 NH4Cl

68. Describe the structure and bonding in ferrocene, Fe(C5H5)2.

Solution:

Structure: - Sandwich compound with Fe between two parallel cyclopentadienyl

spectroscopy.

Fluxional behavior of PF5: - Trigonal bipyramidal structure - Rapid Berry

pseudorotation exchanges axial and equatorial F atoms

NMR study: - At low temperatures: two 19F NMR signals (axial and equatorial F) - At

high temperatures: single averaged 19F NMR signal - Variable temperature NMR can

determine the energy barrier for this process - 31P NMR shows coupling to all F

atoms, appearing as a quintet

71. Consider the reaction of XeF4 with water:

a) Draw the Lewis structure of XeF4 and XeO2F2. b) Determine the hybridization of

Xe in both molecules.

72. Explain the lanthanide contraction and discuss two significant consequences of this

phenomenon in inorganic chemistry.

Solution:

Lanthanide contraction: - Poor shielding of 4f electrons - Increased effective nuclear

charge across the series - Decrease in atomic and ionic radii across the series

Two significant consequences: 1. Similar sizes of later lanthanides and early

similar chemical properties

whether the complex is high-spin or low-spin. c) Calculate its magnetic moment in

Bohr magnetons.

Solution:

a) Oxidation state of Fe: 2+ (CN− is -1, so 6 × (-1) + x = -4; x = +2)

= 0 B.M.

Solution:

Similarities: - Both are Lewis acids due to empty p orbital on boron - Both form

Differences: - BF3 is a weaker Lewis acid than BCl3 - BF3 has stronger B-F bonds due

to p𝜋-p𝜋 back-bonding - BCl3 has weaker B-Cl bonds, making it more reactive - BF3

is planar, while BCl3 is slightly pyramidal due to larger Cl atoms

76. Explain why [Co(NH3)6]3+ is diamagnetic while [CoF6]3− is paramagnetic.

Solution:

[Co(NH3)6]3+: - Co3+ is d6 - NH3 is a strong-field ligand - Low-spin configuration:

[CoF6]3−: - Co3+ is d6 - F− is a weak-field ligand - High-spin configuration: (t2𝑔)4

Bonding: - 18-electron rule satisfied: Fe(0) + 2(C5H5−) = 8 + 2(5) = 18 e− - Fe 3d

79. Explain the concept of isolobal analogy and provide an example.

Solution:

Isolobal analogy: - Fragments with similar frontier orbitals in terms of symmetry,

energy, and electron occupation - These fragments can often replace each other in

molecules

Example: CH3+ and Mn(CO)5+ are isolobal - CH3+ has an empty sp3 hybrid orbital -

similar types of bonds with ligands

80. Describe the fluxional behavior of PF5 and explain how it can be studied using NMR

Solution:

Fluxional behavior of PF5: - Trigonal bipyramidal structure - Rapid Berry

pseudorotation exchanges axial and equatorial F atoms

Xe in both molecules.

Solution:

geometry)

82. Explain the lanthanide contraction and discuss two significant consequences of this

phenomenon in inorganic chemistry.

Lanthanide contraction: - Poor shielding of 4f electrons - Increased effective nuclear

charge across the series - Decrease in atomic and ionic radii across the series

Two significant consequences: 1. Similar sizes of later lanthanides and early

transition metals (e.g., Lu3+ and Y3+) 2. Difficulty in separating lanthanides due to

similar chemical properties

Bohr magnetons.

stability of the linear geometry

Differences: - BF3 is a weaker Lewis acid than BCl3 - BF3 has stronger B-F bonds due

Isolobal analogy: - Fragments with similar frontier orbitals in terms of symmetry,

stability of the linear geometry

97. Propose a synthetic route for the preparation of [Ru(bpy)3](PF6)2 from RuCl3 ·

nH2O.

Solution:

1. RuCl3 · nH2O + 3 bpy → [Ru(bpy)3]Cl3 (in ethanol, reflux) 2. [Ru(bpy)3]Cl3 + Zn

dust → [Ru(bpy)3]Cl2 (reduction) 3. [Ru(bpy)3]Cl2 + 2 NH4PF6 → [Ru(bpy)3](PF6)2

+ 2 NH4Cl

98. Describe the structure and bonding in ferrocene, Fe(C5H5)2.

Solution:

Structure: - Sandwich compound with Fe between two parallel cyclopentadienyl

spectroscopy.

Fluxional behavior of PF5: - Trigonal bipyramidal structure - Rapid Berry

pseudorotation exchanges axial and equatorial F atoms

NMR study: - At low temperatures: two 19F NMR signals (axial and equatorial F) - At

high temperatures: single averaged 19F NMR signal - Variable temperature NMR can

determine the energy barrier for this process - 31P NMR shows coupling to all F

atoms, appearing as a quintet

101. Consider the reaction of XeF4 with water:

a) Draw the Lewis structure of XeF4 and XeO2F2. b) Determine the hybridization of

Xe in both molecules.

102. Explain the lanthanide contraction and discuss two significant consequences of this

phenomenon in inorganic chemistry.

Solution:

Lanthanide contraction: - Poor shielding of 4f electrons - Increased effective nuclear

charge across the series - Decrease in atomic and ionic radii across the series

Two significant consequences: 1. Similar sizes of later lanthanides and early

similar chemical properties

whether the complex is high-spin or low-spin. c) Calculate its magnetic moment in

Bohr magnetons.

Solution:

a) Oxidation state of Fe: 2+ (CN− is -1, so 6 × (-1) + x = -4; x = +2)

= 0 B.M.

Solution:

Similarities: - Both are Lewis acids due to empty p orbital on boron - Both form

Differences: - BF3 is a weaker Lewis acid than BCl3 - BF3 has stronger B-F bonds due

to p𝜋-p𝜋 back-bonding - BCl3 has weaker B-Cl bonds, making it more reactive - BF3

is planar, while BCl3 is slightly pyramidal due to larger Cl atoms

106. Explain why [Co(NH3)6]3+ is diamagnetic while [CoF6]3− is paramagnetic.

Solution:

[Co(NH3)6]3+: - Co3+ is d6 - NH3 is a strong-field ligand - Low-spin configuration:

[CoF6]3−: - Co3+ is d6 - F− is a weak-field ligand - High-spin configuration: (t2𝑔)4

Two significant consequences: 1. Similar sizes of later lanthanides and early

similar chemical properties

whether the complex is high-spin or low-spin. c) Calculate its magnetic moment in

Bohr magnetons.

Solution:

a) Oxidation state of Fe: 2+ (CN− is -1, so 6 × (-1) + x = -4; x = +2)

= 0 B.M.

Solution:

Similarities: - Both are Lewis acids due to empty p orbital on boron - Both form

Differences: - BF3 is a weaker Lewis acid than BCl3 - BF3 has stronger B-F bonds due

to p𝜋-p𝜋 back-bonding - BCl3 has weaker B-Cl bonds, making it more reactive - BF3

is planar, while BCl3 is slightly pyramidal due to larger Cl atoms

136. Explain why [Co(NH3)6]3+ is diamagnetic while [CoF6]3− is paramagnetic.

Solution:

[Co(NH3)6]3+: - Co3+ is d6 - NH3 is a strong-field ligand - Low-spin configuration:

[CoF6]3−: - Co3+ is d6 - F− is a weak-field ligand - High-spin configuration: (t2𝑔)4

Bonding: - 18-electron rule satisfied: Fe(0) + 2(C5H5−) = 8 + 2(5) = 18 e− - Fe 3d

139. Explain the concept of isolobal analogy and provide an example.

Solution:

Isolobal analogy: - Fragments with similar frontier orbitals in terms of symmetry,

energy, and electron occupation - These fragments can often replace each other in

molecules

Example: CH3+ and Mn(CO)5+ are isolobal - CH3+ has an empty sp3 hybrid orbital -

similar types of bonds with ligands

140. Describe the fluxional behavior of PF5 and explain how it can be studied using NMR

Solution:

Fluxional behavior of PF5: - Trigonal bipyramidal structure - Rapid Berry

pseudorotation exchanges axial and equatorial F atoms

Xe in both molecules.

Solution:

geometry)

142. Explain the lanthanide contraction and discuss two significant consequences of this

phenomenon in inorganic chemistry.

Lanthanide contraction: - Poor shielding of 4f electrons - Increased effective nuclear

charge across the series - Decrease in atomic and ionic radii across the series

Two significant consequences: 1. Similar sizes of later lanthanides and early

transition metals (e.g., Lu3+ and Y3+) 2. Difficulty in separating lanthanides due to

similar chemical properties

Bohr magnetons.

stability of the linear geometry

Differences: - BF3 is a weaker Lewis acid than BCl3 - BF3 has stronger B-F bonds due

Isolobal analogy: - Fragments with similar frontier orbitals in terms of symmetry,

stability of the linear geometry

157. Propose a synthetic route for the preparation of [Ru(bpy)3](PF6)2 from RuCl3 ·

nH2O.

Solution:

1. RuCl3 · nH2O + 3 bpy → [Ru(bpy)3]Cl3 (in ethanol, reflux) 2. [Ru(bpy)3]Cl3 + Zn

dust → [Ru(bpy)3]Cl2 (reduction) 3. [Ru(bpy)3]Cl2 + 2 NH4PF6 → [Ru(bpy)3](PF6)2

+ 2 NH4Cl

158. Describe the structure and bonding in ferrocene, Fe(C5H5)2.

Solution:

Structure: - Sandwich compound with Fe between two parallel cyclopentadienyl

spectroscopy.

Fluxional behavior of PF5: - Trigonal bipyramidal structure - Rapid Berry

pseudorotation exchanges axial and equatorial F atoms

NMR study: - At low temperatures: two 19F NMR signals (axial and equatorial F) - At

high temperatures: single averaged 19F NMR signal - Variable temperature NMR can

determine the energy barrier for this process - 31P NMR shows coupling to all F

atoms, appearing as a quintet

161. Consider the reaction of XeF4 with water:

a) Draw the Lewis structure of XeF4 and XeO2F2. b) Determine the hybridization of

Xe in both molecules.

162. Explain the lanthanide contraction and discuss two significant consequences of this

phenomenon in inorganic chemistry.

Solution:

Lanthanide contraction: - Poor shielding of 4f electrons - Increased effective nuclear

charge across the series - Decrease in atomic and ionic radii across the series

Two significant consequences: 1. Similar sizes of later lanthanides and early

similar chemical properties

whether the complex is high-spin or low-spin. c) Calculate its magnetic moment in

Bohr magnetons.

Solution:

a) Oxidation state of Fe: 2+ (CN− is -1, so 6 × (-1) + x = -4; x = +2)

= 0 B.M.

Solution:

Similarities: - Both are Lewis acids due to empty p orbital on boron - Both form

Differences: - BF3 is a weaker Lewis acid than BCl3 - BF3 has stronger B-F bonds due

to p𝜋-p𝜋 back-bonding - BCl3 has weaker B-Cl bonds, making it more reactive - BF3

is planar, while BCl3 is slightly pyramidal due to larger Cl atoms

166. Explain why [Co(NH3)6]3+ is diamagnetic while [CoF6]3− is paramagnetic.

Solution:

[Co(NH3)6]3+: - Co3+ is d6 - NH3 is a strong-field ligand - Low-spin configuration:

[CoF6]3−: - Co3+ is d6 - F− is a weak-field ligand - High-spin configuration: (t2𝑔)4

Bonding: - 18-electron rule satisfied: Fe(0) + 2(C5H5−) = 8 + 2(5) = 18 e− - Fe 3d

169. Explain the concept of isolobal analogy and provide an example.

Solution:

Isolobal analogy: - Fragments with similar frontier orbitals in terms of symmetry,

energy, and electron occupation - These fragments can often replace each other in

molecules

Example: CH3+ and Mn(CO)5+ are isolobal - CH3+ has an empty sp3 hybrid orbital -

similar types of bonds with ligands

170. Describe the fluxional behavior of PF5 and explain how it can be studied using NMR

Solution:

Fluxional behavior of PF5: - Trigonal bipyramidal structure - Rapid Berry

pseudorotation exchanges axial and equatorial F atoms

Xe in both molecules.

Solution:

geometry)

172. Explain the lanthanide contraction and discuss two significant consequences of this

phenomenon in inorganic chemistry.

Lanthanide contraction: - Poor shielding of 4f electrons - Increased effective nuclear

charge across the series - Decrease in atomic and ionic radii across the series

Two significant consequences: 1. Similar sizes of later lanthanides and early

transition metals (e.g., Lu3+ and Y3+) 2. Difficulty in separating lanthanides due to

similar chemical properties

Bohr magnetons.

stability of the linear geometry