CHEM 471 - Trans Effect in Square Planar Complexes.

CHEM 471 - Trans Effect in Square Planar Complexes

Problems and Solutions

1. Explain the trans effect in square planar complexes and provide an example of how

it can be used in synthesis.

Solution:

The trans effect is the labilization of ligands trans to certain other ligands in square

planar complexes. Ligands with a strong trans effect (e.g., CO, CN-, C2H4) weaken the

metal-ligand bond trans to them.

valence electrons, achieving a noble gas configuration.

Calculation: metal valence e− + donor e− from ligands = 18

Violations occur due to: 1. Steric effects (e.g., Pt(PPh3)3 with 16 e−) 2. Electronic

effects (e.g., d0 complexes like Ti(CH3)4 with 8 e−) 3. Square planar d8 complexes

(e.g., [PtCl4]2− with 16 e−) 4. High oxidation state compounds (e.g., WF6 with 12 e−)

3. Compare and contrast the bonding and properties of CO and N2 as ligands in

transition metal complexes.

Solution:

Similarities: - Both are strong field ligands - Both can act as σ-donors and π-

acceptors

stability of metal complexes.

Solution:

HSAB principle: - Hard acids/bases: small, highly charged, weakly polarizable - Soft

acids/bases: large, low charge, highly polarizable

Hard acids prefer hard bases, soft acids prefer soft bases.

Predictions: - Hard acid (Ti4+) + Hard base (F−) → Stable complex [TiF6]2− - Soft

acid (Pt2+) + Soft base (PR3) → Stable complex [Pt(PR3)4]2+ - Hard acid (Fe3+) + Soft

base (RS−) → Less stable complex

5. Describe the structure and bonding in the [Cu(NH3)4]2+ complex, explaining any

distortions from ideal geometry.

6. Discuss the differences in reactivity between octahedral and square planar d8

complexes.

Solution:

Octahedral d8 complexes: - Usually high-spin - Kinetically labile - Undergo

associative substitution - Example: [Ni(H2O)6]2+

Square planar d8 complexes: - Always low-spin - Kinetically inert - Undergo

associative substitution via a 5-coordinate intermediate - Example: [PtCl4]2−

Differences in reactivity due to: - Electronic configuration (high-spin vs. low-spin) -

Ligand field stabilization energy - Availability of coordination sites for incoming

ligands

catalysis and materials science

8. Describe the bonding in metallocenes, using ferrocene as an example. How does this

change for bent metallocenes?

Solution:

Ferrocene bonding: - Two parallel cyclopentadienyl (Cp) rings sandwich Fe - Each

Cp ring contributes 6 π electrons - Fe uses 3d, 4s, and 4p orbitals for bonding - 18-

electron rule satisfied: 6 (Fe) + 2(6) (Cp) = 18

Molecular orbitals: - Cp π orbitals interact with Fe d orbitals - Strongest interaction:

Cp e1 with Fe d𝑥𝑧 and d𝑦𝑧

Bent metallocenes (e.g., Cp2TiCl2): - Cp rings tilted - Reduced symmetry - Additional

ligands in equatorial positions - Changes in MO interactions due to tilting - Often do

not follow 18-electron rule

magnetic properties 5. Impacts thermodynamic and kinetic stability

Example: [Fe(H2O)6]2+ (high-spin) vs. [Fe(CN)6]4− (low-spin) - Different colors

(pale green vs. yellow) - Different magnetic properties (paramagnetic vs.

diamagnetic)

10. Describe the Wade-Mingos rules for predicting the structures of borane and

carborane clusters.

Solution:

Wade-Mingos rules: - Predict structures of electron-deficient cluster compounds -

Based on the number of skeletal electron pairs (SEP)

Rules: 1. closo structures: n + 1 SEP (n = number of vertices) 2. nido structures: n +

it can be used in synthesis.

Solution:

The trans effect is the labilization of ligands trans to certain other ligands in square

Example in synthesis: To synthesize cis-[Pt(NH3)2Cl2] from [PtCl4]2−: 1. [PtCl4]2− +

The strong trans effect of Cl− ensures the second NH3 enters cis to the first.

12. Describe the 18-electron rule and explain why some compounds violate this rule.

Solution:

The 18-electron rule states that stable transition metal complexes often have 18

valence electrons, achieving a noble gas configuration.

Calculation: metal valence e− + donor e− from ligands = 18

Violations occur due to: 1. Steric effects (e.g., Pt(PPh3)3 with 16 e−) 2. Electronic

effects (e.g., d0 complexes like Ti(CH3)4 with 8 e−) 3. Square planar d8 complexes

Differences: - CO is a better π-acceptor than N2 - CO forms stronger bonds with

metals - N2 complexes are generally less stable - CO can bond through C or O (rare),

while N2 usually bonds through one N - N2 can act as a bridging ligand more readily

than CO

Examples: CO complex: [Ni(CO)4] N2 complex: [Ru(NH3)5(N2)]2+

14. Explain the concept of hard and soft acids and bases (HSAB) and use it to predict the

stability of metal complexes.

Solution:

HSAB principle: - Hard acids/bases: small, highly charged, weakly polarizable - Soft

acids/bases: large, low charge, highly polarizable

Solution:

Structure: Distorted square planar Bonding: - Cu2+ is d9 - Four σ bonds between Cu

and NH3 ligands - Jahn-Teller distortion elongates two trans Cu-N bonds

Explanation of distortion: - d9 configuration has uneven occupation of e𝑔 orbitals -

Elongation along z-axis lowers energy of d𝑧2 orbital - This distortion reduces overall

energy of the complex

16. Discuss the differences in reactivity between octahedral and square planar d8

complexes.

Solution:

Octahedral d8 complexes: - Usually high-spin - Kinetically labile - Undergo

17. Explain the concept of metal-metal multiple bonding and provide an example.

Solution:

Metal-metal multiple bonding: - Direct bonding between two metal centers - Can

involve σ, π, and δ bonds - Common in low oxidation state complexes

Example: [Re2Cl8]2− - Re-Re quadruple bond - Bond order: 1σ + 2π + 1δ = 4 - Short

catalysis and materials science

18. Describe the bonding in metallocenes, using ferrocene as an example. How does this

change for bent metallocenes?

Solution:

Wade-Mingos rules: - Predict structures of electron-deficient cluster compounds -

Based on the number of skeletal electron pairs (SEP)

Rules: 1. closo structures: n + 1 SEP (n = number of vertices) 2. nido structures: n +

2 SEP 3. arachno structures: n + 3 SEP 4. hypho structures: n + 4 SEP

Calculation of SEP: SEP = (total valence e− - e− in external bonds) / 2

Ligand field stabilization energy - Availability of coordination sites for incoming

of coordination compounds.

Solution:

spin vs. low-spin) 3. Influences electronic spectra and color of complexes 4. Affects

magnetic properties 5. Impacts thermodynamic and kinetic stability

metal-ligand bond trans to them.

Example in synthesis: To synthesize cis-[Pt(NH3)2Cl2] from [PtCl4]2−: 1. [PtCl4]2− +

The strong trans effect of Cl− ensures the second NH3 enters cis to the first.

32. Describe the 18-electron rule and explain why some compounds violate this rule.

Solution:

The 18-electron rule states that stable transition metal complexes often have 18

valence electrons, achieving a noble gas configuration.

while N2 usually bonds through one N - N2 can act as a bridging ligand more readily

than CO

Examples: CO complex: [Ni(CO)4] N2 complex: [Ru(NH3)5(N2)]2+

stability of metal complexes.

Solution:

Predictions: - Hard acid (Ti4+) + Hard base (F−) → Stable complex [TiF6]2− - Soft

acid (Pt2+) + Soft base (PR3) → Stable complex [Pt(PR3)4]2+ - Hard acid (Fe3+) + Soft

55. Describe the structure and bonding in the [Cu(NH3)4]2+ complex, explaining any

distortions from ideal geometry.

Solution:

complexes.

Solution:

Octahedral d8 complexes: - Usually high-spin - Kinetically labile - Undergo

associative substitution - Example: [Ni(H2O)6]2+

Square planar d8 complexes: - Always low-spin - Kinetically inert - Undergo

associative substitution via a 5-coordinate intermediate - Example: [PtCl4]2−

Differences in reactivity due to: - Electronic configuration (high-spin vs. low-spin) -

Ligand field stabilization energy - Availability of coordination sites for incoming

ligands

57. Explain the concept of metal-metal multiple bonding and provide an example.

Metal-metal multiple bonding: - Direct bonding between two metal centers - Can

involve σ, π, and δ bonds - Common in low oxidation state complexes

Importance: - Unusual electronic and magnetic properties - Potential applications in

catalysis and materials science

change for bent metallocenes?

Solution:

Ferrocene bonding: - Two parallel cyclopentadienyl (Cp) rings sandwich Fe - Each

Cp ring contributes 6 π electrons - Fe uses 3d, 4s, and 4p orbitals for bonding - 18-

2 SEP 3. arachno structures: n + 3 SEP 4. hypho structures: n + 4 SEP

Calculation of SEP: SEP = (total valence e− - e− in external bonds) / 2

= 6 SEP = (26 - 6) / 2 = 10 n + 1 = 6 + 1 = 7, confirming closo structure (octahedral)

61. Explain the trans effect in square planar complexes and provide an example of how

it can be used in synthesis.

Solution:

The trans effect is the labilization of ligands trans to certain other ligands in square

Example in synthesis: To synthesize cis-[Pt(NH3)2Cl2] from [PtCl4]2−: 1. [PtCl4]2− +

while N2 usually bonds through one N - N2 can act as a bridging ligand more readily

stability of metal complexes.

Solution:

HSAB principle: - Hard acids/bases: small, highly charged, weakly polarizable - Soft

acids/bases: large, low charge, highly polarizable

Hard acids prefer hard bases, soft acids prefer soft bases.

Predictions: - Hard acid (Ti4+) + Hard base (F−) → Stable complex [TiF6]2− - Soft

acid (Pt2+) + Soft base (PR3) → Stable complex [Pt(PR3)4]2+ - Hard acid (Fe3+) + Soft

base (RS−) → Less stable complex

65. Describe the structure and bonding in the [Cu(NH3)4]2+ complex, explaining any

distortions from ideal geometry.

66. Discuss the differences in reactivity between octahedral and square planar d8

complexes.

Solution:

Octahedral d8 complexes: - Usually high-spin - Kinetically labile - Undergo

associative substitution - Example: [Ni(H2O)6]2+

Square planar d8 complexes: - Always low-spin - Kinetically inert - Undergo

associative substitution via a 5-coordinate intermediate - Example: [PtCl4]2−

Differences in reactivity due to: - Electronic configuration (high-spin vs. low-spin) -

Ligand field stabilization energy - Availability of coordination sites for incoming

ligands

catalysis and materials science

68. Describe the bonding in metallocenes, using ferrocene as an example. How does this

change for bent metallocenes?

Solution:

Ferrocene bonding: - Two parallel cyclopentadienyl (Cp) rings sandwich Fe - Each

Cp ring contributes 6 π electrons - Fe uses 3d, 4s, and 4p orbitals for bonding - 18-

electron rule satisfied: 6 (Fe) + 2(6) (Cp) = 18

Molecular orbitals: - Cp π orbitals interact with Fe d orbitals - Strongest interaction:

Cp e1 with Fe d𝑥𝑧 and d𝑦𝑧

Bent metallocenes (e.g., Cp2TiCl2): - Cp rings tilted - Reduced symmetry - Additional

ligands in equatorial positions - Changes in MO interactions due to tilting - Often do

not follow 18-electron rule

magnetic properties 5. Impacts thermodynamic and kinetic stability

Example: [Fe(H2O)6]2+ (high-spin) vs. [Fe(CN)6]4− (low-spin) - Different colors

(pale green vs. yellow) - Different magnetic properties (paramagnetic vs.

diamagnetic)

70. Describe the Wade-Mingos rules for predicting the structures of borane and

carborane clusters.

Solution:

Wade-Mingos rules: - Predict structures of electron-deficient cluster compounds -

Based on the number of skeletal electron pairs (SEP)

Rules: 1. closo structures: n + 1 SEP (n = number of vertices) 2. nido structures: n +

it can be used in synthesis.

Solution:

The trans effect is the labilization of ligands trans to certain other ligands in square

Example in synthesis: To synthesize cis-[Pt(NH3)2Cl2] from [PtCl4]2−: 1. [PtCl4]2− +

The strong trans effect of Cl− ensures the second NH3 enters cis to the first.

72. Describe the 18-electron rule and explain why some compounds violate this rule.

Solution:

The 18-electron rule states that stable transition metal complexes often have 18

valence electrons, achieving a noble gas configuration.

Calculation: metal valence e− + donor e− from ligands = 18

Violations occur due to: 1. Steric effects (e.g., Pt(PPh3)3 with 16 e−) 2. Electronic

effects (e.g., d0 complexes like Ti(CH3)4 with 8 e−) 3. Square planar d8 complexes

Differences: - CO is a better π-acceptor than N2 - CO forms stronger bonds with

metals - N2 complexes are generally less stable - CO can bond through C or O (rare),

while N2 usually bonds through one N - N2 can act as a bridging ligand more readily

than CO

Examples: CO complex: [Ni(CO)4] N2 complex: [Ru(NH3)5(N2)]2+

74. Explain the concept of hard and soft acids and bases (HSAB) and use it to predict the

stability of metal complexes.

Solution:

HSAB principle: - Hard acids/bases: small, highly charged, weakly polarizable - Soft

acids/bases: large, low charge, highly polarizable

Solution:

Structure: Distorted square planar Bonding: - Cu2+ is d9 - Four σ bonds between Cu

and NH3 ligands - Jahn-Teller distortion elongates two trans Cu-N bonds

Explanation of distortion: - d9 configuration has uneven occupation of e𝑔 orbitals -

Elongation along z-axis lowers energy of d𝑧2 orbital - This distortion reduces overall

energy of the complex

76. Discuss the differences in reactivity between octahedral and square planar d8

complexes.

Solution:

Octahedral d8 complexes: - Usually high-spin - Kinetically labile - Undergo

77. Explain the concept of metal-metal multiple bonding and provide an example.

Solution:

Metal-metal multiple bonding: - Direct bonding between two metal centers - Can

involve σ, π, and δ bonds - Common in low oxidation state complexes

Example: [Re2Cl8]2− - Re-Re quadruple bond - Bond order: 1σ + 2π + 1δ = 4 - Short

catalysis and materials science

78. Describe the bonding in metallocenes, using ferrocene as an example. How does this

change for bent metallocenes?

Solution:

Wade-Mingos rules: - Predict structures of electron-deficient cluster compounds -

Based on the number of skeletal electron pairs (SEP)

Rules: 1. closo structures: n + 1 SEP (n = number of vertices) 2. nido structures: n +

2 SEP 3. arachno structures: n + 3 SEP 4. hypho structures: n + 4 SEP

Calculation of SEP: SEP = (total valence e− - e− in external bonds) / 2

Ligand field stabilization energy - Availability of coordination sites for incoming

carborane clusters.

Solution:

Wade-Mingos rules: - Predict structures of electron-deficient cluster compounds -

Based on the number of skeletal electron pairs (SEP)

Rules: 1. closo structures: n + 1 SEP (n = number of vertices) 2. nido structures: n +

2 SEP 3. arachno structures: n + 3 SEP 4. hypho structures: n + 4 SEP

Example: B6H62− (closo) Total valence e− = (6 × 3) + (6 × 1) + 2 = 26 External bonds

= 6 SEP = (26 - 6) / 2 = 10 n + 1 = 6 + 1 = 7, confirming closo structure (octahedral)

111. Explain the trans effect in square planar complexes and provide an example of how

it can be used in synthesis.

Solution:

The trans effect is the labilization of ligands trans to certain other ligands in square

planar complexes. Ligands with a strong trans effect (e.g., CO, CN-, C2H4) weaken the

metal-ligand bond trans to them.

Example in synthesis: To synthesize cis-[Pt(NH3)2Cl2] from [PtCl4]2−: 1. [PtCl4]2− +

112. Describe the 18-electron rule and explain why some compounds violate this rule.

valence electrons, achieving a noble gas configuration.

Solution:

Similarities: - Both are strong field ligands - Both can act as σ-donors and π-

acceptors

while N2 usually bonds through one N - N2 can act as a bridging ligand more readily

than CO

Examples: CO complex: [Ni(CO)4] N2 complex: [Ru(NH3)5(N2)]2+

stability of metal complexes.

Solution:

Predictions: - Hard acid (Ti4+) + Hard base (F−) → Stable complex [TiF6]2− - Soft

acid (Pt2+) + Soft base (PR3) → Stable complex [Pt(PR3)4]2+ - Hard acid (Fe3+) + Soft

115. Describe the structure and bonding in the [Cu(NH3)4]2+ complex, explaining any

distortions from ideal geometry.

Solution:

complexes.

Solution:

Octahedral d8 complexes: - Usually high-spin - Kinetically labile - Undergo

associative substitution - Example: [Ni(H2O)6]2+

Square planar d8 complexes: - Always low-spin - Kinetically inert - Undergo

associative substitution via a 5-coordinate intermediate - Example: [PtCl4]2−

Differences in reactivity due to: - Electronic configuration (high-spin vs. low-spin) -

Ligand field stabilization energy - Availability of coordination sites for incoming

ligands

117. Explain the concept of metal-metal multiple bonding and provide an example.

Metal-metal multiple bonding: - Direct bonding between two metal centers - Can

involve σ, π, and δ bonds - Common in low oxidation state complexes

Importance: - Unusual electronic and magnetic properties - Potential applications in

catalysis and materials science

change for bent metallocenes?

Solution:

Ferrocene bonding: - Two parallel cyclopentadienyl (Cp) rings sandwich Fe - Each

Cp ring contributes 6 π electrons - Fe uses 3d, 4s, and 4p orbitals for bonding - 18-

2 SEP 3. arachno structures: n + 3 SEP 4. hypho structures: n + 4 SEP

Calculation of SEP: SEP = (total valence e− - e− in external bonds) / 2

= 6 SEP = (26 - 6) / 2 = 10 n + 1 = 6 + 1 = 7, confirming closo structure (octahedral)

121. Explain the trans effect in square planar complexes and provide an example of how

it can be used in synthesis.

Solution:

The trans effect is the labilization of ligands trans to certain other ligands in square

Example in synthesis: To synthesize cis-[Pt(NH3)2Cl2] from [PtCl4]2−: 1. [PtCl4]2− +

while N2 usually bonds through one N - N2 can act as a bridging ligand more readily

stability of metal complexes.

Solution:

HSAB principle: - Hard acids/bases: small, highly charged, weakly polarizable - Soft

acids/bases: large, low charge, highly polarizable

Hard acids prefer hard bases, soft acids prefer soft bases.

Predictions: - Hard acid (Ti4+) + Hard base (F−) → Stable complex [TiF6]2− - Soft

acid (Pt2+) + Soft base (PR3) → Stable complex [Pt(PR3)4]2+ - Hard acid (Fe3+) + Soft

base (RS−) → Less stable complex

125. Describe the structure and bonding in the [Cu(NH3)4]2+ complex, explaining any

distortions from ideal geometry.

126. Discuss the differences in reactivity between octahedral and square planar d8

complexes.

Solution:

Octahedral d8 complexes: - Usually high-spin - Kinetically labile - Undergo

associative substitution - Example: [Ni(H2O)6]2+

Square planar d8 complexes: - Always low-spin - Kinetically inert - Undergo

associative substitution via a 5-coordinate intermediate - Example: [PtCl4]2−

Differences in reactivity due to: - Electronic configuration (high-spin vs. low-spin) -

Ligand field stabilization energy - Availability of coordination sites for incoming

ligands

catalysis and materials science

128. Describe the bonding in metallocenes, using ferrocene as an example. How does this

change for bent metallocenes?

Solution:

Ferrocene bonding: - Two parallel cyclopentadienyl (Cp) rings sandwich Fe - Each

Cp ring contributes 6 π electrons - Fe uses 3d, 4s, and 4p orbitals for bonding - 18-

electron rule satisfied: 6 (Fe) + 2(6) (Cp) = 18

Molecular orbitals: - Cp π orbitals interact with Fe d orbitals - Strongest interaction:

Cp e1 with Fe d𝑥𝑧 and d𝑦𝑧

Bent metallocenes (e.g., Cp2TiCl2): - Cp rings tilted - Reduced symmetry - Additional

ligands in equatorial positions - Changes in MO interactions due to tilting - Often do

not follow 18-electron rule

magnetic properties 5. Impacts thermodynamic and kinetic stability

Example: [Fe(H2O)6]2+ (high-spin) vs. [Fe(CN)6]4− (low-spin) - Different colors

(pale green vs. yellow) - Different magnetic properties (paramagnetic vs.

diamagnetic)

130. Describe the Wade-Mingos rules for predicting the structures of borane and

carborane clusters.

Solution:

Wade-Mingos rules: - Predict structures of electron-deficient cluster compounds -

Based on the number of skeletal electron pairs (SEP)

Rules: 1. closo structures: n + 1 SEP (n = number of vertices) 2. nido structures: n +

it can be used in synthesis.

Solution:

The trans effect is the labilization of ligands trans to certain other ligands in square

Example in synthesis: To synthesize cis-[Pt(NH3)2Cl2] from [PtCl4]2−: 1. [PtCl4]2− +

The strong trans effect of Cl− ensures the second NH3 enters cis to the first.

132. Describe the 18-electron rule and explain why some compounds violate this rule.

Solution:

The 18-electron rule states that stable transition metal complexes often have 18

valence electrons, achieving a noble gas configuration.

Calculation: metal valence e− + donor e− from ligands = 18

Violations occur due to: 1. Steric effects (e.g., Pt(PPh3)3 with 16 e−) 2. Electronic

effects (e.g., d0 complexes like Ti(CH3)4 with 8 e−) 3. Square planar d8 complexes

Differences: - CO is a better π-acceptor than N2 - CO forms stronger bonds with

metals - N2 complexes are generally less stable - CO can bond through C or O (rare),

while N2 usually bonds through one N - N2 can act as a bridging ligand more readily

than CO

Examples: CO complex: [Ni(CO)4] N2 complex: [Ru(NH3)5(N2)]2+

134. Explain the concept of hard and soft acids and bases (HSAB) and use it to predict the

stability of metal complexes.

Solution:

HSAB principle: - Hard acids/bases: small, highly charged, weakly polarizable - Soft

acids/bases: large, low charge, highly polarizable

Solution:

Structure: Distorted square planar Bonding: - Cu2+ is d9 - Four σ bonds between Cu

and NH3 ligands - Jahn-Teller distortion elongates two trans Cu-N bonds

Explanation of distortion: - d9 configuration has uneven occupation of e𝑔 orbitals -

Elongation along z-axis lowers energy of d𝑧2 orbital - This distortion reduces overall

energy of the complex

136. Discuss the differences in reactivity between octahedral and square planar d8

complexes.

Solution:

Octahedral d8 complexes: - Usually high-spin - Kinetically labile - Undergo

137. Explain the concept of metal-metal multiple bonding and provide an example.

Solution:

Metal-metal multiple bonding: - Direct bonding between two metal centers - Can

involve σ, π, and δ bonds - Common in low oxidation state complexes

Example: [Re2Cl8]2− - Re-Re quadruple bond - Bond order: 1σ + 2π + 1δ = 4 - Short

catalysis and materials science

138. Describe the bonding in metallocenes, using ferrocene as an example. How does this

change for bent metallocenes?

Solution:

Wade-Mingos rules: - Predict structures of electron-deficient cluster compounds -

Based on the number of skeletal electron pairs (SEP)

Rules: 1. closo structures: n + 1 SEP (n = number of vertices) 2. nido structures: n +

2 SEP 3. arachno structures: n + 3 SEP 4. hypho structures: n + 4 SEP

Calculation of SEP: SEP = (total valence e− - e− in external bonds) / 2