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Why is Ni[(PPh₃)₂Cl₂] tetrahedral?

What kind of isomerism does Ni(PPh3)2Cl2 have and what is their IUPAC name?Tetrahedral or Square PlanarWhy does Fe(CO)₄ adopt a tetrahedral, as opposed to square planar, geometry?Why is [PdCl4]2- square planar whereas [NiCl4]2- is tetrahedral?Crystal Field Splitting of d-Orbitals in Octahedral and Tetrahedral Ligand FieldsWhy do tetrahedral complexes have approximately 4/9 the field split of octahedral complexes?Why does steric hindrance cause a d8 complex to have a tetrahedral geometry rather than a square planar geometry?

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$begingroup$

Since PPh₃ is strong field ligand and, the famous Wilkinson's catalyst, which also possess this ligand is square planar, then what makes the above complex tetrahedral?

coordination-compounds transition-metals crystal-field-theory organotransition-metal-chemistry ligand-field-theory

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edited May 19 at 12:02

andselisk♦

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8

2

$begingroup$

Since PPh₃ is strong field ligand and, the famous Wilkinson's catalyst, which also possess this ligand is square planar, then what makes the above complex tetrahedral?

coordination-compounds transition-metals crystal-field-theory organotransition-metal-chemistry ligand-field-theory

share|improve this question

edited May 19 at 12:02

andselisk♦

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asked May 19 at 11:56

user226375user226375

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$endgroup$

add a comment
 | 

$begingroup$

Since PPh₃ is strong field ligand and, the famous Wilkinson's catalyst, which also possess this ligand is square planar, then what makes the above complex tetrahedral?

coordination-compounds transition-metals crystal-field-theory organotransition-metal-chemistry ligand-field-theory

share|improve this question

edited May 19 at 12:02

andselisk♦

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asked May 19 at 11:56

user226375user226375

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$endgroup$

share|improve this question

edited May 19 at 12:02

andselisk♦

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asked May 19 at 11:56

user226375user226375

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share|improve this question

edited May 19 at 12:02

andselisk♦

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asked May 19 at 11:56

user226375user226375

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edited May 19 at 12:02

andselisk♦

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edited May 19 at 12:02

andselisk♦

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asked May 19 at 11:56

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asked May 19 at 11:56

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2 Answers
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oldest

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12

$begingroup$

We sometimes call this type of complex 'pseudotetrahedral' since there is an isomerism from a tetrahedral to a square planar complex possible. I was unable to find the original work here but this link gives some information. As you already mentioned there are two strong and two weak ligands so it's hard to tell how strong the ligand field splitting will be. For your particular complex it seems to be right on the spot where it would change from one to the other so depending on what you do you can influence the equilibrium. From what I read this may depend on the ability of the solvent to coordinate to the complex as well, the temperature, etc.

This is also mentioned in Earnshaw's Chemistry of the elements

1. 
   Planar-tetrahedral equilibria. Compounds
   such as $ce[NiBr2(PEtPh2)2]$ mentioned above as
   well as a number of sec-alkylsalicylaldiminato
   derivatives (i.e. Me in Fig. 27.6b replaced by
   a sec-alkyl group) dissolve in non-coordinating
   solvents such as chloroform or toluene to give
   solutions whose spectra and magnetic properties
   are temperature-dependent and indicate the presence
   of an equilibrium mixture of diamagnetic
   planar and paramagnetic tetrahedral molecules.

share|improve this answer

answered May 19 at 12:40

JustanotherchemistJustanotherchemist

2,31977 silver badges2323 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Can it be justified by saying that due to steric repulsions of the bulky PPh3, to minimise repulsions, the equilibrium is on the side of tetrahedral form?
  $endgroup$
  – user226375
  May 19 at 13:52
  

  
  
  
  

add a comment
 | 

9

$begingroup$

Dichlorobis(triphenylphosphine)nickel(II), or $ceNiCl2[P(C6H5)3]2$ in square planar form is red and diamagnetic. The blue form is paramagnetic and features tetrahedral Ni(II) centers. Both tetrahedral and square planar isomers coexist in solutions. Weak field ligands, favor tetrahedral geometry and strong field ligands favor the square planar isomer. Both weak field ($ceCl−$) and strong field ($cePPh3$) ligands comprise $ceNiCl2(PPh3)2$, hence this compound is borderline between the two geometries.

Steric effects also affect the equilibrium; larger ligands favoring the less crowded tetrahedral geometry.[1]

Reference

1. Greenwood, Norman N.; Earnshaw, Alan (1997). Chemistry of the Elements (2nd ed.).

share|improve this answer

edited May 19 at 15:11

Gaurang Tandon

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answered May 19 at 14:45

Chakravarthy KalyanChakravarthy Kalyan

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12

$begingroup$

We sometimes call this type of complex 'pseudotetrahedral' since there is an isomerism from a tetrahedral to a square planar complex possible. I was unable to find the original work here but this link gives some information. As you already mentioned there are two strong and two weak ligands so it's hard to tell how strong the ligand field splitting will be. For your particular complex it seems to be right on the spot where it would change from one to the other so depending on what you do you can influence the equilibrium. From what I read this may depend on the ability of the solvent to coordinate to the complex as well, the temperature, etc.

This is also mentioned in Earnshaw's Chemistry of the elements

1. 
   Planar-tetrahedral equilibria. Compounds
   such as $ce[NiBr2(PEtPh2)2]$ mentioned above as
   well as a number of sec-alkylsalicylaldiminato
   derivatives (i.e. Me in Fig. 27.6b replaced by
   a sec-alkyl group) dissolve in non-coordinating
   solvents such as chloroform or toluene to give
   solutions whose spectra and magnetic properties
   are temperature-dependent and indicate the presence
   of an equilibrium mixture of diamagnetic
   planar and paramagnetic tetrahedral molecules.

share|improve this answer

answered May 19 at 12:40

JustanotherchemistJustanotherchemist

2,31977 silver badges2323 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Can it be justified by saying that due to steric repulsions of the bulky PPh3, to minimise repulsions, the equilibrium is on the side of tetrahedral form?
  $endgroup$
  – user226375
  May 19 at 13:52
  

  
  
  
  

add a comment
 | 

12

$begingroup$

We sometimes call this type of complex 'pseudotetrahedral' since there is an isomerism from a tetrahedral to a square planar complex possible. I was unable to find the original work here but this link gives some information. As you already mentioned there are two strong and two weak ligands so it's hard to tell how strong the ligand field splitting will be. For your particular complex it seems to be right on the spot where it would change from one to the other so depending on what you do you can influence the equilibrium. From what I read this may depend on the ability of the solvent to coordinate to the complex as well, the temperature, etc.

This is also mentioned in Earnshaw's Chemistry of the elements

1. 
   Planar-tetrahedral equilibria. Compounds
   such as $ce[NiBr2(PEtPh2)2]$ mentioned above as
   well as a number of sec-alkylsalicylaldiminato
   derivatives (i.e. Me in Fig. 27.6b replaced by
   a sec-alkyl group) dissolve in non-coordinating
   solvents such as chloroform or toluene to give
   solutions whose spectra and magnetic properties
   are temperature-dependent and indicate the presence
   of an equilibrium mixture of diamagnetic
   planar and paramagnetic tetrahedral molecules.

share|improve this answer

answered May 19 at 12:40

JustanotherchemistJustanotherchemist

2,31977 silver badges2323 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Can it be justified by saying that due to steric repulsions of the bulky PPh3, to minimise repulsions, the equilibrium is on the side of tetrahedral form?
  $endgroup$
  – user226375
  May 19 at 13:52
  

  
  
  
  

add a comment
 | 

$begingroup$

We sometimes call this type of complex 'pseudotetrahedral' since there is an isomerism from a tetrahedral to a square planar complex possible. I was unable to find the original work here but this link gives some information. As you already mentioned there are two strong and two weak ligands so it's hard to tell how strong the ligand field splitting will be. For your particular complex it seems to be right on the spot where it would change from one to the other so depending on what you do you can influence the equilibrium. From what I read this may depend on the ability of the solvent to coordinate to the complex as well, the temperature, etc.

This is also mentioned in Earnshaw's Chemistry of the elements

1. 
   Planar-tetrahedral equilibria. Compounds
   such as $ce[NiBr2(PEtPh2)2]$ mentioned above as
   well as a number of sec-alkylsalicylaldiminato
   derivatives (i.e. Me in Fig. 27.6b replaced by
   a sec-alkyl group) dissolve in non-coordinating
   solvents such as chloroform or toluene to give
   solutions whose spectra and magnetic properties
   are temperature-dependent and indicate the presence
   of an equilibrium mixture of diamagnetic
   planar and paramagnetic tetrahedral molecules.

share|improve this answer

answered May 19 at 12:40

JustanotherchemistJustanotherchemist

2,31977 silver badges2323 bronze badges

$endgroup$

share|improve this answer

answered May 19 at 12:40

JustanotherchemistJustanotherchemist

2,31977 silver badges2323 bronze badges

answered May 19 at 12:40

JustanotherchemistJustanotherchemist

2,31977 silver badges2323 bronze badges

answered May 19 at 12:40

JustanotherchemistJustanotherchemist

2,31977 silver badges2323 bronze badges

9

$begingroup$

Dichlorobis(triphenylphosphine)nickel(II), or $ceNiCl2[P(C6H5)3]2$ in square planar form is red and diamagnetic. The blue form is paramagnetic and features tetrahedral Ni(II) centers. Both tetrahedral and square planar isomers coexist in solutions. Weak field ligands, favor tetrahedral geometry and strong field ligands favor the square planar isomer. Both weak field ($ceCl−$) and strong field ($cePPh3$) ligands comprise $ceNiCl2(PPh3)2$, hence this compound is borderline between the two geometries.

Steric effects also affect the equilibrium; larger ligands favoring the less crowded tetrahedral geometry.[1]

Reference

1. Greenwood, Norman N.; Earnshaw, Alan (1997). Chemistry of the Elements (2nd ed.).

share|improve this answer

edited May 19 at 15:11

Gaurang Tandon

5,50488 gold badges3030 silver badges7272 bronze badges

answered May 19 at 14:45

Chakravarthy KalyanChakravarthy Kalyan

3,03211 gold badge55 silver badges2525 bronze badges

$endgroup$

add a comment
 | 

9

$begingroup$

Dichlorobis(triphenylphosphine)nickel(II), or $ceNiCl2[P(C6H5)3]2$ in square planar form is red and diamagnetic. The blue form is paramagnetic and features tetrahedral Ni(II) centers. Both tetrahedral and square planar isomers coexist in solutions. Weak field ligands, favor tetrahedral geometry and strong field ligands favor the square planar isomer. Both weak field ($ceCl−$) and strong field ($cePPh3$) ligands comprise $ceNiCl2(PPh3)2$, hence this compound is borderline between the two geometries.

Steric effects also affect the equilibrium; larger ligands favoring the less crowded tetrahedral geometry.[1]

Reference

1. Greenwood, Norman N.; Earnshaw, Alan (1997). Chemistry of the Elements (2nd ed.).

share|improve this answer

edited May 19 at 15:11

Gaurang Tandon

5,50488 gold badges3030 silver badges7272 bronze badges

answered May 19 at 14:45

Chakravarthy KalyanChakravarthy Kalyan

3,03211 gold badge55 silver badges2525 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Dichlorobis(triphenylphosphine)nickel(II), or $ceNiCl2[P(C6H5)3]2$ in square planar form is red and diamagnetic. The blue form is paramagnetic and features tetrahedral Ni(II) centers. Both tetrahedral and square planar isomers coexist in solutions. Weak field ligands, favor tetrahedral geometry and strong field ligands favor the square planar isomer. Both weak field ($ceCl−$) and strong field ($cePPh3$) ligands comprise $ceNiCl2(PPh3)2$, hence this compound is borderline between the two geometries.

Steric effects also affect the equilibrium; larger ligands favoring the less crowded tetrahedral geometry.[1]

Reference

1. Greenwood, Norman N.; Earnshaw, Alan (1997). Chemistry of the Elements (2nd ed.).

share|improve this answer

edited May 19 at 15:11

Gaurang Tandon

5,50488 gold badges3030 silver badges7272 bronze badges

answered May 19 at 14:45

Chakravarthy KalyanChakravarthy Kalyan

3,03211 gold badge55 silver badges2525 bronze badges

$endgroup$

share|improve this answer

edited May 19 at 15:11

Gaurang Tandon

5,50488 gold badges3030 silver badges7272 bronze badges

answered May 19 at 14:45

Chakravarthy KalyanChakravarthy Kalyan

3,03211 gold badge55 silver badges2525 bronze badges

edited May 19 at 15:11

Gaurang Tandon

5,50488 gold badges3030 silver badges7272 bronze badges

edited May 19 at 15:11

Gaurang Tandon

5,50488 gold badges3030 silver badges7272 bronze badges

answered May 19 at 14:45

Chakravarthy KalyanChakravarthy Kalyan

3,03211 gold badge55 silver badges2525 bronze badges

answered May 19 at 14:45

Chakravarthy KalyanChakravarthy Kalyan

3,03211 gold badge55 silver badges2525 bronze badges