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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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?










share|improve this question












$endgroup$





















    8














    $begingroup$


    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?










    share|improve this question












    $endgroup$

















      8












      8








      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?










      share|improve this question












      $endgroup$




      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
















      share|improve this question













      share|improve this question




      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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          2 Answers
          2






          active

          oldest

          votes


















          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










          $endgroup$










          • 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


















          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.



          enter image description here



          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












          $endgroup$
















            Your Answer








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            2 Answers
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            active

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            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










            $endgroup$










            • 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















            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










            $endgroup$










            • 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













            12














            12










            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










            $endgroup$



            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













            share|improve this answer




            share|improve this answer










            answered May 19 at 12:40









            JustanotherchemistJustanotherchemist

            2,3197 silver badges23 bronze badges




            2,3197 silver badges23 bronze badges










            • 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












            • 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







            1




            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




            $begingroup$
            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













            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.



            enter image description here



            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












            $endgroup$



















              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.



              enter image description here



              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












              $endgroup$

















                9














                9










                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.



                enter image description here



                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












                $endgroup$



                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.



                enter image description here



                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















                share|improve this answer




                share|improve this answer








                edited May 19 at 15:11









                Gaurang Tandon

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                5,5048 gold badges30 silver badges72 bronze badges










                answered May 19 at 14:45









                Chakravarthy KalyanChakravarthy Kalyan

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                3,0321 gold badge5 silver badges25 bronze badges































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