
In Wilkinson’s catalyst, the hybridization of central metal ion and its shape are respectively:
A.\[s{p^3}d - trigonalbipyramidal\]
B.\[{d^2}s{p^3} - octahedral\]
C.\[ds{p^2} - squareplanar\]
D.\[s{p^3} - tetrahedral\]
Answer
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Hint: Wilkinson's catalyst is an important organic reagent. The formula of Wilkinson’s catalyst is \[RhCl{(PP{h_3})_3}\]where Rh is rhodium and Ph is the phenyl group.
Rhodium is a d block element having an electronic configuration of \[5{s^2},4{d^7}\].
Step-by-step explanation:
Step 1:
Coordination compounds are those molecular compounds in which a central metal atom or ion is permanently attached to certain atoms or groups of atoms called ligands. The ligands are capable of donating at least a pair of electrons to the central metal atom or ion and get attached to it by coordinate or dative bonds.
Step 2:
Coordination compounds are of two major types: Four coordination compounds, where the number of ligands attached to the metal atom are four and six coordination compounds, where the number of ligands attached to the metal atom are six.
Step 3:
The four coordination compounds are again classified into two types:
Square planar complexes: The complex where the central metal atom possesses \[ds{p^2}\]hybridisation. In these complexes, the four ligands and the central atom lie in the same plane and thereby possess a plane or an axis of symmetry.
Tetrahedral complexes: The complex where the central metal atom possesses \[s{p^3}\] hybridisation. They possess a tetrahedral geometry.
Step 4:
The IUPAC name of Wilkinson’s catalyst is chloridotris(triphenylphosphine)rhodium(I). Here the central metal atom is Rhodium which is a transition metal ion having a \[ds{p^2}\] hybridisation with a coordination number of four. As it has a \[ds{p^2}\] hybridisation so the geometry of the complex is square planar.
Hence, the central metal of Wilkinson’s catalyst has a \[ds{p^2}\]hybridisation and a square planar geometry, so option (C ) is the correct option.
Note: The square planar and the tetrahedral complexes both have 4 coordination numbers (number of ligands attached to the central metal ion) but their geometry and optical activity is different.
Rhodium is a d block element having an electronic configuration of \[5{s^2},4{d^7}\].
Step-by-step explanation:
Step 1:
Coordination compounds are those molecular compounds in which a central metal atom or ion is permanently attached to certain atoms or groups of atoms called ligands. The ligands are capable of donating at least a pair of electrons to the central metal atom or ion and get attached to it by coordinate or dative bonds.
Step 2:
Coordination compounds are of two major types: Four coordination compounds, where the number of ligands attached to the metal atom are four and six coordination compounds, where the number of ligands attached to the metal atom are six.
Step 3:
The four coordination compounds are again classified into two types:
Square planar complexes: The complex where the central metal atom possesses \[ds{p^2}\]hybridisation. In these complexes, the four ligands and the central atom lie in the same plane and thereby possess a plane or an axis of symmetry.
Tetrahedral complexes: The complex where the central metal atom possesses \[s{p^3}\] hybridisation. They possess a tetrahedral geometry.
Step 4:
The IUPAC name of Wilkinson’s catalyst is chloridotris(triphenylphosphine)rhodium(I). Here the central metal atom is Rhodium which is a transition metal ion having a \[ds{p^2}\] hybridisation with a coordination number of four. As it has a \[ds{p^2}\] hybridisation so the geometry of the complex is square planar.
Hence, the central metal of Wilkinson’s catalyst has a \[ds{p^2}\]hybridisation and a square planar geometry, so option (C ) is the correct option.
Note: The square planar and the tetrahedral complexes both have 4 coordination numbers (number of ligands attached to the central metal ion) but their geometry and optical activity is different.
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