Wilkinson's catalyst react with `H_(2)` to from an octahedral complex in which Rh (Z= 45) has the following electronic configuration in the ligand field `t_(2g)^(2.2.2,e_(g)^(0.0).` Then which of the following is (are) correct about this new complex ?

Options

A. It is diamagnetic
B. Its IUPAC name is chlorodihydridotris (triphenylphosphine) rohidium (III)
C. Hybridisation of Rh (I) is `d^(2)sp^(3)`
D. It is a paramagnetic complex

Explanation

(A,B)
(A) `Delta _(o)prop` strength of ligand. `CN^(-)` is stronger field ligand than `H_(2)O.So Delta_(o) of [Co(CN)_(6)]^(3-) gt [Co(H_(2)O)_(6)]^(2+) [Co(CN_(6))]^(3)`
(B) `Delta_(o)prop` oxidation state of central metal ion. So `Delta_(o)` o`[Rh^(1)Cl(PPh_(3))_(3)]+H_(2) rarr [Rh^(III) CiH_(2)(PPh_(3))_(3)][Rh^(III)ClH_(2)(PPh_(3))_(3)]` gives electronic congiguration of `t_(2g)^(2,2,2,),e_(g)^(0.0),` inner orbital complex and thus to have two d-orbitals empty, electrons wil pair up and give `d^(2)sp^(3)` hydridisation. `Rh^(1)` with C.N = 4 has `dap^(2)` hybridisation (4d series element)