The refluxing of RhCl₃ · 3H₂O with an excess of PPh₃, in ethanol gives a complex A. Complex A and the valence… · ABC26IN0248
Subject: Inorganic Chemistry · Chapter: Organometallic Compounds · Topic: 18 Electron Rule · Exam: CSIR NET 2015 · Marks: 2 · Difficulty: Hard
The refluxing of $\mathrm{RhCl_3 \cdot 3H_2O}$ with an excess of $\mathrm{PPh_3}$, in ethanol gives a complex A. Complex A and the valence electron count on rhodium are, respectively.
(a)$\mathrm{[RhCl(PPh_3)_3]}$, 16
(b)$\mathrm{[RhCl(PPh_3)_5]}$, 16
(c)$\mathrm{[RhCl(PPh_3)_3]}$, 18
(d)$\mathrm{[RhCl(PPh_3)_5]}$, 18 The refluxing of $\mathrm{RhCl_3 \cdot 3H_2O}$ with an excess of $\mathrm{PPh_3}$, in ethanol gives a complex $\mathrm{RhCl(PPh_3)_3}$ is Wilkinson catalyst and common method for preparation of $\mathrm{RhCl(PPh_3)_3}$ is refluxing of $\mathrm{RhCl_3 \cdot 3H_2O}$ with $\mathrm{PPh_3}$. Moreover, on counting the total valence electrons, $\mathrm{RhCl(PPh_3)_3} = [9+3\times2+1] = 16e$. {Cl atom cannot be bridging, its contribution is 1}. Wilkinson catalyst is $16e^-$ species.It is a square planer 16-electron complex. Valence electron counting on rhodium metal centre. (i) there is no overall charge on complex (ii) there is one anionic ligand ($\mathrm{Cl^-}$) (iii) Rh metal atom must have +1 charge to compensate for the one negatively charged ligand. So, the oxidation state of Rh is +1.
Answer
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