Q76 · CSIR NET Chemistry, December 2013
Paper: CSIR NET December 2013 · Subject: Inorganic Chemistry · Chapter: Organometallic Compounds · Topic: Organometallic Compounds – General · Marks: 2 · Difficulty: Hard
Reaction of $\mathrm{Ph_2PCH_2CH_2PPh_2}$ with $\mathrm{[RhCl(CO)_2]_2}$ in a 2:1 molar ratio gives a crystalline solid A. The IR spectrum of complex A shows $\nu_{\mathrm{CO}}$ at $1985\,\mathrm{cm^{-1}}$. The $^{31}\mathrm{P}(^{1}\mathrm{H})$ NMR spectrum of A consists of two doublets of doublets of equal intensities ($^{103}\mathrm{Rh}$ is 100% abundant and I =1/2 ). The structure of complex A is
(a)
(b)
(c)
(d)
Answer
Answer: A ✓ checked by 4AB · confidence medium
Explanation
Two dppe per Rh₂ dimer is one dppe per Rh, giving square-planar RhCl(CO)(dppe) with a terminal CO (1985 cm⁻¹). Its two P atoms are inequivalent (trans to Cl and trans to CO), and each couples to ¹⁰³Rh and to the other P: two doublets of doublets of equal intensity (a).
Study loop for Organometallic Compounds