ABC26IN0247 · Organometallic Compounds

Subject: Inorganic Chemistry · Chapter: Organometallic Compounds · Topic: 18 Electron Rule · Exam: CSIR-NET 2014 · Marks: 2 · Difficulty: Hard

Complexes $\mathrm{HM(CO)_5}$ and $\mathrm{[((\eta^{5}\text{-}C_5H_5)M'(CO)_3]_2}$ obey the 18-electron rule. Identify M and $\mathrm{M'}$ and their $^{1}\mathrm{H}$ NMR chemical shifts relative to TMS.
(a)M = Mn, $-7.5$; $\mathrm{M'}$= Cr, 4.10
(b)M= Cr, $-4.10$; $\mathrm{M'}$= Mn, $-7.5$
(c)M=V, $-7.5$; $\mathrm{M'}$= Cr, 4.10
(d)M= Mn, $-10.22$ $\mathrm{M'}$= Fe, 2.80
Answer
Answer (as printed): A
Explanation
Complex (A) $\mathrm{HM(CO)_5}=1+7+10=18$ electron $M=\mathrm{Mn}$ [7 electron] $1+7+10=18$ electron Complex (B) $[(\eta^5\text{-}\mathrm{C_5H_5})M'(\mathrm{CO})_3]_2$ $[(\eta^5\text{-}\mathrm{C_5H_5})\mathrm{Cr(CO)_3}]_2$ $M'=\mathrm{Cr}=6$ electron $10+12+12+2$ (M–M bond) $=36$ electron.

Open in whiteboard · Browse this chapter in the app