ABC26IN0282 · Organometallic Compounds

Subject: Inorganic Chemistry · Chapter: Organometallic Compounds · Topic: Metal Carbonyls · Exam: CSIR-NET 2019 · Marks: 2 · Difficulty: Hard

The correct order of metal-carbon distance is

(a)$\mathrm{Fe(\eta^{5}\text{-}Cp)_2 > Co(\eta^{5}\text{-}Cp)_2 > Ni(\eta^{5}\text{-}\,Cp)_2}$

(b)$\mathrm{Fe(\eta^{5}\text{-}Cp)_2 > Ni(\eta^{5}\text{-}\,Cp)_2 > Co(\eta^{5}\text{-}Cp)_2}$

(c)$\mathrm{Ni(\eta^{5}\text{-}\,Cp)_2 > Fe(\eta^{5}\text{-}Cp)_2 > Co(\eta^{5}\text{-}Cp)_2}$

(d)$\mathrm{Ni(\eta^{5}\text{-}Cp)_2 > Co(\eta^{5}\text{-}Cp)_2 > Fe(\eta^{5}\text{-}Cp)_2}$

Answer

SELF-PRACTICE — the source book printed no answer.

Nothing is invented here, so this question has no answer on record.

Explanation

$\mathrm{Cp_2Fe}$ $10e^{-}+8e^{-}=18e^{-}$; number of electron in anti-bonding orbital = 0 $\mathrm{Cp_2Co}$ $10e^{-}+9e^{-}=19e^{-}$; number of electron in anti-bonding orbital = 1 $\mathrm{Cp_2Ni}$ $10e^{-}+10e^{-}=20e^{-}$; number of electron in anti-bonding orbital = 2 As the number of electron present in the anti-bonding orbital increases, M–ligand bond becomes weak and hence, bond length increases. Therefore, the correct order of M–C distance is $\mathrm{Cp_2Ni} > \mathrm{Cp_2Co} > \mathrm{Cp_2Fe}$

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