Q66 · CSIR-NET Chemistry, December 2011

Paper: CSIR-NET December 2011 · Subject: Inorganic Chemistry · Chapter: Organometallic Compounds · Topic: Cluster Compounds Wade Rules · Marks: 2 · Difficulty: Medium

Reaction of $\mathrm{Fe}(\mathrm{CO})_{5}$ with OH leads to complex A which on oxidation with $\mathrm{MnO}_{2}$ gives B . Compounds A and B respectively are
(a)$\left[\mathrm{HFe}(\mathrm{CO})_{4}\right]^{-}$and $\mathrm{Fe}_{3}(\mathrm{CO})_{12}$
(b)$\left[\mathrm{Fe}(\mathrm{CO})_{5}(\mathrm{OH})\right]^{-}$and $\mathrm{Fe}_{2}(\mathrm{CO})_{10}$
(c)$\left[\mathrm{Fe(CO)_4}\right]^{2-}$ and $\mathrm{Mn_2(CO)_{10}}$
(d)$\left[\mathrm{HFe}(\mathrm{CO})_{4}\right]^{-}$, and $\mathrm{Fe}_{2} \mathrm{O}_{3}$
Answer
Answer: A ✓ checked by 4AB · confidence high
Explanation
Hydroxide attacks a CO of Fe(CO)₅ (Hieber base reaction) to give [HFe(CO)₄]⁻ + CO₂; oxidation of this hydride with MnO₂ yields Fe₃(CO)₁₂.

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