Q77 · CSIR-NET Chemistry, December 2013

Paper: CSIR-NET December 2013 · Subject: Inorganic Chemistry · Chapter: Organometallic Compounds · Topic: Cluster Compounds Wade Rules · Marks: 2 · Difficulty: Medium

On reducing $\mathrm{Fe}_{3}(\mathrm{CO})_{12}$ with an excess of sodium, a carbonylate ion is formed. The iron is isoelectronic with

(a)$\left[\mathrm{Mn}(\mathrm{CO})_{5}\right]^{-}$

(b)$\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$

(c)$\left[\mathrm{Mn}(\mathrm{CO})_{5}\right]^{+}$

(d)$\left[\mathrm{V}(\mathrm{CO})_{6}\right]^{-}$

Answer

Answer: B ✓ checked by 4AB · confidence high

Explanation

Reduction gives [Fe(CO)₄]²⁻ (Collman's reagent), an 18-electron d¹⁰ species isoelectronic with Ni(CO)₄.

Study loop for Organometallic Compounds