Q80 · CSIR NET Chemistry, December 2015
Paper: CSIR NET December 2015 · Subject: Inorganic Chemistry · Chapter: Organometallic Compounds · Topic: Cluster Compounds Wade Rules · Marks: 2 · Difficulty: Hard
Reaction of $[\mathrm{Mn_2(CO)_{10}}]$ with $\mathrm{I_2}$, results in A without loss of CO. Compound A, on heating of $120^{\circ}\mathrm{C}$ loses a CO ligand to give B, which does not have a Mn-Mn bond. Compound B reacts with pyridine to give 2 equivalents of C. Compounds A, B and C from the following respectively, are (I) 
(II) 
(III) 
(IV) 
(V) 
(II) (III) (IV) (V) (a)II, V and IV
(b)II, III and IV
(c)V, III and IV
(d)II, V and III
Answer
Answer: A ✓ checked by 4AB · confidence high
Explanation
I₂ cleaves the Mn–Mn bond, giving Mn(CO)₅I (II). Heating loses CO, giving the iodide-bridged [Mn(CO)₄I]₂ (V) with no Mn–Mn bond. Pyridine splits it into 2 Mn(CO)₄(py)I (IV) (a).
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