Subject: Inorganic Chemistry · Chapter: Organometallic Compounds · Topic: Metal Carbonyls · Exam: CSIR-NET JUNE 2011 · Marks: 2 · Difficulty: Medium
A and F 3. A and D 4. C and E
(a)Ti(Ch2Ph) 4 <Ti(i-pr) 4 <TiEt4 < TiMe4
(b)TiEt4 <TiMe4 <Ti(i-Pr) 4 <Ti(CH2Ph)4
(c)Ti(i-Pr)4 <TiEt4 <TiMe4 <Ti(CH2Ph)4
(d)TiMe4 < TiEt4 <Ti(i-Pr) 4 <Ti(CH2Ph)4
Answer
Answer (as printed): C
Explanation
This is the correct stability order because $\mathrm{Ti(CH_2Ph)_4}$ has no hydrogen. Hence no elimination, and the compound is stable, whereas $\mathrm{Ti(i\text{-}Pr)_4}$ has the maximum number of hydrogens, followed by $\mathrm{TiEt_4}$ and $\mathrm{TiMe_4}$. So $\mathrm{Ti(i\text{-}Pr)_4}$ undergoes fast elimination, hence least stable.$\mathrm{Ti(i\text{-}Pr)_4} < \mathrm{TiEt_4} < \mathrm{TiMe_4} < \mathrm{Ti(CH_2Ph)_4}$