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In Zeigler-Natta polymerization of ethylene, the active species is

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${E t}_{3} A l$

$A l {C l}_{3}$

$H_{2} C = {C H}_{2}$

${T i}^{3 +}$

answer is D.

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Detailed Solution

Terminal 1-alkenes (ethylene and alkenes with the vinyl double bond) are polymerized as follows using Ziegler Natta catalysts:

${n C H}_{2} = C H R \to {{[C H}_{2} - C H R]}_{n}$ .

Zeigler-Natta catalyst is formed by the action of (C₂H₅)₃Al and TiCl₄. Active species is Ti³⁺ as (C₂H₅)₃Al) reduces TiCl₄ to TiCl₃.

Ti³⁺ has active site vacant and thus, can accommodate one alkyl group.

Hence the correct option is (D).

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