Question

In Zeigler Natta polymerisation of ethylene, the active species is:

• A. $AlC{l}_3$
• B. $E{t}_{3}Al$
• C. $C{H}_{2}C{H}_2$
• D. $T{i}^{3+}$

Answer 1

The correct option is D $T{i}^{3+}$

$(C_2{H}_5{)}_{3}Al+TiC{l}_4\to$ Active species.

$T{i}^{3+}$ has one active site vacant and thus accommodate one alkyl group. [as $(C_2{H}_5{)}_{3}Al$ reduces $TiC{l}_4$ to $TiC{l}_3$].