Question

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

• A. $AlC{l}_3$
• B. $(C_2{H}_5{)}_{3}Al$
• C. $C_2{H}_5$
• D. $T{i}^{3+}$

Answer 1

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

When $TiC{l}_4$ and trialkyl aluminium react to produce a fibrous form of $TiC{l}_3$, which is active metal species in Zeigler-Natta catalyst. Since in $TiC{l}_3$, $Ti$ is in +3 oxidation state.

So, option D is the correct option.