Question

In Ziegler-Natta polymerisation of ethylene, the active species are:

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

Answer 1

Answer: (C)

$C{H}_{2}C{H}_2$

$T{i}^{3+}$ ions from the titanium compound having active site vacant, which can accomodate one alkyl group.So the active species in Ziegler-Natta polymerization of ethylene is $T{i}^{3+}$

Hence option C is correct.