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

A l C l_{3}

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E t_{3} A l

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C_{2} H_{4}

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T i C l_{4}

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Solution

The correct option is D $T i C l_{4}$
Zeigler - Natta polymerisation of ethylene takes place as above. ${TiCl}_{4}$ is the active species.
Hence, the correct option is $D$

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

T i C l_{4} + (C_{2} H_{5})_{3} A l

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