1. Tardigrade
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  3. Chemistry
  4. In the compound CH 2= CH - CH 2- CH 2- C ≡ CH, the ( C 2- C 3) bond is of the type:

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Q. In the compound $CH _2= CH - CH _2- CH _2- C \equiv CH$, the $\left( C _2- C _3\right)$ bond is of the type:

Hydrocarbons

Solution:

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Hence, $C _2$ and $C _3$ are $s p^2$-and $s p^3$-hybridised