A molecule $(X)$ has (i) four sigma bonds formed by the overlap of $s p^2$ and $s$ orbitals (ii) one sigma…
- $\mathrm{C}_2 \mathrm{H}_6$
- $\mathrm{C}_2 \mathrm{H}_3 \mathrm{Cl}$
- $\mathrm{C}_2 \mathrm{H}_2 \mathrm{Cl}_2$
- $\mathrm{C}_2 \mathrm{H}_4$
Solution



So, the compound $(X)$ is $\mathrm{C}_2 \mathrm{H}_4$.
Asked in: AP EAMCET 2006
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