The ratio of sigma-bond and pi-bond in tetrac | vedclass

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(B) The structure of tetracyanoethylene is $(CN)_2C=C(CN)_2$.In this molecule,there is a central $C=C$ double bond (one $\sigma$ and one $\pi$ bond).E

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$1 : 1$

Vedclass · Answer

(B) The structure of tetracyanoethylene is $(CN)_2C=C(CN)_2$.In this molecule,there is a central $C=C$ double bond (one $\sigma$ and one $\pi$ bond).Each of the four $C-CN$ bonds contains one $\sigma$ bond.Each of the four $C \equiv N$ triple bonds contains one $\sigma$ bond and two $\pi$ bonds.Total $\sigma$-bonds = $1$ (central $C-C$) + $4$ $(C-CN)$ + $4$ $(C-N)$ = $9$.Total $\pi$-bonds = $1$ (central $C=C$) + $4 	imes 2$ $(C \equiv N)$ = $9$.The ratio of $\sigma$-bond to $\pi$-bond is $9 : 9$,which simplifies to $1 : 1$.

The ratio of $\sigma$-bond and $\pi$-bond in tetracyanoethylene is

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