Number of sp hybridised carbon atoms in the g | vedclass

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(C) To determine the hybridisation of carbon atoms,we count the number of sigma bonds and lone pairs (if any) attached to each carbon atom.$1$. $CH_2=

Vedclass · Accepted Answer

$3$

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(C) To determine the hybridisation of carbon atoms,we count the number of sigma bonds and lone pairs (if any) attached to each carbon atom.$1$. $CH_2=CH-C\equiv C-CH_2-CN$$2$. The structure is $H_2C=CH-C\equiv C-CH_2-C\equiv N$.$3$. Carbon atoms involved in a triple bond ($C\equiv C$ or $C\equiv N$) are $sp$ hybridised.$4$. In the given molecule,the carbons in the $C\equiv C$ group are $sp$ hybridised ($2$ carbons).$5$. The carbon in the $CN$ group is also $sp$ hybridised ($1$ carbon).$6$. Total number of $sp$ hybridised carbon atoms = $2 + 1 = 3$.

Number of $sp$ hybridised carbon atoms in the given compound $CH_2=CH-C\equiv C-CH_2-CN$ is:

Similar Questions

In which of the compounds given below is there more than one kind of hybridisation $(sp, sp^2, sp^3)$ for carbon?
$(i)$ $CH_3CH_2CH_2CH_3$
$(ii)$ $CH_3-CH=CH-CH_3$
$(iii)$ $CH_2=CH-CH=CH_2$
$(iv)$ $HC\equiv CH$

Among the following compounds,which have more than one type of hybridization for carbon atoms?
$(i) CH_3CH_2CH_2CH_3$
$(ii) CH_3-CH=CH-CH_3$
$(iii) CH_2=CH-C\equiv CH$
$(iv) HC\equiv CH$

The hybridization of $C_2$ and $C_3$ in $CH_3-CH=C=CH-CH_3$ is

What hybrid orbitals are present in the following compound $H_3C - CH = CH - CH_2 - CH_3$?

Number of pure $p-$ orbitals in the given molecule $\text{hexa-}1, 3-\text{dien-}5-\text{yne}$.

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