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(A) In the molecule $H_2C = CH - C \equiv CH$: $1$. The first carbon atom $(CH_2)$ is bonded to one double bond,so it is $sp^2$ hybridized. $2$. The s

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$H_2C = CH - C \equiv CH$

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(A) In the molecule $H_2C = CH - C \equiv CH$: $1$. The first carbon atom $(CH_2)$ is bonded to one double bond,so it is $sp^2$ hybridized. $2$. The second carbon atom $(CH)$ is bonded to one double bond and one single bond,so it is $sp^2$ hybridized. $3$. The third carbon atom $(C)$ is bonded to one single bond and one triple bond,so it is $sp$ hybridized. $4$. The fourth carbon atom $(CH)$ is bonded to one triple bond,so it is $sp$ hybridized. Thus,the hybridization sequence is $sp^2 - sp^2 - sp - sp$.

Which of the following represents the given mode of hybridization $sp^2 - sp^2 - sp - sp$ from left to right?

Similar Questions

Hybridization of $1$ and $2$ carbon atoms in $\mathop {CH_2}\limits^1 = \mathop C\limits^2 = CH_2$

What is the increasing order of stability of carbon ions with the change in hybridization of carbon?

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

The compounds containing $sp$-hybridised carbon atom are:
$(I)$ $H_3C-N(CH_3)-CHO$
$(II)$ Pyridine
$(III)$ $H_3C-CN$
$(IV)$ $H_2C=C=CH-CH_3$

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

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