A molecule (X) has (i) four sigma bonds forme | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The molecule $C_2H_4$ (ethene) has the structure $CH_2=CH_2$. Each carbon atom is $sp^2$ hybridized. $(i)$ There are four $C-H$ sigma bonds formed

Vedclass · Accepted Answer

$C_2H_4$

Vedclass · Answer

(D) The molecule $C_2H_4$ (ethene) has the structure $CH_2=CH_2$. Each carbon atom is $sp^2$ hybridized. $(i)$ There are four $C-H$ sigma bonds formed by the overlap of $sp^2$ orbitals of carbon and $s$ orbitals of hydrogen. $(ii)$ There is one $C-C$ sigma bond formed by the overlap of $sp^2$ and $sp^2$ orbitals. $(iii)$ There is one $\pi$ bond formed by the lateral overlap of unhybridized $p_z$ orbitals on each carbon atom. Thus,the molecule $(X)$ is $C_2H_4$.

$A$ molecule $(X)$ has $(i)$ four sigma bonds formed by the overlap of $sp^2$ and $s$ orbitals,$(ii)$ one sigma bond formed by $sp^2$ and $sp^2$ orbitals,and $(iii)$ one $\pi$ bond formed by $p_z$ and $p_z$ orbitals. Which of the following is $X$?

Similar Questions

Seminal plasma in human males is rich in

Which of the following radiations has the least wavelength?

The protective coat of structural protein in viruses is made up of units called:

$A$ particle moves along a circle of radius $\frac{20}{\pi} \; m$ with constant tangential acceleration. If the velocity of the particle is $80 \; m/s$ at the end of the second revolution after motion has begun,the tangential acceleration is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module