Nodal planes of pi-bond(s) in CH_2=C=C=CH_2 a | vedclass

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

Question

(C) The molecule $CH_2=C=C=CH_2$ is an allene derivative (specifically,buta$-1,2,3-$triene). In this molecule,the terminal $CH_2$ groups are perpendic

Vedclass · Accepted Answer

one in molecular plane and two in plane perpendicular to molecular plane which contains $C-C$ $\sigma$-bonds

Vedclass · Answer

(C) The molecule $CH_2=C=C=CH_2$ is an allene derivative (specifically,buta$-1,2,3-$triene). In this molecule,the terminal $CH_2$ groups are perpendicular to each other. The two $\pi$-bonds are formed by the overlap of $p$-orbitals. One $\pi$-bond is formed by the overlap of $p_y$ orbitals,and the other $\pi$-bond is formed by the overlap of $p_z$ orbitals. The nodal plane of a $\pi$-bond is the plane containing the internuclear axis and perpendicular to the plane of the $p$-orbitals. For the two $\pi$-bonds in this system,one nodal plane lies in the molecular plane (containing the $C-C$ $\sigma$-bonds),and the other nodal plane is perpendicular to the molecular plane.

Nodal planes of $\pi$-bond$(s)$ in $CH_2=C=C=CH_2$ are located in

Similar Questions

$A$ carbon-carbon triple bond in ethyne $(HC \equiv CH)$ consists of:

The correct order of the extent of overlapping is:

Which of the following statements is $NOT$ true regarding the sigma and pi bonds formed between two carbon atoms?

Explain the characteristics of $\pi$ bonds. Or,give the requirements for the formation of $\pi$ bonds and explain the active centers.

Which one is stronger,$\sigma$ bond or $\pi$ bond? Why?

Vedclass Test Series

Exam Paper Generator

Online Exam Module