How many $\sigma$ and $\pi$ bonds are present in each of the following molecules?
$(a)$ $HC \equiv C-CH=CH-CH_3$
$(b)$ $CH_2=C=CH-CH_3$
$(a)$ $HC \equiv C-CH=CH-CH_3$
$(b)$ $CH_2=C=CH-CH_3$
(N/A) For $(a)$ $HC \equiv C-CH=CH-CH_3$:
Total $\sigma$ bonds = $4$ ($C$-$C$) + $6$ ($C$-$H$) = $10$ $\sigma$ bonds.
Total $\pi$ bonds = $2$ ($\pi$ in $C \equiv C$) + $1$ ($\pi$ in $C=C$) = $3$ $\pi$ bonds.
For $(b)$ $CH_2=C=CH-CH_3$:
Total $\sigma$ bonds = $3$ ($C$-$C$) + $6$ ($C$-$H$) = $9$ $\sigma$ bonds.
Total $\pi$ bonds = $2$ ($\pi$ in $C=C=C$) = $2$ $\pi$ bonds.
Total $\sigma$ bonds = $4$ ($C$-$C$) + $6$ ($C$-$H$) = $10$ $\sigma$ bonds.
Total $\pi$ bonds = $2$ ($\pi$ in $C \equiv C$) + $1$ ($\pi$ in $C=C$) = $3$ $\pi$ bonds.
For $(b)$ $CH_2=C=CH-CH_3$:
Total $\sigma$ bonds = $3$ ($C$-$C$) + $6$ ($C$-$H$) = $9$ $\sigma$ bonds.
Total $\pi$ bonds = $2$ ($\pi$ in $C=C=C$) = $2$ $\pi$ bonds.
Explore More
Similar Questions
Which of the following combinations does not form a $\sigma$ bond?
$(i) \ 2p_y - 2p_y$
$(ii) \ 2p_z - 2p_z$
$(iii) \ 2s - 2p_x$
$(iv) \ 2p_x - 2p_x$
$(i) \ 2p_y - 2p_y$
$(ii) \ 2p_z - 2p_z$
$(iii) \ 2s - 2p_x$
$(iv) \ 2p_x - 2p_x$
Easy
View SolutionThe correct order of the extent of overlapping is:
Difficult
View SolutionExplain $\sigma$ and $\pi$ bonds,and mention the orbital overlapping involved in their formation.
In a double bond between two carbon atoms of ethene,there are
Medium
View SolutionIn $N_2$ molecule,the atoms are bonded by
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo