What is the correct order of bond strength?

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

Question

(A) The bond strength is directly proportional to the bond order and the $s$-character of the hybrid orbitals involved in the bond formation.For $C-C$

Vedclass · Accepted Answer

$C-C < C=C < C \equiv C$

Vedclass · Answer

(A) The bond strength is directly proportional to the bond order and the $s$-character of the hybrid orbitals involved in the bond formation.For $C-C$ (alkane),the hybridization is $sp^3$ ($25\% \ s$-character).For $C=C$ (alkene),the hybridization is $sp^2$ ($33.3\% \ s$-character).For $C \equiv C$ (alkyne),the hybridization is $sp$ ($50\% \ s$-character).As the $s$-character increases,the bond becomes shorter and stronger.Therefore,the correct order of bond strength is $C-C < C=C < C \equiv C$.

Similar Questions

Which of the following molecules is electron-deficient?

The interatomic distances in $H_2$ and $Cl_2$ molecules are $74 \, pm$ and $198 \, pm$ respectively. The bond length of $HCl$ is $.............$ $pm$.

The correct increasing order of $C-H(A)$, $C-O(B)$, $C=O(C)$ and $C\equiv N(D)$ bonds in terms of covalent bond length is:

Which of the following bonds has the highest bond enthalpy?

Explain the Valence Bond Theory $(VBT)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module