Question

The amount of energy required to break a bond is same as the amount of energy released when the same bond is formed. In gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy (BDE) or Bond Strength. BDE is affected by s-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below:

Correct match of the $\text{C–H}$ bonds (shown in bold) in Column J with their $\text{BDE}$ in Column $\text{K}$ is

Column $\text{J}$ Molecule	Column $\text{K}$ $BDE\left(kcalmo{l}^{-1}\right)$
(P) $H-CH(C{H}_3{)}_2$	(i) $132$
(Q) $H-C{H}_{2}Ph$	(ii) $110$
(R) $H-CH=C{H}_2$	(iii) $95$
(S) $H-C\equiv CH$	(iv) $88$

• A. P-iii, Q - iv, R - ii, S - I
• B. P - i, Q - ii, R - iii, S - iv
• C. P - iii, Q - ii, R - i, S - iv
• D. P - ii, Q - i, R - iv, S - iii

Answer 1

The correct option is A P-iii, Q - iv, R - ii, S - I

As percentage of S character increases bond dissociation energy$\left(BDE\right)$ increases. Each sp hybridized orbital has an equal amount of s and p character – 50% s and 50% p character and Each of the $s{p}^2$ hybrid orbitals formed has a 33.33% ‘s’ character and 66.66%‘p’ character. Option P and Q can form most stable form by loosing H so the $BDE$ is less. Thus option (a) is correct.