The correct order of bond enthalpy $\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)$ is :

(D) $\mathrm{C}-\mathrm{C}>\mathrm{Si}-\mathrm{Si}>\mathrm{Ge}-\mathrm{Ge}>\mathrm{Sn}-\mathrm{Sn}$

(A) $\mathrm{Si}-\mathrm{Si}>\mathrm{C}-\mathrm{C}>\mathrm{Sn}-\mathrm{Sn}>\mathrm{Ge}-\mathrm{Ge}$

(B) $\mathrm{Si}-\mathrm{Si}>\mathrm{C}-\mathrm{C}>\mathrm{Ge}-\mathrm{Ge}>\mathrm{Sn}-\mathrm{Sn}$

(C) $\mathrm{C}-\mathrm{C}>\mathrm{Si}-\mathrm{Si}>\mathrm{Sn}-\mathrm{Sn}>\mathrm{Ge}-\mathrm{Ge}$

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Chemical Thermodynamics

The correct order of bond enthalpy

(kJ mol^{- 1})

is :

(A) $Si - Si > C - C > Sn - Sn > Ge - Ge$
(B) $Si - Si > C - C > Ge - Ge > Sn - Sn$
(C) $C - C > Si - Si > Sn - Sn > Ge - Ge$
(D) $C - C > Si - Si > Ge - Ge > Sn - Sn$ ✓ Correct

Answer: (D)

Solution

Single-bond enthalpy within a group falls as the atoms grow larger and orbital overlap becomes less effective. For the group-14 elements the carbon–carbon bond is the strongest because the compact

2 p

orbitals overlap most efficiently, and the strength then decreases steadily down the group as the valence orbitals become larger and more diffuse. This places the bond enthalpies in the order

C - C > Si - Si > Ge - Ge > Sn - Sn

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Single correct option

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JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Chemical Thermodynamics

Solution

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