Easy MCQ +4 / -1 PYQ · JEE Mains 2023

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

  1. A $\mathrm{C-C > Si-Si > Sn - Sn > Ge-Ge}$
  2. B $\mathrm{Si-Si > C-C > Ge-Ge > Sn-Sn}$
  3. C $\mathrm{C-C > Si-Si > Ge-Ge > Sn-Sn}$ Correct answer
  4. D $\mathrm{Si-Si > C-C > Sn-Sn > Ge-Ge}$

Solution

C–C bonds bond energy = 348 kJ mol<sup>-1</sup> <br/><br/>Si–Si bonds bond energy = 297 kJ mol<sup>-1</sup> <br/><br/>Ge–Ge bonds bond energy = 260 kJ mol<sup>-1</sup> <br/><br/>Sn–Sn bonds bond energy = 240 kJ mol<sup>-1</sup> <br/><br/>$\therefore$ Correct order is - <br/><br/>$\mathrm{C-C > Si-Si > Ge-Ge > Sn-Sn}$ <br/><br/><b>Note :</b> <br/><br/>The property of forming bonds with atoms of the same element or tendency to self linking called catenation. <br/><br/>Carbon shows maximum catenation. On moving down the group catenation tendency decreases. This because the strength of $\mathrm{C}-\mathrm{C}$ bond is very high and in case of other elements, strength of ${M}-{M}$ (where ${M}={Si}, {G e}, {S n}, {P b}$ ) bond is decreases down the group.

About this question

Subject: Chemistry · Chapter: p-Block Elements · Topic: Group 13: Boron Family

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