Question

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

• A. \(C - C > Si - Si > Ge - Ge > Sn - Sn\)
• B. \(C - C > Si - Si > Sn - Sn > Ge - Ge\)
• C. \(Si - Si > C - C > Ge - Ge > Sn - Sn\)
• D. \(Si - Si > C - C > Sn - Sn > Ge - Ge\)

Hint:

Remember that bond enthalpy generally decreases down a group due to the increase in atomic size and bond length.

Answer 1

The Correct Option is A

Step 1: Consider the Trend Down the Group
Bond enthalpy generally decreases down a group in the periodic table. This is because as the atomic size increases, the bond length increases, and longer bonds are weaker.
Step 2: Analyze the Given Elements
The elements in question are C, Si, Ge, and Sn. They all belong to Group 14. Their atomic size increases down the group in the order:
\[\text{C} < \text{Si} < \text{Ge} < \text{Sn}.\]
Step 3: Determine the Bond Enthalpy Order
Since bond enthalpy decreases with increasing atomic size, the correct order of bond enthalpy is:
\[\text{C–C} > \text{Si–Si} > \text{Ge–Ge} > \text{Sn–Sn}.\]