Question:
Bond enthalpy of Ge-Ge bond is 260 kJ mol\(^{-1}\). The bond enthalpies of Si-Si and Sn-Sn bonds in kJ mol\(^{-1}\) are respectively:
Bond enthalpy of Ge-Ge bond is 260 kJ mol\(^{-1}\). The bond enthalpies of Si-Si and Sn-Sn bonds in kJ mol\(^{-1}\) are respectively:
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Bond enthalpy increases with shorter bond lengths and stronger covalent character. Moving up a group increases bond enthalpy.
Updated On: Jun 26, 2026
- 240, 270
- 297, 297
- 297, 240
- 200, 348
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The Correct Option is C
Solution and Explanation
Step 1: Recall bond enthalpy trend.
Bond enthalpies generally increase as you move up the group in the periodic table due to shorter bond lengths and stronger bonds.
Step 2: Analyze given bonds.
- Ge-Ge bond = 260 kJ mol\(^{-1}\) (middle of group 14)
- Si-Si bond is above Ge in group 14 → stronger bond → higher enthalpy
- Sn-Sn bond is below Ge in group 14 → weaker bond → lower enthalpy
Step 3: Assign approximate values.
- Si-Si bond ≈ 297 kJ mol\(^{-1}\)
- Sn-Sn bond ≈ 240 kJ mol\(^{-1}\)
Step 4: Conclusion.
The bond enthalpies of Si-Si and Sn-Sn are \[ \boxed{297, 240 \text{ kJ mol}^{-1}} \]
Bond enthalpies generally increase as you move up the group in the periodic table due to shorter bond lengths and stronger bonds.
Step 2: Analyze given bonds.
- Ge-Ge bond = 260 kJ mol\(^{-1}\) (middle of group 14)
- Si-Si bond is above Ge in group 14 → stronger bond → higher enthalpy
- Sn-Sn bond is below Ge in group 14 → weaker bond → lower enthalpy
Step 3: Assign approximate values.
- Si-Si bond ≈ 297 kJ mol\(^{-1}\)
- Sn-Sn bond ≈ 240 kJ mol\(^{-1}\)
Step 4: Conclusion.
The bond enthalpies of Si-Si and Sn-Sn are \[ \boxed{297, 240 \text{ kJ mol}^{-1}} \]
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