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.

• A. 240, 270
• B. 297, 297
• C. 297, 240
• D. 200, 348

Hint:

Bond enthalpy for X-X single bonds in Group 14 generally decreases down the group: C-C>Si-Si>Ge-Ge>Sn-Sn>Pb-Pb.
This is due to increasing atomic size, leading to longer bonds and less effective overlap of orbitals.
Use the given Ge-Ge bond enthalpy as a reference point. Si-Si should be higher, and Sn-Sn should be lower.
Check options against this relative trend.

Answer 1

The Correct Option is C

Si, Ge, and Sn are all elements of Group 14 (Carbon group) of the periodic table. The order in the group is: C>Si>Ge>Sn>Pb. Bond enthalpy (or bond strength) for single bonds between identical atoms (X-X) in this group generally decreases down the group as the atomic size increases and the extent of effective orbital overlap decreases. So, the expected trend for X-X bond enthalpies (where X = C, Si, Ge, Sn, Pb) is: C-C>Si-Si>Ge-Ge>Sn-Sn>Pb-Pb. Given: Bond enthalpy of Ge-Ge = $260 \text{ kJ mol}^{-1}$. We need to find the bond enthalpies for Si-Si and Sn-Sn. Based on the trend: Bond enthalpy (Si-Si) should be greater than Bond enthalpy (Ge-Ge). So, BE(Si-Si)>$260 \text{ kJ mol}^{-1}$. Bond enthalpy (Sn-Sn) should be less than Bond enthalpy (Ge-Ge). So, BE(Sn-Sn)<$260 \text{ kJ mol}^{-1}$. Let's examine the options: (Si-Si value, Sn-Sn value) % Option (a) (240, 270): BE(Si-Si) = 240 (which is<260, incorrect). BE(Sn-Sn) = 270 (which is>260, incorrect). % Option (b) (297, 297): BE(Si-Si) = 297 (which is>260, possible). BE(Sn-Sn) = 297 (which is>260, incorrect). % Option (c) (297, 240): BE(Si-Si) = 297 (which is>260, possible). BE(Sn-Sn) = 240 (which is<260, possible). This option fits the expected trend. % Option (d) (200, 348): BE(Si-Si) = 200 (which is<260, incorrect). BE(Sn-Sn) = 348 (which is>260, incorrect). Option (c) is the only one that follows the expected trend: Si-Si bond is stronger than Ge-Ge bond, and Ge-Ge bond is stronger than Sn-Sn bond. BE(Si-Si) = $297 \text{ kJ mol}^{-1}$ BE(Ge-Ge) = $260 \text{ kJ mol}^{-1}$ (given) BE(Sn-Sn) = $240 \text{ kJ mol}^{-1}$ This order $297>260>240$ is consistent with the trend. Typical literature values for these bond enthalpies are approximately: C-C: ~347 kJ/mol Si-Si: ~226 kJ/mol (or values up to ~300 kJ/mol in some sources, e.g., for bulk silicon ~222-226) Ge-Ge: ~188 kJ/mol (or ~260 kJ/mol as given in problem) Sn-Sn: ~146 kJ/mol (or higher in some contexts) The values in options can sometimes deviate from precise literature averages due to different sources or specific molecular contexts. However, the relative trend is usually robust. The question gives Ge-Ge as 260 kJ/mol. Option (c) gives Si-Si = 297 kJ/mol and Sn-Sn = 240 kJ/mol. This fits the order: Si-Si (297)>Ge-Ge (260)>Sn-Sn (240). This is the correct trend. \[ \boxed{\text{297, 240}} \]