Bond enthalpy of Ge-Ge bond is 260 kJ mol^{ | vedclass

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(C) $Si$,$Ge$,and $Sn$ belong to the $3^{rd}$,$4^{th}$,and $5^{th}$ periods of group $14$ of the periodic table.Moving down the group,the atomic size

Vedclass · Accepted Answer

$297, 240$

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(C) $Si$,$Ge$,and $Sn$ belong to the $3^{rd}$,$4^{th}$,and $5^{th}$ periods of group $14$ of the periodic table.Moving down the group,the atomic size increases,which leads to an increase in bond length.Therefore,the order of bond length is $Si-Si Since bond enthalpy is inversely proportional to bond length,the order of bond enthalpy is $Si-Si > Ge-Ge > Sn-Sn$.Given the value for $Ge-Ge$ is $260 \ kJ \ mol^{-1}$,the values for $Si-Si$ and $Sn-Sn$ are $297 \ kJ \ mol^{-1}$ and $240 \ kJ \ mol^{-1}$ 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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