The correct order of bond enthalpy \(\left( kJ mol ^{-1}\right)\) is :
Remember that bond enthalpy generally decreases down a group due to the increase in atomic size and bond length.
\(C - C > Si - Si > Ge - Ge > Sn - Sn\)
\(C - C > Si - Si > Sn - Sn > Ge - Ge\)
\(Si - Si > C - C > Ge - Ge > Sn - Sn\)
\(Si - Si > C - C > Sn - Sn > Ge - Ge\)
To determine the correct order of bond enthalpy for the given bonds, we need to understand the concept of bond enthalpy. Bond enthalpy is the energy required to break one mole of a bond in a molecule in the gaseous state.
The bond enthalpies of single bonds involving elements from the carbon group, like C-C, Si-Si, Ge-Ge, and Sn-Sn, are influenced by factors such as bond length, bond order, and atomic size. In general, as the atomic size increases down a group, the bond length increases, leading to weaker bonds and thus lower bond enthalpies.
Let's analyze each option:
The correct answer is option 1: \(C - C > Si - Si > Ge - Ge > Sn - Sn\), as it accurately reflects the decreasing bond enthalpy with increasing atomic size down the group.
\(O - O\) bond length in \(H _2 O _2\) is X than the \(O - O\) bond length in \(F _2 O _2\)The \(O - H\) bond length in \(H _2 O _2\)is Y than that of the\(O - F\) bond in \(F _2 O _2\)Choose the correct option for \(\underline{X} and \underline{Y}\) from those given below :