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

We need to arrange the bond enthalpies of $$C-C$$, $$Si-Si$$, $$Ge-Ge$$, and $$Sn-Sn$$ in the correct order.

Bond enthalpy (or bond energy) is the energy required to break one mole of bonds in the gaseous state. For single bonds between identical atoms in the same group of the periodic table, the bond enthalpy generally decreases as we go down the group.

As we move down Group 14 (C, Si, Ge, Sn):

- The atomic size increases due to the addition of new electron shells.

- The bond length increases because larger atoms cannot approach each other as closely.

- Longer bonds have weaker orbital overlap, resulting in lower bond enthalpy.

- The bonding electrons are farther from the nuclei, so the electrostatic attraction holding the bond together is weaker.

The approximate bond enthalpies are:

- $$C-C$$: 348 kJ/mol

- $$Si-Si$$: 226 kJ/mol

- $$Ge-Ge$$: 188 kJ/mol

- $$Sn-Sn$$: 151 kJ/mol

Therefore, the decreasing order of bond enthalpy is:

$$C-C > Si-Si > Ge-Ge > Sn-Sn$$

The correct answer is Option (4): $$C-C > Si-Si > Ge-Ge > Sn-Sn$$.