The correct order of bond enthalpy `(kJ mol^(–1))` is :

Options

A. `C – C gt Si – Si gt Ge – Ge gt Sn – Sn `
B. `C – C gt Si – Si gt Sn – Sn gt Ge – Ge`
C. `Si – Si gt C – C gt Ge – Ge gt Sn – Sn`
D. `Si – Si gt C – C gt Sn – Sn gt Ge – Ge`

Answer

To determine the correct order of bond enthalpy for the elements mentioned in the question (carbon, silicon, germanium, tin), we need to analyze the bond enthalpy values and the trends in the periodic table.

### Step-by-Step Solution:

1. **Identify the Elements:**
The elements in question are carbon (C), silicon (Si), germanium (Ge), and tin (Sn). All these elements belong to Group 14 of the periodic table.

2. **Understand Bond Enthalpy:**
Bond enthalpy (or bond dissociation enthalpy) is the energy required to break a bond in a molecule. Higher bond enthalpy indicates a stronger bond.

3. **List Known Bond Enthalpy Values:**
- Carbon-Carbon (C-C) bond enthalpy: approximately 348 kJ/mol
- Silicon-Silicon (Si-Si) bond enthalpy: approximately 297 kJ/mol
- Germanium-Germanium (Ge-Ge) bond enthalpy: approximately 260 kJ/mol
- Tin-Tin (Sn-Sn) bond enthalpy: approximately 240 kJ/mol

4. **Analyze the Trend:**
As we move down Group 14 from carbon to tin, the size of the atoms increases. This increase in atomic size leads to a decrease in bond strength due to the longer bond length and weaker overlap of orbitals. Therefore, the bond enthalpy decreases down the group.

5. **Order the Bond Enthalpy Values:**
Based on the values provided:
- C-C: 348 kJ/mol (highest)
- Si-Si: 297 kJ/mol
- Ge-Ge: 260 kJ/mol
- Sn-Sn: 240 kJ/mol (lowest)

6. **Write the Correct Order:**
The correct order of bond enthalpy from highest to lowest is:
\[ \text{C-C} > \text{Si-Si} > \text{Ge-Ge} > \text{Sn-Sn} \]

### Conclusion:
The correct order of bond enthalpy is C-C > Si-Si > Ge-Ge > Sn-Sn.