Given:

ΔH_atom (CH₄) = x kJ mol⁻¹

ΔH_atom (C₂H₆) = y kJ mol⁻¹

Find out bond energy (C − C) in kJ/mol.

To find the bond energy of the carbon-carbon (C - C) single bond, we can relate the enthalpy of atomization of methane (${\mathrm{CH}}_4$) and ethane ($C_2{H}_6$) to their respective bond energies.

First, let us analyze methane, ${\mathrm{CH}}_4$:
Methane consists of 4 identical carbon-hydrogen (C - H) single bonds.
Therefore, the enthalpy of atomization of ${\mathrm{CH}}_4$ is equal to the energy required to break 4 moles of C - H bonds:
${\mathrm{\Delta H}}_{\mathrm{atom}}\left({\mathrm{CH}}_4\right)=4\times \mathrm{BE}(C-H)$
Given that ${\mathrm{\Delta H}}_{\mathrm{atom}}\left({\mathrm{CH}}_4\right)=x\mathrm{kJ}{\mathrm{mol}}^{-1}$, we have:
$4\times \mathrm{BE}(C-H)=x$
This gives the bond energy of a C - H bond as:
$\mathrm{BE}(C-H)=\frac{x}{4}\mathrm{kJ}{\mathrm{mol}}^{-1}$

Next, let us analyze ethane, $C_2{H}_6$:
Ethane consists of 1 carbon-carbon (C - C) single bond and 6 carbon-hydrogen (C - H) single bonds.
Therefore, the enthalpy of atomization of $C_2{H}_6$ is the sum of the bond energies of 1 C - C bond and 6 C - H bonds:
${\mathrm{\Delta H}}_{\mathrm{atom}}\left(C_2{H}_6\right)=\mathrm{BE}(C-C)+6\times \mathrm{BE}(C-H)$
Given that ${\mathrm{\Delta H}}_{\mathrm{atom}}\left(C_2{H}_6\right)=y\mathrm{kJ}{\mathrm{mol}}^{-1}$, we can write:
$y=\mathrm{BE}(C-C)+6\times \mathrm{BE}(C-H)$

Now, substitute the value of $\mathrm{BE}(C-H)$ from the methane equation into the ethane equation:
$y=\mathrm{BE}(C-C)+6\times \left(\frac{x}{4}\right)$
Simplify the term representing the C - H bond contributions:
$y=\mathrm{BE}(C-C)+\frac{3x}{2}$

Rearranging the equation to solve for the carbon-carbon bond energy, $\mathrm{BE}(C-C)$, we get:
$\mathrm{BE}(C-C)=y-\frac{3x}{2}\mathrm{kJ}{\mathrm{mol}}^{-1}$

Thus, the bond energy of the C - C bond is $y-3x/2$.