How to calculate the bond energy of C-Cl?
The bond dissociation enthalpy* $D$ is the enthalpy change for conversion of bonded into dissociated molecules following homolytic cleavage of the bond in question:
$$\ce{A-B (g)-&gt;A(g) + B(g)},\qquad D(\ce{A-B})=\Delta H^\circ_f (\ce{A(g)})+\Delta H^\circ_f (\ce{B(g)})-\Delta H^\circ_f (\ce{A-B(g)})$$
This is the property usually discussed as representing the &quot;strength&quot; of a bond, and is positive because the process of breaking a bond requires an input of heat (is endothermic). Breaking stronger bonds requires a greater positive enthalpy change.
The enthalpy change for formation of the bond by the dissociated molecular constituents is negative the dissociation enthalpy:
$$\Delta H^\circ (\ce{A-B})=-D (\ce{A-B})$$
and is a negative quantity because heat is released when bonds are formed (the process is exothermic).
The enthalpy of formation of a $\ce{Cl-Cl}$ bond is the enthalpy change for the following process:
$$\ce{2Cl(g)-&gt; Cl2(g)}$$
and therefore can be expressed as
$$\Delta H^\circ (\ce{Cl-Cl})=\Delta H^\circ_f (\ce{Cl-Cl(g)})-2\Delta H^\circ_f (\ce{Cl(g)})$$
and since $\Delta H^\circ_f (\ce{Cl2(g)})=0$,
$$\Delta H^\circ (\ce{Cl-Cl})=-2\Delta H^\circ_f (\ce{Cl(g)})=-D(\ce{Cl-Cl})$$
In the process
$$\ce{C(g) + 4 Cl (g)  -&gt; CCl4 (g)}$$
four $\ce{C-Cl}$ bonds are formed, so the average heat of formation of one bond is
$$\Delta H^\circ(\ce{C-Cl}) = \frac{1}{4}\left(\Delta H^\circ_\mathrm{f}(\ce{CCl_4(g))}-\Delta H^\circ_\mathrm{f}(\ce{C(g)}) - 4 \Delta H^\circ_\mathrm{f}(\ce{Cl(g)}) \right) $$
which is a negative quantity (exothermic).
Note again that the corresponding bond dissociation enthalpy is then a positive quantity.

*Sources (from IUPAC Gold Book website):

PAC, 1990, 62, 2167. (Glossary of atmospheric chemistry terms (Recommendations 1990)) on page 2176.
PAC, 1994, 66, 1077. (Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)) on page 1089.