Formation of radicals from C-H and C-C bonds
Note that most radical mechanisms do not explicitly feature homolysis of $\ce{C-H}$ bonds or $\ce{C-C}$ bonds. If one were to compare these two processes, say for ethane $\ce{CH3CH3}$, we would find that homlysis of $\ce{C-C}$ is favored over homolysis of $\ce{C-H}$. Obligatory bond dissociation energy references for the organics and here for the halogens.


  Homolysis 


$\ce{C-C}$

$$\ce{H3C-CH3 -&gt; 2H3C.}\ \ \Delta H^\circ=377\ \text{kJ/mol}$$

$\ce{C-H}$

$$\ce{H3CCH2-H -&gt; H3CCH2. + H.}\ \ \Delta H^\circ=423\ \text{kJ/mol}$$

However, $\ce{C-C}$ and $\ce{C-H}$ homolysis steps are not common steps in radical mechanisms of alkanes. For example, the chlorination of ethane begins with the homolysis of $\ce{Cl2}$, which is much more favorable than homolysis of $\ce{C-C}$ or $\ce{C-H}$.

$$\ce{Cl2 -&gt; 2Cl.} \ \ \Delta H^\circ = 243\ \text{kJ/mol}$$

The chlorine radicals then react with the alkane by abstraction, meaning that bonds are being formed as well as being broken.


  Abstraction 


$\ce{C-C}$   

$$\ce{Cl. + H3C-CH3 -&gt; Cl-CH3 +H3C.}$$

$$\begin{array}{c|c|c|} 
 &amp; \text{Broken} &amp; \text{Formed} \\ \hline
\text{bond} &amp; \ce{H3C-CH3}  &amp; \ce{Cl-CH3} \\ \hline
\Delta H^\circ &amp; 377\ \text{kJ/mol} &amp; -350 \ \text{kJ/mol} \\ \hline
\end{array}$$

$$\Delta_r H^\circ = +27 \ \text{kJ/mol}$$

$\ce{C-H}$

$$\ce{Cl. + H3CCH2-H -&gt; Cl-H +H3CCH2.}$$

$$\begin{array}{c|c|c|} 
 &amp; \text{Broken} &amp; \text{Formed} \\ \hline
\text{bond} &amp; \ce{H3CCH2-H}  &amp; \ce{Cl-H} \\ \hline
\Delta H^\circ &amp; 423\ \text{kJ/mol} &amp; -432 \ \text{kJ/mol} \\ \hline
\end{array}$$
$$\Delta_r H^\circ = -9 \ \text{kJ/mol}$$

Thus, abstraction of $\ce{H}$ by $\ce{Cl.}$ is more favored because it is coupled to the formation of the $\ce{H-Cl}$ bond.