Ionic Mechanism for Wurtz Reaction Instead of going for the normal free radical mechanism for Wurtz reaction, What reasons prevent us from believing in the following ionic mechanism?

\begin{align}
\ce{R-X + 2Na &amp;-&gt; R^--Na+ + NaX}\\
\ce{R-X + R^--Na+ &amp;-&gt; R-R + NaX}\\
\ce{2R-X + 2Na &amp;-&gt; R-R  + 2NaX}
\end{align}

Related but doesn't answer my question here
The actual mechanism is believed to be strongly dependent on the nature of halide, metal (not only $\ce{Na}$ is used in this reaction), and solvent. The 'ionic' mechanism you have written here is the part of the most accepted mechanism which involves both, an 'ionic' and a radical reaction.
First step. Through the single-electron transfer from $\ce{Na}$ to the halogen atom, an alkyl radical $\ce{R^.}$ and sodium halide is formed:
$$\ce{Na + R-X -&gt; Na+X^- + R^.}$$
Alkyl radical further acts as an electron acceptor from another $\ce{Na}$ atom to form highly nucleophilic intermediate (can be isolated in some cases):
$$\ce{R^. + Na -&gt; R^-Na+}$$
This alkyl sodium intermediate reacts with another alkyl halide in the aliphatic nucleophilic substitution:
$$\ce{R^-Na+ + RX -&gt; R-R + NaX}$$