5. Let $E$, $F$, and $C$ be the midpoints of the edges $\overline{AB}$, $BC$, and $CD$ of a regular tetrahedron $ABCD$, respectively. Then the size of the dihedral angle $C$ $F G-E$ is ( ).
(A) $\arcsin \frac{\sqrt{6}}{3}$
(B) $\frac{\pi}{2}+\arccos \frac{\sqrt{3}}{3}$
(C) $\frac{\pi}{2}-\operatorname{arctg} \sqrt{2}$
(D) $\pi-\operatorname{arcctg} \frac{\sqrt{2}}{2}$
5.D.

As shown in the figure, construct $E E_{1} \perp$
plane $B C D$, then $E E_{1} \perp$
$B G$.
$\because A B C D$ is a regular

tetrahedron,
$$
\begin{array}{l}
\therefore A C \perp B D . \\
\text { Also, since } E F / / A C, \\
F G / / B D, \\
\therefore E F \perp F G .
\end{array}
$$

Thus, $E_{1} F \perp F G$. Therefore, $\angle E F E_{1}$ is the plane angle of the dihedral angle $E-F G$ $-B$, denoted as $\alpha$.
$$
\begin{array}{l}
A H=\sqrt{A B^{2}-B H^{2}}:=\sqrt{1^{2}-\left(\frac{\sqrt{3}}{3}\right)^{2}}=\frac{\sqrt{6}}{3} \text {. } \\
\therefore E E_{1}=\frac{1}{2} A H=\frac{\sqrt{6}}{6} \text {. } \\
\text { Also, } E F=\frac{1}{2} A C=\frac{1}{2} \text {, } \\
\therefore \alpha=\arcsin \frac{\frac{\sqrt{6}}{6}}{\frac{1}{2}}=\arcsin \frac{\sqrt{6}}{3} \text {. } \\
\end{array}
$$

Therefore, the size of the dihedral angle $C-F G-E$ is
$$
\pi-\arcsin \frac{\sqrt{6}}{3}=\pi-\operatorname{arcctg} \frac{\sqrt{2}}{2} .
$$