5. Let $E$, $F$, and $G$ be the midpoints of the edges $AB$, $BC$, and $CD$ of a regular tetrahedron $ABCD$, respectively. Then the dihedral angle $C-FG-E$ is
A. $\arcsin \frac{\sqrt{6}}{3}$
B. $\frac{\pi}{2}+\arccos \frac{\sqrt{3}}{3}$
C. $\frac{\pi}{2}-\arctan \sqrt{2}$
D. $\pi-\operatorname{arccot} \frac{\sqrt{2}}{2}$
Let the midpoint of $B D$ be $P$, connect $P A, P C$ intersecting $F G, E H$ at $M, N$.
Then $E F G H$ is a square $\Rightarrow M N \perp F G \Rightarrow \angle C M N$
is the plane angle of the dihedral angle $C-F G-E$ and $\angle P C A=\angle P M N$ $=\pi-\angle C M N$. Let $A C=2 \Rightarrow P A=P C=\sqrt{3}$
$\Rightarrow \cos \angle P C A=\frac{\sqrt{3}}{3} \Rightarrow \cot \angle P C A=\frac{\sqrt{2}}{2} \Rightarrow \angle C M N$ $=\pi-\operatorname{arccot} \frac{\sqrt{2}}{2}$. Therefore, the answer is $D$.