6. Let points $E, F, G$ be the midpoints of edges $A B, B C, C D$ of a regular tetrahedron $A B C D$, respectively. Then the size of the dihedral angle $C-F G-E$ is ( ).
A. $\arcsin \frac{\sqrt{6}}{3}$
B. $\frac{\pi}{2}-\arctan \sqrt{2}$
C. $\frac{\pi}{2}+\arccos \frac{\sqrt{3}}{3}$
D. $\pi-\operatorname{arccot} \frac{\sqrt{2}}{2}$
6. (D). Draw $E E_{1} \perp$ plane $B C D$ at $E_{1}$, then $E E_{1} \perp B G$. Also, $A C \perp B D, E F / / A C, F G / / B D$, so $E F \perp F G$, which means $E_{1} F \perp F G$, thus $\angle E F E_{1}$ is the plane angle of the dihedral angle $E-F G-B$.

Assume the edge length of the regular tetrahedron $A B C D$ is 1, then its height $A H=\frac{\sqrt{6}}{3}, E E_{1}=\frac{1}{2} A H=\frac{\sqrt{6}}{6}$. Also, $E F=\frac{1}{2} A C=\frac{1}{2}$, then $\sin \angle E F E_{1}=\frac{\sqrt{6}}{3}$, so the size of the dihedral angle $C-F G-E$ is $\pi-\arcsin \frac{\sqrt{6}}{3}=\pi-\operatorname{arccot} \frac{\sqrt{2}}{2}$.