5. If two congruent regular tetrahedrons are glued together by overlapping their bases, they form a hexahedron with all dihedral angles equal. If the shortest edge length of this hexahedron is 2, then the distance between the farthest two vertices is $\qquad$ .
5. 3 As shown in the figure, the hexahedron $A-C D F-B$, let the side length of $\triangle C D F$ be $2 a$, and the length of edge $A D$ be $b$. Construct $C E \perp A D$ at $E$, connect $E F$, it is easy to know that $E F \perp A D$, thus $\angle C E F$ is the plane angle of the dihedral angle formed by faces $A D F$ and $A D C$.
Let $G$ be the midpoint of $C D$,
Similarly, $\angle A G B$ is the plane angle of the dihedral angle formed by faces $A C D$ and $B C D$.
Given that $\angle C E F=\angle A G B$,
In $\triangle A C D$, $A G=\sqrt{b^{2}-a^{2}}, C E \cdot b=A G \cdot 2 a$,
Therefore, $C E=\frac{A G \cdot 2 a}{b}=\frac{2 a \sqrt{b^{2}-a^{2}}}{b}$.
In $\triangle A B C$, $A B=2 \sqrt{b^{2}-\left(\frac{2}{3} \sqrt{3} a\right)^{2}}=2 \sqrt{b^{2}-\frac{4}{3} a^{2}}$.
Since isosceles $\triangle C E F \backsim$ isosceles $\triangle A G B$, thus $\frac{A B}{C F}=\frac{A G}{C E}$,
That is, $\frac{2 \sqrt{b^{2}-\frac{4}{3} a^{2}}}{2 a}=\frac{\sqrt{b^{2}-a^{2}}}{\frac{2 a \sqrt{b^{2}-a^{2}}}{b}}$, solving this gives $b=\frac{4}{3} a<2 a$,
Since the shortest edge length is 2, thus $b=2$,
Therefore, $2 a=3$, and $A B=2 \sqrt{b^{2}-\frac{4}{3} a^{2}}=2$,
Thus, the distance between the farthest two vertices is 3.