11. The four sides of the quadrilateral pyramid are isosceles triangles with leg lengths of $\sqrt{7}$ and base lengths of 2. The maximum possible volume of this quadrilateral pyramid is $\qquad$.
11.3 The tetrahedron that meets the requirements has the following three scenarios:
(1) The lengths of the four lateral edges $O A, O B, O C, O D$ are all $\sqrt{7}$, and the lengths of the four base edges are all 2. In this case, it is easy to find that the volume of the tetrahedron $V_{1}=\frac{4}{3} \sqrt{5}$;
(2) As shown in Figure (1), the lengths of the four base edges and the two lateral edges $O A, O C$ are all $\sqrt{7}$.

Draw $O E \perp$ plane $A B C D$ at $E$, and let $A E=y, O E=x$.
Thus, $x^{2}+y^{2}=7$, and $4-x^{2}=7-y^{2}$.
Solving these, we get $x=\sqrt{2}, y=\sqrt{5}$.
At this point, the volume of the tetrahedron
$$
\begin{aligned}
V_{2} & =\frac{1}{3} \times S_{A B C D} \times O E=\frac{1}{3} \times \frac{1}{2} A C \times B D \times O E \\
& =\frac{1}{3} \times \sqrt{2} \times 2 \sqrt{5} \times \sqrt{2}=\frac{4}{3} \sqrt{5}=V_{1} ;
\end{aligned}
$$
(3) As shown in Figure (2), the lengths of the three lateral edges $O B, O C, O D$ are $\sqrt{7}$, and one lateral edge $O A=2$. Let $A C$ and $B D$ intersect at point $E$, and let $B E=x$.
Since $A B=A D=\sqrt{7}, B C=D C=2$,
we have $C E^{2}=4-x^{2}, O E^{2}=7-x^{2}, A E^{2}=7-x^{2}$.
Because $\cos \angle O E C=-\cos \angle O E A$,
thus $\frac{O E^{2}+E C^{2}-O C^{2}}{2 O E \cdot E C}=-\frac{O E^{2}+E A^{2}-O A^{2}}{2 O E \cdot E A}$.
Substituting (1) into (2), we solve to get $x=\sqrt{3}$ or $2 \sqrt{2}$.
Clearly, $x \neq 2 \sqrt{2}$, so $x=\sqrt{3}$, at this point the volume of the tetrahedron $V_{3}=\frac{1}{3} \sqrt{3} \times 3 \sqrt{3}=3$. In summary, $V_{3}>V_{2}=V_{1}$, so the maximum volume of the tetrahedron is $V_{3}=3$.