5. The four sides of the quadrilateral pyramid are isosceles triangles with a leg length of $\sqrt{7}$ and a base length of 2. Then the maximum possible volume of this quadrilateral pyramid is
5. 3 .

There are three scenarios for the quadrilateral pyramid that meet the requirements.
(1) As shown in Figure 4, all four lateral edges are $\sqrt{7}$, and the volume is calculated as $V_{1}=\frac{4}{3} \sqrt{5}$.
(2) As shown in Figure 5, two lateral edges are $\sqrt{7}$.

Construct $O E \perp$ plane $A B C D$, and let $A E$ $=y, O E=x$. Thus,
$$
x^{2}+y^{2}=7 \text {, and } 4-x^{2}=7-y^{2} \text {. }
$$

Solving these equations, we get $x=\sqrt{2}, y=\sqrt{5}$. At this point,
$$
V_{2}=\frac{1}{3} \times \sqrt{2} \times 2 \sqrt{5} \times \sqrt{2}=\frac{4}{3} \sqrt{5}=V_{1} \text {. }
$$
(3) As shown in Figure 6, three lateral edges $(O B, O C, O D)$ are $\sqrt{7}$, and one lateral edge $O A=2$.
Let $A D$ and $B C$ intersect at point $E$, and let $B E=x$.
Since $A B=A C=\sqrt{7}, B D=D C=2$, we have
$$
D E^{2}=4-x^{2}, O E^{2}=7-x^{2}, A E^{2}=7-x^{2} \text {. }
$$

Because $\cos \angle O E D=-\cos \angle O E A$, we have
$$
\frac{O E^{2}+E D^{2}-O D^{2}}{20 E \cdot E D}=-\frac{O E^{2}+E A^{2}-O A^{2}}{20 E \cdot E A} \text {. }
$$

Substituting the previous results into the above equation, we solve to get $x=\sqrt{3}$ or $2 \sqrt{2}$. Clearly, $x \neq 2 \sqrt{2}$, so $x=\sqrt{3}$. At this point, $V_{3}=\frac{1}{3} \sqrt{3} \times 3 \sqrt{3}=3$.
In summary, $V_{3}>V_{2}=V_{1}$. Therefore, $V_{\text {max }}=3$.