7. Given a tetrahedron $ABCD$ where one edge length is 3, and the other five edge lengths are all 2. Then the radius of its circumscribed sphere is $\qquad$
7. $\frac{\sqrt{21}}{3}$.

As shown in Figure 2, let $B C=3, A B=A C=A D=B D=$ $C D=2, E, F$ be the midpoints of $B C, A D$ respectively.

Then the projection of point $D$ on plane $A B C$, $H$, should be the circumcenter of $\triangle A B C$.

Since any point on $D H$ is equidistant from points $A, B, C$, the center of the circumscribed sphere $O$ lies on $D H$.

Also, $\triangle A B C \cong \triangle D B C \Rightarrow A E=D E$, thus, $E F$ is the perpendicular bisector of $A D$.

Therefore, the center of the sphere $O$ is the intersection of $D H$ and $E F$, and is the orthocenter of isosceles $\triangle E A D$. Let the radius of the sphere be $r$.
$$
\begin{array}{l}
\text { By } \triangle D O F \backsim \triangle E A F \Rightarrow r=\frac{A E \cdot D F}{E F} . \\
\text { And } A E^{2}=2^{2}-\left(\frac{3}{2}\right)^{2}=\frac{7}{4}, D F=A F=1, \text { so } \\
E F^{2}=A E^{2}-A F^{2}=\frac{3}{4} \\
\Rightarrow r=\sqrt{\frac{7}{3}}=\frac{\sqrt{21}}{3} .
\end{array}
$$