2. If the two foci of the hyperbola $C$ are exactly the two vertices of the ellipse $\Gamma$ : $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$, and the two vertices of the hyperbola $C$ are exactly the two foci of the ellipse $\Gamma$, then the equation of the hyperbola $C$ is $\qquad$
2. $\frac{x^{2}}{7}-\frac{y^{2}}{9}=1$.

According to the conditions, we know that the center of the hyperbola $C$ is at the origin, and its real axis of symmetry is the $x$-axis. Let its equation be $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.
Then its vertices are $( \pm a, 0)$, and its foci are $( \pm c, 0)$.
The vertices of the ellipse on the major axis are $( \pm 4,0)$, and the foci are $( \pm \sqrt{7}, 0)$, thus,
$$
a=\sqrt{7}, c=4 \Rightarrow b=\sqrt{c^{2}-a^{2}}=3 \text {. }
$$

Therefore, the equation of the hyperbola is $\frac{x^{2}}{7}-\frac{y^{2}}{9}=1$.