Answer: (D)

Solution: By the property of the diagonals of a rectangle, we know
$$
S_{\text {rectangle } G C H O}=S_{\text {rectangle } F A E O}=6,
$$

Let $G(0, a), H(b, 0)$, then $Q\left(\frac{2}{a}, a\right), P\left(b, \frac{2}{b}\right)$,
so $S_{\triangle O P Q}=S_{\text {rectangle } G C H O}-S_{\triangle G O Q}-S_{\triangle P H O}-S_{\triangle P Q C}=6-\frac{1}{2} a \cdot \frac{2}{a}-\frac{1}{2} b \cdot \frac{2}{b}-\frac{1}{2}\left(b-\frac{2}{a}\right) \cdot\left(a-\frac{2}{b}\right)$, given $a b=6$, substituting yields $S_{\triangle O P Q}=\frac{8}{3}$.
23. As shown in the figure, the diagonal $BD$ of rectangle $ABCD$ passes through the origin $O$, and the two sides of the rectangle are parallel to the coordinate axes. The graph of the inverse proportion function $y=\frac{2}{x}$ intersects the rectangle at points $P$ and $Q$. Given that point $A(-3,-2)$, the area of $\triangle OPQ$ is ( ).
(A) $\frac{3}{2}$
(B) $\frac{7}{2}$
(C) $\frac{7}{3}$
(D) $\frac{8}{3}$