2. The clock points to a certain moment between 9 a.m. and 10 a.m., and at this moment, the hour hand 2 minutes earlier and the minute hand 2 minutes later are in a straight line (not considering the overlapping case). Then this moment is ( ).
(A) 9 a.m. $13 \frac{7}{11}$ minutes
(B) 9 a.m. 16 minutes
(C) 9 a.m. 12 minutes
(D) 9 a.m. 14 minutes
2.D.

On the clock, there are 60 small divisions, starting from the mark 12, dividing the circumference into 60 equal parts. The minute hand moves 60 divisions per hour, while the hour hand moves 5 divisions per hour. The speed of the minute hand is 12 times that of the hour hand.
As shown in Figure 2, points $A$, $O$, and $B$ are collinear.
Let the required time be 9:$x$.
At 9:$(x-2)$, the hour hand moves from $A$ to $C$, covering $\frac{x-2}{12}$ divisions; at 9:$(x+2)$, the minute hand moves from $B$ to $D$, covering $(x+2)-3 \times 5$ divisions.
According to the problem, $\overparen{A C}=\overparen{B D}$.
Therefore, $\frac{x-2}{12}=(x+2)-3 \times 5$.
Solving for $x$, we get $x=14$.
Thus, the time is 9:14.