2. As shown in Figure 1, given the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, points $M$ and $N$ are on $A B_{1}$ and $B C_{1}$ respectively, and $A M=B N$. Then,
(1) $A A_{1} \perp M N$;
(2) $A_{1} C_{1} / / M N$;
(3) $M N / /$ plane $A_{1} B_{1} C_{1} D_{1}$;
(4) $M N$ and $A_{1} C_{1}$ are skew lines.
Among the above 4 conclusions, the number of incorrect conclusions is $(\quad)$.
(A) 1
(B) 2
(C) 3
(D) 4
2. (B).

As shown in Figure 3, draw $M P \perp A B$ at $P$, and draw $N Q \perp B C$ at $Q$, then connect $P Q$. Then $P M \parallel Q N$, and $\triangle A M P$ and $\triangle B N Q$ are both isosceles right triangles. Also, $A M = B N$, so $P M = Q N$. It is easy to see that $P M \perp$ plane $A B C D$, hence $P M \perp P Q$, and quadrilateral $P M N Q$ is a rectangle. Therefore, $A A_{1} \perp M N$, and $M N \parallel$ plane $A B C D$, i.e., $M N \parallel$ plane $A_{1} B_{1} C_{1} D_{1}$. Conclusions (1) and (3) are correct.

When $B P \neq B Q$, $P Q$ is not parallel to $A C$, i.e., $M N \parallel A_{1} C_{1}$ is not correct; and when $M$ and $N$ are the midpoints of $A B_{1}$ and $B C_{1}$ respectively, $M N \parallel P Q \parallel A C \parallel A_{1} C_{1}$, i.e., $M N$ and $A_{1} C_{1}$ are coplanar. Therefore, conclusions (2) and (4) are not correct.