2. Let $0<a, b, c, d<1$. Then the relationship between $a(1-b)+b(1-c)+c(1-d)+d(1-a)$ and 2 is ( ).
(A) greater than
(B) less than
(C) equal to
(D) cannot be determined
2.B.

As shown in Figure 4, for a square
with side length 1, then
$$
\begin{array}{c}
\frac{1}{2}[a(1-b)+ \\
b(1-c)+ \\
c(1-d)+ \\
d(1-a)] \\
=S_{1}+S_{2}+S_{3}+S_{4} \\
<S_{\text {square }}(B C D=1 .
\end{array}
$$
Therefore, $a(1-b)+b(1-c)+c(1-d)+d(1-a)<2$.