2. Given $a^{2}+b^{2}=1, b^{2}+c^{2}=c^{2}+a^{2}=2$. Then the minimum value of $a b+b c+a c$ is ( ).
(A) $\sqrt{3}-\frac{1}{2}$
(B) $\frac{1}{2}-\sqrt{3}$
(C) $-\frac{1}{2}-\sqrt{3}$
(D) $\frac{1}{2}+\sqrt{3}$
2. B.
$$
\begin{array}{l}
\text { Given } a^{2}+b^{2}=1, b^{2}+c^{2}=2, a^{2}+c^{2}=2 \\
\Rightarrow(a, b, c)=\left( \pm \frac{\sqrt{2}}{2}, \pm \frac{\sqrt{2}}{2}, \pm \frac{\sqrt{6}}{2}\right) .
\end{array}
$$

Then $a b+b c+c a$
$$
\begin{array}{l}
\geqslant \frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2} \times\left(-\frac{\sqrt{6}}{2}\right)+\left(-\frac{\sqrt{6}}{2}\right) \times \frac{\sqrt{2}}{2} \\
=\frac{1}{2}-\sqrt{3} .
\end{array}
$$