5. Given real numbers $a, b, c, d$ satisfy
$$
a^{2}+b^{2}=1, c^{2}+d^{2}=1, a c+b d=0 \text {. }
$$

The following conclusions are given:
(1) $a^{2}+c^{2}=1$;
(2) $b^{2}+d^{2}=1$;
(3) $a b+c d=0$;
(4) $a d+b c=0$.

The number of correct conclusions is ( ) .
(A) 1
(B) 2
(C) 3
(D) 4
5. C.

Notice
$$
\begin{array}{l}
a^{2}+b^{2}=c^{2}+d^{2}=1 \\
\Rightarrow\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right) \\
\quad=a^{2} c^{2}+a^{2} d^{2}+b^{2} c^{2}+b^{2} d^{2}=1 ; \\
a c+b d=0 \Rightarrow a^{2} c^{2}=b^{2} d^{2} .
\end{array}
$$

Then $a^{2} d^{2}+2 a^{2} c^{2}+b^{2} c^{2}=1$, that is
$$
a^{2}\left(c^{2}+d^{2}\right)+c^{2}\left(a^{2}+b^{2}\right)=1 \Rightarrow a^{2}+c^{2}=1 \text {. }
$$

Similarly, $b^{2}+d^{2}=1$.
Therefore, conclusions (1) and (2) hold.
$$
\begin{array}{l}
\text { By } a b+c d=\left(c^{2}+d^{2}\right) a b+\left(a^{2}+b^{2}\right) c d \\
=a b c^{2}+a b d^{2}+c d a^{2}+c d b^{2} \\
=\left(a b c^{2}+c d a^{2}\right)+\left(a b d^{2}+c d b^{2}\right) \\
=(a d+c b)(a c+b d)=0,
\end{array}
$$

Thus, conclusion (3) also holds.
When $a=\frac{\sqrt{3}}{2}, b=-\frac{1}{2}, c=\frac{1}{2}, d=\frac{\sqrt{3}}{2}$ satisfy the conditions, we have $a d+b c \neq 0$, so conclusion (4) does not hold.