Step 1: Understand what an irrational number is and the target range.
An irrational number is a real number that cannot be expressed as a simple fraction of two integers (i.e., $\frac{p}{q}$ where $p, q$ are integers and $q \ne 0$). Its decimal representation is non-terminating and non-repeating. Examples include $\sqrt{2}$, $\pi$.
We need to find an irrational number that lies strictly between 2 and 2.5.
To check if a number is between 2 and 2.5, it's often easiest to square the boundary values:
$2^2 = 4$
$(2.5)^2 = 6.25$
So, we are looking for a number $x$ such that $\sqrt{x}$ is irrational and $4 < x < 6.25$.
Step 2: Evaluate each option to find its approximate value and check if it's irrational.
(1) $\sqrt{11}$:
Since $3^2 = 9$ and $4^2 = 16$, $\sqrt{11}$ lies between 3 and 4.
$\sqrt{11} \approx 3.317$.
This value is not between 2 and 2.5. (11 is not a perfect square, so $\sqrt{11}$ is irrational).
(2) $\sqrt{5$:}
Since $2^2 = 4$ and $3^2 = 9$, $\sqrt{5}$ lies between 2 and 3.
More precisely, $\sqrt{5} \approx 2.236$.
This value is indeed between 2 and 2.5 ($2 < 2.236 < 2.5$). Also, 5 is not a perfect square, so $\sqrt{5}$ is an irrational number.
(3) $\sqrt{22.5$:}
Since $4^2 = 16$ and $5^2 = 25$, $\sqrt{22.5}$ lies between 4 and 5.
$\sqrt{22.5} \approx 4.743$.
This value is not between 2 and 2.5.
(4) $\sqrt{12.5$:}
Since $3^2 = 9$ and $4^2 = 16$, $\sqrt{12.5}$ lies between 3 and 4.
$\sqrt{12.5} \approx 3.535$.
This value is not between 2 and 2.5.
Step 3: Conclude the correct option.
From the evaluation, $\sqrt{5}$ is the only irrational number that falls within the specified range of 2 and 2.5.
\[
\mathbf{(2)} \quad \sqrt{5}
\]