Step 1: Understand what is given.
An ion $X^{3+}$ has a magnetic moment of $\sqrt{15}$ BM. We need the atomic number of $X$.
Step 2: Use the spin-only formula.
The magnetic moment links to unpaired electrons by $\mu = \sqrt{n(n+2)}$, where $n$ is the number of unpaired electrons.
Step 3: Find the number of unpaired electrons.
Set $\sqrt{n(n+2)} = \sqrt{15}$. Squaring gives $n(n+2)=15$. Trying $n=3$: $3\times5 = 15$. So $n = 3$.
Step 4: Decide the d configuration of the ion.
Three unpaired electrons in a transition metal ion fit a $d^3$ arrangement, since three orbitals each hold one electron. So $X^{3+}$ is $d^3$.
Step 5: Build back the neutral atom.
The ion lost 3 electrons. A $d^3$ ion of type $M^{3+}$ comes from chromium. For Cr, the $3+$ ion is $[Ar]3d^3$, which has exactly 3 unpaired electrons.
Step 6: Read off the atomic number.
Chromium has atomic number 24. This matches our $d^3$ result.
Step 7: State the final answer.
The atomic number of $X$ is:
\[ \boxed{24} \]