The magnetic moment of a transition metal ion is √(24) B.M — JEE Chemistry

Using spin-only formula:

$$\mu = \sqrt{n(n+2)} B.M.$$

$$\sqrt{n(n+2)} = \sqrt{24} \Rightarrow n(n+2) = 24 \Rightarrow n^2+2n-24 = 0$$

$$n = \frac{-2 + \sqrt{4+96}}{2} = \frac{-2+10}{2} = 4 unpaired electrons$$

(e.g., $Fe^{2+}$: $[Ar] 3d^6$ with 4 unpaired electrons in high-spin)