To determine which metal ions have a calculated spin-only magnetic moment of 4.9 Bohr Magnetons (B.M.), we can use the formula:

$ \text{Magnetic Moment (M.M)} = \sqrt{n(n+2)} \, \text{B.M.} $

where $ n $ is the number of unpaired electrons.

Given that the magnetic moment is 4.9 B.M., we can set up the equation:

$ 4.9 = \sqrt{n(n+2)} $

Solving this equation, we find that $ n = 4 $.

Let's analyze each metal ion:

(A) ${}_{24} \mathrm{Cr}^{2+}$: Electronic configuration: $[\mathrm{Ar}] 3d^4$. This ion has 4 unpaired electrons.

(B) ${}_{26} \mathrm{Fe}^{2+}$: Electronic configuration: $[\mathrm{Ar}] 3d^6$. This ion also has 4 unpaired electrons.

(C) ${}_{26} \mathrm{Fe}^{3+}$: Electronic configuration: $[\mathrm{Ar}] 3d^5$. This ion has 5 unpaired electrons, which does not match the required $ n $ value of 4.

(D) ${}_{27} \mathrm{Co}^{2+}$: Electronic configuration: $[\mathrm{Ar}] 3d^7$. This ion has 3 unpaired electrons.

(E) ${}_{25} \mathrm{Mn}^{3+}$: Electronic configuration: $[\mathrm{Ar}] 3d^4$. This ion has 4 unpaired electrons.

Based on the number of unpaired electrons, the metal ions with a spin-only magnetic moment of approximately 4.9 B.M. are $\mathrm{Cr}^{2+}$, $\mathrm{Fe}^{2+}$, and $\mathrm{Mn}^{3+}$.