Use the spin-only magnetic moment formula

For a transition metal ion, $\mu_{\text{spin only}} = \sqrt{n(n+2)}\ \text{B.M.}$ where $n$ is the number of unpaired electrons.

Find the value of $n$ for $\mu = 4.9$ B.M.

Check: $\sqrt{4(4+2)} = \sqrt{24} \approx 4.90$

So, ions having 4 unpaired electrons will have spin-only magnetic moment approximately $4.9$ B.M.

Evaluate each ion

(A) $\mathrm{Cr}^{2+}$

Chromium: $Z=24$ $\mathrm{Cr} = [Ar]3d^5 4s^1$ $\mathrm{Cr}^{2+} = [Ar]3d^4$ A free/high-spin $d^4$ ion has 4 unpaired electrons.

So, $\mathrm{Cr}^{2+}$ gives: $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9\ \text{B.M.}$ A is correct.

(B) $\mathrm{Fe}^{2+}$

Iron: $Z=26$ $\mathrm{Fe} = [Ar]3d^6 4s^2$ $\mathrm{Fe}^{2+} = [Ar]3d^6$ For a high-spin $d^6$ ion, number of unpaired electrons $=4$.

Hence, $\mu = \sqrt{24} \approx 4.9\ \text{B.M.}$ B is correct.

(C) $\mathrm{Fe}^{3+}$

$\mathrm{Fe}^{3+} = [Ar]3d^5$ A high-spin $d^5$ ion has 5 unpaired electrons.

Then, $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\ \text{B.M.}$ C is not correct.

(D) $\mathrm{Co}^{2+}$

Cobalt: $Z=27$ $\mathrm{Co} = [Ar]3d^7 4s^2$ $\mathrm{Co}^{2+} = [Ar]3d^7$ A high-spin $d^7$ ion has 3 unpaired electrons.

Then, $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\ \text{B.M.}$ D is not correct.

(E) $\mathrm{Mn}^{3+}$

Manganese: $Z=25$ $\mathrm{Mn} = [Ar]3d^5 4s^2$ $\mathrm{Mn}^{3+} = [Ar]3d^4$ A high-spin $d^4$ ion has 4 unpaired electrons.

So, $\mu = \sqrt{24} \approx 4.9\ \text{B.M.}$ E is correct.

Collect the correct ions

The ions with magnetic moment $4.9$ B.M. are: $\mathrm{Cr}^{2+},\ \mathrm{Fe}^{2+},\ \mathrm{Mn}^{3+}$

Match with given options

This corresponds to A, B and E only, i.e. Option B.