The metal ions that have the calculated spin only magnetic moment value of 4.9 B.M. are
A. $$Cr^{2+}$$
B. $$Fe^{2+}$$
C. $$Fe^{3+}$$
D. $$Co^{2+}$$
E. $$Mn^{3+}$$

Choose the correct answer from the options given below

To determine which metal ions have a spin-only magnetic moment of $$4.9\text{ B.M.}$$, we use the spin-only magnetic moment formula

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

where $$n$$ is the number of unpaired electrons.

For

$$\mu = 4.9\text{ B.M.},$$

the corresponding value of $$n$$ is

$$n = 4.$$

Hence, we need to identify the ions containing 4 unpaired electrons.

The electronic configurations and corresponding numbers of unpaired electrons are summarized below:

For $$Cr^{2+}$$ with configuration $$3d^4$$, there are four unpaired electrons. Therefore,

$$\mu = \sqrt{4(4+2)}=\sqrt{24}\approx 4.9\text{ B.M.}$$

For $$Fe^{2+}$$ with configuration $$3d^6$$, the electrons are distributed as

$$\uparrow\downarrow,\ \uparrow,\ \uparrow,\ \uparrow,\ \uparrow,$$

giving four unpaired electrons. Hence,

$$\mu = \sqrt{24}\approx 4.9\text{ B.M.}$$

For $$Fe^{3+}$$ with configuration $$3d^5$$, there are five unpaired electrons, so

$$\mu = \sqrt{5(5+2)}=\sqrt{35}\approx 5.9\text{ B.M.}$$

For $$Co^{2+}$$ with configuration $$3d^7$$, there are three unpaired electrons, giving

$$\mu = \sqrt{3(3+2)}=\sqrt{15}\approx 3.87\text{ B.M.}$$

For $$Mn^{3+}$$ with configuration $$3d^4$$, there are four unpaired electrons. Therefore,

$$\mu = \sqrt{4(4+2)}=\sqrt{24}\approx 4.9\text{ B.M.}$$

Hence, the ions exhibiting a spin-only magnetic moment of approximately $$4.9\text{ B.M.}$$ are

$$\boxed{Cr^{2+},\ Fe^{2+},\ \text{and}\ Mn^{3+}}.$$

Therefore, the correct answer is A, B and E only.