The spin-only magnetic moment value of $$M^{3+}$$ ion (in gaseous state) from the pairs $$Cr^{3+}/Cr^{2+}$$, $$Mn^{3+}/Mn^{2+}$$, $$Fe^{3+}/Fe^{2+}$$ and $$Co^{3+}/Co^{2+}$$ that has negative standard electrode potential, is ______ B.M.

We need to find the spin-only magnetic moment of the $$M^{3+}$$ ion (from the given pairs) that has a negative standard electrode potential for the $$M^{3+}/M^{2+}$$ couple.

Identify the standard electrode potentials of the given pairs

The standard electrode potentials $$E°(M^{3+}/M^{2+})$$ are:

- $$Cr^{3+}/Cr^{2+}$$: $$E° = -0.41 \text{ V}$$ (negative)

- $$Mn^{3+}/Mn^{2+}$$: $$E° = +1.51 \text{ V}$$ (positive)

- $$Fe^{3+}/Fe^{2+}$$: $$E° = +0.77 \text{ V}$$ (positive)

- $$Co^{3+}/Co^{2+}$$: $$E° = +1.81 \text{ V}$$ (positive)

Identify the ion with negative $$E°$$

Only $$Cr^{3+}/Cr^{2+}$$ has a negative standard electrode potential. So we need the magnetic moment of $$Cr^{3+}$$.

Find the electronic configuration of $$Cr^{3+}$$

$$Cr$$: $$[Ar] 3d^5 4s^1$$ (anomalous configuration)

$$Cr^{3+}$$: $$[Ar] 3d^3$$ (3 unpaired electrons)

Calculate the spin-only magnetic moment

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

where $$n$$ = number of unpaired electrons = 3

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

Round to the nearest integer

$$\mu \approx 4 \text{ B.M.}$$

The answer is 4 B.M.