Animated Solution for Chemistry - d and f-Block Elements: The correct order of $E_{M^{2+}/M}^{\circ }$ values with negative sign for the four successive elements Cr, Mn, Fe and Co is

Electrode potentials can often feel like a maze of arbitrary numbers, but they are actually a beautiful reflection of atomic structure and thermodynamics. In this problem, we are asked to arrange four successive 3d transition elements—Chromium (Cr), Manganese (Mn), Iron (Fe), and Cobalt (Co)—based on their $E_{M^{2+}/M}^{\circ }$ values.

But there is a crucial catch in the wording: "values with negative sign".

This phrasing often trips students up. It doesn't mean we are arranging the actual algebraic values from most negative to least negative. Instead, it means we are looking at the magnitude of the negative potential. If an element has an $E^{\circ }$ of $-1.18\text{ V}$, the "value with negative sign" is simply $1.18\text{ V}$. We are essentially comparing $-E^{\circ }$.

To understand the order, we first need to look at the general trend of standard reduction potentials across the 3d series. The $E^{\circ }$ value is determined by the sum of three energy terms: the enthalpy of sublimation, the ionization enthalpy (to remove two electrons), and the hydration enthalpy of the resulting $M^{2+}$ ion.

Generally, as we move from left to right across the 3d series, the nuclear charge increases, making it harder to remove electrons. Consequently, the sum of the first and second ionization enthalpies increases. Because it costs more energy to form the $M^{2+}$ ion, the standard reduction potential $E^{\circ }$ becomes less negative.

Following this general rule, we would expect the magnitude of the negative potential to decrease smoothly: $\text{Cr}>\text{Mn}>\text{Fe}>\text{Co}$.

However, chemistry is rarely without its elegant exceptions. Enter Manganese.

Manganese has an atomic number of 25, and its neutral electron configuration is $[\text{Ar}]3d^{5}4s^2$. When it loses two electrons to form the ${\text{Mn}}^{2+}$ ion, it achieves a perfectly half-filled $3d^5$ configuration.

In the quantum world, half-filled subshells offer exceptional exchange energy and symmetrical stability. Because the ${\text{Mn}}^{2+}$ state is so thermodynamically favorable, Manganese is highly eager to oxidize into this $+2$ state. This eagerness translates into an exceptionally negative standard reduction potential ($E^{\circ }\approx -1.18\text{ V}$).

Because of this $3d^5$ stability, Manganese breaks the general trend. Its potential is significantly more negative than Chromium's ($E^{\circ }\approx -0.91\text{ V}$).

Let's look at the approximate magnitudes ($-E^{\circ }$) for clarity: Mn: $1.18\text{ V}$ Cr: $0.91\text{ V}$ Fe: $0.44\text{ V}$ Co: $0.28\text{ V}$

Arranging these magnitudes in decreasing order, we get: $\text{Mn}>\text{Cr}>\text{Fe}>\text{Co}$

This perfectly matches option (a). The beauty of this question lies not just in knowing the trend, but in recognizing how the quantum mechanical stability of a half-filled d-orbital manifests in macroscopic thermodynamic properties like electrode potentials.