The key idea is that $E^\circ(M^{2+}/M)$ reflects the combined energy cost of converting solid metal into hydrated $M^{2+}$ — sublimation (atomisation) enthalpy, the first two ionisation enthalpies, and hydration enthalpy — and these terms do not vary regularly across the series.

Reasoning:

1. The summed ionisation enthalpies ($\Delta_iH_1 + \Delta_iH_2$) vary irregularly from V to Cu because of the varying stability of the different $3d$ configurations (for example, the stable half-filled $3d^5$ of $Mn^{2+}$).
2. The sublimation (atomisation) enthalpies of manganese and vanadium are relatively much lower than those of their neighbours, which makes these metals easier to oxidise to $M^{2+}(aq)$ — so their $E^\circ$ values ($-1.18$ V each) sit more negative than the general trend predicts.
3. Where the overall oxidation costs less energy, $E^\circ$ is more negative; where it costs more (as for Cu, with its very high second ionisation enthalpy), $E^\circ$ is less negative or even positive.

Why This Happens: The Concept

$E^\circ(M^{2+}/M)$ measures the overall ease of the process $M(s) \rightarrow M^{2+}(aq) + 2e^-$, which is the sum of three steps:

1. Atomisation ($\Delta_aH^\circ$) — solid metal → gaseous atoms.
2. Ionisation ($\Delta_iH_1 + \Delta_iH_2$) — gaseous atom → gaseous $M^{2+}$.
3. Hydration ($\Delta_{hyd}H^\circ$) — gaseous $M^{2+}$ → $M^{2+}(aq)$.

If all three terms varied smoothly across V→Cu, $E^\circ$ would too. They don't, because electronic configuration (specifically, the stability of half-filled/fully-filled $d$-subshells) affects atomisation and ionisation enthalpies in a non-uniform way.

Step-by-Step Reasoning, Using Table 4.4's Real Values

1. The general trend. Going from V to Cu, the summed ionisation enthalpy rises fairly steadily (2064 → 2703), which is why $E^\circ$ generally becomes less negative — larger ionisation enthalpies make it harder to remove electrons and reach $M^{2+}$, so more of that unfavourable energy must be repaid by atomisation + hydration, pushing $E^\circ$ upward... except where atomisation enthalpy itself breaks the pattern.

2. Manganese and vanadium have unusually low atomisation enthalpies. Mn's $\Delta_aH^\circ$ (279 kJ mol$^{-1}$) is the lowest of this group — far below its neighbours Cr (398) and Fe (418) — because Mn's half-filled $3d^5 4s^2$ configuration contributes comparatively few unpaired electrons to interatomic metallic bonding, weakening the metal lattice. This unusually low atomisation enthalpy makes the overall process $M(s)\rightarrow M^{2+}(aq)$ for Mn cost less than the smooth trend predicts — and the stability of half-filled $3d^5$ $Mn^{2+}$ reinforces this — so Mn is oxidised more readily and its $E^\circ$ ($-1.18$ V) is more negative than its neighbours Cr and Fe. V's atomisation enthalpy, while numerically higher in absolute terms, is likewise relatively low set against the size of its ionisation-enthalpy contribution, keeping V's $E^\circ$ ($-1.18$ V) anomalously negative too, matching Mn instead of continuing the rising trend from Ti.

3. Copper is the opposite case. Cu has the highest summed ionisation enthalpy (2703, driven by the very high $\Delta_iH_2=1958$ needed to break into the stable, fully-filled $3d^{10}$ core of $\text{Cu}^+$) — normally this alone would make Cu hard to oxidise. Cu's low atomisation enthalpy (339) helps, but the only genuinely energy-releasing step — hydration — is not large enough to repay the combined atomisation + ionisation cost — the high energy of transformation of Cu(s) to Cu2+(aq) is not balanced by its hydration enthalpy — so the overall process $Cu(s) \rightarrow Cu^{2+}(aq)$ stays energetically uphill, giving Cu the series' only positive $E^\circ$ ($+0.34$ V): it is reluctant to be oxidised at all, behaving as a "noble" metal.

Method: Break $E^\circ$ into Its Enthalpy Terms

$E^\circ(M^{2+}/M)$ reflects the overall process $M(s) \rightarrow M^{2+}(aq) + 2e^-$, which is the sum of three steps. The smaller the net cost of this process, the more negative the $E^\circ$.

Step 1 – Write the thermochemical cycle

1. Sublimation/atomisation: $M(s) \rightarrow M(g)$ — costs $\Delta_aH^\circ$
2. Ionisation: $M(g) \rightarrow M^{2+}(g)$ — costs $\Delta_iH_1 + \Delta_iH_2$
3. Hydration: $M^{2+}(g) \rightarrow M^{2+}(aq)$ — releases $\Delta_{hyd}H^\circ$

Step 2 – Tabulate the terms (Table 4.4, kJ mol$^{-1}$)

Step 3 – Find which terms break the smooth trend

• The summed ionisation enthalpies vary irregularly (2064 → 2245 → 2226 → 2323 …), reflecting the varying stability of the different $3d$ configurations.
• The sublimation enthalpies of Mn (279) and V are relatively much lower than their neighbours' demands would suggest — Mn's half-filled $3d^5 4s^2$ configuration contributes few unpaired d-electrons to metallic bonding.

Step 4 – Conclude

• A metal whose overall oxidation to $M^{2+}(aq)$ is cheap has a more negative $E^\circ$: the relatively low sublimation enthalpies keep Mn and V at $-1.18$ V, more negative than the general trend.
• Cu's very high summed ionisation enthalpy (2703, dominated by $\Delta_iH_2 = 1958$ into the stable $3d^{10}$ of $Cu^+$) is not repaid by atomisation + hydration, giving Cu the only positive value ($+0.34$ V).

Key result: the irregularity in $E^\circ(M^{2+}/M)$ comes from the irregular variation of $\Delta_iH_1 + \Delta_iH_2$ together with the relatively low sublimation enthalpies of Mn and V — not from any single half-filled-shell rule alone.

Here are the common mistakes students make when explaining the irregularity in $E^\circ (M^{2+}/M)$ values for first-row transition metals, and how to avoid each.

Mistake 1: Ignoring the "Why" and Just Listing Values

• What students do: They simply state that Cu has a positive $E^\circ$ while others are negative, or that Mn and V have the same value ($-1.18$ V), without explaining the reason.
• Why it's wrong: The question asks you to explain the irregularity. Listing data is not an explanation.
• How to avoid: Always connect the $E^\circ$ value to the ease of oxidation. A more negative $E^\circ$ means the reduction $M^{2+} + 2e^- \rightarrow M$ is less favourable — the metal is easily oxidised and $M^{2+}(aq)$ is the more stable form. A less negative (or positive) $E^\circ$ means the metal itself resists oxidation.

Mistake 2: Forgetting the Role of Hydration Enthalpy

• What students do: They only talk about ionisation enthalpy (IE) or sublimation enthalpy, ignoring that $E^\circ$ depends on the sum of three factors:

• Why it's wrong: The trend in $E^\circ$ is not simply the trend in IE. Hydration enthalpy varies significantly across the series and often dominates the irregularity.
• How to avoid: Always mention that the high hydration enthalpy of $Cu^{2+}$ (due to its small size and high charge density) partially offsets — but cannot overcome — copper's huge ionisation cost, leaving its $E^\circ$ positive. For Mn, the anomaly is driven mainly by its relatively low sublimation enthalpy together with the stable half-filled $d^5$ of $Mn^{2+}$ — its hydration enthalpy is actually the least negative of its neighbours (Table 4.4), so it slightly offsets, rather than causes, the anomaly.

Mistake 3: Misapplying the "Half-Filled Stability" Argument

• What students do: They say "Mn has a half-filled $d^5$ subshell, so $Mn^{2+}$ is very stable, therefore $E^\circ$ is very negative." This is correct but often incomplete.
• Why it's wrong: They forget that the metal (Mn) also has a configuration ($[Ar] 3d^5 4s^2$). The stability of $Mn^{2+}$ is relative to the metal. The half-filled stability makes $Mn^{2+}$ more stable than expected, which lowers its tendency to get reduced back to Mn, making $E^\circ$ more negative.
• How to avoid: Explicitly state: "The $d^5$ configuration of $Mn^{2+}$ gives it extra exchange energy and symmetry stability. This makes the $Mn^{2+}$ ion unusually stable in solution, so the reduction potential $Mn^{2+} \rightarrow Mn$ is less favourable (more negative)."

Mistake 4: Confusing the Trend with the "Dip" at Cr and Cu

• What students do: They try to explain the entire trend (V to Cu) using only the $d^5$ or $d^{10}$ argument, missing the specific irregularity at Cr and Cu.
• Why it's wrong: The question specifically asks about the irregularity in the given values. The main irregularity is:
  • Cu has a positive $E^\circ$ (unlike all others).
  • Mn and V have the same value ($-1.18$ V) despite being far apart.
  • Cr has a less negative value ($-0.91$ V) than expected from its position.
• How to avoid: Focus on the exceptions:
  • Cu: $Cu^{2+}$ has high hydration enthalpy + high IE₂ (due to $d^9$ configuration) → overall $E^\circ$ positive.
  • Cr: $Cr^{2+}$ has a $d^4$ configuration (less stable than $d^5$), so it is easier to reduce → less negative $E^\circ$.
  • Mn: $Mn^{2+}$ has $d^5$ (extra stable) → harder to reduce → more negative $E^\circ$.

Mistake 5: Using the Wrong Sign Convention

• What students do: They say "Cu has a positive $E^\circ$, so it is easily oxidised." This is backwards.
• Why it's wrong: $E^\circ (M^{2+}/M)$ is the reduction potential. A positive value means the reduction $M^{2+} + 2e^- \rightarrow M$ is spontaneous (favoured). So $Cu^{2+}$ is easily reduced to Cu metal, meaning Cu metal is resistant to oxidation (noble).
• How to avoid: Always interpret $E^\circ$ as a reduction potential: more positive means the reduction $M^{2+} + 2e^- \rightarrow M$ is more favoured, i.e. the metal is the stable form. For copper, $Cu^{2+}$ is easily reduced, so Cu metal resists oxidation (noble character).

Quick Summary Table: Mistakes & Fixes

Final Tip: When answering, structure your explanation as:

1. Identify the irregularity (Cu positive, Mn & V same, Cr less negative).
2. State the three factors (sublimation, ionisation, hydration) — and name the printed grounds: the irregular variation of $\Delta_iH_1 + \Delta_iH_2$, plus the relatively low sublimation enthalpies of Mn and V.
3. Explain each exception using electronic configuration and hydration enthalpy.