Q.How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements?

The key idea is that first ionisation enthalpy rises only mildly and irregularly across the series (d-electrons shield the 4s electrons somewhat, but not perfectly), while the real, sharp irregularity shows up in the second ionisation enthalpy.

Reasoning:

1. From Sc to Zn, nuclear charge increases steadily, so first ionisation enthalpy ($\Delta_iH_1$) generally rises — but only slightly and with minor bumps, because the added $3d$ electrons shield the outer $4s$ electron from the growing nuclear charge almost as effectively as it shields itself, keeping the rise gentle (unlike the steep rise across a normal period of non-transition elements).
2. The second ionisation enthalpy ($\Delta_iH_2$) shows the real irregularity: it is unusually high for Cr and Cu, because their singly-charged ions ($\text{Cr}^+ = 3d^5$, $\text{Cu}^+ = 3d^{10}$) already have an extra-stable, half-filled or fully-filled $d$-subshell — removing a second electron means breaking into that stable subshell, which costs much more energy. …

The Real Data (Table 4.2)

Step-by-Step Reasoning

1. First ionisation enthalpy — a gentle, only mildly irregular rise.

Across Sc→Zn, nuclear charge rises by one unit each step, so $\Delta_iH_1$ generally increases (631 → 906). But the rise is much gentler than across a normal (non-transition) period, because each new electron is added to an inner $3d$ orbital rather than the outer shell — a $3d$ electron shields the $4s$ electrons from the nucleus almost as effectively as another $4s$ electron would, so the effective nuclear charge felt by the valence electron increases only slowly. This is why the chapter describes the first-ionisation-enthalpy trend as "irregular... though of little chemical significance," without pinning the irregularity to any one specific element.

2. Second ionisation enthalpy — the real, well-defined break.

$\Delta_iH_2$ removes an electron from the singly-charged ion $M^+$, and here the electronic configuration of $M^+$ matters directly:

• Chromium: neutral Cr is $3d^5 4s^1$ (the well-known half-filled-stability exception), so $\text{Cr}^+$ is already $3d^5$ — a stable, half-filled $d$-subshell. Removing a second electron means breaking into this stable arrangement, so $\Delta_iH_2$ for Cr (1592) is unusually high.
• Copper: neutral Cu is $3d^{10}4s^1$, so $\text{Cu}^+$ is $3d^{10}$ — a stable, fully-filled subshell. Breaking into it likewise makes Cu's $\Delta_iH_2$ (1958) the highest in the row.
• Manganese: neutral Mn is $3d^5 4s^2$, so $\text{Mn}^+$ is $3d^5 4s^1$ — the stable $3d^5$ core is already intact, with one "spare" $4s$ electron still to remove. Taking that easy $4s$ electron gives Mn a comparatively low $\Delta_iH_2$ (1509) — a dip just below Cr's spike, not a peak. …

Method: Configuration-and-Data Analysis (Table 4.2)

This method explains the irregular variation by pairing each element's electronic configurations (atom and $M^+$ ion) with the actual printed data, instead of relying on remembered "dip" rules.

Steps

Step 1: Write the configurations that matter

$\Delta_iH_1$ removes an electron from the neutral atom M; $\Delta_iH_2$ removes one from the $M^+$ ion. So write both configurations for each element (Table 4.2's M and $M^+$ rows).

Step 2: Check the first ionisation enthalpy against the data

Table 4.2 (kJ/mol): Sc 631, Ti 656, V 650, Cr 653, Mn 717, Fe 762, Co 758, Ni 736, Cu 745, Zn 906. The rise is gentle and only mildly irregular — each added $3d$ electron shields the $4s$ electrons from the growing nuclear charge, so the effective nuclear charge climbs slowly. Note there is no clean dip at Cr or Cu: Cr (653) is slightly above V (650), and Cu (745) is slightly above Ni (736).

Step 3: Explain the second ionisation enthalpy — where the sharp irregularity lives

• Cr⁺ is $3d^5$ (stable, half-filled) — removing another electron breaks into it, so $\Delta_iH_2$ is unusually high (1592).
• Cu⁺ is $3d^{10}$ (stable, fully filled) — breaking into it makes Cu's $\Delta_iH_2$ the highest in the row (1958).
• Mn⁺ is $3d^5 4s^1$ — the second electron comes from the spare $4s$, leaving the stable $3d^5$ intact, so $\Delta_iH_2$ is comparatively low (1509).
• Zn⁺ is $3d^{10} 4s^1$ — same pattern, so Zn's $\Delta_iH_2$ (1734) sits below Cu's.

Step 4: Summarise (all values from Table 4.2, kJ/mol)

Here is a breakdown of the common mistakes students make when explaining the irregular variation of ionisation enthalpies in the first transition series, along with how to avoid each.

The Core Concept (The "Why")

The irregular variation is attributed to the varying degrees of stability of the different $3d$ configurations (e.g., $d^0$, $d^5$, $d^{10}$ are exceptionally stable). Two things follow from the real data (Table 4.2):

1. First ionisation enthalpy ($\Delta_i H_1$): rises gently and only mildly irregularly across Sc → Zn (631 → 906 kJ/mol), because each added $3d$ electron partially shields the $4s$ electrons from the growing nuclear charge. There is no clean dip at Cr or Cu.
2. Second ionisation enthalpy ($\Delta_i H_2$): shows the sharp, well-defined irregularity — unusually high for Cr (1592) and Cu (1958), whose $M^+$ ions are the stable $3d^5$ and $3d^{10}$; comparatively low for Mn (1509) and Zn (1734), whose $M^+$ ions still carry one spare, easily-removed $4s$ electron.

Common Mistake #1: Claiming the First Ionisation Enthalpy "Dips" at Cr and Cu

The Mistake: Students reason "Cr and Cu each have only one $4s$ electron, and removing it gives a stable $d^5$/$d^{10}$ ion — so their $\Delta_i H_1$ must dip below their neighbours'."

Why it's wrong: Table 4.2's own numbers refute it: Cr's $\Delta_i H_1$ (653) is slightly higher than V's (650), and Cu's (745) is slightly higher than Ni's (736). The stable-configuration effect shows up cleanly in $\Delta_i H_2$, not as first-ionisation dips.

How to Avoid: Quote the data, not the remembered rule. If asked about $\Delta_i H_1$, say the rise is gentle and only mildly irregular; save the $d^5$/$d^{10}$ stability argument for $\Delta_i H_2$.

Common Mistake #2: Telling the Same Story for $\Delta_i H_1$ and $\Delta_i H_2$

The Mistake: Students say "Cr has low $\Delta_i H_1$ and also low $\Delta_i H_2$" (or high for both), applying one blanket rule to both quantities.

The Correct Logic: The two quantities remove electrons from different species:

• $\Delta_i H_1$ (M → M⁺) depends on the neutral atom.
• $\Delta_i H_2$ (M⁺ → M²⁺) depends on the $M^+$ ion:
  • Cr⁺ ($3d^5$) → Cr²⁺ ($3d^4$): destroys stable $d^5$ → high $\Delta_i H_2$.
  • Cu⁺ ($3d^{10}$) → Cu²⁺ ($3d^9$): destroys stable $d^{10}$ → high $\Delta_i H_2$.
  • Mn⁺ ($3d^5 4s^1$) → Mn²⁺ ($3d^5$): removes the spare $4s$ electron, leaving stable $d^5$ → low $\Delta_i H_2$.
  • Zn⁺ ($3d^{10} 4s^1$) → Zn²⁺ ($3d^{10}$): same pattern → $\Delta_i H_2$ below Cu's.

How to Avoid: Write the configurations of M, M⁺, and M²⁺ for each element. Ask: "Is the electron being removed a spare $4s$ electron (easy), or does it break a stable $d^5$/$d^{10}$ core (hard)?"

Common Mistake #3: Writing Wrong Ground-State Configurations

The Mistake: Students write Cr as $3d^4 4s^2$ or Cu as $3d^9 4s^2$, and then cannot explain the irregularities.

The Correct Logic: The actual ground-state configurations are the two exceptions:

• Cr: $[Ar]\, 3d^5 4s^1$ (not $3d^4 4s^2$)
• Cu: $[Ar]\, 3d^{10} 4s^1$ (not $3d^9 4s^2$)

How to Avoid: Memorise the two exceptions (Cr and Cu). For every other 3d element, the general $3d^n 4s^2$ rule holds.

Common Mistake #4: Blaming Only "Nuclear Charge" or "Shielding"

The Mistake: Students say "ionisation enthalpy increases because nuclear charge increases" and stop there. That explains the general rise, not the irregularities. …