What ionisation enthalpy means
A quantitative measure of an element's tendency to lose an electron is its ionisation enthalpy ($\Delta_i H$). NCERT defines it as the energy required to remove an electron from an isolated gaseous atom (X) in its ground state. The first ionisation enthalpy is the enthalpy change for the process in which the single most loosely bound electron departs:
$$\ce{X(g) -> X+(g) + e-} \qquad \Delta_i H_1$$
Because work must be done against the attraction between the negative electron and the positive nucleus, energy is always absorbed; ionisation enthalpy is therefore always positive. It is reported in $\text{kJ mol}^{-1}$. The qualifier "isolated gaseous atom" is deliberate — it ensures that the value reflects only the atom itself, free from the interactions an atom would feel in a liquid or solid. When the term ionisation enthalpy is used without qualification, it always means the first ionisation enthalpy.
The first ionisation step on an energy axis
Successive ionisation enthalpies
Once the first electron has gone, a second electron can be removed from the resulting cation. The energy needed for this is the second ionisation enthalpy, the energy required to detach the second most loosely bound electron:
$$\ce{X+(g) -> X^2+(g) + e-} \qquad \Delta_i H_2$$
NCERT states the governing rule plainly: the second ionisation enthalpy is always higher than the first, because it is more difficult to remove an electron from a positively charged ion than from a neutral atom. In the same way the third ionisation enthalpy exceeds the second, and so on. The reason is electrostatic — after one electron leaves, the nucleus still carries its full charge but now acts on fewer electrons, so each remaining electron is held more tightly. The general inequality is therefore:
$$\Delta_i H_1 < \Delta_i H_2 < \Delta_i H_3 < \dots$$
| Quantity | Process | Comparison |
|---|---|---|
| First IE, $\Delta_i H_1$ | X(g) → X⁺(g) + e⁻ | Smallest of the set |
| Second IE, $\Delta_i H_2$ | X⁺(g) → X²⁺(g) + e⁻ | Larger than the first |
| Third IE, $\Delta_i H_3$ | X²⁺(g) → X³⁺(g) + e⁻ | Larger than the second |
Factors that control IE
Because ionisation enthalpy measures how firmly the outermost electron is held, it is fixed by the balance between two opposing influences identified in NCERT: the attraction of the electrons towards the nucleus and the repulsion of the electrons from one another. Working through these, four operational factors emerge.
| Factor | Effect on IE | Why |
|---|---|---|
| Nuclear charge (Z) | Higher Z → higher IE | A larger positive charge pulls the valence electron in more strongly. |
| Atomic size / distance | Larger atom → lower IE | The farther the electron, the weaker the nuclear pull; IE and atomic radius are closely related properties. |
| Shielding (screening) | More shielding → lower IE | Inner-core electrons screen the valence electron, so it feels a reduced effective nuclear charge. |
| Penetration of the orbital | Better penetration → higher IE | An s electron penetrates closer to the nucleus than a p electron of the same shell, so it is held more tightly. |
The unifying idea is effective nuclear charge. NCERT explains that the charge a valence electron actually experiences is less than the full nuclear charge because the intervening core electrons shield it. In lithium, for example, the 2s electron is screened by the inner $\ce{1s^2}$ core, so it experiences a net pull noticeably smaller than $+3$. Shielding is most effective when the inner shells are completely filled — the situation in the alkali metals, which carry a single $ns$ electron outside a noble-gas core and consequently have the lowest ionisation enthalpies in their periods.
The two extremes of the periodic landscape make the same point. The alkali metals, with their single, well-shielded valence electron, sit at the minima of any plot of ionisation enthalpy; their low values correlate directly with their high chemical reactivity. The noble gases, with closed shells and very stable configurations, sit at the maxima. Every periodic trend in ionisation enthalpy is, at heart, a contest between rising nuclear charge on one side and growing size plus shielding on the other.
Trends across a period and down a group
Two broad trends govern the first ionisation enthalpy: it generally increases across a period and decreases down a group. Each follows from the same factors, but with a different factor winning.
Moving from left to right across a period, successive electrons enter orbitals of the same principal quantum level. The shielding by the inner core barely increases, yet the nuclear charge keeps climbing. NCERT puts it crisply: across a period, increasing nuclear charge outweighs the shielding. The outermost electrons are therefore held more and more tightly and the ionisation enthalpy rises. Descending a group, the valence electron lies in a shell increasingly far from the nucleus, and shielding by the filled inner shells grows. Here the increase in shielding outweighs the rising nuclear charge, the outermost electron is held more loosely, and less energy is needed to remove it.
Direction of increasing first ionisation enthalpy
Ionisation enthalpy and atomic size move in opposite directions. To see why the radius shrinks across a period and grows down a group, read Atomic and Ionic Radius.
Second-period data and the saw-tooth
The general "increase across a period" rule is only the backbone of the trend. Plotting the actual first ionisation enthalpies of the second period reveals a saw-tooth, not a straight climb: the rise from lithium to neon is interrupted at boron and again at oxygen. The values from NCERT Fig. 3.6(a) make the dips explicit.
| Element | Valence config. | $\Delta_i H_1$ / kJ mol⁻¹ | Note |
|---|---|---|---|
| Li | 2s¹ | 520 | Period minimum |
| Be | 2s² | 899 | Filled 2s gives a local peak |
| B | 2s² 2p¹ | 801 | Dip — below Be |
| C | 2s² 2p² | 1086 | Trend resumes |
| N | 2s² 2p³ | 1402 | Half-filled 2p peak |
| O | 2s² 2p⁴ | 1314 | Dip — below N |
| F | 2s² 2p⁵ | 1681 | Sharp rise |
| Ne | 2s² 2p⁶ | 2080 | Period maximum |
First ionisation enthalpy of Li → Ne against atomic number
The Be > B and N > O anomalies
The two dips in Figure 3 are not measurement quirks — they follow directly from the penetration factor and from electron configuration stability. Both must be reasoned, not memorised, because the examiner phrases them in many disguises.
Beryllium above boron. Boron (Z = 5) has a higher nuclear charge than beryllium (Z = 4), yet its first ionisation enthalpy is the lower of the two. The reason lies in which electron is removed. In beryllium ($\ce{2s^2}$) the electron removed is a 2s electron; in boron ($\ce{2s^2 2p^1}$) it is the 2p electron. A 2s electron penetrates closer to the nucleus than a 2p electron, so the 2p electron of boron is more shielded by the inner core and is held less tightly. It is therefore easier to remove the 2p electron of boron than the 2s electron of beryllium, and boron has the smaller first ionisation enthalpy.
Nitrogen above oxygen. Nitrogen ($\ce{2s^2 2p^3}$) carries its three 2p electrons in three separate orbitals, in line with Hund's rule. Oxygen ($\ce{2s^2 2p^4}$) must place two of its four 2p electrons in the same orbital, and the resulting electron–electron repulsion makes that paired electron easier to remove. So although oxygen has the greater nuclear charge, its fourth 2p electron is less firmly held than any of nitrogen's three unpaired electrons, and the ionisation enthalpy of nitrogen exceeds that of oxygen. The extra stability of the exactly half-filled 2p³ shell is the same idea, viewed from nitrogen's side.
The two dips that overturn the "left-to-right increase" rule
Candidates write the period-2 order as a clean increase Li < Be < B < C < N < O < F. That is wrong on two counts. Boron drops below beryllium (a 2p electron is removed, not a 2s), and oxygen drops below nitrogen (the half-filled 2p³ of N is extra stable). The correct increasing order is Li < B < Be < C < O < N < F < Ne.
The same two effects repeat in period 3: Mg > Al (3s versus 3p removal, the basis of NCERT's Problem 3.6) and P > S (half-filled 3p³ stability). Watch the wording — a question phrased as "decreasing order" or "incorrect arrangement" hides the same anomaly.
Reason from configuration, not from atomic number. A higher Z does not guarantee a higher ionisation enthalpy when the comparison straddles an s-to-p or a half-filled-shell boundary.
Working out IE orders
Most NEET items reduce to one task: arrange a handful of elements in order of first ionisation enthalpy. The method is to apply the general period and group trend first, then check whether any pair crosses an s-to-p boundary or a half-filled-shell boundary, and finally insert the anomaly. Two worked cases show the routine.
Q. Arrange Li, Be, B, C, N in increasing order of first ionisation enthalpy.
All five lie in period 2, so the baseline trend is an increase from Li to N. Now check the anomaly: Be ($\ce{2s^2}$) versus B ($\ce{2s^2 2p^1}$) crosses the 2s-to-2p boundary, so B falls below Be. No half-filled-shell pair appears here (the N versus O dip is not in this set). Inserting the single anomaly gives Li < B < Be < C < N.
Q. Predict whether the first $\Delta_i H$ of Al is closer to 575 or 760 kJ mol⁻¹, given Na = 496, Mg = 737, Si = 786 kJ mol⁻¹. (NCERT Problem 3.6)
Aluminium ($\ce{3s^2 3p^1}$) loses a 3p electron, whereas magnesium ($\ce{3s^2}$) loses a 3s electron. The 3p electron is shielded by the 3s electrons, so Al's value should fall below Mg's 737, not rise toward Si. The answer is therefore closer to 575 kJ mol⁻¹ — the Mg > Al anomaly in action.
Ionisation enthalpy at a glance
- $\Delta_i H$ is the energy to remove the most loosely held electron from a gaseous atom, $\ce{X(g) -> X+(g) + e-}$; it is always positive and measured in kJ mol⁻¹.
- Successive values rise: $\Delta_i H_1 < \Delta_i H_2 < \Delta_i H_3$, because removing an electron from a cation is harder than from a neutral atom.
- Four factors: nuclear charge (↑ IE), atomic size (↓ IE), shielding (↓ IE), orbital penetration (s > p ⇒ ↑ IE).
- IE increases across a period (Z outweighs shielding) and decreases down a group (shielding outweighs Z).
- Anomalies: Be > B and N > O in period 2; Mg > Al and P > S in period 3.
- Period-2 increasing order: Li < B < Be < C < O < N < F < Ne.