Ionization enthalpy is a fundamental concept in atomic structure that describes the energy required to remove an electron from an isolated gaseous atom. It plays a crucial role in understanding periodic trends, chemical reactivity, and the stability of elements. The study of factors affecting ionization enthalpy and successive ionization enthalpies helps explain why different elements exhibit varying tendencies to lose electrons.
- What is Ionization Enthalpy?
- What are the common units of Ionization Enthalpy?
- What is meant by Successive Ionization Enthalpies?
- Why does Ionization Enthalpy increase successively (IE1 < IE2 < IE3 <…) ?
- What are the factors affecting Ionization Enthalpy (Ionization Energy)
- 1. How does atomic size affect ionization enthalpy?
- 2. How does nuclear charge affect ionization enthalpy?
- 3. How does Screening effect of the inner electrons influence ionization enthalpy ?
- 4. How does penetration effect of electrons affect ionization enthalpy?
- 5. How does electronic arrangement of the valence electrons of the atom affect ionization enthalpy?
What is Ionization Enthalpy?
Electrons in an atom are attracted towards the positively charged nucleus. To remove an electron, energy must be supplied to overcome this attraction. This energy is called Ionization Enthalpy.
What is the definition of Ionization Enthalpy?
Ionization enthalpy is defined as the energy required to remove the most loosely bound electron from an isolated gaseous atom of an element in its ground state.
Why is ionization enthalpy also called ionization potential?
Ionization Enthalpy is also called Ionization Potential because it is the minimum potential difference (in a discharge tube) required to remove the most loosely bound electron and form a gaseous cation.
What are the common units of Ionization Enthalpy?
The ionization enthalpy is expressed in units of kJ mol–1 or electron volt (eV) per atom or kcal mol–1.
What is the conversion relation between electron volt and kJ/mol?
The conversion relation between electron volt and kJ/mol are in the following table.
| Unit | Symbol | Conversion |
|---|---|---|
| Kilojoule per mole | kJ mol⁻¹ | – |
| Electron volt per atom | eV/atom | 1 eV per atom = 96.49 kJ/mol |
| Kilocalorie per mole | kcal mol⁻¹ | 1 eV per atom = 23.06 kcal/mol |
For example in hydrogen atom :
H (g) → H+(g) + e−
Ionization enthalpy = 1312 kJ/mol.
What is meant by Successive Ionization Enthalpies?
The atom may not only lose one electron but can lose more than one electrons also. It may be noted that if the gaseous atom is to lose more than one electron, these are removed one after the other. The energies required to remove subsequent electrons from an atom in the gaseous state are known as successive ionization enthalpies. The term first, second, third. ………. ionisation enthalpy refers to the removal of first, second, third…………..electron respectively.
Why are electrons removed one by one from an atom?
Removing electrons one by one rather than all at once is primarily due to the increasing electrostatic grip the nucleus exerts on the remaining electrons. This process is measured as successive ionization energies ($IE_1, IE_2, IE_3$, etc.).
Practically It is statistically and physically improbable for an atom to be hit by a single massive blast of energy that perfectly synchronizes the release of all its electrons simultaneously. Instead, the atom absorbs energy and loses electrons sequentially as each energy threshold is met. Hence an atom can lose more than one electron, but they are removed one by one in steps :
What is the first ionization enthalpy (IE1) ?
Energy required to remove the most loosely bound electron from a neutral gaseous atom is called the first ionization enthalpy (IE1).
Reaction :
X(g) + IE1 → X+(g) + e−
When remove the first electron from a neutral isolated gaseous atom, the number of protons (positive charge) stays the same, but the number of electrons (negative charge) decreases.
In a neutral atom, the positive and negative charges are balanced. Once the first electron is removed, the atom becomes a cation (positively charged). The remaining electrons feel a stronger net pull from the nucleus because there is less “electron-electron repulsion” pushing them away.
What is the second ionization enthalpy (IE2) ?
Energy required to remove the next most loosely bound electron from the singly charged cation is called the second ionization enthalpy (IE2).
Reaction :
X+(g) + IE2 → X2+(g) + e−
Electrons naturally repel each other. This repulsion actually helps “push” outer electrons away from the nucleus, making them easier to remove. As electrons remove one by one there are fewer negative charges to repel the remaining electrons. The remaining electrons “settle” closer to the nucleus. Because they are now closer and less shielded, the energy required to pull the next one away ($IE_2$) is always higher than the energy required for the first ($IE_1$).
What is the third ionization enthalpy (IE3) ?
Energy required to remove the next most loosely bound electron from the doubly charged cation is called the third ionization enthalpy (IE3).
Reaction :
X2+(g) + IE3 → X3+(g) + e−
and so on.
Why are ionization enthalpies always positive?
Because energy must be supplied to overcome the electrostatic attraction between the positively charged nucleus and the negatively charged electron to remove electrons from an isolated gaseous atom and therefore, ionization enthalpies are always positive.
How does ionization enthalpy indicate the ease of electron removal?
Thus, the ionization enthalpy gives the ease with which electron can be removed from an atom. Evidently, the smaller the value of ionization enthalpy, the easier it is to remove the electron from the atom.
Why does Ionization Enthalpy increase successively (IE1 < IE2 < IE3 <…) ?
Successive ionization enthalpies increases because of three main reasons :
How does electron–electron repulsion affect ionization enthalpy?
Decrease in Electron–Electron Repulsion : When the first electron is removed, the number of electrons decreases, but the nuclear charge remains the same. This reduces electron–electron repulsion and allows the nucleus to pull the remaining electrons closer. Hence due to decrease electron – electron repulsion, ionization enthalpy required to remove electron is more.
What is the role of effective nuclear charge in ionization enthalpy?
Increase in Effective Nuclear Charge (Zeff) : After each electron removal, the positive charge of the ion increases. The remaining electrons experience a greater effective nuclear charge, making them more strongly bound to the nucleus. Therefore, more energy is required to remove the next electron.
What happens when electrons are removed from inner shells? or Why are core electrons more difficult to remove than valence electrons?
Removal of electrons from Inner Shells (Core Electrons) : After valence electrons are removed, further electrons come from inner shells. Inner-shell electrons are much closer to the nucleus and are tightly bound due to strong electrostatic attraction. This causes a sudden and large jump in ionization enthalpy values.
Why is more energy required to remove electrons after each step?
Successive ionization enthalpies increases as follows :
IE1 < IE2 < IE3 <….…
because (a) Less shielding effect after each removal of electron (b) Increased effective nuclear charge after each removal of electron. (c) Remaining electrons are more strongly bound (d) Sometimes removal from inner shells cause a sudden large jump.
Why is there a sudden jump in ionization enthalpy after removal of valence electrons?
While the energy required always increases with each electron removed, there are massive “jumps” in energy when you move from one shell to another.
For example, in Sodium ($Na$) : $IE_1$ is relatively low because removing one valence electron from the outermost shell (3rd shell). $IE_2$ is massive (nearly 10x higher) because there is emptied the outer shell and are now trying to remove an electron from a stable electronic configuration of noble gas, full inner shell (2nd shell) that is much closer to the nucleus. That’s why there a sudden jump in ionization enthalpy after removal of valence electrons.
What does a large jump in successive ionization enthalpy values indicate?
It indicates that all valence electrons have been removed and the next electron is being removed from an inner (core) shell.
What are the factors affecting Ionization Enthalpy (Ionization Energy)
Ionization enthalpy is the minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state. The magnitude of ionization enthalpy depends on the following five major factors :
Atomic size, Nuclear charge, Screening effect, Penetration power, Stability of electronic configuration.
1. How does atomic size affect ionization enthalpy?
How does atomic size affect ionization enthalpy?
The ionization enthalpy decreases with increase in size of an atom and vice versa. This is because the attractive force between the electron and the nucleus is inversely proportional to the distance between them. The attractive force between the nucleus and the electron is given by Coulomb’s Law : $$F \propto \frac{Z_{\text{eff}}}{r^2}$$ where :
r = Distance between the nucleus and the outermost electron
Zeff = Effective nuclear charge
Because distance ($r$) is squared in the denominator, the force of attraction drops off very quickly as electrons move away from the nucleus. This is why valence electrons are much easier to remove than inner-shell electrons; they are simply too far away to feel the full “grip” of the protons.
Consequently, as the size of the atom increases, the outermost electrons are less tightly held by the nucleus. As a result, it becomes easier to remove the electron and therefore, ionization enthalpy decreases with increase in atomic size.
Why does ionization enthalpy decrease with increase in atomic size? Why does ionization enthalpy increase with decrease in atomic size ?
Larger atomic radius => Outer electrons are farther from the nucleus => Weaker attraction => Easier to remove => Lower ionization enthalpy.
Smaller atomic radius => Electrons are closer to the nucleus => Stronger attraction => Harder to remove => Higher ionization enthalpy.
For example :
Li (small size) : High ionization enthalpy (~520 kJ/mol)
Cs (large size) : Low ionization enthalpy (~375 kJ/mol)
“Learn More about Atomic Radius“
2. How does nuclear charge affect ionization enthalpy?
How does nuclear charge (Z) affect ionization enthalpy?
Charge on the nucleus : The ionization enthalpy increases with increase in nuclear charge. This is because the attractive force between the nucleus and the electron increases with the increase in nuclear charge.
How is the force between nucleus and electron related to distance and effective nuclear charge?
Coulomb’s Law describes the electrostatic force of attraction that holds the electrons in orbit around the nucleus. The formula for Coulomb’s Law is :
$$F = k \frac{q_1 q_2}{r^2}$$
$F$ (Electrostatic Force) : The pull between the nucleus and the electron.
$q_1$ (Charge of the Nucleus) : This is the total positive charge, represented by the number of protons (Atomic Number, $Z$).
$q_2$ (Charge of the Electron) : The negative charge of a single electron.
$r$ (Distance) : The radius or the distance between the nucleus and the valence shell.
$k$ : Coulomb’s constant.
The force of attraction is directly proportional to the product of charges on the nucleus and the electron. Therefore, with the increase in nuclear charge, it becomes more difficult to remove an electron and hence ionization enthalpy increases.
Therefore a greater nuclear charge makes electron removal more difficult, resulting in higher ionization enthalpy.
For Example in Boron (Z = 5) and Fluorine (Z = 9) : Fluorine has a much higher nuclear charge that is Its outer electrons are pulled more strongly and hence have higher ionization enthalpy.
3. How does Screening effect of the inner electrons influence ionization enthalpy ?
How does screening effect influence ionization enthalpy?
The ionization enthalpy decreases with increase in screening effect of the inner electrons. In multielectron atoms, the outermost electrons are shielded or screened from the nucleus by the inner electrons. This is known as shielding or screening effect. As a result of this, the outermost electron does not feel the full charge of the nucleus. The actual charge felt by an electron is termed as effective nuclear charge. Therefore, the effective nuclear charge (Zeff) is
(Zeff) = Total nuclear charge (Z) – Screening constant (σ)
where screening constant takes into account the screening effect of the inner electrons. If the number of electrons in the inner shell is large, the screening effect will be large. As a result, the nuclear electron attraction will be less. Consequently, ionization enthalpy will decrease.
“Learn more about Screening Effect“
Why does higher shielding lead to lower ionization enthalpy or lower shielding lead to higher ionization enthalpy ?
High screening effect means outer electrons feel less attraction and are easier to remove and hence have lower ionization enthalpy.
Low screening effect means outer electrons feel more attraction and are harder to remove and hence higher ionization enthalpy.
What is the order of screening power of orbitals?
The Screening (or Shielding) refers to the ability of inner-shell electrons to reduce the electrostatic attraction between the positively charged nucleus and the outer-shell electrons.
The order of screening power for orbitals within the same principal energy level is:
s > p > d > f
This means electrons in s-orbitals are less shielded compared to those in d or f orbitals.
The screening power depends entirely on how close an electron can get to the nucleus based on the shape of its orbital.
s-orbitals : These are spherical and have the highest probability of being found very close to the nucleus. Because they “hang out” near the center more often, they are very effective at forming a shield around the nucleus.
p-orbitals : These are dumbbell-shaped and are slightly more spread out than s-orbitals, meaning they spend less time near the nucleus and offer less protection.
d and f-orbitals : These have complex, diffused shapes with many “lobes.” They are spread far apart in space, which makes them very “leaky” shields. They allow the nuclear charge to “peek through” to the outer electrons.
This hierarchy has a massive impact on the behavior of atoms :
Effective Nuclear Charge (Zeff) : Because d and f electrons screen poorly, the outer electrons feel a much stronger pull from the nucleus than expected. This is why atoms with filling d or f subshells (like transition metals or lanthanides) often have smaller atomic radii than predicted.
Energy Levels : The high screening power of s electrons is part of the reason why the 4s orbital is filled before the 3d orbital; the s electron’s ability to get closer to the nucleus makes it more stable (lower energy).
Why are outer electrons in alkali metals easy to remove?
Due to largest size of alkali metals, in alkali metals, the outermost s-electron is shielded by all the inner-core electrons, making it easy to remove, hence low ionization enthalpy.
4. How does penetration effect of electrons affect ionization enthalpy?
What is penetration effect of electrons ?
Penetration refers to the extent to which an electron’s probability density is found close to the nucleus.
How does penetration affect ionization enthalpy?
The ionization enthalpy increases with increase in penetration power of the electrons. It is well known that in case of multielectron atoms, the electrons in the s-orbital have the maximum probability of being found near the nucleus hence they experience the strongest attraction and are harder to remove and this probability goes on decreasing in case of p, d and f-orbitals. Hence, f-electrons are the farthest and are least penetrated, hence easiest to remove. Therefore removing a 2s electron requires more energy than removing a 2p electron from the same atom. Therefore
Greater penetration => Higher attraction => Higher ionization enthalpy.
Lesser penetration => Lower attraction => Lower ionization enthalpy.
What is the order of penetration of orbitals in the same shell?
According to the radial probability curves, the s-orbital has a small peak (a “node” or “inner lobe”) very close to the nucleus, even for higher energy levels.
s penetrates closest to the nucleus.
p penetrates less than s.
d penetrates even less.
f has the least penetration.
In other words, s-electrons are more penetrating towards the nucleus than p– electrons and the penetration power decreases in a given shell (same value of n) in the order
s > p > d > f
Why are s-electrons more difficult to remove than p, d, or f electrons ?
Now, if the penetration power of the electron is more, it will be closer to the nucleus and will be held firmly. Consequently, ionization energy will be high. Thus, for the same shell, the ionization enthalpy would be more to remove a s-electron than the energy required to remove a p-electron, which in turn will be more than that for the removal of a d-electron and so on.
5. How does electronic arrangement of the valence electrons of the atom affect ionization enthalpy?
How does electronic configuration affect ionization enthalpy?
The ionization enthalpy also depends upon the electronic configuration of the atom. It has been observed that certain electronic configurations are more stable than others.
For example, half-filled and completely-filled shells have extra stability associated with them. Consequently, it is difficult to remove electron from these stable configurations and ionization enthalpy is high.
Why do atoms with completely filled and half-filled subshells have high ionization enthalpy?
Ionization enthalpy depends on the stability of the atom’s electronic configuration. Atoms with half-filled or fully filled subshells are exceptionally stable due to exchange energy and symmetry and making electron removal more difficult. These stable configurations have unusually high ionization enthalpies.
This may be illustrated by the following examples :
Why do noble gases have very high ionization enthalpy ?
Noble gases have very small size (in their respective periods), very high nuclear charge and extremely stable ns² np⁶ electronic configuration and have no tendency to lose electrons and and thus have very high ionization enthalpies.
Helium has the highest ionization enthalpy of all elements (~2372 kJ/mol).
Why do Be and Mg have relatively high ionization enthalpy?
The elements like Be (1s² 2s²) and Mg (1s² 2s² 2p⁶ 3s²) have completely filled orbitals, leading to high ionization enthalpies.
Why do N and P show higher ionization enthalpy than expected?
The elements like N (1s² 2s² 2pₓ¹ 2pᵧ¹ 2pz¹) and P (1s² 2s² 2p⁶ 3s² 3pₓ¹ 3pᵧ¹ 3pz¹) have the electronic configurations in which the orbitals of same subshell are exactly half-filled and are also stable. Hence, they need large energy to remove the electron i.e., their ionization enthalpies are high.
Thus, the more stable the electronic configuration, the greater is the ionization enthalpy.
Important Chapter Interlinks
This section provides a complete and interconnected study of Classification of Elements and Periodicity in Properties, starting with detailed theory and notes for Class 11 Chemistry to build a strong conceptual foundation. You can explore atomic radius and its types including covalent, van der Waals, metallic, and ionic radii to understand periodic trends in atomic size. It also includes Screening Effect (Shielding Effect) : Calculation of Effective or Reduced Nuclear Charge (Slater’s Rules), which explains how inner electrons reduce the nuclear attraction on outer electrons and influence periodic trends. In addition, topics like Radius of Cation is Less and Anion is More Than Its Parent Atom, Size Variation in Isoelectronic Series help explain how ionic size changes due to gain or loss of electrons and how nuclear charge affects size in species with the same number of electrons. The causes of periodicity explain why elements show repeating properties based on electronic configuration, which is further supported by the modern periodic law and structure of the modern periodic table including groups, periods, and blocks for elements even beyond atomic number 100. The historical development is covered through Mendeleev’s periodic law and table, leading to the modern classification of elements into s, p, d, and f blocks with prediction of period, group, and block. To strengthen exam preparation, you can practice JEE Main PYQs, IMU CET PYQs and Merchant Navy sponsorship exam MCQs, and other previous year questions with solutions, along with solved examples, conceptual questions, and practice problems on the modern periodic table. Learn more in this section also to radius of cation is less and anion is more than its parent atom and size variation in Isoelectronic Series. Additionally, complete study material, mock tests, and guidance are provided under Anand Classes Chemistry notes, along with expert support from Er Neeraj Anand, making this section a comprehensive resource for competitive exam preparation.