Q.Ionisation enthalpies of Ce, Pr and Nd are higher than Th, Pa and U. Why?

The key idea is that Ce, Pr, and Nd are lanthanoids (4f elements), while Th, Pa, and U are actinoids (5f elements). The difference in ionisation enthalpy arises from the poorer shielding of 4f electrons compared to 5f electrons.

Reasoning:

1. In lanthanoids, the 4f orbitals are more compact and penetrate closer to the nucleus than the 5f orbitals of actinoids.
2. This means 4f electrons shield the outer electrons less effectively from the nuclear charge, resulting in a higher effective nuclear charge ($Z_{\text{eff}}$) experienced by the valence electrons in lanthanoids. …

The Core Idea: Why Compare 4f and 5f Elements?

You’re comparing two series of inner transition elements: the lanthanides (Ce, Pr, Nd — filling the 4f subshell) and the actinides (Th, Pa, U — filling the 5f subshell). Both series add electrons to an inner f-orbital, but the energy required to remove an electron (ionisation enthalpy) is not simply a function of atomic number. The key lies in how tightly the f-electrons are held.

The 4f orbitals are smaller and closer to the nucleus than the 5f orbitals. They also provide poor shielding of the nuclear charge from one another. This means that as you move across the lanthanide series, the effective nuclear charge ($Z_{\text{eff}}$) experienced by each added electron increases sharply. The 4f electrons are therefore pulled in very tightly.

In contrast, the 5f orbitals are more diffuse and extended. They are also subject to relativistic effects (electrons moving at speeds near light, which contracts s and p orbitals but expands and destabilises d and f orbitals). This makes the 5f electrons easier to remove.

Step-by-Step Reasoning

1. The role of orbital size and penetration

The 4f orbitals have no radial nodes and are poorly shielded by the filled 4d and 5s,5p subshells. They penetrate closer to the nucleus than the 5f orbitals do. This means the 4f electron in Ce, Pr, or Nd feels a much larger effective nuclear charge than the 5f electron in Th, Pa, or U.

2. Relativistic destabilisation of 5f orbitals

For heavy elements like thorium ($Z=90$) and uranium ($Z=92$), relativistic effects become significant. The inner 1s electrons move at speeds close to $c$, causing the s and p orbitals to contract. This contraction indirectly expands and destabilises the outer d and f orbitals (the "indirect relativistic effect"). The 5f orbitals in actinides are therefore higher in energy and more loosely bound than a non-relativistic calculation would predict. No such effect is strong enough in the lanthanides ($Z \approx 58-60$) to significantly alter 4f binding.

3. What the actual data shows

NCERT notes that although the ionisation enthalpies of the early actinoids are not accurately known with great precision, the available evidence shows they are lower than those of the corresponding early lanthanoids. For example, the first ionisation enthalpy of thorium is appreciably lower than that of cerium, and the same trend holds for Pa vs Pr and U vs Nd. …

Method: Electronic Configuration Analysis with the Transition Element Definition

This method uses the IUPAC definition of transition elements (elements that have partially filled $d$ or $f$ orbitals in their ground state or common oxidation states) and compares electronic configurations to explain trends in ionisation enthalpy.

Steps

1. Write the ground state electronic configurations of the given elements.

   • Ce (Cerium, Z=58): $[\text{Xe}]\,4f^1\,5d^1\,6s^2$

   • Pr (Praseodymium, Z=59): $[\text{Xe}]\,4f^3\,6s^2$

   • Nd (Neodymium, Z=60): $[\text{Xe}]\,4f^4\,6s^2$

   • Th (Thorium, Z=90): $[\text{Rn}]\,6d^2\,7s^2$

   • Pa (Protactinium, Z=91): $[\text{Rn}]\,5f^2\,6d^1\,7s^2$

   • U (Uranium, Z=92): $[\text{Rn}]\,5f^3\,6d^1\,7s^2$

2. Identify the type of orbitals involved in each set.

   • Ce, Pr, Nd belong to the 4f-series (Lanthanoids). Their differentiating electron enters the 4f orbital.
   • Th, Pa, U belong to the 5f-series (Actinoids). Their differentiating electron enters the 5f orbital.

3. Apply the concept of orbital penetration and shielding.

   • 4f orbitals are more compact and lie closer to the nucleus than 5f orbitals.
   • This means 4f electrons experience greater effective nuclear charge ($Z_{\text{eff}}$) and are more tightly held.
   • 5f orbitals are more diffuse and less penetrating — they are better shielded from the nucleus by inner electrons.

4. Relate to ionisation enthalpy.

   • Higher ionisation enthalpy means more energy is needed to remove an electron. …

Let’s break it down step-by-step.

🔍 The Core Concept

The question compares ionisation enthalpies of:

• Lanthanides: Ce (Cerium), Pr (Praseodymium), Nd (Neodymium)
• Actinides: Th (Thorium), Pa (Protactinium), U (Uranium)

Key fact: Ionisation enthalpy generally decreases down a group because atomic size increases and shielding increases.

But here, the lanthanides (Ce, Pr, Nd) have higher ionisation enthalpies than the actinides (Th, Pa, U) — even though they are in the period above.

Why?

The 5f orbitals in actinides are less tightly held by the nucleus than the 4f orbitals in lanthanides.

• 4f orbitals are more penetrating and closer to the nucleus → electrons are harder to remove.
• 5f orbitals are more diffuse and less shielded → electrons are easier to remove.

Thus, ionisation enthalpy decreases from 4f to 5f series, not increases.

✗ Common Mistake #1: Assuming “Down the Group = Always Higher IE”

The error: Students think that because ionisation enthalpy generally increases across a period, it must also increase down the group for f-block elements.

Why it’s wrong:

• The trend down a group is decrease in IE (due to increased size and shielding).
• The f-block contraction (lanthanoid contraction) makes 4f elements smaller, but the 5f orbitals are more diffuse — so the decrease is even more pronounced.

How to avoid:

• Always check which orbitals are being filled (4f vs 5f).
• Remember: 5f electrons are less tightly bound than 4f electrons.

✗ Common Mistake #2: Confusing “Ionisation Enthalpy” with “Stability of Oxidation States”

The error: Students think that because Ce(IV) and Pr(IV) are stable, their IE must be low — so they expect actinides to have even lower IE.

Why it’s wrong:

• Stability of higher oxidation states in actinides (e.g., U(VI)) is due to 5f electrons being easier to remove, not because IE is lower in a simple trend.
• The ionisation enthalpy is the energy to remove the first electron — not the energy to reach a high oxidation state.

How to avoid:

• Separate the concept of first IE from oxidation state stability.
• First IE depends on orbital penetration and shielding; oxidation state stability depends on relativistic effects and bonding.

✗ Common Mistake #3: Ignoring the Role of Shielding and Penetration

The error: Students treat all f-orbitals as identical — they don’t compare 4f vs 5f.

Why it’s wrong:

• 4f orbitals have greater penetration into the core electron cloud → stronger attraction to nucleus → higher IE. …