Atomic and Ionic Radii of d-Block
Decode the tug-of-war between Nuclear Charge and Shielding Effect. Master the 'constant size' anomaly and the massive implications of Lanthanoid Contraction.
Module Focus: The Tug-of-War
In s- and p-block elements, atomic radius strictly decreases across a period. In the d-block, the trend is much less pronounced and quite anomalous. The size is determined by a fierce competition between the increasing effective nuclear charge ($Z_{eff}$) pulling electrons inward, and the increasing shielding effect of the added inner $(n-1)d$ electrons pushing the outer $ns$ electrons outward.
1. Trend Along a Period (e.g., 3d Series)
As we move from left to right across a transition series, the atomic radius does not decrease uniformly. It exhibits three distinct phases.
Nuclear charge increases by +1 at each step. The newly added $(n-1)d$ electrons provide poor shielding initially. Thus, the increased $Z_{eff}$ pulls the outer $ns$ electrons closer, decreasing the radius.
As more d-electrons are added, their shielding effect becomes significant. The increase in shielding effect almost perfectly counterbalances the increase in nuclear charge. Thus, size remains roughly same.
At the very end ($d^{10}$), electron-electron repulsion between the paired d-electrons outweighs the increased nuclear charge. This repulsion forces the electron cloud to expand slightly.
Ionic Radii
The trend in ionic radii is much more predictable than atomic radii.
- For ions of the same oxidation state (e.g., $M^{2+}$), the ionic radius decreases steadily across the series as the nuclear charge increases.
- For the same element, the ionic radius decreases as oxidation state increases (e.g., $Fe^{3+}$ is significantly smaller than $Fe^{2+}$ because higher $Z_{eff}$ pulls remaining electrons tighter).
2. Trend Down a Group & Lanthanoid Contraction
Normally, as you move down a group, atomic radius increases due to the addition of new principal electron shells. However, between the 4d and 5d series, a massive anomaly occurs.
Why are the elements of the 4d and 5d series almost identical in size (e.g., $Zr \approx Hf$, $Nb \approx Ta$, $Mo \approx W$)?
- Before filling the 5d orbitals, the 4f orbitals are filled with 14 electrons (the Lanthanoids).
- The shape of f-orbitals is highly diffused, resulting in extremely poor shielding of the outer electrons from the nuclear charge.
- Because the nuclear charge increases heavily (+14) but shielding does not increase proportionally, the outer electrons are pulled sharply inward. This shrinkage is called the Lanthanoid Contraction.
- This contraction perfectly cancels out the expected size increase of adding a new shell (n=5), making 4d and 5d elements remarkably similar in size and chemical properties.
Density Consequences
Because the volume of 5d elements is essentially the same as 4d elements, but their atomic mass is nearly double, the density of 5d elements is drastically higher.
For example, the density of Osmium ($Os$) and Iridium ($Ir$) is extremely high (around $22.6 \text{ g/cm}^3$), making them the densest naturally occurring elements.
NEET Grand Test: d-Block Radii
15 High-Yield Questions testing shielding effects, the Zr/Hf anomaly, and ionic radii trends.
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