Why do Zirconium and Hafnium exhibit similar properties ?

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Q. Why do Zirconium and Hafnium exhibit similar properties ?

Puducherry TnboardTamil Nadu HSC (DGE) Board 2025Subjective· 2mImportance★★★★★

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Concept understanding — Lanthanoid Contraction

Lanthanoid Contraction – From Intuition to Precision

Imagine you are walking across a row of the periodic table — from lanthanum (atomic number 57) to lutetium (71). Your first instinct might be: as we add more protons and more electrons, the atom should get bigger. But the opposite happens. The atoms actually shrink, steadily and stubbornly, across these 15 elements. That is the lanthanoid contraction.

Why? The answer lies in the 4f orbitals.

The core intuition: a bad shield

Every new electron you add across the lanthanoid series goes into a 4f orbital. These 4f orbitals are shaped like cloverleaves, but they are tucked deep inside the atom — very close to the nucleus. They are also notoriously poor at shielding the outer electrons from the pull of the nucleus.

Here is the key: each step adds one proton to the nucleus. That proton yanks harder on all the electrons. Normally, the new electron you add would partly cancel that pull (shielding). But 4f electrons are so diffuse and so poorly penetrating that they do a terrible job of shielding. So the net effect is that the effective nuclear charge felt by the outer electrons increases steadily across the series. The outer electrons get pulled inward, and the whole atom shrinks.

Note

The 4f orbitals are "inside" the atom — they lie closer to the nucleus than the 5d and 6s orbitals. So adding electrons there does not push the outer shell outward; instead, the increasing nuclear charge dominates.

The precise statement

Lanthanoid contraction is the progressive and regular decrease in atomic and ionic radii of the lanthanoid elements (Ce to Lu) as atomic number increases. The contraction is about 1 pm per element, totalling roughly 15 pm from La to Lu.

This is not a small effect. It is so consistent that the radii of the later lanthanoids are almost identical to those of the 4d transition metals directly above them in the periodic table. For example, zirconium (Zr) and hafnium (Hf) have nearly the same atomic radius — a direct consequence of the lanthanoid contraction.

\text{Atomic radius (pm)} \approx 187 - 0.9 \times (Z - 57) \quad \text{(rough linear fit for trivalent ions)}

Why it matters

The lanthanoid contraction explains several important patterns in chemistry:

Similarity of 4d and 5d transition metals: Elements like Zr and Hf, Nb and Ta, Mo and W are nearly identical in size and chemical behaviour. Without the lanthanoid contraction, the 5d metals would be much larger. This is why separating Hf from Zr is famously difficult — they are chemical twins.
Trend in basicity: Across the lanthanoid series, the ionic radius decreases. This increases the charge density on the ion, making it more polarising. As a result, the basicity of the hydroxides decreases from La(OH)₃ (strong base) to Lu(OH)₃ (weak base). …

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Hafnium comes right after the 14 lanthanoid elements in the periodic table, and the cumulative lanthanoid contraction in radius almost exactly offsets the size increase that would normally occur on descending from period 5 (Zr) to period 6 (Hf), leaving the two elements with nearly identical radii and hence very similar chemistry.

Normally, on descending a group of transition elements (e.g. from the 4d series to the 5d series), atomic and ionic radii increase due to the addition of an extra electron shell. However, in Group 4, hafnium (5d series, period 6) is preceded by the 14 lanthanoid elements (La to Lu), across which the atomic/ionic radii undergo a steady, cumulative decrease — the lanthanoid contraction — caused by the poor shielding of the diffuse 4f electrons. This contraction is large enough to almost exactly cancel out the expected radius increase from zirconium (Zr, 4d series) to hafnium (Hf, 5d series). As a result, Zr and Hf end up with nearly identical atomic radii ( $\approx145$ pm) and ionic radii ( $Zr^{4+}\approx72$ pm, $Hf^{4+}\approx71$ pm), giving them almost indistinguishable ch …

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Q.Why do Zirconium and Hafnium exhibit similar properties ?

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