Q.(a) Complete the following reactions:

Zirconium (period 5) and hafnium (period 6) have almost identical atomic/ionic radii because of lanthanide contraction, which offsets the expected increase in size on descending a group.

Normally, atomic/ionic radius increases on going down a group as a new shell is added. But between Zr (Z=40) and Hf (Z=72), the 4f orbitals are filled across the lanthanides (Z=58–71). The 4f electrons shield the nuclear charge poorly, so the effective nuclear charge felt by the outer electrons increases steadily across the lanthanide series, causing a steady contraction in atomic/ionic size — the lanthanide contraction.

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The lanthanoid contraction is caused by the poor ability of the diffuse, deeply-buried 4f orbitals to shield one another from the increasing nuclear charge, so effective nuclear charge rises steadily and atomic/ionic radii shrink across the series.

Mechanism: Going from $Ce$ to $Lu$, the atomic number (and hence nuclear charge $Z$) increases by one unit at each step as an electron is added to the inner $4f$ subshell. The $4f$ orbitals are radially compact and have poor overlap/penetration properties, so one $4f$ electron shields another $4f$ electron (or the outer $5d^16s^2$ electrons) from the nucleus very inefficiently — far less efficiently than, say, $s$ or $p$ electrons shield each other. Because this poor screening does not keep pace with the steadily rising nuclear charge, the effective nuclear charge ($Z_{eff}=Z-S$) experienced by the outer electrons increases continuously across the series, pulling the electron cloud in and steadily shrinking the atomic and ionic radii. The cumulative shrinkage across all 14 lanthanoids is the lanthanoid contraction, and it is large enough that the third-row transition elements following the lanthanoids (e.g. $Zr$ and $Hf$, $Nb$ and $Ta$) end up with almost identical atomic radii to their second-row counterparts.

Why the other options are wrong/incomplete: …

Lanthanoid contraction is the steady decrease in atomic/ionic radii across the lanthanoid series, caused by an increase in effective nuclear charge that is imperfectly screened by the diffuse 4f electrons.

Across the lanthanoid series (Ce to Lu), each successive element adds one proton to the nucleus and one electron to the inner 4f subshell. The 4f orbitals have poor shielding ability (diffuse, non-directional shapes), so they do not effectively screen the outer electrons from the increasing nuclear charge.

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Zr4+ (4d series) and Hf4+ (5d series) end up nearly identical in size because lanthanoid contraction cancels out the size increase normally expected on adding a new shell.

Going down a group, ionic radius normally increases as a new principal shell is added. But between the 4d series (e.g. Zr, Z = 40) and the 5d series (e.g. Hf, Z = 72), the 14 lanthanoid elements (57-71) intervene, each adding a poorly-shielding 4f electron. The resulting steady contraction in radius across the lanthanoids ('lanthanoid contraction') almost exactly cancels the expected size increase from Zr to Hf. The net result is that Zr4+ and Hf4+ have almost identical ionic radii (~72 pm) …

Across the lanthanoid series, each added electron goes into the deeply buried, diffuse 4f subshell, which shields the increasing nuclear charge poorly — so the effective nuclear charge felt by outer electrons rises steadily, causing atomic/ionic size to shrink steadily from La to Lu.

Definition

Lanthanoid contraction refers to the steady and cumulative decrease in the atomic and ionic (especially $Ln^{3+}$) radii of the lanthanoid elements as one moves from lanthanum ($Z=57$) to lutetium ($Z=71$) across the 4f series.

Cause

As atomic number increases across the lanthanoids, each successive electron is added to an inner $4f$ orbital. The $4f$ orbitals are radially very diffuse and penetrate poorly, so electrons within the same $4f$ subshell shield each other from the nuclear charge very inefficiently (poor mutual shielding). Consequently, as protons are added to the nucleus, the effective nuclear charge ($Z_{eff}$) experienced by the outer ($5s$, $5p$, $6s$) electrons increases steadily across the series (since the extra positive charge is not being cancelled out by the added, poorly-shielding $4f$ electron). This progressively stronger pull draws the outer electron cloud inward, so the atomic/ionic radius contracts steadily with each additional element.

Consequences

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The 4f orbitals fill across the lanthanides (Ce → Lu).

The lanthanides (the first inner-transition series, elements 58 Ce to 71 Lu) are characterised by the gradual filling of the 4f energy level.

• Actinides fill the 5f orbitals. …