Atomic and Ionic Radii Trends in Transition Elements
Chemistry · D And F Block Elements · NEET
Across a transition series (like Sc to Zn), the atomic radius first decreases, then stays almost the same in the middle, and slightly increases at the end. This happens because added d-electrons shield the nucleus poorly at the start, but electron-electron repulsion balances the pull in the middle. Memory hook: "Drop, flat, tiny rise" - like a plane that dives, cruises level, then lifts a little before landing.
Atomic radius across the 3d series: a clear decrease at first, an almost flat middle (Fe, Co, Ni), and a small rise at Cu and Zn. Poor d-shielding causes the early drop; growing d-d electron repulsion balances the nuclear pull in the middle and end.
Your doubts, answered
Why does the atomic radius first decrease as we move across a transition series?
As you go from Sc to the middle of the series, the nuclear charge (number of protons) increases by one each time. The new electrons go into the inner (n-1)d subshell, not the outermost shell. Because d-electrons shield poorly, the growing nuclear charge pulls the outer 4s electrons in more strongly. So the atom gets smaller at first.
Why does the radius stay almost constant in the middle of the series?
In the middle, two effects balance out. The rising nuclear charge tries to pull electrons in (shrinks the atom), but the extra d-electrons repel each other more and more (pushes the atom out). These two opposite effects roughly cancel, so the size stays nearly the same for elements like Fe, Co, Ni.
Why does the radius slightly increase near the end (Cu, Zn)?
Near the end of the series, the d-orbitals are almost full. Electron-electron repulsion between the many d-electrons becomes strong enough to slightly beat the pull of the nucleus. So the atom expands a little, giving a small increase in radius at Cu and Zn.
Why do Zr (4d) and Hf (5d) have almost the same atomic radius?
Going down a group, the lower element should be much bigger because it has an extra shell. But between the 4d and 5d series lie the 14 lanthanoids. Across the lanthanoids the size shrinks steadily - this is the lanthanoid contraction. This shrinking almost exactly cancels the expected size increase, so Zr and Hf end up nearly identical in size (about 160 pm each).
How is this trend different from a normal period like Na to Cl?
In a normal period, electrons add to the outermost shell, which is poorly shielded, so the radius drops sharply and steadily. In a transition series, electrons add to the inner (n-1)d shell, which shields better, so the decrease is much smaller and the size soon flattens out. That is why transition metals have similar sizes.
How does ionic radius change for the same metal in different oxidation states?
For the same element, higher positive charge means smaller ionic radius. Removing more electrons reduces electron-electron repulsion and lets the nucleus pull the remaining electrons closer. So the size order is: atom > M2+ > M3+. For example, Fe > Fe2+ > Fe3+.
⚠️ The NEET trap ✗ Zr and Hf have the same size because they are in the same group and have similar chemical properties. ✓ Zr and Hf have nearly the same atomic and ionic radii because of the lanthanoid contraction - the 14 lanthanoids between them cause a shrinkage that cancels the normal down-group size increase. 🧠 Same group does NOT explain equal size (Ti is much smaller than Zr, yet all three are in the same group). The special reason is lanthanoid contraction - always pick that for the Zr = Hf question.
Real NEET questions
NEET 2021
Zr (Z = 40) and Hf (Z = 72) have similar atomic and ionic radii because of:
A · lanthanoid contraction ✓
B · having similar chemical properties
C · belonging to same group
D · diagonal relationship
Solution: Zr belongs to the 4d series and Hf to the 5d series, and both are in the same group. Normally the 5d element (Hf) should be clearly larger. But the 14 lanthanoids that come between La and Hf cause a steady decrease in size called the lanthanoid contraction. This contraction almost exactly cancels the expected increase in size on going down the group. So Zr and Hf end up with nearly identical atomic and ionic radii. Correct answer: (A) lanthanoid contraction.
NEET 2016 Phase 2
Which one of the following statements related to lanthanons is incorrect?
A · Europium shows +2 oxidation state.
B · The basicity decreases as the ionic radius decreases from Pr to Lu.
C · All the lanthanons are much more reactive than aluminium. ✓
D · Ce4+ solutions are widely used as oxidizing agent in volumetric analysis.
Solution: Because of the lanthanoid contraction, the ionic radius falls steadily from Pr to Lu, so the hydroxides become less basic - statement (b) is correct and links directly to the size (radius) trend. Eu shows +2 (stable 4f7) and Ce4+ is a known oxidising agent in cerimetry, so (a) and (d) are correct. The over-generalised statement (c) is the incorrect one. This question tests that you connect decreasing ionic radius to decreasing basicity.
Solved D And F Block Elements NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the overall atomic radius trend in the 3d series?
It decreases from Sc to about Cr/Mn, stays almost constant through Fe, Co, Ni, and then slightly increases at Cu and Zn. The overall change is small compared with a normal period.
Which is bigger, Fe2+ or Fe3+?
Fe2+ is bigger. For the same element, a higher positive charge gives a smaller ion, because fewer electrons feel a stronger effective pull from the nucleus. So Fe > Fe2+ > Fe3+.
Are 4d and 5d transition elements bigger than 3d elements?
The 4d elements are larger than the 3d elements. But 4d and 5d elements of the same group are almost the same size because of the lanthanoid contraction (for example, Zr and Hf).
Why is this topic important for NEET?
NEET regularly asks the Zr = Hf lanthanoid-contraction question and links radius to basicity of hydroxides. Knowing the decrease-flat-rise pattern and the lanthanoid contraction lets you answer these directly without confusion.
