Question: Why is there not much change in the atomic radius of the transition metals with the increase in atomic number?

Step 1 (Governing Principles): [1.0 Mark]

The atomic radius of transition metals is governed by the competition between two opposing effects as we move across a period:

• Effective Nuclear Charge ($Z_{eff}$): As the atomic number increases, the number of protons increases, which pulls the electron cloud closer to the nucleus, causing a decrease in atomic radius.
• Shielding Effect (Screening Effect): The electrons being added enter the $(n-1)d$ subshell. These $d$-electrons provide poor shielding for the outer $ns$ electrons against the increasing nuclear charge.

Step 2 (Step-by-Step Logical Deduction): [1.5 Marks]

The net change in atomic radius is determined by the balance between these two factors:

• As we move from left to right in a transition series, the increase in nuclear charge ($Z$) is partially offset by the poor shielding provided by the $d$-electrons.
• Mathematically, the effective nuclear charge experienced by the valence electrons can be represented as: $$Z_{eff} = Z - \sigma$$ where $\sigma$ is the shielding constant.
• Because the $d$-orbitals have a diffuse shape, the increase in $Z$ is almost perfectly balanced by the increase in $\sigma$ (due to the added $d$-electrons).

Step 3 (Final Concluded Answer): [0.5 Mark]

Consequently, the increase in nuclear attraction is compensated by the increase in electron-electron repulsion (shielding). This results in a very small decrease in atomic radius across the series, leading to the observed trend of near-constancy in atomic size for transition elements.

💡 EL10 Concept Insight: Imagine the nucleus is a magnet pulling on the outer electrons; as you add more protons, the pull gets stronger (shrinking the atom), but adding $d$-electrons is like putting a "leaky" shield between the magnet and the electrons, which cancels out that extra pull, keeping the atom's size mostly the same.