A
Inversely proportional to square of effective nuclear charge
B
Directly proportional to effective nuclear charge
C
Directly proportional to square of effective nuclear charge
D
Inversely proportional to effective nuclear charge
The answer is D.
Solution
The size of an ion depends on how strongly its outer electrons are attracted by the nucleus. This attraction is measured by the effective nuclear charge $Z_{eff} = Z - \sigma$ ($\sigma$ = screening constant). The greater the effective nuclear charge, the more strongly the electron cloud is pulled in, and the smaller the ion. Pauling's treatment of ionic radii takes the radius as $r = \dfrac{C_n}{Z - \sigma}$, i.e. inversely proportional to the effective nuclear charge. Example: in the isoelectronic series $O^{2-} \gt F^- \gt Na^+ \gt Mg^{2+}$, Z (and $Z_{eff}$) increases while the number of electrons stays 10, so the radius decreases. So ionic radii are inversely proportional to the effective nuclear charge.
ConceptIonic radius and effective nuclear charge
Classification of Elements and Periodicity in Properties in past papers
41 questions from this chapter have appeared across 23 exam years.