valence
Practice
Atomic structure and periodicityNational 2019

In which list are the ions ranked in order of increasing ionic radius?

17% of national qualifiers got this right

Explanation

draft

Ionic radius depends on both electron configuration (isoelectronic comparisons) and position in the group (period).

  • KX+\ce{K+} and CaX2+\ce{Ca^2+} are isoelectronic (both have the Ar\ce{Ar} core, 18 electrons). Since CaX2+\ce{Ca^2+} has more protons (20) pulling on the same electron shell than KX+\ce{K+} (19 protons), CaX2+\ce{Ca^2+} is smaller: CaX2+<KX+\ce{Ca^2+} < \ce{K+}.
  • Similarly, RbX+\ce{Rb+} and SrX2+\ce{Sr^2+} are isoelectronic (both have the Kr\ce{Kr} core, 36 electrons). SrX2+\ce{Sr^2+} has more protons (38) than RbX+\ce{Rb+} (37), so SrX2+<RbX+\ce{Sr^2+} < \ce{Rb+}.
  • Down a group, ionic radius increases due to added electron shells, so CaX2+<SrX2+\ce{Ca^2+} < \ce{Sr^2+} (Group 2) and KX+<RbX+\ce{K+} < \ce{Rb+} (Group 1).

Putting these relationships together using actual measured radii (CaX2+≈100\ce{Ca^2+} \approx 100 pm, SrX2+≈118\ce{Sr^2+} \approx 118 pm, KX+≈138\ce{K+} \approx 138 pm, RbX+≈152\ce{Rb+} \approx 152 pm):

CaX2+<SrX2+<KX+<RbX+\ce{Ca^2+} < \ce{Sr^2+} < \ce{K+} < \ce{Rb+}

This matches choice (C).

A

Assumes ionic radius depends only on group position (same charge groups compared separately) and ignores that higher charge for the same electron configuration shrinks the ion; incorrectly ranks RbX+\ce{Rb+} before the 2+2+ ions and swaps CaX2+\ce{Ca^2+}/SrX2+\ce{Sr^2+} position relative to KX+\ce{K+}.

B

Mixes up the isoelectronic pairing—compares KX+\ce{K+} to CaX2+\ce{Ca^2+} correctly (smaller charge = larger radius) but then incorrectly places RbX+\ce{Rb+} before SrX2+\ce{Sr^2+}, forgetting that SrX2+\ce{Sr^2+}'s extra proton makes it smaller than RbX+\ce{Rb+}, not larger.

C

Correctly uses isoelectronic comparisons (higher charge = smaller radius for same electron count) and group trends (radius increases down a group) to get CaX2+<SrX2+<KX+<RbX+\ce{Ca^2+} < \ce{Sr^2+} < \ce{K+} < \ce{Rb+}.

D

Incorrectly interleaves the ions by treating 2+2+ and 1+1+ charges as if they alternate by period only, placing KX+\ce{K+} between CaX2+\ce{Ca^2+} and SrX2+\ce{Sr^2+} without properly comparing isoelectronic species; ignores that SrX2+\ce{Sr^2+} (Kr core) is smaller than KX+\ce{K+}... actually reverses the correct SrX2+<KX+\ce{Sr^2+}<\ce{K+} relationship, showing confusion about which ions share electron configurations.

Concept · Periodic trends: ionic radius (isoelectronic series and group trends)

USNCO 2019 National Part I exam, question 44. © American Chemical Society; reproduced for non-commercial practice.