Among the given transition elements, the divalent ion of which metal has smallest ionic size?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Nickel
Across the 3d series, the radii of ions with the same charge decrease as the atomic number increases. Each step adds one proton to the nucleus and one electron to the 3d subshell. The d electrons shield each other poorly, so the effective nuclear charge increases and the electron cloud is pulled closer to the nucleus.
Ionic radii of the divalent ions (pm):
\( \displaystyle \ce{Mn^2+}\ (82) > \ce{Fe^2+}\ (77) > \ce{Co^2+}\ (74) > \ce{Ni^2+}\ (70) \)
Nickel (Z = 28) comes last of the four in the series, so Ni2+ has the smallest ionic size.
NCERT (2026-27 edition):
Variation in Atomic and Ionic Sizes of Transition Metals – …
