Q4 / 160 · AP EAMCET 2021 (23 Aug Shift 1) · Chemistry

The ions $\mathrm{S}^{2-}, \mathrm{Cl}^{-}, \mathrm{K}^{+}, \mathrm{Ca}^{2+}$ are isoelectronic. Their ionic radii show

  • A
    a decrease from $\mathrm{S}^{2-}$ to $\mathrm{Cl}^{-}$and then increase from $\mathrm{K}^{+}$to $\mathrm{Ca}^{2+}$
  • B
    an increase from $\mathrm{S}^{2-}$ to $\mathrm{Cl}^{-}$and then decrease from $\mathrm{K}^{+}$to $\mathrm{Ca}^{2+}$
  • Correct answer
    C
    a significant decrease from $\mathrm{S}^{2-}$ to $\mathrm{Ca}^{2+}$
  • D
    a significant increase from $\mathrm{S}^{2-}$ to $\mathrm{Ca}^{2+}$

Solution

For isoelectronic species, the ionic radii decrease with increase in nuclear charge. So, the cation with greater + ve charge i.e. $\mathrm{Ca}^{2+}$ will have smaller radius and the anion with greater - ve charge i.e. $\mathrm{S}^{2-}$ will have a larger radius. $\because$ The ionic radii show a significant decrease from $\mathrm{S}^{2-}$ to $\mathrm{Ca}^{2+}$.

Hence, the correct answer is option: 'C'.

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