QuestionPlace the following in order of increasing radius. $\mathrm{Ca}^{2+}$ $s^{2-}$ $\mathrm{Cl}^{-}$ $\mathrm{Cl}^{-}<\mathrm{S}^{2-}<\mathrm{Ca}^{2+}$ $\mathrm{S}^{2-}<\mathrm{Cl}^{-}<\mathrm{Ca}^{2+}$ $\mathrm{Cl}^{-}<\mathrm{Ca}^{2+}<\mathrm{S}^{2-}$ $\mathrm{Ca}^{2+}<\mathrm{S}^{2-}<\mathrm{Cl}^{-}$ $\mathrm{Ca}^{2+}<\mathrm{Cl}^{-}<\mathrm{S}^{2-}$ Submit Request Answer

Studdy Solution
Order the ions by increasing radius:
- $\mathrm{Ca}^{2+}$ has the smallest radius due to the highest effective nuclear charge. - $\mathrm{Cl}^{-}$ has a larger radius than $\mathrm{Ca}^{2+}$ but smaller than $\mathrm{S}^{2-}$. - $\mathrm{S}^{2-}$ has the largest radius due to the lowest effective nuclear charge.
The order of increasing radius is:
$\mathrm{Ca}^{2+} < \mathrm{Cl}^{-} < \mathrm{S}^{2-}$
The correct order is:
$\boxed{\mathrm{Ca}^{2+} < \mathrm{Cl}^{-} < \mathrm{S}^{2-}}$