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Arrange the following in the decreasing order of radius.
$$
\mathrm{S}^{2-}, \mathrm{P}^{3-}, \mathrm{Cl}^{-}, \mathrm{Ca}^{2+}, \mathrm{Ar}^{+} \mathrm{K}^{+}
$$
Options:
$$
\mathrm{S}^{2-}, \mathrm{P}^{3-}, \mathrm{Cl}^{-}, \mathrm{Ca}^{2+}, \mathrm{Ar}^{+} \mathrm{K}^{+}
$$
Solution:
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Verified Answer
The correct answer is:
$\mathrm{P}^3>\mathrm{S}^2>\mathrm{Cl}>\mathrm{Ar}>\mathrm{K}^{+}>\mathrm{Ca}^{2+}$
Among isoelectronic species, when negative charge present at atomic species than radii of species increase in comparison of parent atom. On the other hand positive charge at atomic species decreases the atomic radii. In other words, more be the value of negative charge and less the value of positive charge, more is its size and vice-versa. Hence, atomic radii
$$
\mathrm{P}^{3-}>\mathrm{S}^{2-}>\mathrm{CI}^{-}>\mathrm{Ar}>\mathrm{K}^{+}>\mathrm{Ca}^{2+}
$$
$$
\mathrm{P}^{3-}>\mathrm{S}^{2-}>\mathrm{CI}^{-}>\mathrm{Ar}>\mathrm{K}^{+}>\mathrm{Ca}^{2+}
$$
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