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(A) The ionic radii of species depend on the number of shells and the effective nuclear charge.For species in the same period,the ionic radius decreas

Vedclass · Accepted Answer

$O^{2-} < N^{3-} < S^{2-} < P^{3-}$

Vedclass · Answer

(A) The ionic radii of species depend on the number of shells and the effective nuclear charge.For species in the same period,the ionic radius decreases as the atomic number increases (e.g.,$N^{3-} > O^{2-}$).For species in the same group,the ionic radius increases as we move down the group due to the addition of a new shell (e.g.,$P^{3-} > N^{3-}$ and $S^{2-} > O^{2-}$).Comparing the given species:$1$. $O^{2-}$ and $N^{3-}$ are in the second period,so $N^{3-} > O^{2-}$.$2$. $S^{2-}$ and $P^{3-}$ are in the third period,so $P^{3-} > S^{2-}$.$3$. Since $S^{2-}$ and $P^{3-}$ have more shells than $O^{2-}$ and $N^{3-}$,they are larger.Combining these,the order of increasing ionic radii is $O^{2-} < N^{3-} < S^{2-} < P^{3-}$.

Which of the following arrangements represents the increasing order (smallest to largest) of ionic radii of the given species $O^{2-}, S^{2-}, N^{3-}, P^{3-}$?

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