The decreasing order of size of the isoelectr | vedclass

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Question

(D) For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number,$Z$) increases because the effective nuclear attraction

Vedclass · Accepted Answer

$S^{2-} > Cl^{-} > K^{+} > Ca^{2+}$

Vedclass · Answer

(D) For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number,$Z$) increases because the effective nuclear attraction on the electrons increases.The atomic numbers are: $S (Z=16)$,$Cl (Z=17)$,$K (Z=19)$,and $Ca (Z=20)$.Since all these ions have the same number of electrons ($18$ electrons),the size follows the order of increasing atomic number: $S^{2-} (Z=16) > Cl^{-} (Z=17) > K^{+} (Z=19) > Ca^{2+} (Z=20)$.Therefore,the correct decreasing order is $S^{2-} > Cl^{-} > K^{+} > Ca^{2+}$.

The decreasing order of size of the isoelectronic series $K^{+}$,$Ca^{2+}$,$Cl^{-}$,and $S^{2-}$ is:

Similar Questions

The ion with the smallest radius among the following is:

Between $Mg$ and $Mg^{2+}$,which has a smaller size? Why?

Consider the following species: $N^{3-}, O^{2-}, F^{-}, Na^{+}, Mg^{2+}$ and $Al^{3+}$
$(a)$ What is common in them?
$(b)$ Arrange them in the order of increasing ionic radii.

$Na^{+}$,$Mg^{2+}$,$Al^{3+}$,and $Si^{4+}$ are isoelectronic. Their ionic size follows the order:

The correct order of atomic radii of $B, Be, N$ and $C$ is

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