The ionic sizes decrease in the order :-

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Question

(A) The species $S^{2-}, K^{+}, Sc^{3+}, V^{5+},$ and $Mn^{7+}$ are isoelectronic,as they all contain $18$ electrons.For isoelectronic species,the ion

Vedclass · Accepted Answer

$S^{2-} > K^{+} > Sc^{3+} > V^{5+} > Mn^{7+}$

Vedclass · Answer

(A) The species $S^{2-}, K^{+}, Sc^{3+}, V^{5+},$ and $Mn^{7+}$ are isoelectronic,as they all contain $18$ electrons.For isoelectronic species,the ionic size is inversely proportional to the effective nuclear charge $(Z_{eff})$.As the atomic number $(Z)$ increases,the effective nuclear charge increases,which leads to a decrease in ionic size.Since the atomic numbers are $S(16), K(19), Sc(21), V(23),$ and $Mn(25)$,the effective nuclear charge increases in the order $S^{2-} Therefore,the ionic size decreases in the order: $S^{2-} > K^{+} > Sc^{3+} > V^{5+} > Mn^{7+}$.

Similar Questions

The size of isoelectronic species $F^{-},$ $Ne$ and $Na^{+}$ is affected by
$(a)$ Nuclear charge $(Z)$
$(b)$ Valence principal quantum number $(n)$
$(c)$ Electron-electron interaction in the outer orbitals
$(d)$ None of the factors because their size is the same.

If the ionic radii of $K^{+}$ and $F^{-}$ are nearly the same (i.e.,$1.34 \ \mathring{A}$),then the atomic radii of $K$ and $F$ respectively are

Which has the highest value of ionic radius?

Which of the following is arranged in decreasing order of size?

The correct order of increasing ionic radii is .... .

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