MCQ
The ionic sizes decreases in the order :-
- A$K^+ > S^{2-} > Sc^{3+} > V^{5+} > Mn^{7+}$
- ✓$S^{2-} > K^+ > Sc^{3+} > V^{5+} > Mn^{7+}$
- C$Mn^{7+} > V^{5+} > Sc^{3+} > K^+ > S^{2-}$
- D$Mn^{7+} > V^{5+} > Sc^{3+} > S^{2-} > K^+$
✓
Answer
Correct option: B.
$S^{2-} > K^+ > Sc^{3+} > V^{5+} > Mn^{7+}$
b
$S ^{2-}, K ^{+}, Sc ^{3+}, V ^{5+}$ and $Mn ^{7+}$ have equal no. of electrons. So, they are isoelectronic. For iso-electronic species, size is inversely proportional to the effective nuclear charge. Effective nuclear charge is directly proportional to the positive charge.
$S ^{2-}, K ^{+}, Sc ^{3+}, V ^{5+}$ and $Mn ^{7+}$ have equal no. of electrons. So, they are isoelectronic. For iso-electronic species, size is inversely proportional to the effective nuclear charge. Effective nuclear charge is directly proportional to the positive charge.
Thus more the positive charge less the size. Therefore, order of size is : $S ^{2-}\,> \,K ^{+}\,> \,Sc ^{3+}\,>\, V ^{5+}\,> \,Mn ^{7+}$
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