A, B, C, D and E are elements with atomic num | vedclass

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(A) The elements and their atomic numbers are: $A(13)$,$B(11)$,$C(9)$,$D(7)$,$E(16)$.To achieve the nearest inert gas configuration,these elements for

Vedclass · Accepted Answer

$D, A$

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(A) The elements and their atomic numbers are: $A(13)$,$B(11)$,$C(9)$,$D(7)$,$E(16)$.To achieve the nearest inert gas configuration,these elements form the following ions:$A^{3+}$ ($10$ electrons),$B^+$ ($10$ electrons),$C^-$ ($10$ electrons),$D^{3-}$ ($10$ electrons),$E^{2-}$ ($18$ electrons).For isoelectronic ions $(A^{3+}, B^+, C^-, D^{3-})$,the ionic size increases as the nuclear charge $(Z)$ decreases.$Z$ values are: $A=13, B=11, C=9, D=7$.Since $D$ has the lowest $Z$ $(7)$,$D^{3-}$ has the largest size among the isoelectronic series.Comparing $D^{3-}$ $(Z=7)$ and $E^{2-}$ $(Z=16)$: $D^{3-}$ has $10$ electrons in the $n=2$ shell,while $E^{2-}$ has $18$ electrons in the $n=3$ shell. Thus,$D^{3-}$ is larger than $E^{2-}$.Therefore,$X = D$.For the smallest size,we compare the ions. $A^{3+}$ has the highest nuclear charge $(Z=13)$ among the $10$-electron species,making it the smallest among them. $E^{2-}$ $(Z=16)$ has more shells $(n=3)$,making it larger than $A^{3+}$.Therefore,$Y = A$.Thus,$X$ and $Y$ are $D$ and $A$ respectively.

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