Triad-I [N^{3-}, O^{2-}, Na^{+}]Triad-II [N^{ | vedclass

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(A) For Triad-$I$ $[N^{3-}, O^{2-}, Na^{+}]$: These are isoelectronic species with $10$ electrons. The $IP$ increases as the nuclear charge $(Z)$ incr

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$N^{3-}, O^{+}$

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(A) For Triad-$I$ $[N^{3-}, O^{2-}, Na^{+}]$: These are isoelectronic species with $10$ electrons. The $IP$ increases as the nuclear charge $(Z)$ increases. The $Z$ values are $N=7, O=8, Na=11$. Thus,the $IP$ order is $N^{3-} For Triad-$II$ $[N^{+}, C^{+}, O^{+}]$: These are ions of elements in the same period. $IP$ generally increases from left to right across a period. The order of $Z$ is $C(6) < N(7) < O(8)$. Thus,the $IP$ order is $C^{+} < N^{+} < O^{+}$. The species with the highest $IP$ is $O^{+}$.

Triad-$I$ $[N^{3-}, O^{2-}, Na^{+}]$
Triad-$II$ $[N^{+}, C^{+}, O^{+}]$
Choose the species of lowest $IP$ from triad-$I$ and highest $IP$ from triad-$II$ respectively.

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Triad-$I$ $[N^{3-}, O^{2-}, Na^{+}]$Triad-$II$ $[N^{+}, C^{+}, O^{+}]$Choose the species of lowest $IP$ from triad-$I$ and highest $IP$ from triad-$II$ respectively.