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(B) $(a) \, IP$ of $O_{(g)}$ is the energy required to remove an electron from $O_{(g)}$,while $IP$ of $O_{(g)}^{-}$ is the energy required to remove

Vedclass · Accepted Answer

$F F T F$

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(B) $(a) \, IP$ of $O_{(g)}$ is the energy required to remove an electron from $O_{(g)}$,while $IP$ of $O_{(g)}^{-}$ is the energy required to remove an electron from $O_{(g)}^{-}$. Since $O_{(g)}^{-}$ is an anion,it is easier to remove an electron from it than from a neutral $O$ atom. Thus,$IP(O) > IP(O^{-})$. Statement $(a)$ is $F$.$(b) \, IP$ of $Ne_{(g)}$ is the energy to remove an electron from neutral $Ne$. $IP$ of $Ne_{(g)}^{+}$ is the energy to remove an electron from $Ne^{+}$. Removing an electron from a positively charged ion is much harder due to higher effective nuclear charge. Thus,$IP(Ne^{+}) > IP(Ne)$. Statement $(b)$ is $F$.$(c) \, EA$ of $O_{(g)}^{+}$ is the energy released when an electron is added to $O^{+}$. Since $O^{+}$ is positive,it attracts an electron strongly. $EA$ of $O_{(g)}$ is the energy released when an electron is added to $O$. $EA(O^{+}) > EA(O)$. Statement $(c)$ is $T$.$(d) \, IP$ of $N_{(g)}$ is the energy to remove an electron from $N$. $IP$ of $N_{(g)}^{+}$ is the energy to remove an electron from $N^{+}$. As established,$IP(N^{+}) > IP(N)$. Statement $(d)$ is $F$.Therefore,the sequence is $F F T F$.

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