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(A) First,identify the elements based on their electronic configurations:$I. 1s^2 2s^2 2p^6 3s^2 3p^5$ is Chlorine $(Cl)$,which has a high electron af

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$II < IV < III < I$

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(A) First,identify the elements based on their electronic configurations:$I. 1s^2 2s^2 2p^6 3s^2 3p^5$ is Chlorine $(Cl)$,which has a high electron affinity.$II. 1s^2 2s^2 2p^3$ is Nitrogen $(N)$,which has a half-filled $p$-orbital,making it very stable and resulting in a very low (near zero) electron affinity.$III. 1s^2 2s^2 2p^5$ is Fluorine $(F)$. Although $F$ is more electronegative than $Cl$,its electron affinity is slightly lower than $Cl$ due to high inter-electronic repulsion in the small $2p$ subshell.$IV. 1s^2 2s^2 2p^6 3s^1$ is Sodium $(Na)$,which is an alkali metal with a low electron affinity,but higher than Nitrogen $(N)$ because it can form a stable $3s^2$ configuration.Comparing these: Nitrogen $(II)$ has the lowest electron affinity due to its stable half-filled configuration. Sodium $(IV)$ is next. Fluorine $(III)$ is higher than Sodium,and Chlorine $(I)$ has the highest electron affinity among these.Therefore,the increasing order is $II < IV < III < I$.

The increasing order of electron affinity for the given electronic configurations of elements is:
$I. 1s^2 2s^2 2p^6 3s^2 3p^5$
$II. 1s^2 2s^2 2p^3$
$III. 1s^2 2s^2 2p^5$
$IV. 1s^2 2s^2 2p^6 3s^1$

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The increasing order of electron affinity for the given electronic configurations of elements is:$I. 1s^2 2s^2 2p^6 3s^2 3p^5$$II. 1s^2 2s^2 2p^3$$III. 1s^2 2s^2 2p^5$$IV. 1s^2 2s^2 2p^6 3s^1$