Search any question & find its solution
Question:
Answered & Verified by Expert
Which one of the following arrangements represents the correct order of electron gain enthalpy (with negative sign) of the given atomic species?
Options:
Solution:
2056 Upvotes
Verified Answer
The correct answer is:
$\mathrm{O} < \mathrm{S} < \mathrm{F} < \mathrm{Cl}$
The electron gain enthalpy of elements increases along a period towards right and decreases down the group. However, the $\mathrm{EA}$ of $\mathrm{Cl}$ is higher than that of $F$ and EA of $S$ is higher than that of $\mathrm{O}$ as $\mathrm{F}$ and $\mathrm{O}$ are very small in size. Thus, the correct order of EA is:
$$
\mathrm{O} < \mathrm{S} < \mathrm{F} < \mathrm{Cl}
$$
Caution
Oxygen molecule has a less negative electron gain enthalpy than sulphur. This is on the grounds that Oxygen, because of its compressed nature encounter more repulsion between the electrons effectively present and the approaching electron.
$$
\mathrm{O} < \mathrm{S} < \mathrm{F} < \mathrm{Cl}
$$
Caution
Oxygen molecule has a less negative electron gain enthalpy than sulphur. This is on the grounds that Oxygen, because of its compressed nature encounter more repulsion between the electrons effectively present and the approaching electron.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.