For electron gain enthalpies of the elements denoted as $\Delta_{\mathrm{eg}} \mathrm{H}$, the incorrect option is :

<p>First, recall that <strong>electron-gain enthalpy</strong> $\Delta_{\mathrm{eg}}H$ is often quoted as a (negative) numerical value. A “more negative” electron-gain enthalpy means that an atom gains an electron more favorably.</p> <h2>1. Known Trends in the Periodic Table</h2> <p><strong>Halogens (Group 17):</strong> </p> <p><p>Although one might expect the electron-gain enthalpy to become steadily <em>less</em> negative as we go down the group (due to increasing size and shielding), there is an anomaly at the top: </p> <p>$ \text{Most negative: } \mathrm{Cl} > \mathrm{F} > \mathrm{Br} > \mathrm{I} > \mathrm{At} \quad(\text{in terms of magnitude of negativity}). $ </p></p> <p><p>Numerically (since these values are negative), that ordering translates to: </p> <p>$ \Delta_{\mathrm{eg}}H(\mathrm{Cl}) < \Delta_{\mathrm{eg}}H(\mathrm{F}) < \Delta_{\mathrm{eg}}H(\mathrm{Br}) < \Delta_{\mathrm{eg}}H(\mathrm{I}) < \Delta_{\mathrm{eg}}H(\mathrm{At}). $</p> <p>The smaller (more negative) number on a number line corresponds to a <em>larger</em> (more exothermic) electron-gain enthalpy.</p></p> <p><strong>Chalcogens (Group 16):</strong> </p> <p><p>A similar pattern holds, with an anomaly near oxygen. Going down from S to Se to Te to Po, the electron-gain enthalpy becomes <em>less</em> negative: </p> <p>$ \text{Most negative: } \mathrm{S} > \mathrm{Se} > \mathrm{Te} > \mathrm{Po} \quad(\text{in magnitude}). $ </p></p> <p><p>Numerically (all negative values), that means:</p> <p>$ \Delta_{\mathrm{eg}}H(\mathrm{S}) < \Delta_{\mathrm{eg}}H(\mathrm{Se}) < \Delta_{\mathrm{eg}}H(\mathrm{Te}) < \Delta_{\mathrm{eg}}H(\mathrm{Po}). $</p></p> <h2>2. Checking Each Option</h2> <p>We interpret each statement as a comparison of numerical values (remembering they are negative).</p> <h3>(A) $\,\Delta_{\mathrm{eg}}H(\mathrm{I}) < \Delta_{\mathrm{eg}}H(\mathrm{At})$</h3> <p><p>Iodine’s electron-gain enthalpy is <em>more negative</em> than astatine’s. </p></p> <p><p>Numerically, $\Delta_{\mathrm{eg}}H(\mathrm{I})$ is indeed a “smaller” (i.e. more negative) number than $\Delta_{\mathrm{eg}}H(\mathrm{At})$. </p></p> <p><p>So this statement is <strong>correct</strong>.</p></p> <h3>(B) $\,\Delta_{\mathrm{eg}}H(\mathrm{Te}) < \Delta_{\mathrm{eg}}H(\mathrm{Po})$</h3> <p><p>Tellurium (Te) is above polonium (Po) in Group 16, so Te typically has a <em>more negative</em> electron-gain enthalpy than Po. </p></p> <p><p>Numerically, that means $\Delta_{\mathrm{eg}}H(\mathrm{Te})$ is smaller (more negative) than $\Delta_{\mathrm{eg}}H(\mathrm{Po})$. </p></p> <p><p>This statement is <strong>correct</strong>.</p></p> <h3>(C) $\,\Delta_{\mathrm{eg}}H(\mathrm{Cl}) < \Delta_{\mathrm{eg}}H(\mathrm{F})$</h3> <p><p>Among halogens, $\mathrm{Cl}$ actually has the most negative $\Delta_{\mathrm{eg}}H$. Fluorine’s value, while also negative, is a bit less so (because of small-orbital repulsion in F). </p></p> <p><p>Numerically, that means $\Delta_{\mathrm{eg}}H(\mathrm{Cl})$ is indeed a smaller (more negative) number than $\Delta_{\mathrm{eg}}H(\mathrm{F})$. </p></p> <p><p>So this statement is <strong>correct</strong>.</p></p> <h3>(D) $\,\Delta_{\mathrm{eg}}H(\mathrm{Se}) < \Delta_{\mathrm{eg}}H(\mathrm{S})$</h3> <p><p>In Group 16, sulfur (S) has a <em>more negative</em> electron-gain enthalpy than selenium (Se). </p></p> <p><p>Numerically, that implies</p> <p>$ \Delta_{\mathrm{eg}}H(\mathrm{S}) < \Delta_{\mathrm{eg}}H(\mathrm{Se}), $</p> <p>not the other way around. </p></p> <p><p>The statement says $\Delta_{\mathrm{eg}}H(\mathrm{Se}) < \Delta_{\mathrm{eg}}H(\mathrm{S})$, which would mean Se is <em>more</em> negative than S—but that is <strong>not</strong> correct.</p></p> <p>Hence, <strong>(D) is the incorrect statement</strong>.</p>