Given below are two statements:

Statement I: H₂Se is more acidic than H₂Te

Statement II: H₂Se has higher bond enthalpy for dissociation than H₂Te

In the light of the above statements, choose the correct answer from the options given below:

<p>To determine the relative acidity and bond enthalpy of H₂Se and H₂Te, let's examine both:</p> <p><p><strong>Acidic Strength</strong>: </p> <p>The acidity of hydrides increases as we move down the group in the periodic table. This is because the bond strength between hydrogen and the central atom decreases, making it easier for the hydrogen ions to dissociate. Therefore, H₂Te, being further down the group than H₂Se, is more acidic:</p> <p>$ \mathrm{H}_2\mathrm{Se} < \mathrm{H}_2\mathrm{Te} $</p></p> <p><p><strong>Bond Enthalpy</strong>: </p> <p>Bond enthalpy refers to the energy required to dissociate a bond. The bond enthalpy decreases down the group due to weaker bonds forming as the atomic size increases. Hence, H₂Se has a higher bond enthalpy compared to H₂Te:</p> <p>$ \Delta_{\text{dis}} \mathrm{H}: \mathrm{H}_2\mathrm{Se} > \mathrm{H}_2\mathrm{Te} $</p> <p>$ \text{H}_2\mathrm{Se} \, [276 \, \text{KJ/mol}] > \text{H}_2\mathrm{Te} \, [238 \, \text{KJ/mol}] $</p></p> <p>Thus, H₂Se is indeed less acidic than H₂Te, but it has a higher bond enthalpy for dissociation.</p>