Consider the following statements: Statement I: $H_2Se$ is more acidic than $H_2Fe$ Statement II: $H_2Se$ has higher bond dissociation enthalpy In light of the above statements, choose the correct option.
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When comparing the acidities of compounds, consider the size of the central atom and the bond dissociation energies. Larger atoms tend to form weaker bonds with hydrogen, increasing acidity.
Evaluate each statement individually: Statement I: \( H_2Se \) exhibits greater acidity than \( H_2Fe \): The acidity of \( H_2Se \) surpasses that of \( H_2Fe \) due to selenium's position in the periodic table, located in the same group as oxygen but below sulfur. Acidity increases as bond strength between hydrogen and the element (e.g., Se-H) diminishes down a group, facilitating proton (H\(^+\)) release. Conversely, \( H_2Fe \) features a significantly stronger bond between iron and hydrogen, resulting in lower acidity. Consequently, Statement I is accurate. Statement II: \( H_2Se \) possesses a higher bond dissociation enthalpy: Bond dissociation enthalpy is directly proportional to bond strength. Given that the Se-H bonds in \( H_2Se \) are less robust than the Fe-H bonds in \( H_2Fe \), \( H_2Se \) exhibits a lower bond dissociation enthalpy. Therefore, Statement II is inaccurate. Accordingly, the correct conclusion is (1) Statement-I is true & Statement-II is false.
