Question

A \(p\)-block element \(E\) and hydrogen form a binary cation \( (EH_x)^+ \), while \(EH_3\) on treatment with \(K_2HgI_4\) in alkaline medium gives a precipitate of basic mercury(II) amido-iodide. Given below are first ionisation enthalpy values (kJ mol\(^{-1}\)) for the first elements each from groups 13, 14, 15 and 16. Identify the correct first ionisation enthalpy value for element \(E\).

• A. 1402
• B. 801
• C. 1312
• D. 1086

Hint:

Ammonia-like behavior and formation of ammonium-type ions are strong indicators of group 15 elements.

Answer 1

The Correct Option is A

To solve this question, we need to identify the correct first ionization enthalpy value for the element \( E \). The problem states that a \( p \)-block element \( E \) and hydrogen form a binary cation \((EH_x)^+\), and \(EH_3\) reacts with \(K_2HgI_4\) in an alkaline medium to form a precipitate of basic mercury(II) amido-iodide. This indicates that the compound \(EH_3\) is likely an amine derivative, suggesting that the element \( E \) forms hydrides similar to ammonia \((NH_3)\).

The key to solving the question involves understanding which group in the periodic table forms such hydrides and reacts under the described conditions:

• Ammonia (\(NH_3\)) reacts with \(K_2HgI_4\) to form a similar type of compound; thus, we should look for an element in Group 15 (nitrogen group).
• The ionization enthalpy values for the first elements in Groups 13, 14, 15, and 16 are given. We need to identify which value corresponds to nitrogen-like behavior.

Now, let's evaluate the given ionization enthalpy values:

1. \(1402 \text{ kJ/mol} \rightarrow \text{This value is the ionization enthalpy for nitrogen, the first element in Group 15.}\)
2. \(801 \text{ kJ/mol} \rightarrow \text{This is the ionization enthalpy for boron from Group 13.}\)
3. \(1312 \text{ kJ/mol} \rightarrow \text{This is the ionization enthalpy for carbon from Group 14.}\)
4. \(1086 \text{ kJ/mol} \rightarrow \text{This is the ionization enthalpy for oxygen from Group 16.}\)

From the options, the ionization enthalpy value of \(1402 \text{ kJ/mol}\) is correct for nitrogen, which aligns with the properties of forming an amine-like hydride (\(NH_3\)) and reacting to form the indicated product. Thus, the correct choice for element \( E \) that fits all the conditions is from Group 15, with the first ionization enthalpy of nitrogen matching the required characteristics. Therefore, the correct answer is 1402 kJ/mol.