Question:medium

The IUPAC name of the complex ion formed when gold dissolves in aqua regia is:

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In IUPAC naming of coordination compounds, the ligands are named first using prefixes like "tetra-" for four ligands, and the central metal is named with its oxidation state in Roman numerals. For example, \( [\text{AuCl}_4]^- \) becomes Tetrachloridoaurate (III) because gold has an oxidation state of +3.
Updated On: Jan 13, 2026
  • \( \text{Tetrachloroaurate (I)} \)
  • \( \text{Dichloridoaurate (III)} \)
  • \( \text{Tetrachloridoaurate (III)} \)
  • \( \text{Tetrachloroaurate (II)} \)
Show Solution

The Correct Option is C

Solution and Explanation

Gold (Au) reacts in aqua regia, a mixture of concentrated nitric and hydrochloric acids, to form a complex ion. The reaction is: \[ \text{Au} + 4 \text{HCl} + \text{HNO}_3 \rightarrow \text{H[AuCl}_4] + 2 \text{H}_2\text{O} + \text{NO}. \] The resulting complex ion is \( [\text{AuCl}_4]^- \), characterized by chlorido ions (Cl-) coordinated to the central gold ion (Au). This is a typical coordination compound.
Step 1: Determine Gold's Oxidation State. To find the oxidation state of gold (Au) in \( [\text{AuCl}_4]^- \), follow these steps: Assign \( x \) as the oxidation state of gold. Chlorine atoms in chloride ligands have an oxidation state of \( -1 \). For the complex ion: \[ x + 4(-1) = -1. \] Solving the equation: \[ x - 4 = -1 \quad \Rightarrow \quad x = +3. \] Therefore, gold in this complex has an oxidation state of \( +3 \).
Step 2: Establish the IUPAC Name. The complex ion formed is \( [\text{AuCl}_4]^- \). Adhering to IUPAC nomenclature rules: - "tetra-" indicates four chloride (Cl) ligands. - The central metal, gold (Au), is referred to as "aurate". - The metal's oxidation state, \( +3 \), is denoted by Roman numerals as (III). Consequently, the IUPAC name for the complex ion \( [\text{AuCl}_4]^- \) is Tetrachloridoaurate (III).
Conclusion: The correct IUPAC name for the complex ion generated when gold dissolves in aqua regia is \( \mathbf{Tetrachloridoaurate (III)} \). Thus, the correct option is \( \mathbf{(C)} \).
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