The IUPAC name of `[Ni(CN)_(4)]^(2-)` is

Options

A. Tetracyanonickel (II) ion
B. Tetracyanonickelate (II) ion
C. Tetracyanonickel (O) ion
D. Tetracyanonickelate (O) ion

Answer

To determine the IUPAC name of the coordination compound \([Ni(CN)_{4}]^{2-}\), we can follow these steps:

### Step 1: Identify the central metal ion
The central metal ion in the complex is nickel (Ni).

### Step 2: Identify the ligands
The ligands in this complex are cyanide ions (CN). Since there are four cyanide ligands, we will use the prefix "tetra-" to indicate this.

### Step 3: Determine the charge of the complex
The overall charge of the complex ion is given as \(2-\). Since the cyanide ion (CN) has a charge of \(-1\) and there are four of them, the total negative charge contributed by the ligands is \(4 \times (-1) = -4\). To balance this and achieve an overall charge of \(-2\), the nickel must have a charge of \(+2\).

### Step 4: Name the ligands
The cyanide ligand can be referred to as "cyano" in IUPAC nomenclature. Since there are four of them, we will say "tetra-cyano".

### Step 5: Name the metal ion
Since the nickel ion is in the +2 oxidation state, we will use the term "nickelate" to indicate that it is part of a negatively charged complex ion. The oxidation state of nickel is indicated by the Roman numeral II (as it is +2).

### Step 6: Combine the names
Putting it all together, the IUPAC name of the complex \([Ni(CN)_{4}]^{2-}\) is "tetra-cyano-nickelate(II)".

### Final Answer
The IUPAC name of \([Ni(CN)_{4}]^{2-}\) is **tetra-cyano-nickelate(II)**.

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