To find the value of x in the complex ion \([Ni(CN)_4]^x\), we need to consider the oxidation states of the elements involved in the complex.
The central metal atom is Nickel (Ni). In most complexes, Nickel can exhibit oxidation states of +2 or +3. However, in typical cyanide complexes like \([Ni(CN)_4]^{2-}\), Nickel usually exhibits an oxidation state of +2.
Each cyanide ion (CN-) has a charge of -1. Since there are four cyanide ions, the total charge contributed by the cyanide ions is:
\(4 \times (-1) = -4\)
Let's assume the oxidation state of Nickel in \([Ni(CN)_4]^x\) is +2. Therefore, the expression for the charge of the entire complex ion will be:
\(+2 + (-4) = -2\)
This means the overall charge on the complex ion \([Ni(CN)_4]\) is -2, which corresponds to the given charge of x.
Thus, the oxidation state and the value of x in the \([Ni(CN)_4]^x\) complex ion is -2.
| Element/Ion | Oxidation State | Number | Total Charge Contribution |
|---|---|---|---|
| Ni | +2 | 1 | +2 |
| CN- | -1 | 4 | -4 |
| Total Charge on Complex | -2 | ||
Therefore, the correct answer is -2.
The IUPAC name for the complex \( [\text{Co}(\text{ONO})(\text{NH}_3)_5]\text{Cl}_2 \) is