In the complex ion `[Cu(CN_(4))]^(3-)` the hybridization state, oxidation state and number of unpaired electrons of copper are respectively

Options

A. `dsp^(2)`, +1,1
B. `sp^(3), +1`, zero
C. `sp^(3), +2,1`
D. `dsp^(3),+2,` zero

Explanation

To solve the problem regarding the complex ion \([Cu(CN_4)]^{3-}\), we need to determine the oxidation state of copper, its hybridization state, and the number of unpaired electrons. Here’s a step-by-step breakdown:

### Step 1: Determine the Oxidation State of Copper
1. Let the oxidation state of copper be \(x\).
2. The cyanide ion \((CN^−)\) has a charge of \(-1\) and there are 4 cyanide ions in the complex.
3. The overall charge of the complex ion is \(-3\).
4. The equation can be set up as follows:
\[ x + 4(-1) = -3 \]
Simplifying this gives:
\[ x - 4 = -3 \implies x = +1 \]
5. Therefore, the oxidation state of copper in this complex is \(+1\).

### Step 2: Determine the Electron Configuration of Copper
1. The electron configuration of neutral copper (Cu) is:
\[ [Ar] 3d^{10} 4s^1 \]
2. When copper is in the \(+1\) oxidation state, it loses one electron from the \(4s\) orbital:
\[ [Ar] 3d^{10} \]
3. Thus, in the \(+1\) oxidation state, copper has a filled \(3d\) subshell with 10 electrons.

### Step 3: Determine the Hybridization State
1. The cyanide ion \((CN^−)\) is a strong field ligand and is a monodentate ligand, meaning it can donate one pair of electrons.
2. In this complex, there are 4 cyanide ligands, which means copper will use 4 orbitals to bond with these ligands.
3. The hybridization can be determined by the number of ligands:
- With 4 ligands, the hybridization is \(sp^3\).
4. Therefore, the hybridization state of copper in this complex is \(sp^3\).

### Step 4: Determine the Number of Unpaired Electrons
1. The \(3d\) subshell of copper in the \(+1\) oxidation state is \(3d^{10}\).
2. Since all 10 electrons in the \(3d\) subshell are paired, there are no unpaired electrons.
3. Thus, the number of unpaired electrons is \(0\).

### Final Answer
- Oxidation State: \(+1\)
- Hybridization State: \(sp^3\)
- Number of Unpaired Electrons: \(0\)

### Summary
The hybridization state, oxidation state, and number of unpaired electrons of copper in the complex ion \([Cu(CN_4)]^{3-}\) are respectively \(sp^3\), \(+1\), and \(0\).

---