In which of the following , the central atom does NOT exhibit an oxidation state of `+2` ?

Options

A. `K_2[Ni(CN)_4]`
B. `K_4[Fe(CN)_6]`
C. `[Fe(C_2O_4)_(3)]^(3-)`
D. `[Cu(NH_3)_4]^(2+)`

Answer

To solve the question regarding which central atom does NOT exhibit an oxidation state of +2, we will analyze each option step by step.

### Step 1: Analyze the first option - K2[Ni(CN)4]
1. **Identify the charge on the coordination sphere**:
- There are 2 potassium ions (K⁺), each with a +1 charge. Therefore, the total positive charge from potassium is +2.
- To balance this, the charge on the coordination sphere [Ni(CN)₄] must be -2.

2. **Calculate the oxidation state of nickel (Ni)**:
- Let the oxidation state of Ni be X.
- The cyanide ion (CN⁻) has a -1 charge, and there are 4 cyanide ions, contributing a total of -4.
- The equation is: X + 4(-1) = -2
- Simplifying gives: X - 4 = -2 → X = +2.

### Step 2: Analyze the second option - K4[Fe(CN)6]
1. **Identify the charge on the coordination sphere**:
- There are 4 potassium ions (K⁺), contributing a total of +4 charge.
- Therefore, the charge on the coordination sphere [Fe(CN)₆] must be -4.

2. **Calculate the oxidation state of iron (Fe)**:
- Let the oxidation state of Fe be Y.
- The cyanide ion (CN⁻) has a -1 charge, and there are 6 cyanide ions, contributing a total of -6.
- The equation is: Y + 6(-1) = -4
- Simplifying gives: Y - 6 = -4 → Y = +2.

### Step 3: Analyze the third option - [Fe(C2O4)3]³⁻
1. **Identify the charge on the coordination sphere**:
- The oxalate ion (C2O4²⁻) has a -2 charge, and there are 3 oxalate ions, contributing a total of -6.
- Therefore, the charge on the coordination sphere is -3.

2. **Calculate the oxidation state of iron (Fe)**:
- Let the oxidation state of Fe be Z.
- The equation is: Z + 3(-2) = -3
- Simplifying gives: Z - 6 = -3 → Z = +3.

### Step 4: Analyze the fourth option - [Cu(CN)4]²⁻
1. **Identify the charge on the coordination sphere**:
- The coordination sphere has a charge of -2.

2. **Calculate the oxidation state of copper (Cu)**:
- Let the oxidation state of Cu be W.
- The cyanide ion (CN⁻) has a -1 charge, and there are 4 cyanide ions, contributing a total of -4.
- The equation is: W + 4(-1) = -2
- Simplifying gives: W - 4 = -2 → W = +2.

### Conclusion:
- From our analysis:
- K2[Ni(CN)4] → Ni is +2
- K4[Fe(CN)6] → Fe is +2
- [Fe(C2O4)3]³⁻ → Fe is +3
- [Cu(CN)4]²⁻ → Cu is +2

The central atom that does NOT exhibit an oxidation state of +2 is **[Fe(C2O4)3]³⁻**, where iron has an oxidation state of +3.

### Final Answer:
**[Fe(C2O4)3]³⁻**