`K_(2)[Ni(CN)_(4)] underset(NH_(3))overset("K in liq.")to"X'`
Regarding this reaction correct statement is/are:

Options

A. X' is `K_(4)[Ni(CN)_(4)]`
B. The oxidation state of Ni changed +2 to zero
C. The structure of 'X' is tetrahedral
D. `[Ni(CN)_(4)]^(2-)` is square planar complex

Answer

To solve the question regarding the reaction of \( K_2[Ni(CN)_4] \) in the presence of liquid ammonia and potassium ions, we will follow these steps:

### Step 1: Identify the Reactants
The reactant is \( K_2[Ni(CN)_4] \). This is a coordination compound where nickel (Ni) is coordinated to four cyanide (CN) ligands. The oxidation state of nickel in this complex is +2.

### Step 2: Determine the Reaction Conditions
The reaction occurs in the presence of liquid ammonia (NH₃) and potassium ions (K⁺). The presence of NH₃, a strong field ligand, can lead to the formation of a new complex.

### Step 3: Predict the Product
When \( K_2[Ni(CN)_4] \) reacts with NH₃, it forms a new complex, which we denote as \( X \). The resulting complex is likely \( K_4[Ni(CN)_4] \) due to the addition of NH₃ ligands.

### Step 4: Calculate the Oxidation State of Nickel in the Product
In the new complex \( K_4[Ni(CN)_4] \), we need to determine the oxidation state of nickel:
- The overall charge of the complex is neutral.
- The cyanide ligands (CN⁻) contribute a total charge of -4 (since there are four CN⁻ ligands).
- The potassium ions (K⁺) contribute a total charge of +4 (since there are four K⁺ ions).

Thus, the oxidation state of nickel (Ni) can be calculated as follows:
\[ \text{Oxidation state of Ni} + 4(-1) + 4(+1) = 0 \]
\[ \text{Oxidation state of Ni} - 4 + 4 = 0 \]
\[ \text{Oxidation state of Ni} = 0 \]

### Step 5: Determine the Geometry of the New Complex
The geometry of the new complex \( K_4[Ni(CN)_4] \) can be predicted based on the hybridization of the nickel ion:
- Nickel in the +0 oxidation state has the electronic configuration of \( [Ar] 4s^2 3d^8 \).
- The strong field nature of CN⁻ causes pairing of electrons, leading to the hybridization of \( sp^3 \).
- Therefore, the geometry of the complex is tetrahedral.

### Step 6: Verify the Geometry of the Initial Complex
For the initial complex \( K_2[Ni(CN)_4] \):
- The oxidation state of nickel is +2.
- The electronic configuration for Ni²⁺ is \( [Ar] 3d^8 \).
- The hybridization is \( dsp^2 \), which corresponds to a square planar geometry.

### Conclusion
Based on the analysis:
1. The oxidation state of nickel changes from +2 in \( K_2[Ni(CN)_4] \) to 0 in \( K_4[Ni(CN)_4] \).
2. The geometry of the new complex \( K_4[Ni(CN)_4] \) is tetrahedral.
3. The initial complex \( K_2[Ni(CN)_4] \) has a square planar geometry.

### Final Statements
- The oxidation state of nickel changes from +2 to 0.
- The geometry of the product \( X \) is tetrahedral.
- The geometry of the initial complex is square planar.