Which of the following statement(s) is/are true?

Options

A. In ferrocyanide ion, the effective atomic number is 36.
B. Chelating ligands are atleast bidentate ligand
C. `[CrCl_(2)(CN)_(2)(NH_(3))_(2)]^(Theta) and [CrCl_(3)(NH_(3))_(3)]` both have `d^(2)sp^(3)` hybridization
D. As the number of rings in complex increases, stability of complex (chelate) also increases

Answer

To determine which statements are true regarding coordination compounds, we will analyze each statement step by step.

### Step 1: Analyze the Effective Atomic Number (EAN) of Ferrocyanate Ion
- The ferrocyanate ion is represented as [Fe(CN)6]^(4-).
- To find the EAN, we use the formula:
\[ \text{EAN} = Z - O + (n \times 2) \]
where \(Z\) is the atomic number of the central atom (Fe), \(O\) is the oxidation state of the central atom, and \(n\) is the number of ligands.

- **Calculation:**
- Atomic number of Fe (Z) = 26
- Oxidation state of Fe in [Fe(CN)6]^(4-) = +2 (since CN is -1 and there are 6 CN ligands: \(6 \times -1 = -6\); thus \(x - 6 = -4 \Rightarrow x = +2\))
- Contribution from ligands: Each CN donates 2 electrons, and there are 6 CN ligands: \(6 \times 2 = 12\)

Putting it all together:
\[ \text{EAN} = 26 - 2 + 12 = 36 \]
- **Conclusion:** The first statement is true.

### Step 2: Evaluate the Statement on Chelating Ligands
- Chelating ligands are defined as ligands that can attach to a central metal atom at multiple bonding sites.
- Bidentate ligands have two donor atoms, which allows them to form a stable ring structure with the metal ion.

- **Conclusion:** The second statement is true.

### Step 3: Determine the Hybridization of the Given Complexes
1. **First Complex: [CrCl2(CN)2(NH3)2]**
- Calculate the oxidation state of Cr:
- Cl is -1 (2 Cl = -2), CN is -1 (2 CN = -2), NH3 is neutral (0).
- Total charge = -2 - 2 + 0 = -4. If we let the oxidation state of Cr be \(x\), we have:
\[ x - 4 = 0 \Rightarrow x = +3 \]
- For Cr in +3 state, the electron configuration is:
- Ar (3d^3)
- Hybridization: In an octahedral complex, Cr can use 3d and 4s orbitals to form d^2sp^3 hybridization.

2. **Second Complex: [CrCl3(NH3)3]**
- Calculate the oxidation state of Cr:
- Cl is -1 (3 Cl = -3), NH3 is neutral (0).
- Total charge = -3 + 0 = -3. If we let the oxidation state of Cr be \(x\), we have:
\[ x - 3 = 0 \Rightarrow x = +3 \]
- For Cr in +3 state, the electron configuration remains the same (Ar, 3d^3).
- Hybridization: Similar to the first complex, it will also be d^2sp^3.

- **Conclusion:** The third statement is true.

### Step 4: Analyze the Stability of Chelate Complexes
- The statement claims that as the number of rings in a chelate increases, the stability of the complex also increases.
- This is a known principle in coordination chemistry known as the "chelate effect," where chelating ligands form more stable complexes due to the formation of multiple bonds with the metal ion.

- **Conclusion:** The fourth statement is true.

### Final Conclusion
All four statements provided in the question are true.

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