Match the entries of column I with appropriate entries of column II and choose the correct option out of the four option (a), (b), (c ), (d) given at the end of each question.
`{:(,"Column I",,"Column II"),(,"(Metal ion configuration in strong ligand field)",,("CFSE, "Delta_(0)" value")),(("A"),d^(4),(p),-1.6),(("B"),d^(5),(q),-1.8),(("C"),d^(6),(r),-2.0),(("D"),d^(7),(s),-2.4):}`

Options

A. A-p, B-s, C-r, D-q
B. A-q, B-r, C-p, D-s
C. A-p, B-r, C-s, D-q
D. A-r, B-s, C-p, D-q

Answer

To solve the problem of matching the entries of Column I with appropriate entries of Column II, we need to calculate the Crystal Field Stabilization Energy (CFSE) and the corresponding Δ₀ values for each of the given d-electron configurations in a strong ligand field.

### Step-by-Step Solution:

1. **Understanding CFSE**:
- In an octahedral field, the d-orbitals split into two sets: the lower-energy t2g (three orbitals) and the higher-energy eg (two orbitals).
- The CFSE can be calculated using the formula:
\[ \text{CFSE} = \left( \text{Number of electrons in } t_{2g} \times \left(-\frac{2}{5} \Delta_0\right) \right) + \left( \text{Number of electrons in } e_g \times \left(+\frac{3}{5} \Delta_0\right) \right) \]

2. **Case A: d⁴ Configuration**
- In a strong field, d⁴ will have all four electrons paired in t2g.
- CFSE = \(4 \times \left(-\frac{2}{5} \Delta_0\right) + 0 = -\frac{8}{5} \Delta_0\)
- This gives us a CFSE of \(-1.6 \Delta_0\).
- **Match**: A → p

3. **Case B: d⁵ Configuration**
- In a strong field, d⁵ will have all five electrons in t2g.
- CFSE = \(5 \times \left(-\frac{2}{5} \Delta_0\right) + 0 = -2 \Delta_0\)
- **Match**: B → r

4. **Case C: d⁶ Configuration**
- In a strong field, d⁶ will have six electrons with four in t2g and two in eg.
- CFSE = \(6 \times \left(-\frac{2}{5} \Delta_0\right) + 0 = -\frac{12}{5} \Delta_0\)
- This gives us a CFSE of \(-2.4 \Delta_0\).
- **Match**: C → s

5. **Case D: d⁷ Configuration**
- In a strong field, d⁷ will have six electrons in t2g and one in eg.
- CFSE = \(6 \times \left(-\frac{2}{5} \Delta_0\right) + 1 \times \left(+\frac{3}{5} \Delta_0\right)\)
- CFSE = \(-\frac{12}{5} \Delta_0 + \frac{3}{5} \Delta_0 = -\frac{9}{5} \Delta_0\)
- **Match**: D → q

### Final Matching:
- A → p
- B → r
- C → s
- D → q

### Answer Options:
Now we can choose the correct option based on the matches:
- (a) A → p, B → r, C → s, D → q