Write the increasing order of the value is CFSE `(Delta_(0))` for the following species
I.`[Co(NH_(3))_(6)]^(3+)`
II.`[Rh(NH_(3))_(6)]^(3+)`
III.`[Ir(NH_(3))_(6)]^(3+)`

Options

A. IIIltIIltI
B. IltIIltIII
C. IIltIltIII
D. IltIIIltII

Answer

To determine the increasing order of the Crystal Field Stabilization Energy (CFSE) for the given coordination compounds, we will analyze the electronic configurations of the metal ions involved and their corresponding CFSE values.

### Step-by-Step Solution:

**Step 1: Identify the metal ions and their electronic configurations.**

- **Cobalt in [Co(NH₃)₆]³⁺:**
- Cobalt (Co) has the atomic number 27. The electronic configuration is [Ar] 3d⁷ 4s².
- In the +3 oxidation state, Co loses two 4s electrons and one 3d electron, resulting in the configuration: 3d⁶.

- **Rhodium in [Rh(NH₃)₆]³⁺:**
- Rhodium (Rh) has the atomic number 45. The electronic configuration is [Kr] 4d⁸ 5s¹.
- In the +3 oxidation state, Rh loses one 5s and one 4d electron, resulting in the configuration: 4d⁷.

- **Iridium in [Ir(NH₃)₆]³⁺:**
- Iridium (Ir) has the atomic number 77. The electronic configuration is [Xe] 4f¹⁴ 5d⁷ 6s².
- In the +3 oxidation state, Ir loses two 6s and one 5d electron, resulting in the configuration: 5d⁶.

**Step 2: Determine the CFSE for each complex.**

The CFSE can be calculated using the formula:
\[ \text{CFSE} = (n_t \times \Delta) - (n_e \times \frac{1}{2} \Delta) \]
where:
- \( n_t \) = number of electrons in the t₂g orbitals,
- \( n_e \) = number of electrons in the e_g orbitals,
- \( \Delta \) = crystal field splitting energy.

**For the complexes:**
- **[Co(NH₃)₆]³⁺ (3d⁶):**
- In an octahedral field, the configuration is t₂g⁶ e_g⁰.
- CFSE = \( 6 \times \frac{-2}{5} \Delta = -\frac{12}{5} \Delta \).

- **[Rh(NH₃)₆]³⁺ (4d⁷):**
- In an octahedral field, the configuration is t₂g⁶ e_g¹.
- CFSE = \( 6 \times \frac{-2}{5} \Delta + 1 \times \frac{3}{5} \Delta = -\frac{12}{5} \Delta + \frac{3}{5} \Delta = -\frac{9}{5} \Delta \).

- **[Ir(NH₃)₆]³⁺ (5d⁶):**
- In an octahedral field, the configuration is t₂g⁶ e_g⁰.
- CFSE = \( 6 \times \frac{-2}{5} \Delta = -\frac{12}{5} \Delta \).

**Step 3: Compare the CFSE values.**

- For [Co(NH₃)₆]³⁺: CFSE = -12/5 Δ
- For [Rh(NH₃)₆]³⁺: CFSE = -9/5 Δ
- For [Ir(NH₃)₆]³⁺: CFSE = -12/5 Δ

**Step 4: Establish the increasing order of CFSE.**

Since the CFSE values for [Co(NH₃)₆]³⁺ and [Ir(NH₃)₆]³⁺ are the same, and [Rh(NH₃)₆]³⁺ has a higher CFSE (less negative), the increasing order of CFSE is:
\[ [Co(NH₃)₆]^{3+} \approx [Ir(NH₃)₆]^{3+} < [Rh(NH₃)₆]^{3+} \]

### Final Answer:
The increasing order of CFSE is:
\[ [Co(NH₃)₆]^{3+} \approx [Ir(NH₃)₆]^{3+} < [Rh(NH₃)₆]^{3+} \]