If CFSE increases by 30% and 40% respectively for `Co^(3+) " to " Rh^(3+) " to " Ir^(3+)`, then the total increase in CFSE for `Ir^(3+)` with respect to `Co^(3+)` is ……..

Answer

To solve the problem of calculating the total increase in Crystal Field Stabilization Energy (CFSE) for \( \text{Ir}^{3+} \) with respect to \( \text{Co}^{3+} \), we can follow these steps:

### Step 1: Define CFSE for \( \text{Co}^{3+} \)
Let the CFSE for \( \text{Co}^{3+} \) be represented as \( C_{Co} \).

### Step 2: Calculate CFSE for \( \text{Rh}^{3+} \)
According to the problem, the CFSE increases by 30% when moving from \( \text{Co}^{3+} \) to \( \text{Rh}^{3+} \). Therefore, we can express the CFSE for \( \text{Rh}^{3+} \) as:
\[ C_{Rh} = C_{Co} + 0.3 C_{Co} = 1.3 C_{Co} \]

### Step 3: Calculate CFSE for \( \text{Ir}^{3+} \)
Next, we know that the CFSE increases by 40% when moving from \( \text{Rh}^{3+} \) to \( \text{Ir}^{3+} \). Thus, we can express the CFSE for \( \text{Ir}^{3+} \) as:
\[ C_{Ir} = C_{Rh} + 0.4 C_{Rh} = 1.4 C_{Rh} \]
Substituting \( C_{Rh} \) from Step 2 into this equation, we get:
\[ C_{Ir} = 1.4 (1.3 C_{Co}) = 1.82 C_{Co} \]

### Step 4: Calculate the total increase in CFSE from \( \text{Co}^{3+} \) to \( \text{Ir}^{3+} \)
To find the total increase in CFSE from \( \text{Co}^{3+} \) to \( \text{Ir}^{3+} \):
\[ \text{Increase} = C_{Ir} - C_{Co} = 1.82 C_{Co} - C_{Co} = 0.82 C_{Co} \]
This means the total increase in CFSE is:
\[ \text{Total Increase} = 0.82 C_{Co} \]

### Step 5: Express the total increase as a percentage
To express this increase as a percentage of \( C_{Co} \):
\[ \text{Percentage Increase} = \left( \frac{0.82 C_{Co}}{C_{Co}} \right) \times 100\% = 82\% \]

### Final Answer
The total increase in CFSE for \( \text{Ir}^{3+} \) with respect to \( \text{Co}^{3+} \) is **82%**.

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