If the value of C.F.S.E. for "NI" is `Delta` then it is for Pd should be

Options

A. `1.1Delta_(0)`
B. `0.5Delta_(0)`
C. `1.5Delta_(0)`
D. `2Delta_(0)`

Answer

To solve the question regarding the crystal field stabilizing energy (C.F.S.E.) for palladium (Pd) in relation to nickel (Ni), we can follow these steps:

### Step 1: Understand the C.F.S.E. for Nickel
The problem states that the C.F.S.E. for nickel is represented as `Δ`. This value is the splitting energy for the d-orbitals in the crystal field created by ligands surrounding the metal ion.

### Step 2: Identify the Position of Nickel and Palladium
Nickel (Ni) is a 3d transition metal with an atomic number of 28, while palladium (Pd) is a 4d transition metal with an atomic number of 46.

### Step 3: Understand the Trend in Crystal Field Stabilization Energy
As we move from the 3d series (like Ni) to the 4d series (like Pd), the C.F.S.E. tends to increase. The increase is typically around 50% when moving from one series to the next due to the increased effective nuclear charge and reduced shielding effect from inner electrons.

### Step 4: Calculate the C.F.S.E. for Palladium
Given that the C.F.S.E. for Ni is `Δ`, we can express the C.F.S.E. for Pd as follows:
- If the C.F.S.E. for Ni is `Δ`, then for Pd, it will be increased by 50%.
- Therefore, the C.F.S.E. for Pd can be calculated as:

\[ \text{C.F.S.E. for Pd} = \Delta + 0.5 \Delta = 1.5 \Delta \]

### Step 5: Conclusion
Thus, the value of C.F.S.E. for palladium (Pd) is `1.5Δ`.

### Final Answer
The C.F.S.E. for Pd should be `1.5Δ`.

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