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 splitting energy (C.F.S.E.) for nickel (Ni) and palladium (Pd), we will follow these steps:

### Step 1: Understand the given information
We know that the C.F.S.E. for nickel (Ni) is given as Δ (delta). We need to find out what the C.F.S.E. for palladium (Pd) would be.

### Step 2: Identify the position of Ni and Pd in the periodic table
- Nickel (Ni) is a transition metal located in the 3d series.
- Palladium (Pd) is also a transition metal but is located in the 4d series.

### Step 3: Understand the trend in crystal field splitting energy
As we move down a group in the periodic table from the 3d series to the 4d series, the crystal field splitting energy (Δ) increases. Specifically, it is known that the C.F.S.E. increases by approximately 50% when moving from the 3d series to the 4d series.

### Step 4: Calculate the C.F.S.E. for Pd
Since the C.F.S.E. for Ni is Δ, the C.F.S.E. for Pd can be calculated as follows:
- Increase in C.F.S.E. = 50% of Δ = 0.5Δ
- Therefore, C.F.S.E. for Pd = Δ + 0.5Δ = 1.5Δ

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

### Final Answer:
The value of C.F.S.E. for Pd is 1.5Δ.

---