Pd has exceptional valence shell electronic configuration of `4d^(10)5s^(0)`. It is a member of-

Options

A. `4^(th)` period ,group 11
B. `5^(th)` period ,group 10
C. `6^(th)` period ,group 9
D. `3^(rd)` period ,group 16

Answer

To determine the group and period of Palladium (Pd) based on its electronic configuration, follow these steps:

### Step-by-Step Solution:

1. **Identify the Electronic Configuration**:
- The given electronic configuration of Palladium (Pd) is `4d^(10) 5s^(0)`.

2. **Determine the Period**:
- The period of an element is determined by the highest principal quantum number (n) in its electronic configuration.
- In this case, the highest value of n is 5 (from the `5s` orbital).
- Therefore, Palladium belongs to the **5th period**.

3. **Determine the Group**:
- The group of an element can be calculated using the formula:
\[ \text{Group} = ns + (n-1)d \]
- Here, `ns` refers to the number of electrons in the outermost s subshell, and `(n-1)d` refers to the number of electrons in the d subshell of the previous principal quantum level.
- For Palladium:
- `ns` (from `5s`) = 0 electrons (since `5s^(0)`)
- `n-1` = 4 (for the `4d` subshell)
- The number of electrons in `4d` = 10 (from `4d^(10)`).
- Therefore, the group number is:
\[ \text{Group} = 0 + 10 = 10 \]

4. **Conclusion**:
- Palladium (Pd) is in the **5th period** and **10th group** of the periodic table.

### Final Answer:
Palladium (Pd) is a member of the **5th period and 10th group**.

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