An element has the electronic configuration
`1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(8)4s^(2)`
What will be its position in the periodic table?

Options

A. Period 4,group 10
B. Period 2,Group 2
C. Period 4,Group 2
D. Period 2,Group 8

Answer

To determine the position of the element with the electronic configuration `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(8)4s^(2)` in the periodic table, we can follow these steps:

### Step 1: Identify the Period
The period of an element is determined by the highest principal quantum number (n) in its electronic configuration.

- In the given configuration, the highest principal quantum number is 4 (from the `4s^2`).

**Conclusion:** The element is in the 4th period.

### Step 2: Identify the Group
To identify the group, we need to consider the distribution of electrons in the outermost shell (n) and the d subshell (n-1).

- The outermost shell (n=4) has `4s^2`, which contributes 2 electrons.
- The penultimate shell (n-1=3) has `3d^8`, which contributes 8 electrons.

Now, we can use the formula to find the group number for d-block elements:
\[ \text{Group Number} = \text{Number of electrons in } (n-1)d + \text{Number of electrons in } ns \]
\[ \text{Group Number} = 8 + 2 = 10 \]

**Conclusion:** The element is in Group 10.

### Final Position
Based on the findings:
- Period: 4
- Group: 10

Thus, the element with the electronic configuration `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(8)4s^(2)` is located in **Period 4, Group 10** of the periodic table.