The general electronic configuration of outer most and penultimate shell is given as `(n - 1)s^(2)(n-1)p^(6)(n-1)d^(x)ns^(2)`. Then for an element with n=4 und x = 7.

Options

A. 25
B. 26
C. 27
D. 28

Answer

To solve the problem, we need to determine the atomic number of an element based on the given electronic configuration. The configuration provided is `(n-1)s^2(n-1)p^6(n-1)d^x ns^2`, with `n = 4` and `x = 7`.

### Step-by-Step Solution:

1. **Substitute the values of n and x**:
- Given `n = 4`, we can substitute this into the electronic configuration.
- The configuration becomes:
- `(4-1)s^2` → `3s^2`
- `(4-1)p^6` → `3p^6`
- `(4-1)d^x` → `3d^7`
- `ns^2` → `4s^2`

So, the complete electronic configuration is:
\[ 3s^2 \, 3p^6 \, 3d^7 \, 4s^2 \]

2. **Write the complete electronic configuration**:
- The complete electronic configuration includes the filled shells before `3s`, which are `1s`, `2s`, `2p`, and `3s`, `3p`.
- Thus, the full configuration becomes:
\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^7 \, 4s^2 \]

3. **Count the total number of electrons**:
- Now, we will count the electrons in each subshell:
- `1s^2` contributes 2 electrons
- `2s^2` contributes 2 electrons
- `2p^6` contributes 6 electrons
- `3s^2` contributes 2 electrons
- `3p^6` contributes 6 electrons
- `3d^7` contributes 7 electrons
- `4s^2` contributes 2 electrons

Adding these together:
\[ 2 + 2 + 6 + 2 + 6 + 7 + 2 = 27 \]

4. **Determine the atomic number**:
- The atomic number (Z) of the element is equal to the total number of electrons, which we calculated to be 27.

5. **Identify the element**:
- The element with atomic number 27 is Cobalt (Co).

### Final Answer:
The atomic number of the element is **27**, and the element is **Cobalt (Co)**.