The electronic configuration of the outemost shell of the most electronegative element is

Options

A. `1s^(2).2s^(2).2p^(4).3s^(1)`
B. `1s^(2).2s^(2).2p^(6).3s^(2).3p^(5)`
C. `1s^(2).2s^(2).2p^(5)`
D. `1s^(2).2s^(2).2p^(6).3s^(1). 3p^(8)`

Answer

To determine the electronic configuration of the outermost shell of the most electronegative element, we can follow these steps:

### Step 1: Identify the Most Electronegative Element
The most electronegative element in the periodic table is fluorine (F). It is located in Group 17 (the halogens) and Period 2.

### Step 2: Determine the Atomic Number of Fluorine
Fluorine has an atomic number of 9. This means it has 9 electrons.

### Step 3: Write the Electronic Configuration
The electronic configuration of fluorine can be written as follows:
- The first two electrons fill the 1s orbital: **1s²**
- The next two electrons fill the 2s orbital: **2s²**
- The remaining five electrons fill the 2p orbital: **2p⁵**

Thus, the complete electronic configuration for fluorine is:
**1s² 2s² 2p⁵**

### Step 4: Identify the Outermost Shell Configuration
The outermost shell of fluorine is the second shell (n=2), which contains:
- 2 electrons in the 2s subshell
- 5 electrons in the 2p subshell

Therefore, the total number of electrons in the outermost shell is:
**2 + 5 = 7 electrons**

### Step 5: Conclusion
The electronic configuration of the outermost shell of the most electronegative element (fluorine) is:
**2s² 2p⁵**