The electronic distribution of Mn(25) is

Options

A. 2, 8, 13,2
B. 2, 8, 14, 1
C. 2, 8, 8,7
D. 2, 8, 10,5

Explanation

To determine the electronic distribution of manganese (Mn) with atomic number 25, we can follow these steps:

### Step 1: Understand the Atomic Number
The atomic number of manganese is 25, which means it has 25 electrons.

### Step 2: Fill the Electron Shells
We will fill the electron shells according to the maximum capacity of each shell:
- **K shell**: Can hold a maximum of 2 electrons.
- **L shell**: Can hold a maximum of 8 electrons.
- **M shell**: Can hold a maximum of 18 electrons, but we will fill it according to the number of electrons remaining after filling K and L.
- **N shell**: Can also hold a maximum of 32 electrons, but we will fill it last.

### Step 3: Fill the K Shell
- The K shell can hold 2 electrons.
- Electrons in K: **2**

### Step 4: Fill the L Shell
- The L shell can hold 8 electrons.
- Electrons in L: **8**

### Step 5: Fill the M Shell
- After filling K and L, we have used 2 + 8 = 10 electrons.
- We have 25 - 10 = 15 electrons left to distribute.
- The M shell can hold a maximum of 18 electrons, but since we only have 15 left, we will fill it with 15 electrons.
- Electrons in M: **15**

### Step 6: Fill the N Shell
- The N shell can hold electrons, but since we have filled all 25 electrons (2 in K, 8 in L, and 15 in M), we do not need to fill any electrons in the N shell.
- Electrons in N: **0**

### Step 7: Write the Electronic Configuration
Now, we can write the electronic configuration based on the distribution:
- K: 2
- L: 8
- M: 15
- N: 0

Thus, the electronic configuration of manganese (Mn) is:
**2, 8, 15, 0**

### Final Answer
The electronic distribution of Mn (25) is **2, 8, 15, 0**.

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