Which of the following is the electronic configuration of `H_(3)PO_(4)` ?

Options

A. `[Ne]`
B. `[Ne]3s^(2)3p^(3)3d^(1)`
C. `[Ne]3s^(1)3p^(3)`
D. None of these

Explanation

To determine the electronic configuration of \( H_3PO_4 \), we need to focus on the central atom, which is phosphorus (P). Here’s a step-by-step solution:

### Step 1: Identify the oxidation state of phosphorus in \( H_3PO_4 \)
- The oxidation state of hydrogen (H) is +1. Since there are 3 hydrogen atoms, their total contribution is \( 3 \times (+1) = +3 \).
- The oxidation state of oxygen (O) is -2. Since there are 4 oxygen atoms, their total contribution is \( 4 \times (-2) = -8 \).
- Let the oxidation state of phosphorus be \( X \). The overall charge of the molecule is neutral (0), so we can set up the equation:
\[ X + 3 - 8 = 0 \]
Simplifying this gives:
\[ X - 5 = 0 \implies X = +5 \]

### Step 2: Write the electronic configuration of phosphorus
- The atomic number of phosphorus (P) is 15. The electronic configuration of phosphorus in its neutral state is:
\[ 1s^2 2s^2 2p^6 3s^2 3p^3 \]
This can also be expressed in a simpler form as:
\[ [Ne] 3s^2 3p^3 \]

### Step 3: Adjust the electronic configuration for \( P^{+5} \)
- When phosphorus loses 5 electrons to form \( P^{+5} \), it loses the 3 electrons from the 3p subshell and the 2 electrons from the 3s subshell. Thus, the electronic configuration for \( P^{+5} \) becomes:
\[ [Ne] \]
This indicates that all the valence electrons have been removed.

### Conclusion
- Therefore, the electronic configuration of phosphorus in \( H_3PO_4 \) when it is in the +5 oxidation state is:
\[ [Ne] \]