An electron in an atom can be completely designated with the help of four quantum numbers. Out of these, the first three i.e., principal (n), azimuthal (l) and magnetic (m) quantum number are obtained from the solution of Schrödinger wave equation while the spin(s) quantum number arises from the spin of the electron around its axis clockwise or anticlockwise. Of these principal quantum number tells about the size, azimuthal quantum number about the shape and magnetic quantum signifies the orientation of the electron orbital.
The electronic configuration of P in `H_(3)PO_(4)` is

Options

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

Explanation

To determine the electronic configuration of phosphorus (P) in phosphoric acid (H₃PO₄), we will follow these steps:

### Step 1: Determine the Oxidation State of Phosphorus
To find the oxidation state of phosphorus in H₃PO₄, we can set up the following equation:

Let the oxidation state of phosphorus be \( x \).

- The oxidation state of each hydrogen (H) is +1. Since there are 3 hydrogen atoms, their total contribution is \( 3 \times +1 = +3 \).
- The oxidation state of each oxygen (O) is -2. Since there are 4 oxygen atoms, their total contribution is \( 4 \times -2 = -8 \).

Now, we can write the equation for the overall charge of the molecule, which is neutral (0):

\[ x + 3 + (-8) = 0 \]

### Step 2: Solve for \( x \)
Rearranging the equation gives:

\[ x - 5 = 0 \]

Thus,

\[ x = +5 \]

This means that phosphorus is in the +5 oxidation state in H₃PO₄.

### Step 3: Determine the Electronic Configuration of Neutral Phosphorus
The atomic number of phosphorus is 15. Therefore, the electronic configuration of neutral phosphorus (without any oxidation state) is:

\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^3 \]

### Step 4: Adjust the Configuration for the +5 Oxidation State
Since phosphorus is in the +5 oxidation state in H₃PO₄, it has lost 5 electrons. The electrons are removed from the outermost shells first:

- 3 electrons are removed from the 3p subshell.
- 2 electrons are removed from the 3s subshell.

Thus, the configuration after removing 5 electrons will be:

\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^0 \, 3p^0 \]

### Final Electronic Configuration
Therefore, the electronic configuration of phosphorus in H₃PO₄ is:

\[ 1s^2 \, 2s^2 \, 2p^6 \]

### Summary
The electronic configuration of phosphorus in H₃PO₄ (with +5 oxidation state) is:

\[ 1s^2 \, 2s^2 \, 2p^6 \]

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