Which one is a wrong statement ?

Options

A. The electronic configuration of N-atom is

B. An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers
C. Total orbital angular momentum of electron in 's' orbital is equal to zero
D. The value of m for `d_(z^(2))` is zero

Explanation

To determine which statement is incorrect regarding the nitrogen atom's electronic configuration, we will analyze each statement step by step.

### Step 1: Analyze the Electronic Configuration of Nitrogen
The electronic configuration of a nitrogen atom is given as:
- 1s² 2s² 2p³

This means:
- 2 electrons in the 1s orbital,
- 2 electrons in the 2s orbital,
- 3 electrons in the 2p orbitals (which can be distributed among 2px, 2py, and 2pz).

**Hint:** Check the distribution of electrons in the p orbitals and their spins.

### Step 2: Check the Spin of Electrons
According to Hund's rule, when filling degenerate orbitals (like the three 2p orbitals), each orbital gets one electron with the same spin before any orbital gets a second electron with the opposite spin.

In the given configuration:
- If we assume the 2p electrons are arranged as 2p_x, 2p_y, and 2p_z, they should all have the same spin (let's say +1/2) before pairing up.

If the configuration shows one of the electrons in the opposite spin, it violates Hund's rule.

**Hint:** Remember the importance of Hund's rule in electron configuration.

### Step 3: Verify the Quantum Numbers for Orbitals and Electrons
1. **Orbital Designation**: An orbital is designated by 3 quantum numbers (n, l, m).
2. **Electron Designation**: An electron in an atom is designated by 4 quantum numbers (n, l, m, s).

This statement is true.

**Hint:** Recall the definition of quantum numbers and their significance in atomic structure.

### Step 4: Total Orbital Angular Momentum of s Orbital
The total orbital angular momentum (L) of an electron in an s orbital is indeed 0 because:
- For s orbitals, l = 0.

This statement is also true.

**Hint:** Remember that the shape of the orbital (spherical for s) determines its angular momentum.

### Step 5: Value of m for dz² Orbital
For d orbitals, the magnetic quantum number (m) can take values from -l to +l. For d orbitals, l = 2, so m can be -2, -1, 0, 1, or 2.

The dz² orbital corresponds to m = 0, which is correct.

**Hint:** Recall the relationship between the azimuthal quantum number (l) and the magnetic quantum number (m).

### Conclusion
After analyzing all the statements:
- The incorrect statement is the electronic configuration of the nitrogen atom due to the violation of Hund's rule.

### Final Answer
The wrong statement is regarding the electronic configuration of the nitrogen atom.