The configuration of the valence orbital of an element with atomic number 22 is

Options

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

Explanation

To find the configuration of the valence orbital of an element with atomic number 22, we need to follow these steps:

### Step 1: Identify the Element
The element with atomic number 22 is Titanium (Ti).

### Step 2: Write the Electronic Configuration
We will write the electronic configuration by filling the orbitals according to the Aufbau principle, Hund's rule, and the Pauli exclusion principle.

1. Start filling from the lowest energy level:
- 1s: 2 electrons → 1s²
- 2s: 2 electrons → 2s²
- 2p: 6 electrons → 2p⁶
- 3s: 2 electrons → 3s²
- 3p: 6 electrons → 3p⁶
- 4s: 2 electrons → 4s²
- 3d: 2 electrons → 3d²

Now, let's sum the electrons:
- 1s² = 2
- 2s² = 2
- 2p⁶ = 6
- 3s² = 2
- 3p⁶ = 6
- 4s² = 2
- 3d² = 2

Total = 2 + 2 + 6 + 2 + 6 + 2 + 2 = 22 electrons.

### Step 3: Determine the Valence Orbital Configuration
The valence electrons are those in the outermost shell. For Titanium, the outermost shell includes:
- 4s²
- 3d²

Thus, the configuration of the valence orbital for Titanium (atomic number 22) is:
**4s² 3d²**

### Step 4: Check the Options
Now, we can compare this with the given options:
- Option A: 3d⁵ 4s¹
- Option B: 4s² 3d²
- Option C: 4s¹ 4p¹
- Option D: 3d² 4s¹ 4p¹

The correct option is **Option B: 4s² 3d²**.

### Final Answer
The configuration of the valence orbital of an element with atomic number 22 is **4s² 3d²**.

---