For chromium element, how many orbital can have `n+l=3`

Options

A. `2`
B. `3`
C. `4`
D. `6`

Explanation

To solve the question of how many orbitals can have \( n + l = 3 \) for the element chromium, we will follow these steps:

### Step 1: Understand the Quantum Numbers
- The principal quantum number \( n \) indicates the energy level of the electron.
- The azimuthal quantum number \( l \) indicates the subshell type:
- \( l = 0 \) for s subshell
- \( l = 1 \) for p subshell
- \( l = 2 \) for d subshell
- \( l = 3 \) for f subshell

### Step 2: Write the Electronic Configuration of Chromium
- The atomic number of chromium (Cr) is 24. Its electronic configuration is:
\[ \text{Cr: } 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^1 \, 3d^5 \]

### Step 3: Calculate \( n + l \) for Each Orbital
- For each subshell in the electronic configuration, calculate \( n + l \):
- **1s**: \( n = 1, l = 0 \) → \( n + l = 1 + 0 = 1 \)
- **2s**: \( n = 2, l = 0 \) → \( n + l = 2 + 0 = 2 \)
- **2p**: \( n = 2, l = 1 \) → \( n + l = 2 + 1 = 3 \)
- **3s**: \( n = 3, l = 0 \) → \( n + l = 3 + 0 = 3 \)
- **3p**: \( n = 3, l = 1 \) → \( n + l = 3 + 1 = 4 \)
- **4s**: \( n = 4, l = 0 \) → \( n + l = 4 + 0 = 4 \)
- **3d**: \( n = 3, l = 2 \) → \( n + l = 3 + 2 = 5 \)

### Step 4: Identify the Orbitals with \( n + l = 3 \)
- From the calculations:
- The orbitals with \( n + l = 3 \) are:
- **2p** (from \( 2p^6 \))
- **3s** (from \( 3s^2 \))

### Step 5: Count the Orbitals
- There are **2 orbitals** that satisfy the condition \( n + l = 3 \).

### Final Answer
The number of orbitals that can have \( n + l = 3 \) for chromium is **2**.

---