In all, how many nodal planes are there in the atomic orbitals for the principal quantum number `n = 3`?

Options

A. 10
B. 9
C. 11
D. 2

Explanation

To determine the total number of nodal planes in atomic orbitals for the principal quantum number \( n = 3 \), we can follow these steps:

### Step 1: Identify the possible azimuthal quantum numbers (l)
The azimuthal quantum number \( l \) can take values from \( 0 \) to \( n-1 \). For \( n = 3 \):
- \( l = 0 \) (s orbital)
- \( l = 1 \) (p orbital)
- \( l = 2 \) (d orbital)

### Step 2: Determine the orbitals corresponding to each \( l \)
- For \( l = 0 \): The orbital is \( 3s \)
- For \( l = 1 \): The orbitals are \( 3p_x, 3p_y, 3p_z \)
- For \( l = 2 \): The orbitals are \( 3d_{xy}, 3d_{yz}, 3d_{zx}, 3d_{x^2y^2}, 3d_{z^2} \)

### Step 3: Calculate the number of nodal planes for each type of orbital
- For \( 3s \) (where \( l = 0 \)): There are **0 nodal planes**.
- For \( 3p \) (where \( l = 1 \)): Each of the three \( 3p \) orbitals has **1 nodal plane**. Therefore, for three \( 3p \) orbitals, the total is \( 3 \times 1 = 3 \) nodal planes.
- For \( 3d \) (where \( l = 2 \)): Each of the four \( 3d \) orbitals (except \( 3d_{z^2} \)) has **2 nodal planes**. The \( 3d_{z^2} \) orbital has **0 nodal planes**. Therefore, the total for \( 3d \) orbitals is \( 4 \times 2 + 0 = 8 \) nodal planes.

### Step 4: Sum the total number of nodal planes
Now, we can sum the nodal planes from all orbitals:
- From \( 3s \): \( 0 \)
- From \( 3p \): \( 3 \)
- From \( 3d \): \( 8 \)

Total nodal planes = \( 0 + 3 + 8 = 11 \)

### Final Answer
Thus, the total number of nodal planes for the atomic orbitals when \( n = 3 \) is **11**.

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