The energy of an electron of `2p_(1)` orbital is

Options

A. Greater than `2p` orbital
B. Less than `2p_(x)` orbital
C. Equal to `2s` orbital
D. Sum of that of`2p_(x)` and `2p_(z)` orbital

Explanation

To determine the energy of an electron in the \(2p_1\) orbital, we can use the \(n + l\) rule, which helps us compare the energies of different orbitals. Here’s a step-by-step solution:

### Step 1: Identify the quantum numbers
The \(2p_1\) orbital is characterized by:
- Principal quantum number \(n = 2\)
- Azimuthal quantum number \(l = 1\) (since it is a p orbital)

### Step 2: Calculate \(n + l\) for the \(2p\) orbital
Using the formula \(n + l\):
\[ n + l = 2 + 1 = 3 \]
This means the energy level for the \(2p\) orbital is determined to be 3.

### Step 3: Calculate \(n + l\) for the \(2s\) orbital
For the \(2s\) orbital:
- \(n = 2\)
- \(l = 0\) (since it is an s orbital)

Calculating \(n + l\):
\[ n + l = 2 + 0 = 2 \]
This indicates that the energy level for the \(2s\) orbital is 2.

### Step 4: Compare the energies
Now we can compare the energies:
- The \(2p\) orbital has \(n + l = 3\)
- The \(2s\) orbital has \(n + l = 2\)

Since \(3 > 2\), we conclude that the energy of the \(2p\) orbital is greater than that of the \(2s\) orbital.

### Step 5: Consider the \(2p_x\) and \(2p_z\) orbitals
The \(2p_x\) and \(2p_z\) orbitals are degenerate, meaning they have the same energy. Therefore, the energy of \(2p_1\) is equal to that of \(2p_x\) and \(2p_z\).

### Conclusion
The energy of the electron in the \(2p_1\) orbital is equal to the energy of the \(2p\) orbitals (which includes \(2p_x\) and \(2p_z\)).

Thus, the correct answer is:
**The energy of an electron in the \(2p_1\) orbital is equal to that of the \(2p\) orbitals.**