Maximum pumber of degenerate orbitals in M shell of `Li^(2+)` is _____. 

Answer

To determine the maximum number of degenerate orbitals in the M shell of \( \text{Li}^{2+} \), we can follow these steps:

### Step 1: Understand the Electron Configuration
- Lithium (\( \text{Li} \)) has an atomic number of 3, meaning it has 3 electrons in its neutral state.
- The electron configuration for neutral lithium is \( 1s^2 2s^1 \).

### Step 2: Determine the Electron Configuration of \( \text{Li}^{2+} \)
- The \( \text{Li}^{2+} \) ion has lost 2 electrons.
- Therefore, the electron configuration for \( \text{Li}^{2+} \) will be \( 1s^2 \), as the 2s electron is also removed.

### Step 3: Identify the M Shell
- The M shell corresponds to the principal quantum number \( n = 3 \).
- The subshells in the M shell are \( 3s, 3p, \) and \( 3d \).

### Step 4: Determine the Number of Degenerate Orbitals
- The number of degenerate orbitals in each subshell is as follows:
- \( 3s \) has 1 orbital.
- \( 3p \) has 3 orbitals.
- \( 3d \) has 5 orbitals.
- However, since \( \text{Li}^{2+} \) only has electrons in the \( 1s \) subshell, the M shell does not contain any electrons.

### Step 5: Conclusion
- Since there are no electrons in the M shell of \( \text{Li}^{2+} \), the maximum number of degenerate orbitals in the M shell is 0.

### Final Answer
The maximum number of degenerate orbitals in the M shell of \( \text{Li}^{2+} \) is **0**.

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