Question

The four quantum numbers for the electron in the outermost orbital of potassium \((Z=19)\) are

• A. \(n=4,\ l=2,\ m=-1,\ s=+\frac{1}{2}\)
• B. \(n=4,\ l=0,\ m=0,\ s=+\frac{1}{2}\)
• C. \(n=3,\ l=0,\ m=1,\ s=+\frac{1}{2}\)
• D. \(n=4,\ l=3,\ m=-2,\ s=-\frac{1}{2}\)

Hint:

For an \(s\)-orbital, \(l=0\) and \(m=0\). Potassium has outermost electron in \(4s^1\).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The goal is to determine the exact set of four quantum numbers that describe the high-energy valence electron of Potassium. We must use the Aufbau principle to locate the electron.
Step 2: Detailed Explanation:

Potassium has an atomic number \(Z = 19\). Its ground-state electronic configuration is:
\(1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^1\).

The outermost (valence) electron is the one added last, which resides in the \(4s\) orbital.

Now, we identify the four quantum numbers (\(n, l, m, s\)) for this \(4s^1\) electron:
1. Principal quantum number (\(n\)): This corresponds to the main shell number. For \(4s\), \(n = 4\).
2. Azimuthal quantum number (\(l\)): This defines the subshell shape. For an 's' orbital, \(l\) is always 0. (p=1, d=2, f=3).
3. Magnetic quantum number (\(m\)): This defines the orientation. The allowed values for \(m\) are between \(-l\) and \(+l\). Since \(l = 0\), \(m\) must also be 0.
4. Spin quantum number (\(s\)): This describes the spin direction. It can be either \(+1/2\) or \(-1/2\). Option B provides \(+1/2\).

Looking at the other options:
- Option A (\(l=2\)) would be a d-orbital.
- Option C (\(n=3\)) is an internal shell.
- Option D (\(l=3\)) is an f-orbital.

Only Option B correctly maps to the \(4s\) orbital.

Step 3: Final Answer:
The set of quantum numbers for the outermost electron of Potassium is \(n=4, l=0, m=0, s=+1/2\).