Question

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

• A. $n=4, l=2, m=-1, s=+1/2$
• B. $n=4, l=0, m=0, s=+1/2$
• C. $n=3, l=0, m=1, s=+1/2$
• D. $n=4, l=3, m=-2, s=-1/2$

Hint:

Potassium is in Period 4, Group 1, so its valence electron always ends in $4s^1$.

Answer 1

The Correct Option is B

Step 1: Concept
Determine the electronic configuration of Potassium ($Z=19$): $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$.

Step 2: Meaning
The outermost electron is in the $4s$ orbital.

Step 3: Analysis
For $4s^1$: Principal quantum number $n = 4$. Azimuthal quantum number $l$ for an $s$-orbital is 0. Magnetic quantum number $m$ must be 0 if $l=0$. Spin quantum number $s = +1/2$.

Step 4: Conclusion
The set is (4, 0, 0, +1/2).
Final Answer: (B)