Question:medium

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

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Potassium is in Period 4, Group 1, so its valence electron always ends in $4s^1$.
  • $n=4, l=2, m=-1, s=+1/2$
  • $n=4, l=0, m=0, s=+1/2$
  • $n=3, l=0, m=1, s=+1/2$
  • $n=4, l=3, m=-2, s=-1/2$
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
To find the quantum numbers of the outermost electron, we must first write the electronic configuration of the atom based on the Aufbau principle.
Step 2: Key Formula or Approach:
1. Potassium ($Z=19$) configuration: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$.
2. Identify the outermost electron: $4s^1$.
Step 3: Detailed Explanation:
For the $4s^1$ electron: * The principal quantum number ($n$) is the coefficient: 4. * For an 's' orbital, the azimuthal quantum number ($l$) is 0. * If $l=0$, the magnetic quantum number ($m$) must be 0. * The spin quantum number ($s$) can be $+1/2$ (or $-1/2$). Looking at the options, $n=4, l=0, m=0, s=+1/2$ is the only correct set.
Step 4: Final Answer:
The quantum numbers are $n=4, l=0, m=0, s=+1/2$.
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