JEE Mains · Chemistry · STD 11 - 2. structure of atom
The four quantum numbers for the electron in the outer most orbital of potassium (atomic no. \(19\)) are _______.
- A \(\mathrm{n}=4, l=2, \mathrm{~m}=-1, s=+\frac{1}{2}\)
- B \(\mathrm{n}=4, l=0, \mathrm{~m}=0, s=+\frac{1}{2}\)
- C \(\mathrm{n}=3, l=0, \mathrm{~m}=1, s=+\frac{1}{2}\)
- D \(\mathrm{n}=2, l=0, \mathrm{~m}=0, s=+\frac{1}{2}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{n}=4, l=0, \mathrm{~m}=0, s=+\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
\({ }_{19} \mathrm{K} \quad 1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6, 3 \mathrm{~s}^2, 3 \mathrm{p}^6, 4 \mathrm{~s}^1 \text {. }\) Outermost orbital of potassium is \(4 \mathrm{~s}\) orbital \(\mathrm{n}=4, \mathrm{l}=0, \mathrm{~m}_1=0, \mathrm{~s}= \pm \frac{1}{2}\)
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