Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
The four quantum numbers of the outermost orbital of $K$ (atomic no. $=19$ ) are
ChemistryStructure of AtomJEE Main
Options:
  • A $n=2 l=0, m=0, s=+\frac{1}{2}$
  • B $n=4, l=0, m=0, s=+\frac{1}{2}$
  • C $n=3, l=1, m=1, s=+\frac{1}{2}$
  • D $n=4, I=2 m=-1, s=+\frac{1}{2}$
Solution:
2474 Upvotes Verified Answer
The correct answer is: $n=4, l=0, m=0, s=+\frac{1}{2}$
$K_{19}=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6, 4 s^1$
for $4 s^1$ electrons.
$n=4, I=0, m=0$ and $s=+\frac{1}{2}$.

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.