JEE Advanced · Chemistry · 2. Atomic Structure
The maximum number of electrons that can have principal quantum number, \(n=3\) and spin quantum number, \(m_s=-\frac{1}{2}\), is
- A 3
- B 6
- C 9
- D 12
Answer & Solution
Correct Answer
(C) 9
Step-by-step Solution
Detailed explanation
When \(n=3, l=0,1,2\) i.e., there are \(3 s, 3 p\) and \(3 d\) orbitals. If all these orbitals are completely occupied as
Total 18 electrons, 9 electrons with \(s=+\frac{1}{2}\) and 9 with \(s=-\frac{1}{2}\).
Alternatively In any \(n\)th orbit, there can be a maximum of \(2 n^2\) electrons. Hence, when \(n=3\), number of maximum electrons \(=18\). Out of these 18 electrons, 9 can have spin \(-\frac{1}{2}\) and remaining nine with spin \(=+\frac{1}{2}\).
Total 18 electrons, 9 electrons with \(s=+\frac{1}{2}\) and 9 with \(s=-\frac{1}{2}\).
Alternatively In any \(n\)th orbit, there can be a maximum of \(2 n^2\) electrons. Hence, when \(n=3\), number of maximum electrons \(=18\). Out of these 18 electrons, 9 can have spin \(-\frac{1}{2}\) and remaining nine with spin \(=+\frac{1}{2}\).
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