JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Atomic Structure

We need electrons with:
• Principal quantum number $n=4$
• $|m_l|=1$, meaning $m_l=+1$ or $m_l=-1$
• Spin $m_s=-\frac{1}{2}$ (spin-down)

For $n=4$, possible subshells are 4s, 4p, 4d, 4f.

4s ($l=0$): $m_l=0$ only. Does not satisfy $|m_l|=1$. ✗

4p ($l=1$): $m_l=-1,0,+1$. Two orbitals have $|m_l|=1$ ($m_l=\pm 1$). Each can hold 1 electron with $m_s=-\frac{1}{2}$.
→ 2 electrons

4d ($l=2$): $m_l=-2,-1,0,+1,+2$. Two orbitals have $|m_l|=1$ ($m_l=\pm 1$). Each can hold 1 electron with $m_s=-\frac{1}{2}$.
→ 2 electrons

4f ($l=3$): $m_l=-3,-2,-1,0,+1,+2,+3$. Two orbitals have $|m_l|=1$ ($m_l=\pm 1$). Each can hold 1 electron with $m_s=-\frac{1}{2}$.
→ 2 electrons

Total: 2 + 2 + 2 = 6 electrons