A. There can be 5 electrons for which m_s = +1/2 and 4 electrons for which m_s = −1/2.

B. There is only one electron in p_z orbital.

C. The last electron goes to orbital with n = 2 and l = 1.

D. The sum of angular nodes of all the atomic orbitals is 1.

Choose the correct answer from the options given below:

Atomic number 9 → Fluorine
Electronic configuration: 1s² 2s² 2p⁵

In 2p⁵ configuration, maximum unpaired electrons = 1
Total electrons = 9
Possible distribution: 5 electrons with m_s = +1/2 and 4 with m_s = −1/2
✔ Statement A is correct.

2p subshell has three orbitals (p_x, p_y, p_z).
Electrons distribute according to Hund’s rule → p_z can have 2 electrons.
✘ Statement B is incorrect.

Last electron enters 2p orbital → n = 2, l = 1
✔ Statement C is correct.

Angular nodes = l
1s (l = 0): 0 × 2 electrons
2s (l = 0): 0 × 2 electrons
2p (l = 1): contributes angular node = 1
✔ Statement D is correct.

Correct Answer = A and C Only