Correct statements for an element with atomic number 9 are :
A. There can be 5 electrons for which $$m_s = +\frac{1}{2}$$ and 4 electrons for which $$m_s = -\frac{1}{2}$$
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:

The atomic number $$Z = 9$$ corresponds to fluorine. Its ground-state electronic configuration is
$$1s^2\, 2s^2\, 2p^5$$.

Useful facts:
(i) For every electron the set of quantum numbers $$\{n,\,l,\,m_l,\,m_s\}$$ must be unique (Pauli exclusion principle).
(ii) In a given subshell electrons first occupy different orbitals with parallel spins (Hund’s rule) and pairing begins only after every orbital is singly occupied.

Testing statement A
Fluorine has 9 electrons. Split them according to their spin quantum number $$m_s$$.
• $$1s^2$$ : one $$m_s = +\frac12$$ and one $$m_s = -\frac12$$ ⇒ 1 up, 1 down.
• $$2s^2$$ : one up, one down ⇒ +1 up, +1 down (cumulative 2 up, 2 down).
• $$2p^5$$ : first three electrons occupy the three $$2p$$ orbitals with $$m_s = +\frac12$$, next two pair with $$m_s = -\frac12$$.
⇒ 3 up + 2 down (cumulative 5 up, 4 down).
Possible distribution: 5 electrons with $$m_s = +\frac12$$ and 4 electrons with $$m_s = -\frac12$$. Hence statement A is correct.

Testing statement B
The five $$2p$$ electrons are placed as
$$\uparrow\downarrow \; (p_x),\; \uparrow \; (p_y),\; \uparrow \; (p_z).$$
Only one of the three $$2p$$ orbitals ends up doubly occupied; which one is arbitrary because the three orbitals are degenerate. Therefore it is not mandatory that exactly the $$p_z$$ orbital contains a single electron. Statement B is not universally correct and is rejected.

Testing statement C
For the last (ninth) electron the available orbitals in increasing energy order are
$$1s \lt 2s \lt 2p$$.
The $$1s$$ and $$2s$$ orbitals are already full; hence the ninth electron enters an orbital with $$n = 2,\, l = 1$$ (a $$2p$$ orbital). Statement C is correct.

Testing statement D
Angular nodes of an orbital $$= l$$.
List every occupied orbital:

1s : $$l = 0 \Rightarrow 0$$ angular nodes
2s : $$l = 0 \Rightarrow 0$$ angular nodes
2p_x, 2p_y, 2p_z : each has $$l = 1 \Rightarrow 1$$ angular node

Total angular nodes of all five distinct orbitals
$$0 + 0 + 1 + 1 + 1 = 3$$.
Statement D claims the sum is 1, which is incorrect.

Hence only statements A and C are correct.

Correct option: Option B (A and C Only).