For atomic number $9$, the element is fluorine.

Its ground-state electronic configuration is $1s^2\,2s^2\,2p^5$

Now check each statement.

1. Statement A: There can be 5 electrons with $m_s=+\tfrac{1}{2}$ and 4 electrons with $m_s=-\tfrac{1}{2}$.

For fluorine, total electrons $=9$. In ground state:

• $1s^2$: one electron has $+\tfrac{1}{2}$, one has $-\tfrac{1}{2}$
• $2s^2$: one electron has $+\tfrac{1}{2}$, one has $-\tfrac{1}{2}$
• $2p^5$: by Hund's rule, three $p$ orbitals first get one electron each with parallel spins, then two electrons pair up.

So in $2p^5$, we can have:

• $3$ electrons with $m_s=+\tfrac{1}{2}$
• $2$ electrons with $m_s=-\tfrac{1}{2}$

Total:

• $+\tfrac{1}{2}$ electrons $=1+1+3=5$
• $-\tfrac{1}{2}$ electrons $=1+1+2=4$

So A is correct.

2. Statement B: There is only one electron in $p_z$ orbital.

For $2p^5$, electrons are distributed among three $p$ orbitals: $p_x, p_y, p_z$. According to Hund’s rule, the distribution will be like: $2,2,1$ across the three $p$ orbitals, but which one has the single electron is arbitrary. It need not specifically be $p_z$.

So saying there is only one electron in $p_z$ orbital is not निश्चित/uniquely correct. Hence B is false.

3. Statement C: The last electron goes to orbital with $n=2$ and $l=1$.

The last electron enters a $2p$ orbital. For a $p$ orbital: $n=2,\quad l=1$ So C is correct.

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

Occupied orbitals are:

• one $1s$ orbital
• one $2s$ orbital
• three $2p$ orbitals

Angular nodes for an orbital $=l$. Thus:

• $1s$: $l=0$  angular nodes $=0$
• $2s$: $l=0$  angular nodes $=0$
• each $2p$: $l=1$  angular nodes $=1$

Since there are three $2p$ orbitals, total angular nodes of all atomic orbitals $0+0+1+1+1=3$ not $1$.

So D is false.

Therefore, the correct statements are: $\boxed{A \text{ and } C \text{ only}}$

Hence the correct option is $\boxed{\text{C}}$