1. For a given principal quantum number $n=4$, the allowed values of azimuthal quantum number $l$ are: $l=0,1,2,3$

2. For each value of $l$, the magnetic quantum number $m_l$ can take values: $m_l=-l,-(l-1),\dots,0,\dots,(l-1),l$

3. We need orbitals with $m_l=0$.

4. Check each allowed $l$:

• For $l=0$, $m_l=0$ is allowed
• For $l=1$, $m_l=0$ is allowed
• For $l=2$, $m_l=0$ is allowed
• For $l=3$, $m_l=0$ is allowed

5. Thus, for each of the four possible $l$ values, there is exactly one orbital with $m_l=0$.

So total number of orbitals is: $4$

Therefore, the maximum number of orbitals identified by $n=4$ and $m_l=0$ is $4$.