We are given:

• Principal quantum number: $n = 5$
• Magnetic quantum number: $m_l = +2$

For a given principal quantum number $n$, the azimuthal quantum number $l$ can take values: $l = 0, 1, 2, 3, \dots, n-1$ So for $n=5$: $l = 0,1,2,3,4$

For a given $l$, the magnetic quantum number $m_l$ can take values: $m_l = -l, -(l-1), \dots, 0, \dots, +(l-1), +l$

We need those orbitals for which $m_l = +2$ is possible. This requires: $l \ge 2$ because only then $+2$ lies in the allowed range from $-l$ to $+l$.

Among the allowed $l$ values for $n=5$:

• $l=0$ : $m_l=+2$ not possible
• $l=1$ : $m_l=+2$ not possible
• $l=2$ : $m_l=+2$ possible
• $l=3$ : $m_l=+2$ possible
• $l=4$ : $m_l=+2$ possible

Thus, the possible orbitals correspond to: $l = 2,3,4$ That gives a total of: $3$ orbitals.

Therefore, the required number of orbitals is: $\boxed{3}$

Comparison with stored correct answer:

• Derived answer = $3$
• Stored correct answer = $3$
• They match.