Question

The maximum number of orbitals which can be identified with \( n = 4 \) and \( m_l = 0 \) is ______

Answer 1

Correct Answer: 4

For \( n = 4 \), the possible subshells are :
4s (1 orbital)
4p (3 orbitals)
4d (5 orbitals)
4f (7 orbitals)
Among these, only one orbital from each subshell can have a magnetic quantum number \( m_l = 0 \). Therefore, there is one such orbital from 4s, one from 4p, one from 4d, and one from 4f. 
Thus, the total number of orbitals with \( n = 4 \) and \( m_l = 0 \) is 4.

Answer 2

Step 1: Understanding the given quantum numbers 
We are told that the principal quantum number \( n = 4 \) and the magnetic quantum number \( m_l = 0 \).
We must find the total number of orbitals that can have these quantum numbers simultaneously.

Step 2: Recall quantum number relationships
For any orbital, the possible quantum numbers are related as:
- \( n \): principal quantum number (shell)
- \( l \): azimuthal (subshell) quantum number, values range from \(0\) to \(n - 1\)
- \( m_l \): magnetic quantum number, values range from \(-l\) to \(+l\) in integer steps.

Here, \( n = 4 \Rightarrow l = 0, 1, 2, 3 \) (that is, 4s, 4p, 4d, 4f subshells).

Step 3: Apply the condition \( m_l = 0 \)
For each allowed value of \( l \), \( m_l \) can take one value from \(-l\) to \(+l\). Let’s check if \( m_l = 0 \) exists for each subshell:
 

Subshell	l	Possible ml values	Is ml=0 allowed?
4s	0	0	Yes
4p	1	−1, 0, +1	Yes
4d	2	−2, −1, 0, +1, +2	Yes
4f	3	−3, −2, −1, 0, +1, +2, +3	Yes

Hence, \( m_l = 0 \) exists for all four subshells: 4s, 4p, 4d, and 4f.

Step 4: Count the total number of orbitals
Each valid combination of \( n \), \( l \), and \( m_l \) represents one orbital. Since \( m_l = 0 \) is possible for all four subshells, the total number of such orbitals is:
\[ \text{Number of orbitals} = 4. \]

Final answer
 

4