The number of orbitals with \( n = 5 \) and \( m_l = +2 \) is _________. (Round off to the Nearest Integer)
Detailed Step-by-Step Solution
To find the total number of specific orbitals, we need to apply the rules governing the Principal (\(n\)), Azimuthal (\(l\)), and Magnetic (\(m_l\)) quantum numbers.
Step 1: Determine the possible values of \(l\)
For a given principal quantum number \(n\), the azimuthal quantum number \(l\) can take integer values ranging from \(0\) up to \((n - 1)\).
Given \( n = 5 \), the possible values for \(l\) are:
These correspond to the 5s, 5p, 5d, 5f, and 5g subshells respectively.
Step 2: Check each subshell for \( m_l = +2 \)
For each value of \(l\), the magnetic quantum number \(m_l\) takes integer values from \(-l\) to \(+l\), including zero. We need to find which subshells contain an orbital where \( m_l = +2 \).
-
❌
If \( l = 0 \) (5s subshell):
\( m_l = 0 \). (There is no \(+2\)) -
❌
If \( l = 1 \) (5p subshell):
\( m_l = -1, 0, +1 \). (There is no \(+2\)) -
✅
If \( l = 2 \) (5d subshell):
\( m_l = -2, -1, 0, +1, \mathbf{+2} \). (One orbital found) -
✅
If \( l = 3 \) (5f subshell):
\( m_l = -3, -2, -1, 0, +1, \mathbf{+2}, +3 \). (One orbital found) -
✅
If \( l = 4 \) (5g subshell):
\( m_l = -4, -3, -2, -1, 0, +1, \mathbf{+2}, +3, +4 \). (One orbital found)
Conclusion: We found exactly one orbital with \( m_l = +2 \) in the 5d, 5f, and 5g subshells. Therefore, the total number of orbitals is \( 1 + 1 + 1 = 3 \). The correct integer is 3.
Mastering Quantum Numbers
Quantum number questions in JEE Main are highly scoring if you are careful. A common shortcut for finding the number of orbitals with a specific \(m_l\) value in a given shell (\(n\)) is: if \( |m_l| < n \), the number of such orbitals is exactly \( n - |m_l| \). In this case: \( 5 - |+2| = 3 \).
To learn more of these time-saving tricks and solidify your understanding of subshells, nodes, and probability density, check out our comprehensive notes on the Structure of Atom Class 11 Chemistry.
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