Question:easy

Which quantum number combination is impossible for a hydrogenic atom orbital?

Show Hint

Always check \(l < n\) first — most errors come from this rule.
Updated On: Jun 10, 2026
  • \(n = 3, l = 2, m_l = -2, m_s = +\frac{1}{2}\)
  • \(n = 4, l = 0, m_l = 0, m_s = -\frac{1}{2}\)
  • \(n = 3, l = 3, m_l = +1, m_s = +\frac{1}{2}\)
  • \(n = 2, l = 1, m_l = 0, m_s = -\frac{1}{2}\)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Know what each quantum number means.
Every electron in an atom is described by four labels. The principal number $n$ tells the shell, the azimuthal number $l$ tells the sub-shell shape, the magnetic number $m_l$ tells the orbital direction, and the spin number $m_s$ tells the spin. Each label has its own allowed range.

Step 2: Write the rule for $l$.
The value of $l$ can only go from $0$ up to one less than $n$. In short form: \[ l = 0, 1, 2, \ldots, (n-1) \] So the largest $l$ is always $n-1$.

Step 3: Write the rule for $m_l$ and $m_s$.
For a chosen $l$, the value of $m_l$ runs from $-l$ to $+l$. The spin $m_s$ can only be $+\tfrac{1}{2}$ or $-\tfrac{1}{2}$. These rules must all hold at the same time.

Step 4: Test each given set against the rules.
For $n=3$, the allowed $l$ values are $0, 1, 2$. So $l$ cannot reach $3$ when $n=3$. Any set claiming $n=3$ with $l=3$ breaks the basic rule.

Step 5: Spot the impossible set.
The set $n=3, l=3, m_l=+1, m_s=+\tfrac{1}{2}$ asks for $l=3$ while $n=3$. Since $l$ can be at most $n-1=2$, this set can never exist for a real orbital.

Step 6: Confirm the others are fine.
The sets with $n=3,l=2$, with $n=4,l=0$, and with $n=2,l=1$ all obey $l \le n-1$, and their $m_l$ and $m_s$ values are in range. So only one set is wrong.
\[ \boxed{n = 3,\; l = 3,\; m_l = +1,\; m_s = +\tfrac{1}{2}} \]
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