Quantum numbers and electron designation

Principal quantum number ($n$): $n = 1, 2, 3, \ldots$, labelling the shell (K, L, M, N for $n = 1, 2, 3, 4$). Larger $n$ means higher energy and, on average, greater distance from the nucleus.

Azimuthal quantum number ($l$): for a given $n$, $l$ runs from $0$ to $(n-1)$, labelling the subshell shape — $l = 0, 1, 2, 3$ are the $s, p, d, f$ subshells. A shell with principal number $n$ therefore has exactly $n$ possible subshells.

Magnetic quantum number ($m_l$): for a given $l$, $m_l = -l, \ldots, 0, \ldots, +l$, giving $(2l+1)$ values, one per orbital — the spatial orientation of that orbital shape. An $s$ subshell has $1$ orbital, $p$ has $3$, $d$ has $5$, $f$ has $7$.

Spin quantum number ($m_s$): only two values, $m_s = +\tfrac12$ or $-\tfrac12$, independent of $n$, $l$, $m_l$. Each orbital holds at most two electrons, and by the Pauli exclusion principle those two must have opposite spin. Put together, a full set $(n, l, m_l, m_s)$ names one electron uniquely. Before counting how many electrons a shell can hold, it helps to see the four numbers stacked for a whole shell in one picture.

Question. For $n = 3$, list all possible subshells, the number of orbitals in each, and the total number of orbitals and maximum electrons for the whole shell. Then, for the $2p$ subshell holding exactly three electrons (as in nitrogen's $2p^3$), write the complete set of four quantum numbers for each electron.

Step 1. For $n=3$, $l$ takes values $0, 1, 2$, giving subshells $3s$ ($l=0$), $3p$ ($l=1$), $3d$ ($l=2$).

Step 2. Orbitals per subshell from $(2l+1)$: $3s$ has $2(0)+1=1$, $3p$ has $2(1)+1=3$, $3d$ has $2(2)+1=5$. Total orbitals $=1+3+5=9=n^2=3^2$, and maximum electrons $=2\times9=18=2n^2=2(3)^2$, matching equations (1)–(2).

Step 3. Now the $2p$ subshell: $n=2$, $l=1$, so $m_l = -1, 0, +1$ — three orbitals. By Hund's rule, three electrons entering a partially filled subshell occupy the three orbitals singly, with parallel spin, before any orbital is doubly occupied.

Step 4. The four quantum numbers for each electron: electron 1 is $n=2, l=1, m_l=-1, m_s=+\tfrac12$; electron 2 is $n=2, l=1, m_l=0, m_s=+\tfrac12$; electron 3 is $n=2, l=1, m_l=+1, m_s=+\tfrac12$. Each set differs at least in $m_l$, so no two are identical — consistent with the Pauli exclusion principle — and no orbital is doubly occupied, consistent with Hund's rule.

Final answer: $n=3$ has 9 orbitals and holds 18 electrons; the half-filled $2p^3$ configuration is $(2,1,-1,+\tfrac12)$, $(2,1,0,+\tfrac12)$, $(2,1,+1,+\tfrac12)$.

Competitive stretch

For $n=4$ (shell N), the subshells are $4s, 4p, 4d, 4f$ with $1, 3, 5, 7$ orbitals respectively, summing to $16 = n^2$, so the shell holds $2(16) = 32 = 2n^2$ electrons — the same pattern, one shell further out.

• Forgetting $l$ only runs $0$ to $(n-1)$ — claiming a $2d$ subshell exists is wrong, since $n=2$ permits only $l=0$ ($2s$) and $l=1$ ($2p$).
• Confusing $l$ (which subshell) with $m_l$ (which orbital within that subshell) — $l$ has one value per subshell, $m_l$ has $(2l+1)$ values.
• Treating $m_s$ as dependent on $n$, $l$, or $m_l$ — it is always just $+\tfrac12$ or $-\tfrac12$, regardless of orbital.
• Miscounting orbitals in $d$ as $6$ (confusing the orbital count, $2l+1=5$, with the electron capacity, $2(2l+1)=10$).

Ask the tutor: if you are still unsure which $m_l$ values are allowed for a given $l$, or how to assign spins while filling a partially filled subshell, ask the AI tutor for a few more practice sets.