Groups (1–18), periods (1–7), s - p - d - f blocks: Worked examples

Intuition What this page trains

The parent note told you the rules. Here we use them on real elements — including the awkward ones that break the naive shortcuts. By the end you can take any atomic number and, from its electron filling alone, name its period, group, block, and valence count — and never be caught by a case you haven't seen.

Two words we lean on constantly. A shell is the set of orbitals sharing the same principal quantum number $n$ (think "which ring of the stadium"). A subshell is one orbital type inside a shell — s, p, d, or f — labelled by the number $l$ (the letter tells you the shape section). If either word feels shaky, glance at Quantum numbers (n, l, m_l, m_s) first.

One more term earned up front: each orbital can hold two electrons, distinguished by a property called spin — labelled by the spin quantum number $m_{s} = \pm \frac{1}{2}$ ("spin up" and "spin down"). Picture two seats per orbital, one for each spin. We use $m_{s}$ the moment we count "2 electrons per orbital".

The scenario matrix

Every "place this element" question falls into one of these cells. We will hit all of them.

#	Case class	What makes it tricky	Example element
A	s-block, group 1–2	easiest — outer $n s^{1 - 2}$	Na (Z=11)
B	p-block, groups 13–18	must subtract 10 for the group number	Cl (Z=17)
C	Noble gas / full shell	valence = 8, sits group 18	Ne (Z=10)
D	d-block ("lag" element)	filling electron is $(n - 1) d$ , period ≠ subshell label	Sc (Z=21)
E	Period-1 degenerate case	only $n = 1$ exists → no p possible; H is anomalous	H, He
F	Helium exception	config $1 s^{2}$ looks group 2 but is group 18	He (Z=2)
G	f-block	filling electron is $(n - 2) f$ , pulled out below	Ce (Z=58)
H	Limiting / boundary	end of a period vs start of the next	Ne→Na jump
I	Word problem	given behaviour, deduce group	"forms −1 ion"
J	Exam twist	given period+group, reconstruct config	period 4, group 6
K	Aufbau exceptions	naive $s^{2} d^{4}$ / $s^{2} d^{9}$ is wrong	Cr, Cu

The single engine behind every cell is the fill order from the n+l rule:

$1 s < 2 s < 2 p < 3 s < 3 p < 4 s < 3 d < 4 p < 5 s < 4 d < 5 p < 6 s < 4 f < 5 d < 6 p \dots$

Figure — Groups (1–18), periods (1–7), s - p - d - f blocks

Example A — sodium, the cleanest s-block case

Worked example Cell A. Place

$Na$ , $Z = 11$ . Forecast: guess its period, group and block before reading on.

Step 1 — fill 11 electrons in order. $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$ . Why this step? We simply walk the arrow order until 11 electrons are placed; the last one lands in $3 s$ .

Step 2 — read the period. Highest $n$ occupied $= 3$ → period 3. Why this step? Period number = the biggest shell that has any electron, i.e. the size of the outer ring.

Step 3 — read the block & group. Last electron entered an s subshell → s-block. Outer config is $n s^{1}$ with 1 valence electron → group 1. Why this step? For groups 1–2 the rule is: valence electrons = group number, and s-block valence = the s-count.

Verify: Na is famously a soft, +1-forming alkali metal in period 3, group 1. ✔ Valence count $= 1$ , matching one $3 s$ electron. See Valence electrons and chemical reactivity.

Example B — chlorine, subtract-10 p-block

Worked example Cell B. Place

$Cl$ , $Z = 17$ . Forecast: how many valence electrons — 7, or 17?

Step 1 — configuration. $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}$ . Why this step? Placing 17 electrons, the last lands in $3 p$ .

Step 2 — period. Highest $n = 3$ → period 3. Why this step? As always, the period is the size of the outermost occupied shell; here electrons reach $n = 3$ and no higher, so the outer ring is the 3rd.

Step 3 — group via the $- 10$ rule. Last electron in p → p-block (groups 13–18). Valence $= 2 + 5 = 7$ . Group number $=$ valence $+ 10 = 17$ → group 17. Why this step? p-block group numbers start at 13, not 1, because the 10 d-block columns sit in between. So we add those 10 back on to convert valence into a group number.

Verify: Cl gains 1 electron to reach the noble-gas $3 p^{6}$ , forming $Cl^{-}$ . A group-17 halogen. ✔ ( $7$ valence, group $17$ , $7 = 17 - 10$ .)

Example C — neon, the full-shell case

Worked example Cell C. Place

$Ne$ , $Z = 10$ . Forecast: valence count for a noble gas?

Step 1 — configuration. $1 s^{2} 2 s^{2} 2 p^{6}$ . Why this step? Ten electrons walked through the fill order stop exactly at a filled $2 p$ .

Step 2 — period & block. Highest $n = 2$ → period 2. Last electron in $2 p$ → p-block. Why this step? Period is again the outermost occupied shell ( $n = 2$ ); the block is named after the subshell type that received the last-added electron, which was a p orbital.

Step 3 — group. Outer $2 s^{2} 2 p^{6}$ → valence $= 8$ → group $= 8 + 10 = 18$ . This filled set of eight outer electrons ( $n s^{2} n p^{6}$ ) is called an octet — a complete, especially stable outer shell. Why this step? A completely filled $n s^{2} n p^{6}$ shell means 8 valence electrons (an octet), the maximum for the p-block, which is column 18.

Verify: Ne is inert — full octet, no urge to react. Group 18. ✔ See Noble gases and octet rule.

Example D — scandium, the d-block "lag"

Worked example Cell D. Place

$Sc$ , $Z = 21$ . The trap element. Forecast: is Sc in period 3 (because "3d") or period 4?

Step 1 — fill in energy order. $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{1}$ . Why this step? By the n+l rule $4 s$ ( $n + l = 4$ ) fills before $3 d$ ( $n + l = 5$ ). So the 20th electron goes into $4 s$ , and only the 21st drops into $3 d$ .

Step 2 — period = highest $n$ with electrons. The outermost shell is $n = 4$ (the $4 s^{2}$ ). → period 4, not 3. Why this step? Period counts the outermost shell you actually occupy, which is $4 s$ — the $3 d$ label refers to a subshell tucked inside that shell. Figure 2 draws exactly this: the $4 s$ electrons ride the outer ring while $3 d$ hides deeper in.

Step 3 — block & group. Last electron entered a d subshell → d-block. Sc is the first d-column → group 3. Why this step? d-block spans groups 3–12; the count of d-electrons plus the two $s$ electrons places it: $3 d^{1} 4 s^{2}$ → group $3$ .

Verify: Period 4, group 3, block d. The "3d in the name so period 3" reasoning is wrong precisely because filling-label ≠ period. ✔ (See the mistake callout in the parent.)

Figure 2: Scandium's shells as nested rings. The $4 s$ pair sits on the outermost ring (setting period 4), while the single $3 d$ electron — despite its "3" label — lives on an inner ring. Period is set by the outer ring, not the filling label.

Example E — period-1 degenerate case (and why hydrogen is odd)

Worked example Cell E. Why does period 1 hold only H and He — and where does H belong?

Forecast: why can't period 1 have a $2 p$ -like set of 6 more elements, and is H really a group-1 metal?

Step 1 — list allowed $l$ for $n = 1$ . Rule: $l = 0, 1, \dots, n - 1$ . For $n = 1$ that gives only $l = 0$ (the s subshell). No $p$ exists at $n = 1$ . Why this step? $l$ can never reach or exceed $n$ , so there is physically no $1 p$ orbital to fill.

Step 2 — count capacity. Only $1 s$ → orbitals $= 2 l + 1 = 1$ → electrons $= 2 (2 l + 1) = 2$ . Why this step? One orbital, and two spin slots (the two values $m_{s} = \pm \frac{1}{2}$ we defined at the top). Two slots ⇒ exactly two elements.

Step 3 — place hydrogen ( $Z = 1$ , $1 s^{1}$ ). By configuration it has a lone $s$ electron ( $n s^{1}$ ), so the table draws it in group 1 above Li. Why this step? Group placement follows the outer arrangement, and $1 s^{1}$ matches the alkali $n s^{1}$ pattern.

Step 4 — flag hydrogen's anomaly. Unlike a true alkali metal, H is a diatomic gas; it can lose its electron to make $H^{+}$ (like group 1) but also gain one to reach the full $1 s^{2}$ shell, making $H^{-}$ (like a group-17 halogen). So H sits in group 1 by convention while behaving like neither cleanly. Why this step? With only one shell that is half-full, hydrogen is equidistant from "empty like an alkali" and "full like a noble gas", so it straddles two chemistries — a genuine edge case.

Verify: Exactly 2 elements in period 1: H ( $1 s^{1}$ ) and He ( $1 s^{2}$ ). Period 1 length $= 2$ ; H is drawn in group 1 but is chemically anomalous. ✔

Example F — helium exception

Example G — cerium, into the f-block (full fill sequence shown)

Worked example Cell G. Place

$Ce$ , $Z = 58$ . Forecast: which shell index carries the filling f-electron — $4 f$ , $5 f$ , or $6 f$ ?

Step 1 — fill all 58 electrons by walking Figure 1's arrows in order. Reading the diagonals one by one and tallying the running electron count: $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{2} 4 d^{10} 5 p^{6} 6 s^{2} 4 f^{1} 5 d^{1}$ Running totals as we go: $2, 4, 10, 12, 18, 20, 30, 36, 38, 48, 54, 56, 57, 58$ . The 57th electron opens $4 f$ and the 58th sits in $5 d$ (Ce's real ground state). Why this step? We do not memorize the order — we read it off the n+l chart in Figure 1: each diagonal ( $1 s$ ; $2 s$ ; $2 p, 3 s$ ; $3 p, 4 s$ ; $3 d, 4 p, 5 s$ ; $4 d, 5 p, 6 s$ ; then $4 f, 5 d, 6 p$ ) fills before the next, breaking ties by lower $n$ . Following that gives $6 s$ before $4 f$ , exactly as drawn. (The chart continues into $7 s$ and beyond for the heaviest elements, but we never need past $6 p$ here.)

Step 2 — period. Highest occupied shell is $n = 6$ ( $6 s^{2}$ ) → period 6. Why this step? Once more, the outermost occupied shell sets the period, and here that is $6 s$ — not the $4 f$ that is being filled.

Step 3 — block & the $(n - 2)$ pattern. Filling electron reaches f → f-block (pulled out below the main table). Note the index: period is 6 but the f label is $4$ , i.e. $n - 2 = 6 - 2 = 4$ . Why this step? $4 f$ fills two periods "late" relative to its own $n = 4$ , so the offset is $n - 2$ (mirroring $(n - 1) d$ for d-block).

Verify: Ce is a lanthanide, period 6, f-block. The $4 f$ /period-6 gap equals $6 - 4 = 2$ ; the running total reaches exactly $58$ . ✔ See Electronic configuration of elements.

Example H — the limiting boundary (end of a period → start of the next)

Example I — word problem (behaviour → group)

Example J — exam twist (period + group → config)

Worked example Cell J. "Element X is in

period 4, group 6. Write its outer configuration and give $Z$ ." Forecast: group 6 sits in which block?

Step 1 — locate the block. Groups 3–12 are the d-block; group 6 is the 4th d-column. Why this step? Group 6 minus group 3 (the first d-column) is 3 columns in, i.e. the 4th d-electron ⇒ $d^{4}$ .

Step 2 — apply the $(n - 1) d$ rule for period 4. Period 4 d-block uses $(n - 1) d = 3 d$ with $n = 4$ . Idealized: $4 s^{2} 3 d^{4}$ . Why this step? The parent's rule: d electrons carry index $n - 1$ because $4 s$ fills before $3 d$ .

Step 3 — count $Z$ . $[Ar] 4 s^{2} 3 d^{4} \Rightarrow Z = 18 + 2 + 4 = 24$ . Why this step? Argon core is 18, plus the 6 outer electrons.

Verify: $Z = 24$ is chromium (period 4, group 6, d-block). The position-from-group logic gives $Z = 24$ correctly — but the idealized config $4 s^{2} 3 d^{4}$ is not Cr's true ground state, which is the subject of Example K. ✔

Example K — the Aufbau exceptions (Cr and Cu), worked through

Worked example Cell K. Write the

actual ground states of $Cr (Z = 24)$ and $Cu (Z = 29)$ , and see why the naive rule fails. Forecast: will Cr end in $4 s^{2} 3 d^{4}$ and Cu in $4 s^{2} 3 d^{9}$ , as the plain n+l chant predicts?

Step 1 — write the naive predictions. Following Figure 1 blindly: Cr $= [Ar] 4 s^{2} 3 d^{4}$ , Cu $= [Ar] 4 s^{2} 3 d^{9}$ . Why this step? This is what the raw fill order gives, and it correctly recovers $Z$ ( $18 + 2 + 4 = 24$ , $18 + 2 + 9 = 29$ ).

Step 2 — apply the half-/fully-filled stability nudge. In reality one $4 s$ electron drops into $3 d$ : $Cr : [Ar] 4 s^{1} 3 d^{5}, Cu : [Ar] 4 s^{1} 3 d^{10} .$ Why this step? A half-filled ( $d^{5}$ ) or completely filled ( $d^{10}$ ) d-subshell is extra stable (symmetric, low electron–electron repulsion). Promoting one $4 s$ electron "buys" that stable arrangement, so nature prefers it. The $4 s$ and $3 d$ energies are so close here that this tiny gain wins.

Step 3 — check the placement is unchanged. Both still have outer shell $n = 4$ → period 4; both are d-block. Electron count is identical, so $Z$ , period and block are all unaffected — only the exact split between $4 s$ and $3 d$ changes. Why this step? Placement depends on the outer shell and total count, not on the fine $4 s$ / $3 d$ bookkeeping, so exceptions never move an element on the table.

Verify: Cr $[Ar] 4 s^{1} 3 d^{5}$ sums to $18 + 1 + 5 = 24$ ; Cu $[Ar] 4 s^{1} 3 d^{10}$ sums to $18 + 1 + 10 = 29$ . Both stay period 4, d-block. ✔

Recall One-line placement recipe

Period $:::$ highest $n$ that has any electron. Block $:::$ subshell type (s/p/d/f) of the last-added electron. Group (s-block) $:::$ number of $n s$ electrons (1 or 2). Group (p-block) — value $:::$ valence electrons $+ 10$ . Group (p-block) — why $+ 10$ $:::$ the 10 d-block columns (groups 3–12) sit between the s- and p-blocks, so p-block columns start at 13; adding 10 converts a valence count back into that shifted group number. Group (d-block) $:::$ (count of $s$ + count of $(n - 1) d$ electrons).

Connections

Aufbau principle and n+l rule — the fill order every example rides on.
Quantum numbers (n, l, m_l, m_s) — why $l < n$ forces period 1 to be tiny.
Electronic configuration of elements — the config-writing skill practised here.
Valence electrons and chemical reactivity — turning valence count into ion charge.
Noble gases and octet rule — the full-shell logic behind cells C and F.
Periodic trends — atomic radius, ionization energy — next step once placement is fluent.