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

The vocabulary you need first

WHY the row lengths are 2, 8, 8, 18, 18, 32, 32 (traced, not asserted)

Intuition Read the block widths straight off

$l$ Apply $l = 0 \dots n - 1$ shell by shell, name each subshell, and add its capacity $2 (2 l + 1)$ . The row length is set by which subshells fill during that row, ordered by the $n + l$ rule (lower $n + l$ first; ties broken by lower $n$ ).

Shell $n = 1$ : only $l = 0$ ( $1 s$ ) → capacity 2. Look at the yellow bar on the left of the figure — one tiny block. → period length 2.
Shell $n = 2$ : $l = 0, 1$ ( $2 s, 2 p$ ) → $2 + 6 = 8$ . The first $p$ subshell appears. → 8.
Shell $n = 3$ : $l = 0, 1, 2$ exist, but $3 d$ has $n + l = 5$ , so it waits until after $4 s$ ( $n + l = 4$ ). During period 3 only $3 s, 3 p$ fill → $2 + 6 = 8$ . → 8 (d exists but hasn't got its turn).
Shell $n = 4$ start: $4 s$ fills, then the delayed $3 d$ (10), then $4 p$ (6): $2 + 10 + 6 = 18$ . This is where the d-block first appears. → 18.
Period 5: same pattern $5 s, 4 d, 5 p$ → 18.
Period 6: now $4 f$ ( $n + l = 7$ ) finally gets its turn: $6 s, 4 f, 5 d, 6 p = 2 + 14 + 10 + 6 = 32$ . The f-block first appears. → 32.
Period 7: $7 s, 5 f, 6 d, 7 p$ → 32.

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

True or false — justify

Every element in the same group has the same number of electrons.

False. Same group means same valence arrangement (same outer subshell pattern), not the same total electron count — Na (11 electrons) and K (19 electrons) share

n s^{1}

but differ in inner shells.

The period number equals the highest principal quantum number

n

that is occupied in the ground state.

True. The period is set by the outermost shell's

n

; for Sc that outer shell is

4 s

, so Sc sits in period 4 even though its last electron enters

3 d

The d-block is 10 columns wide because there are 10 d-orbitals.

False. There are only 5 d-orbitals (

2 l + 1 = 5

for

l = 2

); each holds 2 electrons, so

2 \times 5 = 10

columns. The width counts electrons, not orbitals.

A period can never be longer than 32 elements.

True. The widest possible period is

s (2) + p (6) + d (10) + f (14) = 32

; no new subshell type beyond

f

opens up in the seven known periods, so 32 is the ceiling.

Helium's configuration

1 s^{2}

ends in

s^{2}

, so chemically it belongs with the group-2 metals.

False. Grouping follows chemistry, not the raw config ending. He has a complete shell (for

n = 1

1 s^{2}

is full since no

1 p

exists), so it behaves like a noble gas and sits in group 18 above Ne.

Moving down a group, you keep the same valence arrangement but jump to a larger shell.

True. Each step down adds a full new shell while preserving the outer subshell pattern (e.g.

n s^{1}

), which is exactly why chemistry stays similar down a group.

For any element, the last electron added lands in the highest-

n

shell.

False. In the d- and f-blocks the last-added electron enters an inner shell — e.g. Sc's fills

3 d

while its outermost shell is

4 s

. Filling order ≠ outermost shell.

Every subshell of a given period becomes available at the same time.

False. Subshells open by the

n + l

rule: within period 4,

4 s

fills, then

3 d

, then

4 p

. They enter at different energies, which is why the d-block "lags" into the middle of the row.

The Aufbau order (

1 s, 2 s, 2 p, 3 s, \dots

) predicts every ground-state configuration exactly.

False. It predicts most, but there are known anomalies — e.g. Cr is

[Ar] 3 d^{5} 4 s^{1}

(not

3 d^{4} 4 s^{2}

) and Cu is

[Ar] 3 d^{10} 4 s^{1}

(not

3 d^{9} 4 s^{2}

), because a half-filled or fully-filled

d

subshell is extra stable. This is exactly why you rely on the

n + l

trend, not a rigid seat-by-seat fill.

Spot the error

"Scandium is a period-3 element because its filling electron goes into the 3d subshell."

Error: period ≠ subshell label. Period = highest

n

occupied = the

4 s

shell → Sc is period 4. The "3" in 3d is just the orbital's principal number, not the row.

"The d-block appears in period 3 because 3d orbitals exist for

n = 3

Error:

3 d

exists at

n = 3

but does not fill until after

4 s

(since

n + l = 5 > 4

). Filling starts once period 4 has begun, so the first d-row is in period 4.

"Period 1 has only 2 elements because we simply chose to stop there."

Error: it's forced, not chosen. For

n = 1

the only allowed

l

0

(rule

l = 0 \dots n - 1

), so only a

1 s

orbital exists → max 2 electrons → H and He, physically no room for more.

"Group 15 has 15 valence electrons."

Error: for main groups 13–18, valence

=

group

- 10

, so group 15 has 5 valence electrons (

n s^{2} n p^{3}

). Only groups 1–2 have valence equal to the group number.

"All periods have the same length because the table is a neat rectangle."

Error: lengths are 2, 8, 8, 18, 18, 32, 32. New subshell types (p at

n = 2

, d filling at period 4, f at period 6) become available as

n

grows, adding slots and lengthening lower rows.

"The f-block config is

(n - 1) f^{1 - 14}

, matching the d-block's

(n - 1) d

Error: f-block carries index

(n - 2)

, i.e.

(n - 2) f^{1 - 14}

. Because

4 f

fills during period 6, its index lags the period number by 2, not 1.

"A group-2 element like Mg gains 2 electrons to fill its shell."

Error: Mg has 2 valence electrons (

3 s^{2}

) and loses them to reach a noble-gas core, forming

Mg^{2 +}

. Gaining 6 to fill the shell is far less favourable than losing 2.

"Chromium is

[Ar] 3 d^{4} 4 s^{2}

because you fill 4s completely before touching 3d."

Error: Cr is actually

[Ar] 3 d^{5} 4 s^{1}

. One

4 s

electron shifts into

3 d

to give a stable half-filled

d^{5}

. The naive strict fill fails here — a reminder that the

n + l

order is a guide, not an ironclad law.

Why questions

Why is the factor of 2 in

2 (2 l + 1)

present?

Because each orbital holds two electrons of opposite spin (

m_{s} = + \frac{1}{2}

and

- \frac{1}{2}

); the

(2 l + 1)

counts the orbitals and the 2 counts electrons per orbital.

Why does the width

2 + 6 + 10 + 14

equal exactly 32?

These are the max electrons in s, p, d, f subshells; summing them gives the largest possible period once all four subshell types are simultaneously available (periods 6 and 7).

Why does dropping down a group preserve chemical behaviour but not exact reactivity?

The valence count and arrangement stay the same (similar bonding), but the outer shell sits farther from the nucleus and is more shielded, so reactivity shifts gradually (see Periodic trends — atomic radius, ionization energy).

Why do the p-orbitals not exist at

n = 1

A p-orbital needs

l = 1

, but the allowed angular momentum values are

l = 0 \dots n - 1

; for

n = 1

that range is only

{0}

, so

l = 1

is impossible and no

1 p

orbital can form.

Why does the group number connect to reactivity for groups 1, 17, and 18?

Group 1 has 1 loose valence electron (easily lost → +1), group 17 needs just 1 more (easily gained → −1), and group 18 already has a full octet (inert). Valence count sets how badly an atom wants to lose/gain electrons — see Valence electrons and chemical reactivity.

Why is knowing the filling order enough to "draw" the table?

Because the sequence of subshells that fill (via Aufbau principle and n+l rule) fixes both when a new row starts (new highest

n

) and where each block sits; the shape is a direct picture of that order.

Why are Cr and Cu drawn in the middle of the d-block even though their

4 s

is only half-filled?

Placement follows the number of d-block columns crossed (their group), which reflects total valence involvement — the small

4 s \to 3 d

shuffle for extra stability doesn't move them out of the transition-metal region.

Edge cases

For

n = 1

, list every allowed

l

and the resulting period length.

Only

l = 0

(the

1 s

orbital) is allowed, giving 2 electrons → period 1 has exactly 2 elements (H, He).

What is the valence configuration of a group-18 element in period 2, and why is it a boundary case?

Ne is

2 s^{2} 2 p^{6}

— a complete octet. It marks the end of period 2 because there is no

2 d

(

l = 2

needs

n \geq 3

), so the shell is full at 8 and the row must end.

Where does helium fall on the group→valence rule, and why does it break the pattern?

The rule "valence = group − 10" would give He (group 18) a valence of 8, but He only has 2 electrons (

1 s^{2}

). It's a genuine exception: its filled shell (only 2 for

n = 1

) makes it noble despite not having 8 — see Noble gases and octet rule.

What is the outer configuration at the exact start of the d-block (group 3, period 4), and what makes it special?

Sc is

[Ar] 3 d^{1} 4 s^{2}

— the first element whose last-added electron enters an inner

(n - 1) d

shell while the outer shell stays

4 s

. This is where filling order and period number first diverge.

At the join between periods, why does the highest

n

jump by exactly 1?

A new period begins when the next-lowest-energy orbital available is a fresh

n s

subshell (a shell not yet started); starting that shell raises the highest occupied

n

by one, which defines the new row.

Does the noble-gas core ever include a partly filled subshell?

No — a noble-gas core is always a set of completely filled subshells (e.g.

[Ar] = 1 s^{2} \dots 3 p^{6}

). That completeness is exactly what makes noble gases stable reference points for writing configs.

Why are the lanthanides and actinides usually drawn as two detached rows below the main table?

They are the f-block (filling

(n - 2) f

4 f

in period 6,

5 f

in period 7). Inserting all 14 f-columns inline would make the table 32 wide and unwieldy, so by convention the f-block is pulled out and placed below — it still belongs between groups 2 and 3 of periods 6 and 7.

If you inserted the f-block inline, how wide would periods 6 and 7 become, and where exactly would it sit?

Exactly 32 columns wide (

2 + 14 + 10 + 6

). The f-block would slot in right after the group-2 element (between

6 s

and

5 d

for period 6), i.e. between La-position and the start of the d-block — which is precisely the gap the footnote rows are drawn from.

Recall One-line self-test

Cover everything above. Can you state, for any element: (1) its period from the highest $n$ , (2) its block from where the last electron lands, and (3) its valence from the group rule — and name where each rule breaks (He, d-block start, the Cr/Cu anomalies, the pulled-out f-block)? If yes, you own this topic.