Blocks, groups and periods

Periods are the seven horizontal rows. A new period begins exactly when electrons start occupying a new principal shell. Period 1 holds only 2 elements (H, He; $n=1$). Periods 2 and 3 hold 8 each ($n=2,3$, only $s$ and $p$ subshells filling). Periods 4 and 5 hold 18 each, because the $d$-block now joins in. Periods 6 and 7 hold 32 each, because the $f$-block joins too. So the period number is simply the largest $n$ appearing in the configuration — no counting of electrons required.

Groups are the eighteen vertical columns. Elements in one group share the same pattern of valence-shell configuration, and it is that shared pattern — not shared atomic number — that gives them family chemistry. Group 1 (alkali metals) is always $ns^1$; Group 17 (halogens) is always $ns^2np^5$; Group 18 (noble gases) is always $ns^2np^6$ (except He, $1s^2$). Because the pattern repeats every time a shell fills the same way, the table has vertical families rather than just an unbroken sequence of 118 unrelated elements.

Blocks classify elements by where the differentiating electron lands in the Aufbau order:

• $s$-block — Groups 1–2, valence $ns^{1\text{ or }2}$, reactive metals.
• $p$-block — Groups 13–18, valence $ns^2np^{1\text{–}6}$, spans metals, metalloids, non-metals, and the noble gases.
• $d$-block (transition elements) — Groups 3–12, general valence $(n-1)d^{1\text{–}10}ns^{0\text{–}2}$; the differentiating electron enters an inner $(n-1)d$ subshell while the outer shell is already $n$.
• $f$-block (inner-transition elements) — lanthanides and actinides, shown as a separate strip below the main table; general valence $(n-2)f^{1\text{–}14}(n-1)d^{0\text{–}1}ns^2$.

Because group membership is really a statement about the valence-electron count, it can be written as a formula once you know the block. For a $p$-block element with valence configuration $ns^2np^x$ ($x = 1$ to $6$),

where: $x$ is the number of electrons in the outermost $p$ subshell (from 1 to 6), and the constant $12$ accounts for the two $ns$ electrons plus the ten already-filled $d$-electrons counted into the numbering convention for that period. For a $d$-block element with configuration $(n-1)d^{a}ns^{b}$, the rule is simpler still: $\text{Group} = a + b$, the total count of $d$- and $s$-electrons beyond the preceding noble-gas core.

The differentiating-electron rule is the reason this classification is consistent rather than a memorised list. The Aufbau order fills subshells roughly as $1s,2s,2p,3s,3p,4s,3d,4p,\dots$ — note that $4s$ fills before $3d$, even though $3d$ belongs to a lower shell number. This is exactly why the $d$-block "lags behind" by one period: a $d$-block element's outermost occupied shell is $ns$ (giving its period), but its last-added electron sits in the inner $(n-1)d$ subshell (giving its block). The same logic pushes the $f$-block two shells inward, $(n-2)f$, which is why lanthanides and actinides are drawn as a detached strip rather than widening the main table by fourteen more columns. Reading a configuration therefore separates two different questions cleanly: "which shell is outermost" answers the period, and "which subshell just received an electron" answers the block — group number then follows once the block tells you which valence pattern to count.

Question: An element has ground-state configuration $1s^2\,2s^2\,2p^6\,3s^2\,3p^4$. Identify its period, group, and block, and name the element.

Step 1. Count total electrons: $2+2+6+2+4 = 16$, so $Z=16$. This is sulphur.

Step 2. The highest principal shell occupied is $n=3$ (the $3s$ and $3p$ electrons), so the element is in Period 3.

Step 3. The valence configuration is $3s^23p^4$, matching the pattern $ns^2np^x$ with $x=4$. The last electron enters a $3p$ orbital, so the element is in the $p$-block. Applying formula (1): $\text{Group} = 12+4 = 16$.

Step 4. Answer: sulphur sits in Period 3, Group 16, $p$-block — a chalcogen, consistent with its known chemistry (it forms $\text{SO}_2$, $\text{H}_2\text{S}$, and sits directly below oxygen).

Competitive stretch

Question: Chromium ($Z=24$) has the "anomalous" ground-state configuration $[\text{Ar}]\,3d^5\,4s^1$ rather than the Aufbau-predicted $[\text{Ar}]\,3d^4\,4s^2$. Find its period, group, and block.

Step 1. $[\text{Ar}]$ accounts for $Z=18$; adding $3d^54s^1$ reaches $Z=24$.

Step 2. The outermost occupied shell is $n=4$ (the $4s^1$ electron), so chromium is in Period 4. The differentiating electron enters the inner $3d$ subshell, so it is a $d$-block element.

Step 3. Using the $d$-block rule $\text{Group} = a+b$ with $a=5$ ($3d^5$) and $b=1$ ($4s^1$): $\text{Group} = 5+1 = 6$.

Step 4. Answer: chromium is Period 4, Group 6, $d$-block, unchanged by the Hund's-rule anomaly — the half-filled $3d^5$ stability shifts one electron between subshells but the differentiating subshell is still $3d$, so the block assignment is unaffected.

• Placing helium in the $p$-block because it is drawn above the noble gases — its configuration is $1s^2$, so it is genuinely $s$-block; it is only positioned with Group 18 for family resemblance.
• Assuming every period has 8 elements — Periods 4–5 have 18 (the $d$-block joins) and Periods 6–7 have 32 (the $f$-block joins too).
• Swapping "period" (row, set by the highest $n$) with "group" (column, set by the valence pattern) — they answer different questions from the same configuration.
• Applying the simple Aufbau order blindly to $d$- and $f$-block elements — a handful (Cr, Cu, and some lanthanides/actinides) deviate for extra stability of half-filled or fully-filled subshells, but the block is still fixed by where the actual differentiating electron sits, not by the "expected" configuration.

Ask the tutor: if you want practice converting an unfamiliar electronic configuration straight into period, group, and block — or want more anomalous-configuration examples like chromium — ask the AI tutor for extra worked problems.