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

WHAT we do: just read off the two defining counts.

• A group is a vertical column. There are $18$ of them.
• A period is a horizontal row. There are $7$ of them.

Answer: 18 groups, 7 periods.

WHAT we do: a block is named after the orbital type receiving the last electron. The last electron went into a p orbital, so the element is in the p-block. (The "3" tells you it is in period 3, but the letter names the block.)

Width $=$ max electrons the orbital type holds $=2(2l+1)$.

WHY this rule: for groups 13–18 the leading "1" of the tens digit counts the filled inner s+p that don't participate, so we subtract 10. $\text{valence}=\text{group}-10=15-10=\boxed{5}$ Check with config: group 15 outer shell is $n{s}^{2}n{p}^3\Rightarrow 2+3=5$. ✔

WHAT fills in period 4: the subshells that come online are $4s$, then $3d$, then $4p$.

• $4s$: $2$ electrons
• $3d$: $10$ electrons
• $4p$: $6$ electrons

$2+10+6=\boxed{18}$ WHY longer than period 2: the extra $10$ slots from the newly-available $d$ subshell.

Block: the last electron (highest energy added) is in $4p$ → p-block. Period: highest $n$ occupied is $4$ (the $4s$ and $4p$) → period 4. Answer: p-block, period 4. (This is arsenic, group 15.)

Compute $n+l$:

• $4s$: $n=4,l=0\Rightarrow n+l=4$.
• $3d$: $n=3,l=2\Rightarrow n+l=5$.

Lower sum fills first: $4<5$, so $4s$ fills before $3d$. One-sentence why: because $4s$ (the period-4 outer shell) fills first, the atom has already entered period 4 before any $3d$ electron is added — so the first d-block row appears in period 4, and the filling electrons carry index $n-1=3$.

Compute $n+l$:

• $6s$: $6+0=6$
• $4f$: $4+3=7$
• $5d$: $5+2=7$ (tie with $4f$ — broken by lower $n$, so $4f$ before $5d$)

Order: $6s(6)<4f(7,n=4)<5d(7,n=5)$. So an atom enters period 6 (fills $6s$, $n=6$) before the $4f$ shell fills. The filling electron has $n=4$ while the period is $6$, giving index $n-2=6-2=4$. Hence $(n-2)f$.

$2+6+10+14=\boxed{32}$ This is the length of the longest periods (periods 6 and 7, counting the f-block). It confirms the block widths tile perfectly into the maximum row.

WHAT we do: fill 25 electrons in $n+l$ order: $1s2s2p3s3p4s3d\dots$ $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^5$ Count: $2+2+6+2+6+2+5=25$. ✔

• Period: highest $n=4$ → period 4.
• Block: last electron in $3d$ → d-block.
• Group: for a d-block atom, group $=$ (outer $s$) $+$ ($d$ electrons) $=2+5=\boxed{7}$.

Answer: period 4, d-block, group 7.

Group 1 → outer $n{s}^1$. Period 4 → $n=4$, so outer config $=4s^1$ (this is potassium, K). Valence electrons: $1$ → it loses that single electron easily to reach a noble-gas shell. Ion: ${\mathrm{K}}^{+}$ (charge $+1$). Reactivity: very reactive metal — one loosely-held $n{s}^1$ electron is cheap to remove, so it bonds eagerly (see Valence electrons and chemical reactivity).

The trap in the premise: $1s^2$ does end in $s^2$, so a config-only view puts it in group 2. WHY that fails: grouping follows chemical behaviour, and He's $n=1$ shell is completely full ($1s$ holds only 2, since for $n=1$ the only allowed $l=0$). A full outer shell means He is chemically inert, exactly like the noble gases. Answer: group 18, above neon — the filled-shell status outranks the $s^2$ ending.

Step 1 — which subshells exist at $n=2$? Allowed $l=0,1$ (since $l$ runs $0$ to $n-1=1$). Step 2 — orbitals per subshell ($2l+1$):

• $l=0$ (2s): $2(0)+1=1$ orbital
• $l=1$ (2p): $2(1)+1=3$ orbitals

Step 3 — electrons (2 per orbital, from spin $m_s=\pm \frac{1}{2}$): $2\times (1+3)=2\times 4=\boxed{8}$ So period 2 holds Li through Ne — 8 elements. The new $2p$ subshell (6 slots) is exactly what makes it longer than period 1.

WHAT we do: fill 63 electrons. Cumulative counts through the noble-gas cores:

• through $\mathrm{Ar}$ ($Z=18$): $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$
• $+4s^{2}3d^{10}4p^6$ → $Z=36$ (Kr)
• $+5s^{2}4d^{10}5p^6$ → $Z=54$ (Xe)
• remaining: $63-54=9$ electrons → $6s^2$ (now $56$), then $4f$.

After $6s^2$ we have $63-56=7$ electrons left, all entering $4f$: $\Rightarrow 4f^7$. Full outer part: $[\mathrm{Xe}]6s^{2}4f^7$.

• Period: highest $n=6$ → period 6.
• Block: last electron in $4f$ → f-block.

Answer: period 6, f-block (a lanthanide, index $n-2=4$). ✔

New subshell: $l=4$ (g-orbital). Orbitals $=2l+1=2(4)+1=9$ → electrons $=2\times 9=18$. Period 8 capacity (all five types once available): $\underset{s}{\underbrace{2}}+\underset{p}{\underbrace{6}}+\underset{d}{\underbrace{10}}+\underset{f}{\underbrace{14}}+\underset{g}{\underbrace{18}}=\boxed{50}$ So a period containing a g-block would hold 50 elements — the pattern $2,8,8,18,18,32,32,50$ continues by adding the next $2(2l+1)$ slice.