Electronic Configurations & the s-, p-, d-, f-Block Classification

From Quantum Numbers to Periods: The Aufbau Principle

If you want to derive the structure of the periodic table from scratch — not just memorise it — the starting point is the Aufbau Principle. Aufbau is German for "building up", and that is exactly the idea: we imagine constructing each neutral atom by adding electrons one at a time into the lowest-energy orbital that is still available.

Why orbital energies matter

In a hydrogen atom there is only one electron, so orbital energies depend only on the principal quantum number $n$: $1s < 2s = 2p < 3s = 3p = 3d < \dots$ The $2s$ and $2p$ are degenerate, the $3s$, $3p$ and $3d$ are all degenerate, and so on.

But in every multi-electron atom (Z > 1), the electrons screen each other. Orbitals that were degenerate in hydrogen split apart: within a given shell, the $s$ orbital dips below the $p$, which dips below the $d$, which dips below the $f$. Even more dramatically, the $4s$ orbital slides below the $3d$ because it penetrates deeper toward the nucleus. The practical filling order we actually observe is:

$1s \to 2s \to 2p \to 3s \to 3p \to 4s \to 3d \to 4p \to 5s \to 4d \to 5p \to 6s \to 4f \to 5d \to 6p \to 7s \to 5f \to 6d \to 7p$

The $(n + \ell)$ rule — Madelung's rule

A tidy way to remember the order is the $n + \ell$ rule (also called Madelung's rule or the Klechkowski rule):

Rule 1. Orbitals fill in order of increasing $(n + \ell)$, where $\ell$ is the azimuthal quantum number: $\ell = 0$ for $s$, $1$ for $p$, $2$ for $d$, $3$ for $f$.

Rule 2. If two orbitals have the same $(n + \ell)$ value, the one with the lower $n$ fills first.

Worked check: $4s$ has $n + \ell = 4 + 0 = 4$, while $3d$ has $n + \ell = 3 + 2 = 5$. So $4s$ fills before $3d$ — even though 3d seems "lower" by shell number. That is why Period 4 opens with $\text{K}$ ($4s^{1}$) and $\text{Ca}$ ($4s^{2}$) before the 3d block (Sc–Zn) even begins.

Two companion rules you already know

Aufbau is about which orbital gets the next electron. Two partner rules dictate how electrons settle inside a given orbital:

• Pauli's Exclusion Principle: No two electrons in an atom can have identical values of all four quantum numbers $(n, \ell, m_{\ell}, m_{s})$. In practice: a single orbital can hold at most two electrons, and they must have opposite spins.
• Hund's Rule of Maximum Multiplicity: When electrons are filling a set of degenerate orbitals (e.g. the three $2p$ orbitals), they enter one-at-a-time with parallel spins before any orbital is double-occupied.

Put the three rules together and you can predict the ground-state configuration of any atom up to Z ≈ 120 without looking it up.

[JEE Tip] A handful of atoms break the Aufbau order because a half-filled or fully-filled sub-shell is especially stable. The textbook cases are $\text{Cr}$ ($3d^{5}\,4s^{1}$ instead of $3d^{4}\,4s^{2}$) and $\text{Cu}$ ($3d^{10}\,4s^{1}$ instead of $3d^{9}\,4s^{2}$). Similar exceptions occur at $\text{Mo}, \text{Ag}, \text{Pd}$ and in several actinides. These are real configurations — JEE loves to test them.

Why Aufbau Explains Periods and Blocks

The Aufbau order is not just a convenience for remembering atomic configurations — it is the very reason the periodic table has the shape it has.

Period length = shell capacity at that filling stage

A new period begins every time an electron starts a new $ns$ orbital (where $n$ is the next integer). A period ends at the noble gas whose $np$ orbital is full. The length of the period is therefore the total number of orbitals that open up between those two events:

Opens with	Orbitals filled	Elements
$1s$	$1s$	2
$2s$	$2s,\ 2p$	8
$3s$	$3s,\ 3p$	8
$4s$	$4s,\ 3d,\ 4p$	18
$5s$	$5s,\ 4d,\ 5p$	18
$6s$	$6s,\ 4f,\ 5d,\ 6p$	32
$7s$	$7s,\ 5f,\ 6d,\ 7p$	32

The familiar sequence $2, 8, 8, 18, 18, 32, 32$ falls out automatically. The doubling at Periods 4 and 6 is the $d$- and $f$-block's contribution — each doubling corresponds to one new kind of sub-shell coming online.

Block = which sub-shell the last electron enters

Every element lives in exactly one block, determined by the sub-shell type ($s, p, d$ or $f$) that accepts the last electron added in the Aufbau order. This is the rule we hinted at in Section 3; now we can state it crisply:

$$\text{Block}(Z) \;=\; \text{orbital type of the last-filled electron in the ground-state configuration of atom } Z.$$

[Board Level] The rule in words: write the configuration of the neutral atom in Aufbau order; look at the very last orbital with non-zero occupancy; the letter on that orbital is the block.

Group within a block

Within a given block, the group number is read off the number of valence electrons:

• s-block: group = number of valence $ns$ electrons (1 or 2)
• p-block: group = 10 + (valence $ns$ + valence $np$ electrons) → gives 13 to 18
• d-block: group = (valence $ns$ electrons) + ($(n-1)d$ electrons) → gives 3 to 12
• f-block: no single group number is assigned; lanthanides and actinides are all considered to belong to Group 3 of their period (6 or 7 respectively)

Memorising these four one-liners is often enough to answer the whole "place this element" family of JEE/NEET problems in one line.

Predicting an Element's Block, Period and Group from its Electronic Configuration

Given the atomic number $Z$ of any neutral atom, you can place it in the periodic table using exactly the same four-step recipe every time. Once you internalise this recipe, placing an unknown element becomes a 30-second exercise.

The recipe

Step 1. Write the ground-state electronic configuration using the Aufbau diagonal rule. Carry it out digit by digit — do not skip ahead.

Step 2. Identify the period as the largest principal quantum number $n$ that appears in the configuration.

Step 3. Identify the block as the sub-shell type (s / p / d / f) of the last-entering electron.

Step 4. Compute the group number from the block-specific rule of Section 4.2 above.

Example walk-throughs

Let's run the recipe for three different atomic numbers — one from each of the common test regions.

Case A — $Z = 25$

• Configuration: $1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{5}$.
• Largest $n = 4$, so Period 4.
• Last-entering electron is in $3d^{5}$ — the last filled sub-shell written in Aufbau order is $3d$ — so d-block.
• Group = (valence $ns$) + ($(n−1)d$) = 2 + 5 = Group 7.
• The element is manganese (Mn).

Case B — $Z = 52$

• Configuration: $[\text{Kr}]\,4d^{10}\,5s^{2}\,5p^{4}$.
• Largest $n = 5$, so Period 5.
• Last-entering electron is in $5p$ — p-block.
• Group = 10 + (valence $s$ + valence $p$) = 10 + (2 + 4) = Group 16.
• The element is tellurium (Te).

Case C — $Z = 92$

• Configuration: $[\text{Rn}]\,5f^{3}\,6d^{1}\,7s^{2}$.
• Largest $n = 7$, so Period 7.
• Last-entering electron is in $5f$ — f-block (actinide series).
• By IUPAC convention, placed in Group 3 of Period 7.
• The element is uranium (U).

Dealing with the anomalous configurations

When a Cr-type or Cu-type anomaly appears, use the actual ground-state configuration rather than the naive Aufbau configuration — the answer won't change the block or group, only the written form of the configuration:

• $\text{Cr}\ (Z = 24): \;[\text{Ar}]\,3d^{5}\,4s^{1}$ (not $3d^{4}\,4s^{2}$) — still d-block, Group 6.
• $\text{Cu}\ (Z = 29): \;[\text{Ar}]\,3d^{10}\,4s^{1}$ (not $3d^{9}\,4s^{2}$) — still d-block, Group 11.

[NEET Important] The Aufbau-predicted placement and the experimentally-observed placement always agree on block, period and group, even when the anomaly shifts a single electron between $ns$ and $(n{-}1)d$. Examiners exploit this — they give you an "anomalous" element and ask for its group, hoping you will either place it wrongly or skip the question. Use the recipe, ignore the anomaly, and move on.

General Valence-Shell Configurations of the Four Blocks

Every block has a general valence-shell configuration: a single symbolic formula that captures what all its members have in common. Knowing these four formulas (and the example atoms attached to each) is arguably the highest-yield piece of Class 11 chemistry.

s-block — Groups 1 and 2

• General valence config: $ns^{1-2}$.
• Members: H, Li, Na, K, Rb, Cs, Fr (Group 1, alkali metals, $ns^{1}$) and Be, Mg, Ca, Sr, Ba, Ra (Group 2, alkaline-earth metals, $ns^{2}$).
• Typical behaviour: soft, silvery, highly reactive metals. Lose the outer 1–2 electrons easily → low ionisation enthalpies, strongly positive $\text{M}^{+}$ or $\text{M}^{2+}$ ions, exclusively ionic compounds.
• Example: Sodium (Z = 11) has configuration $[\text{Ne}]\,3s^{1}$ — one valence electron in the $3s$ orbital — Period 3, Group 1.

p-block — Groups 13 to 18

• General valence config: $ns^{2}\,np^{1-6}$ (helium is the exception — it is $1s^{2}$ only).
• Members: every group from 13 (Boron family) through 18 (Noble gases) — 36 elements in total spanning Periods 2–7.
• Typical behaviour: the most chemically diverse block — contains non-metals, metalloids, post-transition metals and noble gases. Oxidation states are highly variable (differing by 2 because of the inert-pair effect).
• Example: Bromine (Z = 35) has configuration $[\text{Ar}]\,3d^{10}\,4s^{2}\,4p^{5}$ — last electron in $4p$ — Period 4, Group 17.

d-block — Groups 3 to 12

• General valence config: $(n-1)d^{1-10}\,ns^{0-2}$.
• Members: the 40 "transition metals" filling the middle block of the long form, across Periods 4–7. Formally: elements whose atoms or whose common ions have a partially filled $d$-subshell (IUPAC definition).
• Typical behaviour: hard, high-melting metals; variable oxidation states (e.g. Mn: +2, +3, +4, +6, +7); coloured ions from $d{-}d$ transitions; form complex ions with ligands; catalytic activity.
• Example: Iron (Z = 26) has configuration $[\text{Ar}]\,3d^{6}\,4s^{2}$ — last electron in $3d$ — Period 4, Group 8.

f-block — inner transition metals

• General valence config: $(n-2)f^{1-14}\,(n-1)d^{0-1}\,ns^{2}$.
• Members: the 14 lanthanides (Ce–Lu, filling $4f$) in Period 6 and the 14 actinides (Th–Lr, filling $5f$) in Period 7 — 28 elements in total, shown as two separate rows below the main table.
• Typical behaviour: predominantly show +3 oxidation state (lanthanides); actinides are all radioactive and post-U are man-made. Subtle size-contraction effects (lanthanide contraction, Section 8) have far-reaching consequences.
• Example: Europium (Z = 63) has configuration $[\text{Xe}]\,4f^{7}\,6s^{2}$ — last electron in $4f$ — Period 6, lanthanide series.

Quick count check: 14 + 36 + 40 + 28 = 118 — exactly the number of known elements. ✓

Why Noble Gases Occupy Group 18

Of every column in the periodic table, Group 18 is the easiest to define electronically: its members are the atoms whose outermost electronic configuration is a completely-filled valence shell. Everyone else in the periodic table wants to either lose or gain electrons to reach that stable configuration — which is why noble gases, having already arrived, are so unreactive.

The "stable octet" (and the duet exception)

Every noble gas except helium has the valence configuration $ns^{2}\,np^{6}$ — eight valence electrons arranged as a filled $s$ and a filled $p$ sub-shell. We call this the octet. Helium, sitting in Period 1 where no $2p$ orbital exists yet, is satisfied with $1s^{2}$ — a duet of 2 electrons. In both cases the outermost shell is completely closed, and that closure is the chemical "signal" for extreme stability.

Noble gas	Z	Valence configuration	Octet / duet
He	2	$1s^{2}$	duet
Ne	10	$[\text{He}]\,2s^{2}\,2p^{6}$	octet
Ar	18	$[\text{Ne}]\,3s^{2}\,3p^{6}$	octet
Kr	36	$[\text{Ar}]\,3d^{10}\,4s^{2}\,4p^{6}$	octet
Xe	54	$[\text{Kr}]\,4d^{10}\,5s^{2}\,5p^{6}$	octet
Rn	86	$[\text{Xe}]\,4f^{14}\,5d^{10}\,6s^{2}\,6p^{6}$	octet
Og	118	$[\text{Rn}]\,5f^{14}\,6d^{10}\,7s^{2}\,7p^{6}$	octet

Three observable consequences

1. Very high ionisation enthalpy. Pulling an electron out of a closed shell is energetically costly. $\text{He}$ has IE₁ = 2372 kJ/mol, the highest of any element.
2. Near-zero electron affinity. Adding an electron forces it into a higher, unoccupied shell — so noble gases have weakly positive (endothermic) electron-gain enthalpies.
3. Extreme chemical inertness. The noble gases were historically thought to be completely non-reactive — until Bartlett's $\text{XePtF}_{6}$ in 1962. Even today, their chemistry is essentially confined to Xe and Kr compounds with the most electronegative partners (F and O).

Group 0 or Group 18?

Mendeleev labelled the column Group 0 because it seemed to have zero valency. Once xenon and krypton were shown to form real compounds with fluorine and oxygen, the "zero" label became wrong. IUPAC retired it in 1988 and renamed the column Group 18 — a purely positional label that makes no chemical claim about zero valency. Modern NCERT and all JEE/NEET mark schemes use "Group 18".

Why exactly at the right edge of every period?

Because every period ends when $np$ fills. The last electron of every noble gas (except He) enters $np^{6}$ — and that is also the last orbital available at the current $n$. The next electron must jump to the next shell, i.e. start a new period. That is why the noble-gas column is always the rightmost one, and why it appears once in every row.

[JEE Tip] Any question asking "which element begins/ends a period?" can be reduced to this rule: the one that ends a period always has the generic closed-shell configuration shown above; the one that begins the next period has configuration [noble_gas_core] ns¹.

Exceptions, Caveats and Edge Cases

A handful of atoms break the naive Aufbau filling order, and a handful of column assignments are debated. These are the highest-value "watch-out" items to learn for the exam.

Anomalous configurations: Cr, Cu, Mo, Pd, Ag, Au, and a few actinides

• Chromium (Z = 24): observed $[\text{Ar}]\,3d^{5}\,4s^{1}$, not $3d^{4}\,4s^{2}$. A half-filled d-shell (exactly 5 electrons, one in each orbital with parallel spins) gives extra exchange-energy stability, enough to overcome the tiny $4s$–$3d$ promotion cost.
• Copper (Z = 29): observed $[\text{Ar}]\,3d^{10}\,4s^{1}$, not $3d^{9}\,4s^{2}$. A fully-filled d-shell is similarly stable.
• Molybdenum (Z = 42): observed $[\text{Kr}]\,4d^{5}\,5s^{1}$ — same reason as Cr.
• Silver (Z = 47): observed $[\text{Kr}]\,4d^{10}\,5s^{1}$ — same reason as Cu.
• Palladium (Z = 46): observed $[\text{Kr}]\,4d^{10}\,5s^{0}$ — a double anomaly, giving Pd a noble-gas-like closed $4d^{10}$.
• Gold (Z = 79): observed $[\text{Xe}]\,4f^{14}\,5d^{10}\,6s^{1}$ — fully-filled $5d$.

Where does hydrogen belong?

Hydrogen ($1s^{1}$) is a unique element. Its configuration is identical to the alkali metals ($ns^{1}$), so it is placed at the top of Group 1 in the printed NCERT table. However:

• It also has one electron short of the nearest noble-gas configuration ($\text{He}, 1s^{2}$), just like halogens need one more electron. Under this view, H would sit at the top of Group 17.
• Unlike alkali metals, H is a gas (not a solid) and a non-metal.
• Unlike halogens, H does not form stable -1 ions easily (hydride $\text{H}^{-}$ exists only in ionic hydrides like $\text{NaH}$).

NCERT's official placement: top of Group 1, with a footnote. JEE/NEET treat this dual character as examinable — expect at least one question on "where would you place H and why?" every cycle.

Where does helium belong?

By configuration, $\text{He} = 1s^{2}$ is an s-block element. By chemistry it is a noble gas — completely inert. IUPAC places it in Group 18 of the printed table but remarks that it is formally s-block. Watch for JEE questions that exploit this split: helium is the only element whose block and group classifications disagree on the block.

Where does lanthanum (and actinium) belong?

Strictly by the last-entering-electron rule, $\text{La}\ (Z = 57)$ has configuration $[\text{Xe}]\,5d^{1}\,6s^{2}$ — last electron in $5d$, so La is a d-block element (Group 3, Period 6). But chemically La behaves almost identically to the lanthanides that follow it (Ce–Lu). Most periodic tables — including NCERT's — show La sitting inside the Group 3 cell of Period 6 with an arrow pointing to the lanthanide row; some older tables draw La in the main body and Lu underneath.

The same ambiguity affects $\text{Ac}\ (Z = 89)$. For JEE/NEET, follow the last-entering-electron rule — count La as d-block, and answer "d-block" on multiple-choice questions. Board-level marking schemes accept either.

[Board Level] CBSE tends to award full marks for any internally-consistent answer, as long as you explain your reasoning. JEE/NEET single-answer MCQs follow IUPAC's rule strictly — d-block for La and Ac.

Solved Examples

Example 1: Aufbau order — which orbital fills first?

For each of the following orbital pairs, decide which fills first in the ground-state Aufbau order and justify using the $n + \ell$ rule.

(a) $3d$ vs $4s$ (b) $4f$ vs $6s$ (c) $5d$ vs $6p$ (d) $6s$ vs $5p$

Solution:

Apply the $(n + \ell)$ rule: lower $(n + \ell)$ fills first; ties are broken by smaller $n$.

Pair	$(n + \ell)$ values	Which fills first
(a) $3d$ vs $4s$	5 vs 4	$4s$
(b) $4f$ vs $6s$	7 vs 6	$6s$
(c) $5d$ vs $6p$	7 vs 7 → lower $n$ wins	$5d$
(d) $6s$ vs $5p$	6 vs 6 → lower $n$ wins	$5p$

So the filling order of the orbitals involved is $4s, 5p, 6s, 5d, 4f$ — exactly what the Aufbau diagonal chart predicts.

Example 2: Configuration & placement of Z = 17

Write the ground-state electronic configuration of chlorine (Z = 17) and use it to state its period, block and group.

Solution:

Apply the Aufbau order up to Z = 17:

$$1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{5}$$

• Largest $n = 3$, so Period 3.
• Last-entering electron is in $3p$, so p-block.
• Group (for p-block) = 10 + (valence $s$ + valence $p$) = 10 + (2 + 5) = Group 17 (halogens).

Chlorine is indeed a Period 3, Group 17 halogen. ✓

Example 3: Anomalous configuration — chromium

The naive Aufbau configuration for Z = 24 predicts $[\text{Ar}]\,3d^{4}\,4s^{2}$, but the experimentally observed configuration is $[\text{Ar}]\,3d^{5}\,4s^{1}$. Explain why, and state the block and group of chromium.

Solution:

The observed configuration places an electron from $4s$ into $3d$ to give a half-filled $d$-subshell ($3d^{5}$). Half-filled (and fully-filled) sub-shells are unusually stable because:

• Exchange-energy stabilisation: with five parallel-spin electrons in five separate $d$-orbitals, the number of electron-exchange interactions (each of which lowers energy) is maximised.
• Symmetric charge distribution: a half-filled shell has a relatively symmetric distribution, helping reduce electron-electron repulsion.

The gain from these two effects outweighs the small cost of promoting one $4s$ electron to the slightly higher $3d$. The block/group assignment is unchanged by the anomaly:

• Period = 4
• Block = d-block
• Group = valence $ns$ electrons + $(n{-}1)d$ electrons = $1 + 5 = 6$

So Cr is Period 4, d-block, Group 6.

Example 4: Placing Z = 38 — strontium

Without using a periodic table, determine the period, block, group and likely oxidation state of the element with atomic number 38.

Solution:

Ground-state Aufbau configuration of Z = 38:

$$[\text{Kr}]\,5s^{2}$$

• Largest $n = 5$ → Period 5.
• Last electron in $5s$ → s-block.
• Group = number of valence $ns$ electrons = Group 2 (alkaline-earth metals).
• Likely oxidation state: +2 (losing both $5s$ electrons to attain the Kr noble-gas core).

The element is strontium (Sr), a Period 5 alkaline-earth metal that forms $\text{Sr}^{2+}$ ions.

Example 5: p-block group from valence electrons

An atom has the valence configuration $ns^{2}\,np^{3}$. Which group does it belong to? Name the p-block family with this general configuration.

Solution:

For p-block elements, group number = 10 + (valence $s$ + valence $p$ electrons) = 10 + (2 + 3) = Group 15.

Group 15 is the pnictogens — N, P, As, Sb, Bi, Mc — famous for their $ns^{2}\,np^{3}$ half-filled $p$-subshell and the +3 / +5 oxidation-state pattern.

Example 6: Decoding a d-block configuration

An element has configuration $[\text{Ar}]\,3d^{7}\,4s^{2}$. Identify the element and state its period, group and block.

Solution:

Total electrons = 18 (Ar core) + 7 + 2 = 27 → the element is cobalt (Co, Z = 27).

• Largest $n = 4$ → Period 4.
• Last-entering electron in $3d$ → d-block.
• Group for d-block = valence $ns$ + $(n{-}1)d$ = 2 + 7 = Group 9.

Cobalt sits in Period 4, Group 9 of the modern periodic table. ✓

Example 7: f-block placement — europium

The ground-state configuration of an element is $[\text{Xe}]\,4f^{7}\,6s^{2}$. Identify the element, its period, block and its place in the lanthanide series.

Solution:

Total electrons = 54 + 7 + 2 = 63 → the element is europium (Eu, Z = 63).

• Largest $n = 6$ → Period 6.
• Last-entering electron is in $4f$ → f-block (specifically, the lanthanide series).
• Formally assigned to Group 3 of Period 6.
• $4f^{7}$ is a half-filled $f$-subshell, which gives Eu its famously stable +2 state (unusual for lanthanides, which usually prefer +3).

Eu is the 6th lanthanide if counted from Ce = 58 (Ce, Pr, Nd, Pm, Sm, Eu).

Example 8: Period & group from Z alone (JEE Main style)

Predict the period and group of the element with $Z = 83$ and name it.

Solution:

Ground-state Aufbau configuration of Z = 83:

$$[\text{Xe}]\,4f^{14}\,5d^{10}\,6s^{2}\,6p^{3}$$

• Largest $n = 6$ → Period 6.
• Last electron in $6p$ → p-block.
• Group = 10 + (valence $s$ + valence $p$) = 10 + (2 + 3) = Group 15.

The element is bismuth (Bi), the heavy pnictogen of Period 6.

Example 9: Why noble gases end every period

Show using Aufbau arguments that every period of the periodic table must end at an $np^{6}$ configuration (except Period 1).

Solution:

Periods are organised in the Aufbau order. Look at the sub-shells that fill consecutively within a given period:

• Period 2: $2s, 2p$ — fills $2p^{6}$ last, ending at Ne.
• Period 3: $3s, 3p$ — fills $3p^{6}$ last, ending at Ar.
• Period 4: $4s, 3d, 4p$ — $4p^{6}$ is the very last Aufbau step before $5s$ begins, ending at Kr.
• Period 5: $5s, 4d, 5p$ — last step $5p^{6}$, ending at Xe.
• Period 6: $6s, 4f, 5d, 6p$ — last step $6p^{6}$, ending at Rn.
• Period 7: $7s, 5f, 6d, 7p$ — last step $7p^{6}$, ending at Og.

In every period from 2 onwards, the next Aufbau step after $np^{6}$ is $(n{+}1)s^{1}$ — which by definition starts a new period. Period 1 is the exception because the $1p$ orbital does not exist; Period 1 ends at $1s^{2}$ (He), and the next element (Li) starts $2s^{1}$.

This is the electronic reason that Group 18 always sits at the rightmost edge of the periodic table.

Example 10: Noble-gas core shorthand

Write the electronic configuration of $Z = 35$ (bromine) in full form and then in noble-gas core shorthand. Verify that both forms describe the same atom.

Solution:

Full form, Aufbau order:

$$1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{5}$$

Sum of superscripts: $2 + 2 + 6 + 2 + 6 + 2 + 10 + 5 = 35$. ✓

Noble-gas core shorthand: use the symbol of the preceding noble gas (Ar, $Z = 18$) as the core:

$$[\text{Ar}]\,3d^{10}\,4s^{2}\,4p^{5}$$

Since $[\text{Ar}]$ stands for the full Ar configuration ($1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}$, 18 electrons), the shorthand describes exactly the same 35-electron atom. Either form shows the last-entering electron in $4p$ (p-block) and the highest $n$ as 4 (Period 4).

Example 11: General valence configuration → group

An atom has the general valence configuration $(n{-}1)d^{5}\,ns^{1}$. State its block, its group and identify the element in Period 4.

Solution:

• Block: last-entering electron sits in $(n{-}1)d$, so d-block.
• Group: valence $ns$ + $(n{-}1)d$ = 1 + 5 = Group 6.
• Element in Period 4: $n = 4$ → configuration $[\text{Ar}]\,3d^{5}\,4s^{1}$ → this is the anomalous configuration of chromium (Cr), $Z = 24$.

The same general config in Period 5 would give $[\text{Kr}]\,4d^{5}\,5s^{1}$ → molybdenum (Mo), another anomaly.

Example 12: CBSE Board 3-marker — state the rule & apply it

(a) State the Aufbau Principle and write the order in which atomic orbitals up to $4p$ are filled. (b) Using the Aufbau order, write the ground-state configuration of $Z = 22$ and identify its block, period and group.

Solution:

(a) The Aufbau Principle states: "In the ground state of an atom, electrons occupy atomic orbitals in order of increasing energy, filling the lowest-energy orbital that is still available before any higher-energy orbital is used."

Using the $(n+\ell)$ rule, orbitals up to $4p$ fill in the order:

$$1s \to 2s \to 2p \to 3s \to 3p \to 4s \to 3d \to 4p$$

(b) $Z = 22$ → titanium. Configuration:

$$1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{2}$$

Check: $2+2+6+2+6+2+2 = 22$ ✓.

• Largest $n = 4$ → Period 4.
• Last-entering electron in $3d$ → d-block.
• Group = valence $s$ + $(n{-}1)d$ = 2 + 2 = Group 4.

Titanium sits in Period 4, Group 4 of the modern periodic table. (Board mark-schemes typically award 1 mark for the statement, 1 mark for the filling order, and 1 mark for the correct placement.)