Indicate the number of unpaired electrons in: (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.

Exercises · 2.66

Q. Indicate the number of unpaired electrons in:

(a) P,

(b) Si,

(d) Fe and

(e) Kr.

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Concept understanding — Electron Configuration

Electron Configuration: Where Do Electrons Actually Live?

Imagine a school building. Students don't just wander randomly — they sit in specific classrooms, on specific floors, in specific rows. Electrons in an atom behave similarly. They don't buzz around the nucleus chaotically. They occupy specific energy levels (floors), sublevels (classrooms), and orbitals (seats).

The electron configuration is simply the address system that tells you exactly which "seats" every electron in an atom is sitting in.

The Intuition: Why Can't Electrons Sit Anywhere?

Two big rules force electrons into this orderly arrangement:

Energy matters. Electrons want to be as close to the nucleus as possible (lowest energy). The first "floor" (n=1) is the most comfortable. Higher floors cost more energy.
No crowding. A famous rule called the Pauli Exclusion Principle says: no two electrons in the same atom can have the exact same set of four quantum numbers. In plain language: each orbital (seat) can hold at most two electrons, and they must spin in opposite directions.

So electrons fill up from the bottom floor upward, like students filling a theatre from the front row back.

The Precise Statement

n

For example, the configuration of carbon (6 electrons) is:

1s^2 \; 2s^2 \; 2p^2

This reads: "Two electrons in the 1s orbital, two in the 2s orbital, and two in the 2p orbitals."

The Filling Order: The Aufbau Principle

Electrons don't fill levels in simple numerical order. Here's the actual sequence (memorise this — it's exam gold):

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

Notice: 4s fills before 3d. This is because the 4s orbital is actually lower in energy than 3d. This trips up many students.

Tip

Use the diagonal rule (Madelung's rule) to remember the order: draw arrows diagonally across the $(n+\ell)$ chart. Or just remember the mnemonic: "Silly People Don't Fail" for the sublevel order within each shell.

How to Write Any Configuration (Step-by-Step)

Let's do iron (Fe, atomic number 26).

Step 1: Know the total electrons = 26.

Step 2: Follow the filling order, counting electrons as you go:

$1s^2$ (2 used, 24 left)
$2s^2$ (4 used, 22 left)
$2p^6$ (10 used, 16 left)
$3s^2$ (12 used, 14 left)
$3p^6$ (18 used, 8 left)
$4s^2$ (20 used, 6 left)
$3d^6$ (26 used, 0 left)

Step 3: Write it in order of increasing $n$ (standard notation):

1s^2 \; 2s^2 \; 2p^6 \; 3s^2 \; 3p^6 \; 3d^6 \; 4s^2

Watch out

Many textbooks write configurations in order of filling (4s before 3d), but IUPAC standard lists them by principal quantum number $n$ (3d before 4s). Check which convention your exam uses. For CBSE/ICSE, write in order of increasing $n$ : $1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p...$

The Three Golden Rules (Memorise These)

Rule	What it says	Why it matters
Aufbau Principle	Electrons fill lowest energy orbitals first	Determines the order of filling

Why this formula?

Electron Configuration: Why the Rules Work

Let's build this from the ground up — not just what the rules are, but why they exist.

The Core Question

Why do electrons arrange themselves in specific shells, subshells, and orbitals — and not just pile up anywhere?

The answer lies in three fundamental principles, each rooted in physics and quantum mechanics.

1. The Aufbau Principle: Why "Build Up" in Order?

What it says: Electrons fill orbitals from lowest to highest energy.

Why it holds: Nature seeks the lowest possible energy state (ground state). An atom is most stable when its electrons occupy the lowest available energy levels.

Think of it like water flowing downhill — electrons "fall" into the lowest energy orbitals first.

The energy ordering (for multi-electron atoms) is:

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s...

Why this order? It comes from the $(n + \ell)$ rule:

$n$ = principal quantum number (shell)
$\ell$ = azimuthal quantum number (subshell: $s=0$ , $p=1$ , $d=2$ , $f=3$ )

Orbitals fill in order of increasing $(n + \ell)$ . If two have the same $(n + \ell)$ , the one with lower $n$ fills first.

Example: $4s$ has $(4+0)=4$ , $3d$ has $(3+2)=5$ . So $4s$ fills before $3d$ — even though $4s$ is a higher shell number.

2. Pauli Exclusion Principle: Why Only Two Per Orbital?

What it says: No two electrons in an atom can have the same set of all four quantum numbers.

Why it holds: This is a fundamental law of quantum mechanics — electrons are fermions (spin-1/2 particles). Fermions obey the Pauli exclusion principle, which arises from the antisymmetry of the wavefunction.

The four quantum numbers:

$n$ (shell)
$\ell$ (subshell shape)
$m_\ell$ (orbital orientation)
$m_s$ (spin: $+\frac{1}{2}$ or $-\frac{1}{2}$ )

Since only $m_s$ can differ for electrons in the same orbital, maximum 2 electrons per orbital — one spin-up ( $\uparrow$ ) and one spin-down ( $\downarrow$ ).

Key result: The $s$ subshell ( $\ell=0$ , one orbital) holds 2 electrons. The $p$ subshell ( $\ell=1$ , three orbitals) holds 6 electrons. The $d$ subshell ( $\ell=2$ , five orbitals) holds 10 electrons.

3. Hund's Rule: Why Spread Out First?

What it says: Within a subshell, electrons occupy empty orbitals singly before pairing up — and all unpaired electrons have parallel spins.

Why it holds: Electrons repel each other (Coulomb repulsion). By occupying different orbitals, they stay farther apart, reducing repulsion energy.

The spin alignment (all parallel) comes from exchange energy — a quantum mechanical effect where parallel spins have a slightly lower energy state due to wavefunction symmetry.

Example for carbon ( $1s^2 2s^2 2p^2$ ):

Correct: $\uparrow$ | $\uparrow$ | (two unpaired, parallel)
Wrong: $\uparrow\downarrow$ | | (paired in one orbital — higher repulsion)

The Big Picture: Why These Three Rules Together? …

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The number of unpaired electrons is found by writing the ground-state electron configuration and applying Hund’s rule. The results are: P → 3, Si → 2, Cr → 6, Fe → 4, Kr → 0.

Why electron configuration tells you the number of unpaired electrons

Every atom in its ground state fills orbitals in a specific order — 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, and so on. Within a subshell, electrons occupy orbitals singly before pairing up (Hund’s rule). The unpaired electrons are simply the ones that sit alone in an orbital. So the task reduces to writing the configuration and counting the singles.

A common trap: the 4s orbital fills before 3d, but when writing configurations for atoms beyond calcium, we list 3d before 4s because 3d is lower in energy once occupied. For counting unpaired electrons, the order of writing doesn’t matter — what matters is which orbitals are singly occupied.

Let’s go element by element.

1. Phosphorus (P, atomic number 15)

Configuration: $1s^2 2s^2 2p^6 3s^2 3p^3$

The 3p subshell has three orbitals ( $p_x, p_y, p_z$ ). With three electrons, Hund’s rule says each goes into a separate orbital, all with parallel spins. So all three are unpaired.

Watch out

Do not pair any electrons in the 3p subshell — three electrons in three orbitals means three singles, not one pair plus one single.

Unpaired electrons in P: 3

2. Silicon (Si, atomic number 14)

Configuration: $1s^2 2s^2 2p^6 3s^2 3p^2$

The 3p subshell has two electrons. They occupy two different p orbitals (say $p_x$ and $p_y$ ), with parallel spins. No pairing yet.

Unpaired electrons in Si: 2

3. Chromium (Cr, atomic number 24)

Here is where the textbook exception lives. The expected configuration would be $[Ar] 4s^2 3d^4$ , but that is not the ground state. Half-filled subshells are extra stable, so one electron from the 4s moves into the 3d, giving:

$[Ar] 4s^1 3d^5$

Now the 3d subshell has five orbitals, each with one electron — all unpaired. The 4s has one electron, also unpaired. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 36 on this concept.

AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$X$
›Reveal solutionSolution
This tests electron configuration bookkeeping across period 3 — count s- and p-electrons for each element and match the given ratios.

Concept and Intuition

Every element from Na (Z=11) to Ar (Z=18) fills the $3s$ and $3p$ subshells after the fixed $1s^22s^22p^6$ core. The core always contributes 2 s-electrons ( $1s^2$ ) + 2 s-electrons ( $2s^2$ ) = 4 s-electrons and 6 p-electrons ( $2p^6$ ), fixed for all of them. On top of that, each element adds its own $3s$ and $3p$ electrons.

Step-by-Step Solution

Write full configurations for Z = 11 to 18 and tally total s vs p electrons:

Na (11): $1s^22s^22p^63s^1$ → s = 2+2+1 = 5, p = 6 → ratio $5:6$

Mg (12): $3s^2$ → s = 6, p = 6 → ratio $1:1$

Al (13): $3s^23p^1$ → s = 6, p = 7 → ratio $6:7$

Si (14): $3s^23p^2$ → s = 6, p = 8 → ratio $3:4$

P (15): $3s^23p^3$ → s = 6, p = 9 → ratio $2:3$

S (16): $3s^23p^4$ → s = 6, p = 10 → ratio $3:5$ …
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AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
$[Xe]4f^75d^16s^2$
›Reveal solutionSolution
Gadolinium is a well-known lanthanide exception: it configures as $[Xe]4f^75d^16s^2$ , not $[Xe]4f^85d^06s^2$ , because a half-filled 4f subshell is extra stable.

Concept and Intuition

Across the lanthanide series, electrons are added to the 4f subshell while 5d and 6s stay essentially fixed. But nature strongly favours half-filled ( $f^7$ ) and fully-filled ( $f^{14}$ ) subshells because of the extra exchange stabilisation energy among electrons with parallel spins in degenerate orbitals. When adding the electron that would make $4f^8$ , it is instead energetically cheaper to keep 4f at the stable half-filled $4f^7$ and place the extra electron in 5d. This is exactly analogous to why Cr is $[Ar]3d^54s^1$ instead of $3d^44s^2$ , and Cu is $[Ar]3d^{10}4s^1$ instead of $3d^94s^2$ .

Step-by-Step Solution

Gadolinium, Z = 64, is in the lanthanide series (element after Xe core, filling 4f).

The "expected" regular filling (following Aufbau strictly across lanthanides) would be $[Xe]4f^85d^06s^2$ (since 64 − 54 (Xe) − 2 (6s^2) = 8 electrons distributed to 4f).

However, $4f^7$ (half-filled) is a specially stable configuration (maximum number of unpaired parallel-spin electrons, maximum exchange energy). …
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AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
$4f^7$
›Reveal solutionSolution
Tests recall of lanthanide electronic configurations and which ion retains a $4f^7$ (half-filled, extra-stable) shell. Answer: $\mathrm{Eu^{2+}}$ .

Concept and Intuition

Lanthanides fill the $4f$ subshell after $[\mathrm{Xe}]6s^2$ . A half-filled ( $f^7$ ) or fully-filled ( $f^{14}$ ) $4f$ subshell is extra stable (exchange energy is maximised), so ions that can attain these configurations by losing only their outer $6s$ electrons (instead of also digging into $4f$ ) are especially favoured. $\mathrm{Eu}$ is the classic case: its $+2$ oxidation state is unusually accessible (compared to the "normal" lanthanide $+3$ state) precisely because $\mathrm{Eu^{2+}}$ retains the stable half-filled $4f^7$ core.

Step-by-Step Solution

Write ground-state atomic configurations (using the $[\mathrm{Xe}]$ core, Z=54):

$\mathrm{Pr}$ (Z=59): $[\mathrm{Xe}]4f^3 6s^2$

$\mathrm{Eu}$ (Z=63): $[\mathrm{Xe}]4f^7 6s^2$

$\mathrm{Lu}$ (Z=71): $[\mathrm{Xe}]4f^{14}5d^1 6s^2$

$\mathrm{Ce}$ (Z=58): $[\mathrm{Xe}]4f^1 5d^1 6s^2$

Remove electrons to form the ions in the options, starting with the outermost ( $6s$ , then $5d$ , then $4f$ if needed):

$\mathrm{Pr^{3+}}$ : remove $6s^2$ and one $4f$ electron $\Rightarrow [\mathrm{Xe}]4f^2$

$\mathrm{Eu^{2+}}$ : remove just $6s^2 \Rightarrow [\mathrm{Xe}]4f^7$ …
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AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
$\mathrm{Am^{3+}}$
›Reveal solutionSolution
$(n+l)=8$ singles out the $5f$ subshell; since Am³⁺ has the configuration $[\mathrm{Rn}]5f^6$ , the number of such f-electrons is 6.

Concept and Intuition

The quantum number rule $n+l$ pins down a specific subshell for a given $l$ . For f-electrons, $l=3$ , so $n+l=8$ forces $n=5$ — i.e. these are exactly the $5f$ electrons. So the question reduces to: how many $5f$ electrons does $\mathrm{Am^{3+}}$ have?

Step-by-Step Solution

Americium ( $Z=95$ ) ground-state configuration: $[\mathrm{Rn}]\,5f^7\,7s^2$ (half-filled 5f, like the lanthanide analogue Eu, $[\mathrm{Xe}]4f^7 6s^2$ ).

Forming $\mathrm{Am^{3+}}$ requires removing 3 electrons. For actinide/lanthanide ions, the outer $ns$ electrons are removed first, then (if needed) one from the $f$ subshell: remove both $7s^2$ electrons, then one $5f$ electron.

Resulting configuration: $\mathrm{Am^{3+}} = [\mathrm{Rn}]\,5f^6$ . …
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AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
$\mathrm{Fe^{2+}}(Z=26)$
›Reveal solutionSolution
Fe²⁺ has 6 d-electrons; comparing this against each option's electron count shows only Cl's p-electron count (11) fails to match.

Concept and Intuition

When a transition metal ion forms, electrons are removed from the outermost (4s) orbital first, not from the (n-1)d orbital, even though 4s fills before 3d in the neutral atom. This is a classic conceptual trap in electron configuration problems.

Step-by-Step Solution

Fe (Z=26): $[Ar]3d^64s^2$ . Removing 2 electrons to form Fe²⁺ removes both 4s electrons first: Fe²⁺ = $[Ar]3d^6$ → 6 d-electrons.

(A) Ne (Z=10): $1s^22s^22p^6$ → p-electrons = 6. Equal.

(B) Mg (Z=12): $1s^22s^22p^63s^2$ → s-electrons = 2+2+2=6. Equal.

(C) Fe neutral (Z=26): $3d^6$ → d-electrons = 6. Equal. …
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AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
$m_l = 0$
›Reveal solutionSolution
Counting electrons occupying the $m_l=0$ orbital in each subshell of Cr (Z=24) and Cu (Z=29) — using their actual (exception) configurations — gives 12 and 13 respectively.

Concept and Intuition

Each subshell has $(2l+1)$ orbitals with $m_l$ values from $-l$ to $+l$ ; exactly one of these has $m_l=0$ (the 's'-orbital itself for $l=0$ , or the middle orbital for $p$ / $d$ ). To find how many electrons in an atom have $m_l=0$ , go subshell by subshell: for a completely filled subshell, the $m_l=0$ orbital is full (2 electrons); for a partially filled subshell, use Hund's rule (electrons singly occupy every orbital, including $m_l=0$ , before any pairing).

Step-by-Step Solution

Chromium (Z=24) has the well-known exception configuration $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^5\,4s^1$ (half-filled 3d and 4s for extra stability).

Go through each subshell's $m_l=0$ orbital: $1s$ (2), $2s$ (2), $2p$ full so its $m_l=0$ orbital has 2, $3s$ (2), $3p$ full so 2, $3d^5$ is exactly half-filled — by Hund's rule each of the 5 d-orbitals (including $m_l=0$ ) holds exactly 1 electron, so 1; $4s^1$ contributes 1.

Total for Cr: $2+2+2+2+2+1+1=12$ . …
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AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
$(n + l)$
›Reveal solutionSolution
Writing out Mn's electron configuration and grouping subshells by their $(n+l)$ value shows $2p+3s$ give $x=8$ electrons at $(n+l)=3$ , and $3p+4s$ give $y=8$ electrons at $(n+l)=4$ , so $x+y=16$ .

Concept and Intuition

The Aufbau principle fills orbitals in order of increasing $(n+l)$ , and for equal $(n+l)$ , in order of increasing $n$ (this is exactly why $4s$ fills before $3d$ : $4s$ has $n+l=4+0=4$ while $3d$ has $n+l=3+2=5$ ). To find "how many electrons have a given $(n+l)$ value", we just need the atom's full electron configuration and then tag each occupied subshell with its $(n+l)$ .

Step-by-Step Solution

$Z=25$ is manganese; ground-state configuration: $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^5\,4s^2$ (total $2+2+6+2+6+5+2=25$ electrons, confirming the configuration is right).

Tag each subshell with $(n+l)$ : $1s\to1$ , $2s\to2$ , $2p\to3$ , $3s\to3$ , $3p\to4$ , $3d\to5$ , $4s\to4$ .

Subshells with $(n+l)=3$ : $2p$ (6 electrons) and $3s$ (2 electrons) $\Rightarrow x = 6+2=8$ . …
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AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
$5, 0, 0, +\frac{1}{2}$
›Reveal solutionSolution
Strontium's valence electrons occupy the $5s$ orbital, giving quantum numbers $n=5, l=0, m_l=0, m_s=+\tfrac12$ .

Concept and Intuition

Strontium (Z=38) sits in Group 2, Period 5 — its electron configuration is $[Kr]5s^2$ , so its valence electrons are in a filled $5s$ subshell (an s-orbital always has $l=0$ , $m_l=0$ ).

Step-by-Step Solution

Electron configuration of Sr: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 5s^2 = [Kr]5s^2$ . …
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AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
$Q. Identify the pair of elements in which number of electrons in (n-1) shell is same (A) Fe, Mn (B) Zn, Fe (C) K, Sc (D) Mn, Cr$
›Reveal solutionSolution
Counting electrons in the second-outermost (n−1) shell for each candidate pair shows Mn and Cr

both have 13 electrons there (both being $3d^5$ configurations), making them the matching pair.

Concept and Intuition

For 4th-period transition elements, the outermost shell is n=4 (the 4s electrons), so the

"(n−1) shell" is the 3rd shell, containing the 3s, 3p, and 3d electrons. Counting these for each

candidate element reveals which pairs coincide — this is a good test of knowing the exceptional

configurations of Cr and Cu (half-filled/fully-filled 3d stability).

Step-by-Step Solution

Fe (Z=26): $[Ar]3d^64s^2$ → (n−1)=3rd shell electrons: $2+6+6=14$ .

Mn (Z=25): $[Ar]3d^54s^2$ → 3rd shell: $2+6+5=13$ .

Zn (Z=30): $[Ar]3d^{10}4s^2$ → 3rd shell: $2+6+10=18$ .

K (Z=19): $[Ar]4s^1$ → 3rd shell: $2+6+0=8$ .

Sc (Z=21): $[Ar]3d^14s^2$ → 3rd shell: $2+6+1=9$ . …
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AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
$4f^x\,5d^1\,6s^2$
›Reveal solutionSolution
Only La, Ce, Gd, and Lu among the lanthanoids adopt the $4f^x5d^16s^2$ pattern (extra stability of empty/half-filled/fully-filled 4f); restricting to $x=1$ –14 excludes La, leaving Ce, Gd, Lu.

Concept and Intuition

Most lanthanoids follow the expected $[Xe]4f^{n}6s^2$ filling pattern (no 5d electron), because 4f orbitals are lower in energy and get filled preferentially. However, four lanthanoids break this pattern and instead have one electron in 5d, because that gives their 4f subshell a specially stable configuration — empty ( $4f^0$ , La), exactly half-filled ( $4f^7$ , Gd), or completely filled ( $4f^{14}$ , Lu) — plus Ce, whose $4f^15d^16s^2$ arises from near-degenerate 4f/5d energies at the start of the series.

Step-by-Step Solution

Recall the standard exceptions to lanthanoid configuration: La: $[Xe]5d^16s^2$ ( $4f^0$ ); Ce: $[Xe]4f^15d^16s^2$ ; Gd: $[Xe]4f^75d^16s^2$ ; Lu: $[Xe]4f^{14}5d^16s^2$ .

The question specifies $x=1$ –14, which excludes La (whose $x=0$ ). …
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AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
$m_l = 0$
›Reveal solutionSolution
Fe $^{3+}$ and Cr $^+$ both end up with the identical configuration [Ar] $3d^5$ , so both have exactly 11 electrons with magnetic quantum number $m_l=0$ .

Concept and Intuition

For each subshell, only the orbital with $m_l=0$ contributes to this count (e.g. the $p_z$ orbital in a $p$ subshell, or $d_{z^2}$ in a $d$ subshell); an $s$ orbital is always $m_l=0$ . In a half-filled $d^5$ configuration, Hund's rule places one electron in each of the five $d$ orbitals ( $m_l=-2,-1,0,1,2$ ), so the $m_l=0$ orbital there holds exactly one electron (not two).

Step-by-Step Solution

Fe (Z=26): [Ar] $3d^64s^2$ . Removing 3 electrons to form Fe $^{3+}$ removes both $4s$ electrons first, then one $3d$ electron: Fe $^{3+}$ = [Ar] $3d^5$ .

Cr (Z=24): [Ar] $3d^54s^1$ (anomalous half-filled stability). Removing 1 electron to form Cr $^+$ removes the single $4s$ electron: Cr $^+$ = [Ar] $3d^5$ .

Both ions share the exact configuration $1s^22s^22p^63s^23p^63d^5$ . …
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AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
$4f^7\,6s^0$
›Reveal solutionSolution
Europium's ground state is [Xe]4f⁷6s²; losing the 6s electrons gives Eu²⁺ = [Xe]4f⁷6s⁰, favoured by the extra stability of a half-filled f-subshell.

Concept and Intuition

Most lanthanides prefer the +3 oxidation state, but a few show unusually stable +2 (or +4) states when doing so gives a half-filled ( $f^7$ ) or fully-filled ( $f^{14}$ ) or empty ( $f^0$ ) subshell — extra-stable electronic configurations analogous to the familiar half-filled/fully-filled stability seen in d-block chemistry. Eu²⁺ (f⁷), Yb²⁺ (f¹⁴), and Sm²⁺ (f⁶, less stable) are the classic examples.

Step-by-Step Solution

Europium, Z=63: ground-state electron configuration is $[Xe]4f^76s^2$ (note Eu breaks the "normal" filling trend seen in La–Lu to preserve the extra-stable half-filled 4f⁷).

Removing the two outer 6s electrons (as always happens first upon ionisation) gives $Eu^{2+}: [Xe]4f^76s^0$ — exactly matching the configuration asked for. …
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Q.Indicate the number of unpaired electrons in:

Why electron configuration tells you the number of unpaired electrons

1. Phosphorus (P, atomic number 15)

2. Silicon (Si, atomic number 14)

3. Chromium (Cr, atomic number 24)

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