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.

Telangana TsbieTextbookSubjective· 2mImportance★★★★★

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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.

TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
$Q. In which of the following, elements are correctly arranged in the decreasing order of atomic radii? (A) Eu > Ce > Gd > Ho (B) Ce > Eu > Gd > Ho (C) Gd > Eu > Ce > Ho (D) Ho > Ce > Eu > Gd$
›Reveal solutionSolution
The key idea is that atomic radii in the lanthanide series generally decrease from left to right (lanthanide contraction), but europium (Eu) has an anomalously large radius due to its half-filled 4f⁷ shell. The correct decreasing order is Eu > Ce > Gd > Ho, which corresponds to option (A).

Why This Approach Works

Atomic radii in the lanthanide series (Ce to Lu) follow a well-known trend called the lanthanide contraction: as nuclear charge increases across the series, the 4f electrons are poorly shielded, so the effective nuclear pull on the outer electrons increases, causing a steady decrease in atomic radius. However, there are two important exceptions: europium (Eu) and ytterbium (Yb). Eu has a half-filled 4f⁷ subshell, which is especially stable and causes the atom to be slightly larger than its neighbors. Similarly, Yb has a filled 4f¹⁴ subshell. So, when ordering radii, we must account for these "bumps."

The problem asks for decreasing order among four specific lanthanides: Ce (cerium, atomic number 58), Eu (europium, 63), Gd (gadolinium, 64), and Ho (holmium, 67). We need to place them from largest to smallest radius.

Step-by-Step Reasoning

Recall the general trend across the lanthanide series.

From left to right (increasing atomic number), atomic radii decrease. So, in a "normal" trend, Ce (Z=58) would be the largest, then Eu (Z=63), then Gd (Z=64), then Ho (Z=67) would be the smallest. But this ignores the special stability of Eu.

Identify the anomaly for europium.

Europium has the electron configuration [Xe] 4f⁷ 6s². The half-filled 4f subshell is exceptionally stable, which reduces the effective nuclear charge felt by the outer 6s electrons. This makes the atomic radius of Eu larger than that of its immediate neighbors, including Ce. In fact, Eu’s radius is even larger than that of Ce, despite Ce being to the left.

Compare Eu and Ce directly.

Experimental data (e.g., metallic radii in picometers) show:

Ce: ~181.8 pm

Eu: ~204.2 pm

Gd: ~180.2 pm

Ho: ~176.6 pm Clearly, Eu is the largest of the four, followed by Ce, then Gd, then Ho.

Check the options against this order.

The correct decreasing order is: Eu > Ce > Gd > Ho.

Option (A) says exactly that.

Option (B) says Ce > Eu > Gd > Ho — wrong because Eu is larger than Ce. …
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TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
$l=1$
›Reveal solutionSolution
We first determine the electron configuration of $Zn^{2+}$ by removing electrons from the outermost shell of neutral zinc. Then, we count electrons based on the given quantum numbers $l$ and $m_l$ for p-subshells ( $l=1$ ) and d-subshells ( $l=2$ ). The sum of $x$ and $y$ is $\boxed{6}$ .

The problem asks us to find the total number of electrons in a dipositive zinc ion ( $Zn^{2+}$ ) that satisfy specific conditions on their azimuthal ( $l$ ) and magnetic ( $m_l$ ) quantum numbers. To do this, we need to understand the electron configuration of $Zn^{2+}$ and how quantum numbers describe the state of an electron.

The four quantum numbers ( $n, l, m_l, m_s$ ) uniquely describe an electron in an atom:

$n$ (principal quantum number) defines the electron shell and energy level.

$l$ (azimuthal or angular momentum quantum number) defines the subshell and the shape of the orbital.

$l=0$ corresponds to an s-subshell.

$l=1$ corresponds to a p-subshell.

$l=2$ corresponds to a d-subshell.

$l=3$ corresponds to an f-subshell.

$m_l$ (magnetic quantum number) defines the orientation of the orbital in space. For a given $l$ , $m_l$ can take any integer value from $-l$ to $+l$ , including $0$ .

For $l=0$ (s-subshell), $m_l=0$ (1 orbital).

For $l=1$ (p-subshell), $m_l=-1, 0, +1$ (3 orbitals).

For $l=2$ (d-subshell), $m_l=-2, -1, 0, +1, +2$ (5 orbitals).

$m_s$ (spin quantum number) describes the intrinsic angular momentum (spin) of the electron, which can be $+1/2$ or $-1/2$ . Each orbital can hold a maximum of two electrons, one with $m_s=+1/2$ and one with $m_s=-1/2$ .

We will first determine the electron configuration of $Zn^{2+}$ and then count the electrons based on the given $l$ and $m_l$ values.

Determine the electron configuration of neutral Zinc (Zn).

Zinc (Zn) has an atomic number $Z=30$ , meaning a neutral zinc atom has 30 electrons. Following the Aufbau principle, its electron configuration is:

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

Determine the electron configuration of dipositive Zinc ion ( $Zn^{2+}$ ).

To form a $Zn^{2+}$ ion, two electrons must be removed from the neutral zinc atom. Electrons are always removed from the outermost shell (the shell with the highest principal quantum number, $n$ ) first. In the configuration of neutral Zn, the $4s$ orbital has $n=4$ , which is higher than $n=3$ for the $3d$ orbital. Therefore, the two electrons are removed from the $4s$ orbital.

The electron configuration of $Zn^{2+}$ is:

$1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10}$

Calculate $x$ , the number of electrons with $l=1, m_l=0$ .

The condition $l=1$ refers to p-subshells. In the $Zn^{2+}$ configuration, we have electrons in the $2p^6$ and $3p^6$ subshells. …
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TG EAPCET 2023Set ap-2023-05-11-AN1 markMCQ
$^{-1}$
›Reveal solutionSolution
The pattern of ionization enthalpies — a drop from the first to the second element, then a steady rise — matches the trend across period 2 from Be to O. The correct sequence is Be, B, C, N, O, which is option (B).

The key to this problem is understanding how ionization enthalpy varies across a period. As we move from left to right, the nuclear charge increases, pulling the valence electrons tighter, so ionization enthalpy generally rises. But there are two famous exceptions: the drop from group 2 to group 13 (Be to B) and from group 15 to group 16 (N to O). These dips happen because of changes in electron configuration — from a filled s-orbital to a p-orbital with one electron (easier to remove), and from a half-filled p-subshell to one with a paired electron (also easier to remove).

Here the given values are: 899, 801, 1086, 1402, 1314 kJ mol $^{-1}$ . Notice the second value (801) is lower than the first (899) — that’s the first dip. Then the values climb: 1086, 1402, and then a slight drop to 1314. That second drop matches the N-to-O exception. So the sequence must start with a group 2 element, then group 13, then group 14, then group 15, then group 16. In period 2, that’s Be, B, C, N, O.

Let’s walk through it step by step.

Identify the first dip. The first two values are 899 and 801. Since 801 < 899, the second element has a lower ionization enthalpy than the first. This is the classic Be-to-B drop: Be has a filled 2s $^2$ configuration (stable), while B has 2s $^2$ 2p $^1$ — that single p-electron is easier to remove. So the first element must be Be (group 2) and the second must be B (group 13).

Check the rise after the dip. After 801, the values increase to 1086 and then to 1402. That’s consistent with moving from B (group 13) to C (group 14) to N (group 15). Each step adds a proton, increasing nuclear charge, and the electron is removed from the same shell, so ionization enthalpy rises. …
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TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
$(n+l)$
›Reveal solutionSolution
For chromium (Z=24), the electrons are distributed according to the Aufbau principle with the exception of a half-filled 3d⁵ subshell. Counting electrons by (n+l) values: (n+l)=3 gives 8 electrons, (n+l)=4 gives 7 electrons, (n+l)=5 gives 5 electrons. The correct option is (C).

The key idea is that the sum $n+l$ determines the order of filling orbitals (the Aufbau principle). For a given atom, we need to list all occupied orbitals, group them by their $n+l$ value, and count the electrons in each group. Chromium (Z=24) has a well-known electronic configuration exception: instead of $3d^4 4s^2$ , it is $3d^5 4s^1$ to achieve a half-filled d-subshell.

Let’s work through it step by step.

Write the full electronic configuration of chromium (Z=24). The order of filling is: 1s, 2s, 2p, 3s, 3p, 4s, then 3d. Normally, we would have $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^4$ . But chromium is an exception: one electron from 4s moves to 3d, giving a half-filled 3d⁵. So the actual configuration is:

$1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^1 \, 3d^5$
Total electrons: $2+2+6+2+6+1+5 = 24$ .

List each subshell with its (n, l) and compute $n+l$ .

1s: $n=1, l=0$ → $n+l = 1$

2s: $n=2, l=0$ → $n+l = 2$

2p: $n=2, l=1$ → $n+l = 3$

3s: $n=3, l=0$ → $n+l = 3$

3p: $n=3, l=1$ → $n+l = 4$

4s: $n=4, l=0$ → $n+l = 4$

3d: $n=3, l=2$ → $n+l = 5$

Group the subshells by their $n+l$ value and count electrons.

$n+l = 3$ : includes 2p (6 electrons) and 3s (2 electrons). Total = $6+2 = 8$ .

$n+l = 4$ : includes 3p (6 electrons) and 4s (1 electron). Total = $6+1 = 7$ . …
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TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
$n = 4$
›Reveal solutionSolution
For the $n=4$ energy level, the total number of orbitals is $16$ , and the maximum number of electrons with spin $+\frac{1}{2}$ in those orbitals is $16$ as well — each orbital can hold one electron of each spin. The correct option is (D).

The key idea here is that the number of orbitals in a given energy level is determined by the sum of all possible $m_l$ values across all subshells, while the maximum number of electrons with a specific spin is simply half the total electron capacity — because each orbital holds exactly two electrons of opposite spin.

Let’s break it down.

Number of orbitals in the $n=4$ level

For a principal quantum number $n$ , the possible azimuthal quantum numbers are $l = 0, 1, 2, \dots, n-1$ . So for $n=4$ , we have:

$l=0$ (4s subshell): $m_l = 0$ → 1 orbital

$l=1$ (4p subshell): $m_l = -1, 0, +1$ → 3 orbitals

$l=2$ (4d subshell): $m_l = -2, -1, 0, +1, +2$ → 5 orbitals

$l=3$ (4f subshell): $m_l = -3, -2, -1, 0, +1, +2, +3$ → 7 orbitals

Total orbitals = $1 + 3 + 5 + 7 = 16$ .

A quick formula: the number of orbitals in the $n$ th level is $n^2$ , so $4^2 = 16$ — same result.

Maximum number of electrons with spin $+\frac{1}{2}$ …
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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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