Electronic configuration of a transition element X in +3 oxidation state is $[Ar]3d^5$. What is its atomic number? (i) 25 (ii) 26 (iii) 27 (iv) 24

> [!TLDR] > The +3 oxidation state has lost three electrons from the neutral atom. The $3d^5$ configuration in +3 means the neutral atom had $3d^6 4s^2$ (since 4s electrons are lost first). That gives atomic number 26 — iron (Fe). So the answer is **(ii) 26**. The key to this problem is understanding how transition metals lose electrons when forming ions. Many students memorise the electronic configurations of neutral atoms but forget that when a transition element forms a positive ion, the **4s electrons are lost before the 3d electrons**. This is a classic exam trap. Let’s break it down. 1. **What does $[Ar]3d^5$ in the +3 state tell us?** The ion $\ce{X^3+}$ has the same electron configuration as argon plus five electrons in the 3d subshell. So the total number of electrons in $\ce{X^3+}$ is: $18 \text{ (from Ar)} + 5 = 23$ electrons. 2. **Relating ion electrons to neutral atom electrons** A neutral atom has the same number of electrons as its atomic number $Z$. When it loses 3 electrons to become $\ce{X^3+}$, the number of electrons drops by 3. So: $\text{Electrons in } \ce{X^3+} = Z - 3$ We already know this equals 23, so: $Z - 3 = 23 \implies Z = 26$ That gives atomic number 26 directly — provided the order of electron loss has been accounted for correctly, which the next step verifies. 3. **Why the 4s electrons matter** The neutral atom with $Z=26$ is iron. Its ground state configuration is $[Ar]3d^6 4s^2$. When iron forms $\ce{Fe^3+}$, it loses the two 4s electrons first, then one 3d electron. So: $\ce{Fe}$: $[Ar]3d^6 4s^2$ $\ce{Fe^3+}$: $[Ar]3d^5$ This matches perfectly. > [!WARNING] > A common mistake is to assume the +3 ion’s configuration comes directly from the neutral atom’s configuration by removing 3d electrons first. If you did that, you might think the neutral atom had $3d^8$ (since $3d^5$ in +3 means $3d^8$ in neutral), giving $Z=26$ anyway — but that’s a coincidence here. For other eleme

Electronic configuration of a transition element X in +3 oxidation state is [Ar]3d⁵. What is its atomic number

NCERT Exemplar · Q1

[Ar]3d^5

(i) 25

(ii) 26

(iii) 27

(iv) 24

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Concept understanding — Ionization Energy Trends

Ionization Energy: The First Meeting

Imagine you're holding onto something precious — say, a favourite pen. How hard would someone have to pull to take it from your hand? That's the core idea behind ionization energy. In an atom, the "something precious" is an electron, and the "pulling force" is energy.

Ionization energy (IE) is the minimum energy required to remove the most loosely bound electron from a gaseous atom in its ground state.

Why "gaseous" and "ground state"? Because we want a fair comparison — no extra energy from neighbours or from the atom being already excited. We measure how tightly the atom holds its outermost electron when it's alone and calm.

The unit you'll see most often in exams: kJ/mol (kilojoules per mole of atoms).

The Intuition: What Controls the Grip?

Two factors decide how hard an atom holds its outermost electron:

Nuclear charge — more protons in the nucleus means a stronger pull on the electron. Simple: bigger positive charge, tighter grip.
Distance from the nucleus — the farther the electron is, the weaker the pull. Think of a magnet: it holds a paperclip strongly up close, but barely at arm's length.

But there's a subtle twist: shielding (or screening). Inner electrons partially block the nuclear charge from reaching the outer electron. The outermost electron doesn't "feel" the full nuclear charge — it feels only the effective nuclear charge ( $Z_{\text{eff}}$ ).

Note

$Z_{\text{eff}} = Z - S$ , where $Z$ is the atomic number and $S$ is the shielding constant (roughly the number of inner electrons). This is the net positive charge pulling on the outer electron.

So the real question becomes: How large is $Z_{\text{eff}}$ for the outermost electron, and how far away is it?

The Precise Trend: Across a Period

As you move left to right across a period (say, from Li to Ne in period 2):

Nuclear charge increases steadily (more protons).
Electrons are added to the same shell — no new inner layers.
Shielding stays roughly constant (same number of inner electrons).
Result: $Z_{\text{eff}}$ increases → the outer electron is pulled in tighter → ionization energy increases.

Across a period: IE increases (generally).

Example:

Li (IE = 520 kJ/mol) → Be (900) → B (801) → C (1086) → N (1402) → O (1314) → F (1681) → Ne (2081)

Wait — why does B have lower IE than Be? And O lower than N? That's the exception, not the rule. We'll come back to it.

The Precise Trend: Down a Group

As you move down a group (say, from Li to Cs in group 1):

Nuclear charge increases (more protons).
But electrons are added to new, higher shells — the outermost electron is much farther from the nucleus.
Shielding also increases significantly (more inner electrons).
Result: distance dominates → the outer electron is held more loosely → ionization energy decreases.

Down a group: IE decreases.

Example:

Li (520) → Na (496) → K (419) → Rb (403) → Cs (376) — all in kJ/mol.

The Two Exceptions (and Why They Matter)

Exception 1: Group 13 vs Group 2 (e.g., B vs Be)

Be has a full $2s^2$ subshell. B has $2s^2 2p^1$ . The $2p$ electron is slightly higher in energy and slightly better shielded by the $2s$ electrons than the $2s$ electrons shield each other. So removing the $2p$ electron from B takes less energy than removing a $2s$ electron from Be.

Watch out

Don't memorise "IE increases across a period" blindly. Group 13 always has lower IE than Group 2 in the same period.

Exception 2: Group 16 vs Group 15 (e.g., O vs N)

N has a half-filled $2p^3$ subshell — each $2p$ orbital has one electron. This is an especially stable arrangement (exchange energy stabilisation). O has $2p^4$ — one orbital gets a second electron. That extra electron experiences electron-electron repulsion, making it easier to remove. So O has lower IE than N.

Tip

|--------|--------------|--------------|------------------|

| 2 | N (1402) | O (1314) | N > O |

| 3 | P (1012) | S (1000) | P > S |

| 4 | As (947) | Se (941) | As > Se |

The pattern holds for all periods.

The Big Picture: What You Must Remember

Important

Ionization energy increases across a period (with two dips) and decreases down a group.

The dips occur at Group 13 (lower than Group 2) and Group 16 (lower than Group 15).

The underlying reason is always the same: effective nuclear charge and distance. When $Z_{\text{eff}}$ is high and the electron is close, IE is high. When the electron is far or repulsion helps it leave, IE is low.

A Final Check: First vs Second Ionization Energy

Removing one electron from an atom leaves a positive ion. Removing a second electron from that ion is always harder — the ion has a higher positive charge pulling on the remaining electrons.

Second IE > First IE — always. For example, Na: first IE = 496 kJ/mol, second IE = 4562 kJ/mol. That's nearly 10 times larger. This huge jump tells you that the second electron comes from a different shell (closer to the nucleus).

In exams, this jump is used to identify the group of an element — a sudden large increase in successive ionization energies indicates you've stripped off all valence electrons and are now pulling from a core shell.

"Ionization energy trends periodic table" and "periodicity class 11 chemistry important questions" are extremely common searches, both anchored in the Classification of Elements and Periodicity chapter of the NCERT/CBSE Class 11 Chemistry curriculum. The Group 13 and Group 16 exceptions in particular are a favourite trap question in board exams and JEE Main.

Why this formula?

Ionization Energy Trends: The Why Behind the Trends

Ionization energy (IE) is the minimum energy required to remove the most loosely bound electron from a gaseous atom in its ground state.

It is measured in kJ/mol or eV/atom.

The key trend is:

IE increases across a period (left → right) and decreases down a group (top → bottom).

But why? Let’s break down the reasoning step-by-step.

1. The Core Formula: Coulomb’s Law

The energy needed to remove an electron is fundamentally governed by the electrostatic attraction between the electron and the nucleus.

F = \frac{k \cdot Z_{\text{eff}} \cdot e^2}{r^2}

Where:

$Z_{\text{eff}}$ = effective nuclear charge (net positive charge felt by the electron)
$r$ = distance of the electron from the nucleus
$e$ = charge of electron
$k$ = Coulomb constant

Key insight: The stronger the attraction, the higher the ionization energy.

2. Why IE Increases Across a Period

Reasoning:

As you move left → right, protons increase in the nucleus.
Electrons are added to the same principal energy level (same shell).
Shielding by inner electrons remains roughly constant (same number of inner shells).
Therefore, $Z_{\text{eff}}$ increases — the outer electrons feel a stronger pull.

Result:

\text{IE} \propto Z_{\text{eff}}

So IE increases across a period.

Example:

Na (Z=11): IE = 496 kJ/mol
Mg (Z=12): IE = 738 kJ/mol
Al (Z=13): IE = 578 kJ/mol (slight dip due to p-orbital shielding — see exception below)

3. Why IE Decreases Down a Group

Reasoning:

As you move down a group, principal quantum number $n$ increases.
The outermost electron is farther from the nucleus ( $r$ increases).
Shielding increases because more inner electron shells are present.
$Z_{\text{eff}}$ increases only slightly (not enough to compensate for distance).

Result:

\text{IE} \propto \frac{1}{r^2}

So IE decreases down a group.

Example:

Li (n=2): IE = 520 kJ/mol
Na (n=3): IE = 496 kJ/mol
K (n=4): IE = 419 kJ/mol

4. The Mathematical Expression (Approximation)

For a hydrogen-like atom (single electron), the ionization energy is given by:

\text{IE} = \frac{13.6 \, \text{eV} \cdot Z^2}{n^2}

For multi-electron atoms, we replace $Z$ with $Z_{\text{eff}}$ :

\text{IE} \approx \frac{13.6 \, \text{eV} \cdot Z_{\text{eff}}^2}{n^2}

Why this holds:

$Z_{\text{eff}}$ accounts for shielding by inner electrons.
$n$ is the principal quantum number of the electron being removed.
The $1/n^2$ dependence comes from the Bohr model — energy levels scale as $E_n \propto -Z^2/n^2$ .

5. Exceptions (Why the Trend Isn’t Perfect)

a) Group 13 vs Group 2 (e.g., Al vs Mg)

Al has a p-orbital electron (higher energy, easier to remove) than Mg’s s-orbital.
Also, p-orbitals are more shielded by s- and p-electrons.

b) Group 16 vs Group 15 (e.g., O vs N)

N has a half-filled p-subshell (extra stability).
O has one paired electron — electron-electron repulsion makes removal easier.

6. Summary Table

Factor	Across Period (→)	Down Group (↓)
$Z_{\text{eff}}$	Increases	Increases slightly
$r$ (distance)	Decreases slightly	Increases
Shielding	Constant	Increases
IE	Increases	Decreases

Final Takeaway

Ionization energy is not just a number — it’s a direct consequence of Coulomb’s law, modified by shielding and orbital shape.

The trend is driven by $Z_{\text{eff}}$ (across) and distance + shielding (down).

Always ask: “How strongly is this electron held?” — and the answer lies in the balance of nuclear charge, distance, and shielding.

The +3 oxidation state has lost three electrons from the neutral atom. The $3d^5$ configuration in +3 means the neutral atom had $3d^6 4s^2$ (since 4s electrons are lost first). That gives atomic number 26 — iron (Fe). So the answer is (ii) 26.

The key to this problem is understanding how transition metals lose electrons when forming ions. Many students memorise the electronic configurations of neutral atoms but forget that when a transition element forms a positive ion, the 4s electrons are lost before the 3d electrons. This is a classic exam trap.

Let’s break it down.

What does $[Ar]3d^5$ in the +3 state tell us?

The ion $\ce{X^3+}$ has the same electron configuration as argon plus five electrons in the 3d subshell. So the total number of electrons in $\ce{X^3+}$ is:

$18 \text{ (from Ar)} + 5 = 23$ electrons.
Relating ion electrons to neutral atom electrons

A neutral atom has the same number of electrons as its atomic number $Z$ . When it loses 3 electrons to become $\ce{X^3+}$ , the number of electrons drops by 3. So:

$\text{Electrons in } \ce{X^3+} = Z - 3$

We already know this equals 23, so:

$Z - 3 = 23 \implies Z = 26$

That gives atomic number 26 directly — provided the order of electron loss has been accounted for correctly, which the next step verifies.
Why the 4s electrons matter

The neutral atom with $Z=26$ is iron. Its ground state configuration is $[Ar]3d^6 4s^2$ . When iron forms $\ce{Fe^3+}$ , it loses the two 4s electrons first, then one 3d electron. So:

$\ce{Fe}$ : $[Ar]3d^6 4s^2$

$\ce{Fe^3+}$ : $[Ar]3d^5$

This matches perfectly.

Watch out

A common mistake is to assume the +3 ion’s configuration comes directly from the neutral atom’s configuration by removing 3d electrons first. If you did that, you might think the neutral atom had $3d^8$ (since $3d^5$ in +3 means $3d^8$ in neutral), giving $Z=26$ anyway — but that’s a coincidence here. For other elements, that wrong reasoning would give the wrong answer. Always remember: 4s is higher in energy than 3d for neutral atoms, so 4s electrons are lost first when forming cations.

Checking the options
- (i) 25: Mn — neutral $[Ar]3d^5 4s^2$ ; $\ce{Mn^3+}$ would be $[Ar]3d^4$ (lose two 4s and one 3d). Not correct.
- (ii) 26: Fe — neutral $[Ar]3d^6 4s^2$ ; $\ce{Fe^3+}$ is $[Ar]3d^5$ . Correct.
- (iii) 27: Co — neutral $[Ar]3d^7 4s^2$ ; $\ce{Co^3+}$ is $[Ar]3d^6$ . Not correct.
- (iv) 24: Cr — neutral $[Ar]3d^5 4s^1$ (exception); $\ce{Cr^3+}$ is $[Ar]3d^3$ . Not correct.

Tip

For quick verification: the +3 oxidation state of a first-row transition metal with $3d^5$ configuration is almost always iron. Manganese in +3 gives $3d^4$ , and chromium in +3 gives $3d^3$ . So if you see $[Ar]3d^5$ for a +3 ion, think iron.

✓Final answer

The atomic number is 26, which corresponds to option (ii) .

Common Mistakes & How to Avoid Them

Mistake 1: Forgetting the +3 Oxidation State Means 3 Electrons Removed

Many students directly count electrons from the given configuration $[Ar]3d^5$ and conclude:

Total electrons = 18 (Ar) + 5 = 23
Then assume atomic number = 23 (which isn't even an option)

Why this is wrong:

The configuration $[Ar]3d^5$ is for the +3 ion, not the neutral atom. You must add back the 3 electrons that were removed.

How to avoid:

Always ask: "Is this configuration for the neutral atom or an ion?" If it's an ion, reverse the charge to find the neutral atom's electron count.

Mistake 2: Confusing Atomic Number with Number of Electrons in the Ion

Some students see $3d^5$ and immediately think of Mn (Z=25) because neutral Mn has $[Ar]3d^5 4s^2$ . They pick option (i) 25 without checking the oxidation state.

Why this is wrong:

Neutral Mn has 25 electrons. But here, the +3 ion has 23 electrons (18 from Ar + 5 from 3d). So the neutral atom must have 23 + 3 = 26 electrons, which corresponds to Fe (Z=26).

How to avoid:

Write the neutral configuration first: $[Ar]3d^5 4s^2$ is Mn (Z=25)
Remove 3 electrons (from 4s first, then 3d): $[Ar]3d^4$ — but the question gives $[Ar]3d^5$ , so this doesn't match Mn.
For Fe (Z=26): neutral is $[Ar]3d^6 4s^2$ ; remove 3 electrons → $[Ar]3d^5$ ✓

Mistake 3: Forgetting the 4s Orbital Fills Before 3d (But Empties First)

Students sometimes remove electrons from 3d before 4s, leading to wrong configurations.

Correct order for removal:

When forming positive ions, electrons are removed from the 4s orbital first, even though 3d fills first in the neutral atom.

How to avoid:

Remember the mnemonic: "Last in, first out" for transition metals — 4s fills last but empties first.

Mistake 4: Rushing and Not Checking All Options

Some students calculate 26, see it's an option, and mark it without verifying if other options could also give $[Ar]3d^5$ in +3 state.

Quick verification:

Atomic No.	Neutral Config.	After losing 3e⁻	Matches?
25 (Mn)	$[Ar]3d^5 4s^2$	$[Ar]3d^4$	✗
26 (Fe)	$[Ar]3d^6 4s^2$	$[Ar]3d^5$	✓
27 (Co)	$[Ar]3d^7 4s^2$	$[Ar]3d^6$	✗
24 (Cr)	$[Ar]3d^5 4s^1$	$[Ar]3d^4$	✗

How to avoid:

Always do a quick sanity check — write the neutral configuration for each option and remove electrons in the correct order.

Final Answer

The atomic number is 26 (Option (ii)).

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

KEAM 2026Set eng-2026-04184 marksMCQ
$Q. The correct decreasing order of the first ionization enthalpies among the elements C, N, O, F is (A) N > O > F > C (B) O > F > N > C (C) C > N > O > F (D) F > N > O > C (E) C > O > N > F$
›Reveal solutionSolution

The order is $F>N>O>C$ ; nitrogen's stable half-filled $2p^3$ configuration gives it a higher first IE than oxygen.

Generally first ionization enthalpy increases across a period (C < N < O < F), but the extra-stable half-filled $2p^3$ of nitrogen makes it harder to ionize than oxygen (whose paired $2p^4$ electron is easier to remove).

Approximate values (kJ mol⁻¹): C ≈ 1086, N ≈ 1402, O ≈ 1314, F ≈ 1681.

Decreasing order:
$F>N>O>C$
✓Final answer
$The correct option is (D) .$
KEAM 2026Set eng-2026-04224 marksMCQ
$Q. Which of the following represents the correct increasing order of first ionization enthalpy for Ca, Ba, S, Se and Ar? (A) Ca < S < Ba < Se < Ar (B) Ar < S < Ba < Se < Ca (C) Ba < Ca < Se < S < Ar (D) Ba < Ca < S < Se < Ar (E) Ca < S < Ar < Se < Ba$
›Reveal solutionSolution

First ionization enthalpies rise across a period and fall down a group: Ba < Ca < Se < S < Ar.

Approximate IE $_1$ values (kJ/mol): Ba $\approx$ 503, Ca $\approx$ 590, Se $\approx$ 941, S $\approx$ 1000, Ar $\approx$ 1521. Ba is below Ca in group 2 (lower IE); S is above Se in group 16 (higher IE); Ar, a noble gas, is highest. Increasing order: Ba $<$ Ca $<$ Se $<$ S $<$ Ar.

✓Final answer
$The correct option is (C) .$
KEAM 2025Set eng-2025-04274 marksMCQ
$Q. The alkali metal with the highest first enthalpy of ionization is (A) Cs (B) Rb (C) K (D) Na (E) Li$
›Reveal solutionSolution

Down Group 1, atomic size increases and the outer electron is less tightly held, so ionization enthalpy falls — making Li the highest and Cs the lowest.

Among the alkali metals (Li, Na, K, Rb, Cs), the first ionization enthalpy decreases down the group. As we move down, the atomic radius increases and the valence electron is farther from the nucleus and more shielded, so it is removed more easily.

Lithium, being the smallest, holds its outer electron most tightly and therefore has the highest first ionization enthalpy of the listed alkali metals.

✓Final answer
$The correct option is (E) .$
KEAM 2025Set pha-2025-0429F4 marksMCQ
$Q. The correct order of ionization enthalpy is (A) C < B < O < N (B) B < O < C < N (C) N < C < O < B (D) B < C < O < N (E) C < B < O < N$
›Reveal solutionSolution
[!TLDR]

Ranking the first ionization enthalpies of B, C, N and O gives the increasing order $B < C < O < N$ , so option (D) is correct.

Concept

In this NCERT/CBSE-aligned periodic-trends topic, ionization enthalpy increases across a period as nuclear charge grows. The one exception among these elements is nitrogen: its stable, exactly half-filled $2s^2 2p^3$ subshell makes it harder to ionize than oxygen, whose $2p^4$ configuration has one paired electron that is easier to remove.

Solution

The standard first ionization enthalpies (kJ/mol) are:
$B \approx 801,\quad C \approx 1086,\quad O \approx 1314,\quad N \approx 1402$
Arranging from smallest to largest:
$801 (B) < 1086 (C) < 1314 (O) < 1402 (N)$
Hence $B < C < O < N$ . The reversal of O and N is the tell-tale half-filled-stability effect.

[!ANSWER]

(D) $B < C < O < N$
KEAM 2025Set pha-2025-0429F4 marksMCQ
$Q. The first and second ionization enthalpies of lanthanoids are comparable with the element (A) Chromium (B) Calcium (C) Germanium (D) Cesium (E) Cadmium$
›Reveal solutionSolution

The first and second ionization enthalpies of the lanthanoids are relatively low and lie in the same range as those of calcium, reflecting the ease of forming the common +2/+3 states.

Reasoning

Across the lanthanoid series the first and second ionization enthalpies remain fairly low and change little (poor shielding by 4f electrons keeps them comparable). Their values are of the same order as calcium's, which is why lanthanoids readily attain the +3 (and sometimes +2) oxidation states. This comparison with calcium is a standard NCERT statement.

✓Final answer
$The correct option is (B) .$
KEAM 2024Set eng-2024-06094 marksMCQ
$Q. The decreasing order of first ionisation enthalpy of the following elements is (A) N>O>C>Be (B) O>N>C>Be (C) Be>C>O>N (D) O>N>Be>Ce (E) N>O>Be>C$
›Reveal solutionSolution

Approximate IE $_1$ values: N ( $1402$ ) $>$ O ( $1314$ ) $>$ C ( $1086$ ) $>$ Be ( $899\ \mathrm{kJ\,mol^{-1}}$ ), giving $N>O>C>Be$ .

Ionisation enthalpy generally rises across a period, but nitrogen's extra stable half-filled $2p^3$ configuration makes its IE $_1$ higher than oxygen's (removing an electron from O's paired $2p^4$ is easier). Carbon ( $2p^2$ ) is lower than both, and beryllium ( $2s^2$ ) lowest of these four:
$N (1402) > O (1314) > C (1086) > Be (899)\ \mathrm{kJ\,mol^{-1}}.$
Hence the decreasing order is $N>O>C>Be$ .

✓Final answer
$The correct option is (A) .$
KEAM 2023Set eng-2023-P1-A14 marksMCQ
$\mathrm{Ne} < \mathrm{Xe} > \mathrm{Li} > \mathrm{K} < \mathrm{Cs}$
›Reveal solutionSolution
Ionisation enthalpy order is Ne > Xe > Li > K > Cs.

Concept and Intuition

Noble gases have the highest ionisation enthalpies (stable closed shells), and within a group it falls down the column. So the two noble gases lead, with Ne above Xe, followed by the alkali metals Li > K > Cs (decreasing down group 1).

Step-by-Step Solution

Noble gases first: Ne (approx 2081 kJ/mol) > Xe (approx 1170).

Alkali metals are far lower and decrease down the group: Li (520) > K (419) > Cs (376).

Combined order: Ne > Xe > Li > K > Cs.

Common Mistakes

Placing alkali metals above noble gases, or reversing the down-group trend (thinking Cs > Li).

✓Final answer
$The correct option is (E) — Ne > Xe > Li > K > Cs.$

ANSWER: E
KEAM 2021Set eng-2021-P1-A14 marksMCQ
$Q. The first ionisation enthalpy is the least in (A) Germanium (B) Antimony (C) Tellurium (D) Arsenic (E) Bismuth$
›Reveal solutionSolution
Bismuth has the lowest first ionisation enthalpy of the options.

Concept and Intuition

First ionisation enthalpy generally increases across a period and decreases down a group as the outer electron gets farther from the nucleus and more shielded. Among the options, bismuth sits lowest in the periodic table.

Step-by-Step Solution

Approximate IE $_1$ values (kJ mol $^{-1}$ ): Ge $762$ , As $944$ , Sb $831$ , Te $869$ , Bi $703$ .

The smallest is bismuth at $\approx703$ .

Common Mistakes

Assuming a period-3/4 element like Ge is lowest, ignoring the group trend that favours Bi.

✓Final answer
$The correct option is (E) — Bismuth.$

ANSWER: E

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Q.Electronic configuration of a transition element X in +3 oxidation state is [Ar]3d5[Ar]3d^5[Ar]3d5. What is its atomic number?

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Q.Electronic configuration of a transition element X in +3 oxidation state is $[Ar]3d^5$ . What is its atomic number?

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