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

CBSEMCQ· 1mImportance★★★★★

45% · 59/132 Questions

✓ Free question

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.

CBSE 2024Set 56/1/11 markMCQ
$Q. Which one of the following first row transition elements is expected to have the highest third ionization enthalpy? (A) Iron (Z = 26) (B) Manganese (Z = 25) (C) Chromium (Z = 24) (D) Vanadium (Z = 23)$
›Reveal solutionSolution
The third ionization enthalpy is highest for the element whose +2 ion has the most stable electronic configuration (half-filled or fully filled d-subshell). Among Fe, Mn, Cr, and V, the +2 ion of manganese (Mn²⁺) has a half-filled 3d⁵ configuration, making it exceptionally stable and hardest to remove an electron from. Thus, Mn has the highest third ionization enthalpy.

The key to this question lies not in memorizing numbers but in understanding what the third ionization enthalpy actually measures. It is the energy required to remove the third electron from a gaseous atom — that is, to go from the +2 ion to the +3 ion. So we are really comparing the stability of the M²⁺ ions of these elements. The more stable the M²⁺ ion, the harder it is to pull off another electron, and the higher the third ionization enthalpy.

Now, stability in transition metal ions is heavily influenced by the d-electron configuration. A half-filled d⁵ or fully filled d¹⁰ subshell confers exceptional stability. Let us examine each element’s +2 ion.

Vanadium (Z = 23)

Electronic configuration of V: $[\text{Ar}]\,3d^3\,4s^2$

V²⁺: remove two 4s electrons → $[\text{Ar}]\,3d^3$

This is neither half-filled nor fully filled. It is a relatively ordinary configuration.

Chromium (Z = 24)

Cr has a special ground state: $[\text{Ar}]\,3d^5\,4s^1$ (half-filled d gives extra stability).

Cr²⁺: remove the 4s electron and one 3d electron → $[\text{Ar}]\,3d^4$

This is not half-filled. The half-filled stability of Cr atom is lost in Cr²⁺.

Manganese (Z = 25)

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

Mn²⁺: remove two 4s electrons → $[\text{Ar}]\,3d^5$

This is exactly half-filled! The d⁵ configuration is exceptionally stable. Removing a third electron would break this stable half-filled shell, requiring a large amount of energy.

Iron (Z = 26)

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

Fe²⁺: remove two 4s electrons → $[\text{Ar}]\,3d^6$

This is one electron beyond half-filled. While not as stable as d⁵, it is more stable than d⁴ or d³, but still less stable than the half-filled d⁵ of Mn²⁺.

Watch out

A common mistake is to look at the stability of the neutral atom rather than the +2 ion. For example, Cr has a half-filled d⁵ configuration in its neutral state, but Cr²⁺ is d⁴ — not particularly stable. The third ionization enthalpy depends on the +2 ion, not the atom.

Tip

For first-row transition metals, the third ionization enthalpy often peaks at manganese because Mn²⁺ (d⁵) is the most stable +2 ion. The next highest is usually iron (d⁶), then chromium (d⁴), then vanadium (d³). This pattern is a direct consequence of exchange energy and half-filled shell stability.

Thus, the order of third ionization enthalpy is:

Mn > Fe > Cr > V.

✓Final answer
$The element with the highest third ionization enthalpy is manganese (Mn), corresponding to option (B) .$
CBSE 2024Set 56/1/11 markMCQ
$Q. Assertion (A): Separation of Zr and Hf is difficult. Reason (R): Zr and Hf have similar radii due to lanthanoid contraction. [Codes (A)-(D) as in the Assertion-Reason instruction.]$
›Reveal solutionSolution
Lanthanoid contraction causes Zr and Hf to have nearly identical radii despite being in different periods, making their chemical properties so similar that separation becomes extremely difficult. Both statements are true and R correctly explains A.

The question tests your understanding of how the lanthanoid contraction affects the chemistry of post-lanthanoid elements, particularly the 4d and 5d transition metals.

Why Zr and Hf are Chemical Twins

Zirconium (Zr, atomic number 40) sits in the second transition series (4d block), while hafnium (Hf, atomic number 72) belongs to the third transition series (5d block). Normally, when you move down a group in the periodic table, atomic and ionic radii increase because you're adding entire electron shells. This is why sodium is larger than lithium, and potassium larger than sodium.

But something unusual happens between the 4d and 5d series. Between Zr and Hf, the periodic table inserts the entire lanthanoid series (elements 57–71) — fourteen f-block elements. As electrons fill the poorly shielding 4f orbitals across the lanthanoids, the effective nuclear charge experienced by outer electrons increases steadily. This pulls all the electron shells inward, causing a cumulative contraction in atomic size. By the time we reach Hf, this lanthanoid contraction has almost exactly compensated for the addition of an extra shell.

The result? The atomic radius of Zr is approximately 160 pm, while Hf is about 159 pm — virtually identical. Their ionic radii ( $\text{Zr}^{4+}$ ≈ 72 pm, $\text{Hf}^{4+}$ ≈ 71 pm) are equally similar.

Why Similar Radii Make Separation Difficult

Chemical behavior depends heavily on ionic size and charge. When two elements have:

The same oxidation states (both commonly +4)

Nearly identical ionic radii

The same coordination preferences

...they form compounds with almost indistinguishable properties. Their oxides, halides, and complexes have similar solubilities, crystal structures, and stabilities. Traditional separation methods like fractional crystallization or precipitation rely on differences in these properties, so when the properties are nearly identical, separation becomes extraordinarily challenging.

Historically, chemists struggled for decades to separate Zr and Hf. Even today, industrial separation requires sophisticated techniques like solvent extraction or ion exchange with carefully chosen ligands that can exploit the tiny remaining differences.

Tip

The lanthanoid contraction affects all elements after the lanthanoids. This is why the 5d metals (like Ta, W, Pt) are denser and have higher melting points than you'd predict from simple periodic trends — they're "compressed" by the contraction.

Evaluating the Statements

Assertion (A): "Separation of Zr and Hf is difficult."

This is unequivocally true. The chemical similarity between these two elements is one of the classic examples in inorganic chemistry. For nearly a century after hafnium's discovery in 1923, obtaining pure samples of either element required painstaking effort.

Reason (R): "Zr and Hf have similar radii due to lanthanoid contraction."

This is also true. The lanthanoid contraction is the direct cause of the radius similarity. Without the intervening lanthanoids and their f-orbital contraction, Hf would be significantly larger than Zr (as we see in other groups where no f-block intervenes).

Does R explain A?

Yes, completely. The similar radii lead directly to similar chemical properties, which in turn make separation difficult. The causal chain is: lanthanoid contraction → similar radii → similar chemistry → difficult separation. R provides the fundamental reason for A.

✓Final answer
$The correct option is (A) : Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. The first ionisation enthalpy of Xenon is almost identical with that of:$
(a) Molecular oxygen
(b) Molecular Nitrogen
(c) Molecular Fluorine
(d) Molecular Hydrogen
›Reveal solutionSolution

Xenon's ionisation enthalpy is unusually low for a noble gas — close enough to that of O2 that Xe can be oxidised by the same species that oxidise O2.

Noble gases normally have very high ionisation enthalpies because of their stable, fully-filled valence shells. However, Xenon is a large atom with a valence shell far from the nucleus (weak nuclear hold on outer electrons), so its first ionisation enthalpy (~1170 kJ/mol) is unusually low — almost identical to that of molecular oxygen (~1175 kJ/mol). This numerical coincidence is historically important: Neil Bartlett had made the salt O2+[PtF6]- by oxidising O2 with PtF6; realising Xe's ionisation enthalpy was so close to O2's, he reasoned PtF6 should also oxidise Xe — leading to the first noble-gas compound, Xe+[PtF6]-.

✓Final answer
$Molecular oxygen (option a).$
CBSE 2023Set 56/3/11 markMCQ
$3d^3 4s^2$
›Reveal solutionSolution
The highest oxidation state in transition metals is achieved when all electrons from both the 4s and 3d orbitals are removed. Among the given configurations, $3d^5 4s^2$ (option C) allows the removal of 7 electrons, giving a maximum oxidation state of +7.

The key to this question lies in understanding how transition metals exhibit variable oxidation states. Unlike main group elements where the outermost s and p electrons are the only ones involved, transition metals can use both the ns and (n−1)d electrons for bonding. The highest possible oxidation state for a given configuration is simply the total number of electrons in the outermost s and d orbitals — because in principle, all of them can be lost.

Let’s examine each option carefully.

Option A: $3d^3 4s^2$

Total electrons in the valence shell = $3 + 2 = 5$ . So the maximum oxidation state possible is +5. This is seen in elements like vanadium (V), which indeed shows +5 in compounds like $\text{V}_2\text{O}_5$ .

Option B: $3d^5 4s^1$

Total = $5 + 1 = 6$ . Maximum oxidation state = +6. Chromium (Cr) has this configuration and shows +6 in $\text{CrO}_3$ and dichromates. Notice that chromium’s actual ground state is $3d^5 4s^1$ , not $3d^4 4s^2$ , due to the extra stability of a half-filled d-subshell.

Option C: $3d^5 4s^2$

Total = $5 + 2 = 7$ . Maximum oxidation state = +7. Manganese (Mn) has this configuration and exhibits +7 in permanganate ion ( $\text{MnO}_4^-$ ). This is the highest among the given options.

Option D: $3d^6 4s^2$

Total = $6 + 2 = 8$ . Wait — does that mean +8 is possible? In theory, yes, but in practice, no transition metal with this configuration (iron, for instance) ever reaches +8. The +8 state is known only for ruthenium and osmium (in $\text{RuO}_4$ and $\text{OsO}_4$ ), which belong to the 4d and 5d series, not the 3d series. For 3d metals, the maximum observed oxidation state is +7 (manganese). So while the electron count suggests +8, it is not achievable for the given 3d series.

Watch out

Do not confuse the theoretical maximum (total valence electrons) with the practically observed maximum. For 3d transition metals, the highest oxidation state actually seen is +7 (Mn), not +8. Iron ( $3d^6 4s^2$ ) never reaches +8; its highest is +6 (in ferrates).

Tip

A quick way: count the total number of electrons in the 4s and 3d orbitals. The highest oxidation state among the options will correspond to the largest count, but only if that state is actually observed in the 3d series. Here, +7 is the practical limit.

Thus, among the given configurations, $3d^5 4s^2$ (manganese) shows the highest oxidation state of +7.

✓Final answer
$The correct option is (C), corresponding to manganese with a maximum oxidation state of +7.$
CBSE 2023Set ANNUAL1 mark
$Q. Which is the most electronegative element in periodic table?$
›Reveal solutionSolution

Electronegativity generally increases across a period and decreases down a group; fluorine, at the top-right of the periodic table (excluding noble gases), is the most electronegative element.

Electronegativity is the tendency of an atom to attract a shared pair of electrons towards itself in a bond. It increases left-to-right across a period (smaller atomic size, more nuclear charge) and decreases top-to-bottom down a group (larger atomic size, more shielding). Fluorine, in period 2, group 17, sits at the position of highest electronegativity among all elements (Pauling value ~3.98, usually rounded to 4.0).

✓Final answer
$Fluorine (F) is the most electronegative element.$
CBSE 2022Set HE2181 mark
$Q. Fill in the blank: The highest electron affinity has ______ element.$
›Reveal solutionSolution

Chlorine, not fluorine, has the largest electron affinity, because fluorine's very small atomic size causes strong electron-electron repulsion in its compact valence shell when an extra electron is added.

Electron affinity (electron gain enthalpy) generally becomes more negative (larger magnitude) across a period (increasing effective nuclear charge) and less negative down a group (increasing atomic size, the added electron feels a weaker pull). Naively this predicts fluorine (top of Group 17) should have the highest electron affinity. However, fluorine is anomalous: its atom is so small and its 2p subshell so compact that the incoming electron experiences significant inter-electron repulsion, which partly offsets the favourable nuclear attraction and reduces the energy released. Chlorine, being one size step larger with more room in its 3p subshell, accommodates the extra electron more comfortably and releases more energy — giving chlorine the highest electron affinity of any element (~~-349 kJ/mol) compared to fluorine (~~-328 kJ/mol).

✓Final answer
$Chlorine (Cl) has the highest electron affinity among all elements.$
CBSE 2022Set ANNUAL1 markMCQ
$Q. Which of the following element does not show allotropy?$
(a) Nitrogen
(b) Bismuth
(c) Antimony
(d) Arsenic
›Reveal solutionSolution
Nitrogen exists only as the diatomic $N_2$ molecule and has no allotropes; the heavier Group-15 elements (As, Sb, Bi) do show allotropy. Option (A).

Allotropy is the existence of an element in two or more different physical forms in the same physical state.

Nitrogen — exists as the gas $N_2$ with a strong $N\equiv N$ triple bond; it does not form allotropes.

Arsenic, Antimony, Bismuth — the heavier Group-15 elements are solids that display allotropic forms (e.g. grey/metallic and yellow arsenic; metallic and yellow antimony; different forms of bismuth).

Hence, among the given options, nitrogen is the element that does not show allotropy.

✓Final answer
$(A) Nitrogen.$

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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?

Concept understanding — Ionization Energy Trends

Ionization Energy: The First Meeting

The Intuition: What Controls the Grip?

The Precise Trend: Across a Period

The Precise Trend: Down a Group

The Two Exceptions (and Why They Matter)

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

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

The Big Picture: What You Must Remember

A Final Check: First vs Second Ionization Energy

Why this formula?

Ionization Energy Trends: The Why Behind the Trends

1. The Core Formula: Coulomb’s Law

2. Why IE Increases Across a Period

Reasoning:

Result:

3. Why IE Decreases Down a Group

Reasoning:

Result:

4. The Mathematical Expression (Approximation)

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

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

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

6. Summary Table

Final Takeaway

Method: Electronic Configuration Reconstruction

Steps

Final Answer

Common Mistakes & How to Avoid Them

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

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

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

Mistake 4: Rushing and Not Checking All Options

Final Answer

Why Zr and Hf are Chemical Twins

Why Similar Radii Make Separation Difficult

Evaluating the Statements

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?

Unlock everything free for 14 days