Why Eominus values for Mn, Ni and Zn are more negative than expected?

NCERT Exemplar · Q23

E^\ominus

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Concept understanding — Stability of Oxidation States

Stability of Oxidation States – From Intuition to Precision

Imagine you're holding a ball on a hill. If you place it exactly at the top, it's balanced — but the slightest push sends it rolling down. That's an unstable position. If you place it in a small dip on the hillside, it stays put even if nudged — that's stable. Oxidation states work the same way: some are like the hilltop (easily changed), others like the dip (hard to change).

The Core Intuition

An oxidation state is just a number we assign to an atom to track how many electrons it has gained or lost compared to its neutral state. But atoms don't "want" to stay in arbitrary oxidation states — they want to reach a configuration that minimises their energy.

Stability here means: how reluctant is that oxidation state to change under normal conditions? A stable oxidation state resists being oxidised further or reduced further. An unstable one readily changes into something else.

The Precise Statement

Stability of an oxidation state refers to the tendency of an element to maintain that particular oxidation state under given conditions (temperature, pH, presence of other reagents). A stable oxidation state is one that does not easily undergo redox reactions — it is neither easily oxidised nor easily reduced.

This depends on three key factors:

Electronic configuration – Half-filled and fully-filled d or f subshells confer extra stability (e.g., $\text{Fe}^{3+}$ with $d^5$ is more stable than $\text{Fe}^{2+}$ with $d^6$ in some contexts).
Inert pair effect – Heavier p-block elements (like Tl, Pb, Bi) show lower oxidation states (e.g., +1 for Tl) as more stable than higher ones (+3 for Tl), because the s-electrons become reluctant to participate.
Disproportionation tendency – Some oxidation states are unstable because they spontaneously convert into two other states (e.g., $\text{Cu}^+$ in aqueous solution gives $\text{Cu}^{2+}$ and $\text{Cu}$ ).

Important

Stability is relative — it depends on the environment. $\text{Mn}^{2+}$ is stable in acidic solution but easily oxidised in alkaline medium. Always specify conditions when discussing stability.

Examples That Make It Concrete

Transition metals – $\text{Cr}^{3+}$ ( $d^3$ ) and $\text{Mn}^{2+}$ ( $d^5$ ) are exceptionally stable because half-filled/half-filled-like configurations have low energy. $\text{Cr}^{2+}$ ( $d^4$ ) is easily oxidised to $\text{Cr}^{3+}$ — it's unstable.

p-block elements – $\text{Pb}^{2+}$ is stable, $\text{Pb}^{4+}$ is a strong oxidising agent (unstable). $\text{Sn}^{2+}$ is a reducing agent (easily oxidised to $\text{Sn}^{4+}$ ), so $\text{Sn}^{4+}$ is more stable for tin.

Common pattern – For most elements, the most common oxidation state is the most stable one under standard conditions. But "most common" isn't always "most stable" — e.g., $\text{Fe}^{3+}$ is common but $\text{Fe}^{2+}$ is more stable in acidic solution.

Watch out

Do not confuse "stability" with "occurrence". $\text{Mn}^{7+}$ (as $\text{MnO}_4^-$ ) is common in the lab but is a powerful oxidising agent — it is not stable in the sense of resisting change. It readily accepts electrons.

How to Think About It in Exams …

Why this formula?

Stability of Oxidation States: Why It Works

This concept explains why certain oxidation states of an element are more stable than others — and why some states are never observed at all.

The Core Idea: Energy Minimisation

An oxidation state is stable when the total energy of the system is at a minimum. This depends on three competing factors:

Ionisation energy (energy needed to remove electrons)
Lattice energy (for ionic compounds) or bond energy (for covalent compounds)
Electronic configuration (half-filled / fully-filled subshells)

There is no single formula for stability — instead, we use trends and principles that act as "formulae" for prediction.

Key Principle 1: Inert Pair Effect (for p-block elements)

Why it holds:

For heavier elements (e.g., Tl, Pb, Bi), the 6s² electrons are held very tightly due to poor shielding and relativistic effects. They resist removal.

Result: Lower oxidation state (e.g., +1 for Tl, +2 for Pb) becomes more stable than the higher state (+3, +4).
Example: $\ce{Tl^3+}$ is a strong oxidising agent — it readily gains two electrons to become $\ce{Tl^+}$ .

Derivation logic:

The energy cost to remove the 6s² electrons is greater than the energy gained by forming additional bonds or lattice. So the system stays in the lower state.

Key Principle 2: Half-Filled / Fully-Filled Subshell Stability

Why it holds:

A half-filled ( $d^5$ , $f^7$ ) or fully-filled ( $d^{10}$ , $f^{14}$ ) subshell has extra exchange energy and symmetry — making it unusually stable.

Example: $\ce{Mn^{2+}}$ ( $d^5$ ) is more stable than $\ce{Mn^{3+}}$ ( $d^4$ ). $\ce{Fe^{3+}}$ ( $d^5$ ) is more stable than $\ce{Fe^{2+}}$ ( $d^6$ ).

Derivation logic:

The exchange energy (Hund's rule) is maximum for half-filled configurations. Removing an electron from a half-filled shell costs extra energy — so the half-filled state is favoured.

Key Principle 3: Lattice Energy / Hydration Energy Compensation

For transition metals, stability of a particular oxidation state in aqueous solution depends on:

\Delta G^\circ = \Delta H^\circ_{\text{hydration}} - \Delta H^\circ_{\text{ionisation}}

Why it holds:

Higher oxidation states have higher ionisation energy (harder to remove electrons).
But they also have higher hydration energy (smaller, more charged ions attract water more strongly).
The balance determines which state is stable.

Example:

$\ce{Cu^+}$ is unstable in water because its hydration energy is too low to compensate for the loss of the second electron.
$\ce{Cu^{2+}}$ is stable in water.

Key Principle 4: Disproportionation

Some oxidation states are unstable and spontaneously convert to two other states:

\ce{2Cu^+ -> Cu + Cu^{2+}}

Why it holds: …

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The more negative $E^\ominus$ values for Mn, Ni, and Zn each have a different root cause: Mn and Zn both have an unusually low enthalpy of atomisation (weak metallic bonding) combined with the exceptional stability of their $d^5$ / $d^{10}$ ions, while Ni's anomaly comes from $Ni^{2+}$ having the highest (most exothermic) hydration enthalpy in the series.

The question touches on a beautiful anomaly in the 3d transition series. You'd expect a smooth trend in standard electrode potentials ( $E^\ominus$ ) as you move across the series — but nature throws in surprises. Let's see why.

The Core Idea: What $E^\ominus$ Actually Measures

The standard electrode potential for a metal $M^{2+}/M$ couple is a measure of how easily the metal loses electrons to form its ion in solution. The overall process is:

M(s) \rightarrow M^{2+}(aq) + 2e^-

This involves three energy steps:

Atomisation: $M(s) \rightarrow M(g)$ — enthalpy $\Delta H_{atom}$
Ionisation: $M(g) \rightarrow M^{2+}(g) + 2e^-$ — sum of first two ionisation enthalpies ( $IE_1 + IE_2$ )
Hydration: $M^{2+}(g) \rightarrow M^{2+}(aq)$ — hydration enthalpy $\Delta H_{hyd}$

The overall enthalpy change is:

\Delta H = \Delta H_{atom} + (IE_1 + IE_2) + \Delta H_{hyd}

A more negative $E^\ominus$ means the metal is less willing to form its ion — i.e., the overall $\Delta H$ is less favourable (more positive) than expected.

Step-by-Step: Why Mn, Ni, and Zn Stand Out

1. The Expected Trend

For most 3d metals, as atomic number increases, nuclear charge increases, and you'd expect ionisation enthalpies to rise steadily. This should make $E^\ominus$ values less negative (more positive) across the series. But Mn, Ni, and Zn break this pattern.

2. Mn: The Half-Filled $d^5$ Stability

Manganese has the electronic configuration $[Ar] 3d^5 4s^2$ . The $d^5$ configuration is half-filled — a state of exceptional stability due to exchange energy and symmetrical distribution of electrons.

Important

Contrary to what you might expect, NCERT notes that the enthalpy of atomisation of Mn (and V) is actually lower than the general trend across the series — the half-filled $d^5$ configuration gives a more symmetric, less itinerant arrangement of d-electrons, so the metallic lattice of Mn is held together somewhat more weakly than its neighbours.

This lower $\Delta H_{atom}$ makes it easier to atomise Mn metal, which—combined with the irregular (lower-than-trend) sum of the first and second ionisation enthalpies at this point in the series, itself a consequence of the extra stability gained on reaching the half-filled $Mn^{2+}$ ( $d^5$ ) configuration—makes the overall conversion $Mn(s) \rightarrow Mn^{2+}(aq)$ more exothermic than the general trend would predict.

The net effect: the overall $\Delta H$ for $Mn(s) \rightarrow Mn^{2+}(aq)$ is more favourable (more negative) than the trend would suggest, giving a more negative $E^\ominus$ .

Note

Zn actually has the lowest enthalpy of atomisation in the whole first-row series—its $3d^{10}4s^2$ configuration leaves no unpaired or readily-available d-electrons to contribute to metallic bonding, so the lattice is unusually weak. In addition, the $Zn^{2+}$ ion ( $d^{10}$ ) gains no crystal field stabilisation energy (CFSE) in an octahedral field, making its hydration enthalpy less exothermic than ions that do gain CFSE.

The combination of an unusually low atomisation enthalpy (easy to atomise) and the exceptional stability of the resulting $d^{10}$ ion makes Zn's $E^\ominus$ more negative than the general trend.

4. Ni: The $d^8$ Anomaly

Nickel has $[Ar] 3d^8 4s^2$ . Unlike Mn and Zn, Ni's anomaly is not primarily about atomisation enthalpy or a specially stable ion configuration ( $d^8$ is neither half-filled nor fully filled).

Watch out

The deciding factor for Ni is different from Mn and Zn: $Ni^{2+}$ has the highest (most exothermic) enthalpy of hydration of any ion in the 3d series. A more exothermic hydration step makes $M(s) \rightarrow M^{2+}(aq)$ more energetically favourable overall, which pushes $E^\ominus$ more negative than the general trend predicts. …

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Method: Electronic Configuration & Stability Analysis

This method uses electronic configuration and exchange energy to explain why certain transition metals show unexpectedly negative $E^\ominus$ values.

Step 1 – Recall the expected trend

For 3d-series transition metals, $E^\ominus$ values generally become more negative from left to right, but Mn, Ni, and Zn deviate — they are more negative than the trend predicts.

Step 2 – Write the electronic configurations

Metal	Configuration (M)	Configuration (M²⁺)
Mn	$[\text{Ar}]\,3d^5\,4s^2$	$[\text{Ar}]\,3d^5$
Ni	$[\text{Ar}]\,3d^8\,4s^2$	$[\text{Ar}]\,3d^8$
Zn	$[\text{Ar}]\,3d^{10}\,4s^2$	$[\text{Ar}]\,3d^{10}$

Step 3 – Identify the special stability

Mn²⁺ has a half-filled d⁵ configuration — extra stability due to symmetry and exchange energy.
Zn²⁺ has a fully-filled d¹⁰ configuration — maximum stability.
Ni²⁺ has a d⁸ configuration — gains extra crystal field stabilization energy in octahedral complexes.

Step 4 – Link to $E^\ominus$ negativity

A more negative $E^\ominus$ means the metal is more difficult to reduce (or easier to oxidise).

Why? Because the M²⁺ state is unusually stable — it resists being reduced back to M.

For Mn: $E^\ominus(\text{Mn}^{2+}/\text{M}) = -1.18\ \text{V}$ The half-filled d⁵ makes Mn²⁺ very stable → reduction is unfavourable → $E^\ominus$ is more negative. …

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Here are the common mistakes students make when answering why $E^\ominus$ values for Mn, Ni, and Zn are more negative than expected, along with how to avoid each.

Mistake 1: Confusing "more negative" with "less stable"

What students do: They see a very negative $E^\ominus$ and think the element is unstable or easily reduced.
Why it’s wrong: A more negative $E^\ominus$ (e.g., $\mathrm{Mn^{2+}/Mn}$ = –1.18 V) means the metal is a strong reducing agent — it gives up electrons more readily to form the ion (the reverse reduction, $M^{2+} + 2e^- \rightarrow M$ , is thermodynamically unfavourable). So the metal is actually easier to oxidise, not harder.
How to avoid: Always interpret $E^\ominus$ $E^{⊖}$ as a measure of reducing power:
- More negative → stronger reducing agent → metal is more stable in its zero oxidation state.
- The question asks: Why is the metal so stable that it takes a very negative potential to reduce the ion?

Mistake 2: Forgetting the electronic configuration reason

What students do: They memorise the $E^\ominus$ values but cannot explain why Mn, Ni, and Zn are special.
Why it’s wrong: The answer lies in extra stability of half-filled / fully-filled d-subshells:
- Mn²⁺: $3d^5$ (half-filled) — very stable.
- Zn²⁺: $3d^{10}$ (fully filled) — very stable.
- Ni²⁺: $3d^8$ — not half/fully filled, but Ni has high hydration enthalpy and low sublimation energy, making the ion unusually stable.
How to avoid: For each element, write the d-electron configuration of the +2 ion and note if it’s half-filled ( $d^5$ ) or fully filled ( $d^{10}$ ). For Ni, remember the hydration enthalpy argument.

Mistake 3: Ignoring the three-step enthalpy cycle

What students do: They only talk about ionisation energy or hydration enthalpy in isolation.
Why it’s wrong: $E^\ominus$ depends on the net Gibbs energy of:

\mathrm{M(s) \rightarrow M^{2+}(aq) + 2e^-}

This involves:

Sublimation enthalpy ( $\Delta H_{\text{sub}}$ )
Ionisation enthalpy (first + second)
Hydration enthalpy ( $\Delta H_{\text{hyd}}$ )

How to avoid: Always mention that a more negative $E^\ominus$ means the sum of these three steps is less favourable (i.e., the metal is reluctant to form the ion). For Mn, Ni, Zn, the hydration enthalpy is high enough to compensate, but the ionisation enthalpy is also high — the net effect is a more negative $E^\ominus$ .

Mistake 4: Thinking "more negative" means "more reactive"

What students do: They say “Mn is more reactive because its $E^\ominus$ is more negative.”
Why it’s wrong: Reactivity in displacement reactions depends on the actual $E^\ominus$ value, but the question asks why the value is more negative than expected — i.e., why the trend deviates from the general increase across the series.
How to avoid: Compare with neighbours (e.g., Cr, Fe, Co, Cu). For example:
- Cr²⁺ → Cr: –0.91 V
- Mn²⁺ → Mn: –1.18 V (more negative than expected)
- Fe²⁺ → Fe: –0.44 V
- Ni²⁺ → Ni: –0.25 V
- Zn²⁺ → Zn: –0.76 V (more negative than expected) The anomaly is that Mn and Zn are more negative than their neighbours, not that they are the most negative overall.

Mistake 5: Mixing up Mn and Zn with Cu or Ag

What students do: They say “Zn has a fully filled d¹⁰, so it’s stable” — correct — but then they incorrectly apply the same logic to Cu (d¹⁰ in Cu⁺, but Cu²⁺ is d⁹). …

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

KCET 2025Set D-41 markMCQ
$[\text{NiCl}_4]^{2-}$
(a) The complex has tetrahedral geometry
(b) Co-ordination number of Ni is 2 and oxidation state is +4
(c) The complex is $sp^3$ hybridised
(d) It is a high spin complex
(e) The complex is paramagnetic (A) a, c, d and e (B) a, b, d and e (C) b, c, d and e (D) a, b, c and d
›Reveal solutionSolution
Find the oxidation state (+2, so $d^8$ ), note Cl⁻ is a weak-field ligand so no pairing occurs, giving a high-spin $sp^3$ tetrahedral paramagnetic complex — every statement is true except (b), whose CN and oxidation state are both wrong.

Step 1 — Oxidation state and d-configuration of nickel

Let the oxidation state of Ni be $x$ . Chloride carries $-1$ each, and the overall charge is $-2$ :
$x + 4(-1) = -2 \quad\Longrightarrow\quad x = -2 + 4 = +2$
So the metal is $\ce{Ni^{2+}}$ .

Nickel is $Z = 28$ : $\;\ce{Ni} = [\text{Ar}]\,3d^{8}\,4s^{2}$ . Removing the two $4s$ electrons:
$\ce{Ni^{2+}} = [\text{Ar}]\,3d^{8} \qquad (\text{a } d^8 \text{ ion})$
Step 2 — Coordination number

Chloride is a unidentate ligand (one donor atom, Cl). With four of them:
$\text{Coordination number} = 4$
Step 3 — This already settles statement (b)

(b) "Co-ordination number of Ni is 2 and oxidation state is +4"

Both halves are wrong — the CN is 4 (Step 2) and the oxidation state is +2 (Step 1). (b) is FALSE.

This is decisive: every option containing (b) — namely (B), (C) and (D) — is eliminated at once. Only (A) a, c, d and e remains. Let us verify that all four of those statements are indeed true.

Step 4 — Ligand field strength decides everything else

In the spectrochemical series, chloride sits near the weak-field end:
$\ce{I- < Br- < \underline{Cl-} < F- < H2O < NH3 < en < CN- \approx CO}$
$\ce{Cl-}$ is a weak-field ligand, so the splitting energy $\Delta$ it produces is small — smaller than the electron pairing energy $P$ :
$\Delta < P$
When $\Delta < P$ , it costs less energy for an electron to occupy a higher orbital than to pair up in a lower one. Therefore no pairing occurs — the $3d^8$ configuration is left untouched, with its two unpaired electrons intact.

Since the $3d$ orbitals are not vacated, no inner $3d$ orbital is available for hybridisation.

Step 5 — Hybridisation and geometry ⇒ statements (a) and (c)

With the $3d$ set unavailable, Ni²⁺ must use its outer orbitals: one $4s$ and three $4p$ .
$4s + 4p_x + 4p_y + 4p_z \;\longrightarrow\; \textbf{four } sp^3 \textbf{ hybrid orbitals}$
Four $sp^3$ orbitals point to the corners of a tetrahedron ( $109.5^\circ$ ).

(a) "The complex has tetrahedral geometry" — TRUE ✓

(c) "The complex is $sp^3$ hybridised" — TRUE ✓ …
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COMEDK 2025Set 2025-A1 markMCQ
$Q. Larger number of oxidation states are exhibited by the actinoids than those of lanthanoids. The reason is: (A) Lesser energy difference between 5 f and 6 d than between 4 f and 5 d orbitals (B) More energy difference between 5 f and 6 d than between 4 f and 5 d orbitals (C) 4 f orbitals are more diffused than 5 f orbitals (D) Highly reactive nature of the actinoids$
›Reveal solutionSolution
The key idea is that actinoids show more oxidation states because their 5f and 6d orbitals are closer in energy than the 4f and 5d orbitals of lanthanoids, making it easier to involve f-electrons in bonding. The correct option is (A).

The question asks why actinoids exhibit a larger number of oxidation states than lanthanoids. This is a classic comparison in f-block chemistry, rooted in the electronic structure of the two series.

Concept and Intuition

Oxidation states arise when an atom loses electrons. For f-block elements, the electrons lost can come from both the f and d orbitals. The ease of removing f-electrons depends on how tightly they are held, which is related to the energy gap between the f and d orbitals. A smaller gap means f-electrons can be promoted to d orbitals more readily, allowing a wider range of oxidation states. Actinoids (5f series) have a smaller 5f–6d energy difference than lanthanoids (4f–5d), so they can access more oxidation states.

Let’s break it down step by step.

Understand the orbital energy trends

In lanthanoids, the 4f orbitals are deeply buried inside the atom, shielded by outer electrons. The 5d orbitals are at a significantly higher energy. This large 4f–5d energy gap makes it difficult to remove or promote 4f electrons, so lanthanoids typically show only +3 (and occasionally +2 or +4) oxidation states.

In actinoids, the 5f orbitals are less shielded and more extended (diffuse). The 5f and 6d orbitals are much closer in energy. This small energy difference allows 5f electrons to be easily promoted to 6d orbitals or directly lost, enabling a variety of oxidation states (e.g., +3, +4, +5, +6, and even +7 in some cases like neptunium and plutonium).

Evaluate the options

(A) Lesser energy difference between 5f and 6d than between 4f and 5d orbitals — This matches the explanation above.

(B) More energy difference — This would make it harder to involve f-electrons, reducing oxidation states, so incorrect.

(C) 4f orbitals are more diffused than 5f orbitals — Actually, 5f orbitals are more diffused (less tightly held) due to poorer shielding, so this is false. …
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COMEDK 2024Set 2024-E1 markMCQ
$\mathrm{Cr}(\mathrm{VI})$
›Reveal solutionSolution
The question tests knowledge of transition metal chemistry: magnetic properties of manganate vs. permanganate, properties of interstitial compounds, oxidation states of cerium, and stability of Cr(VI) vs. W(VI). Only statement (C) is correct.

Let’s examine each statement carefully, using chemical principles to decide which one is accurate.

Statement (A): “The green manganate ion shows diamagnetic nature but the permanganate ion exhibits paramagnetic nature.”

The manganate ion is $\text{MnO}_4^{2-}$ , where manganese is in the +6 oxidation state. Electronic configuration of Mn in +6: $[\text{Ar}]\,3d^1$ . That’s one unpaired electron → paramagnetic, not diamagnetic.

The permanganate ion is $\text{MnO}_4^{-}$ , with Mn in +7: $[\text{Ar}]\,3d^0$ . No unpaired electrons → diamagnetic.

So the statement gets both magnetic natures backwards. False.

Statement (B): “Interstitial compounds of transition metals have lower melting points than that of pure transition metals and their compounds are chemically reactive.”

Interstitial compounds (e.g., carbides, nitrides, hydrides) form when small atoms like C, N, or H occupy holes in the metal lattice. This usually increases hardness and melting point (often very high), and they are chemically inert (not reactive).

Both claims here are opposite to reality. False.

Statement (C): “Cerium is a lanthanoid metal which exists in a stable oxidation state of +4, besides exhibiting an oxidation state of +3.”

Cerium (Ce, atomic number 58) has the electron configuration $[Xe]\,4f^1\,5d^1\,6s^2$ . The common +3 state arises from losing the 5d and 6s electrons.

The +4 state is also stable because losing one more electron gives a $4f^0$ configuration (empty f-subshell), which is especially stable. Ce(IV) is well-known in compounds like $\text{CeO}_2$ and ceric ammonium nitrate.

This is a textbook fact. True. …
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COMEDK 2024Set 2024-M1 markMCQ
$\mathrm{Cu}$
›Reveal solutionSolution
The key idea is to evaluate each statement about transition-metal chemistry using periodic trends and redox stability; the incorrect statement is (A) because fluorine cannot exceed oxygen in stabilising high oxidation states.

Let’s go through each option carefully, building the reasoning step by step.

Option (A): “The ability of Fluorine to stabilise the higher oxidation states of transition metals exceeds that of Oxygen”

In transition-metal oxyanions (like $\mathrm{CrO_4^{2-}}$ , $\mathrm{MnO_4^-}$ ), oxygen stabilises high oxidation states via strong $\pi$ -bonding (O donates electron density to the metal, reducing its effective charge).

Fluorine is more electronegative than oxygen, but it is a poor $\pi$ -donor (it has no available d-orbitals for back-bonding) and forms weaker multiple bonds.

For example, $\mathrm{Mn_2O_7}$ (Mn in +7) is stable, but $\mathrm{MnF_7}$ does not exist; the highest fluoride of Mn is $\mathrm{MnF_4}$ (+4).

Conclusion: Oxygen stabilises high oxidation states better than fluorine. So statement (A) is false.

Option (B): “Cu(I) compounds in aqueous medium undergo disproportionation reaction”

Disproportionation: $2\mathrm{Cu^+} \rightarrow \mathrm{Cu} + \mathrm{Cu^{2+}}$ .

In water, the standard reduction potentials: $\mathrm{Cu^+ + e^- \rightarrow Cu}$ ( $E^\circ = +0.52\,\mathrm{V}$ ) and $\mathrm{Cu^{2+} + e^- \rightarrow Cu^+}$ ( $E^\circ = +0.16\,\mathrm{V}$ ).

The net cell potential for disproportionation is $E^\circ_{\text{cell}} = 0.52 - 0.16 = +0.36\,\mathrm{V} > 0$ , so it is spontaneous.

Conclusion: Statement (B) is true.

Option (C): “ $\mathrm{Cr^{2+}}$ is a stronger reducing agent than $\mathrm{Fe^{2+}}$ ”

Standard reduction potentials: $\mathrm{Cr^{3+} + e^- \rightarrow Cr^{2+}}$ ( $E^\circ = -0.41\,\mathrm{V}$ ) $\mathrm{Fe^{3+} + e^- \rightarrow Fe^{2+}}$ ( $E^\circ = +0.77\,\mathrm{V}$ )

A more negative reduction potential means the reduced form (here $\mathrm{Cr^{2+}}$ ) is more easily oxidised — i.e., a stronger reducing agent. …
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KCET 2023Set D-21 markMCQ
$(n-1)d^{10}ns^2$
›Reveal solutionSolution
The configuration $(n-1)d^{10}ns^2$ is the ground-state pattern of group-12 elements. Copper is the exception — it is $(n-1)d^{10}ns^1$ , not $ns^2$ — so the pair in which the configuration fails is (D) Cu, Zn.

The pattern $(n-1)d^{10}ns^2$ means a filled $(n-1)d$ subshell together with a filled $ns$ subshell. This is the hallmark of the group-12 elements — Zn, Cd, Hg and Cn — in their ground state. To find the pair that breaks the pattern, we check each element's configuration.

Option (A): Zn, Cd.

Zn ( $Z=30$ ) is $[\text{Ar}]3d^{10}4s^2$ ; Cd ( $Z=48$ ) is $[\text{Kr}]4d^{10}5s^2$ . Both group 12 — both fit the pattern.

Option (B): Cd, Hg.

Cd fits; Hg ( $Z=80$ ) is $[\text{Xe}]4f^{14}5d^{10}6s^2$ . Both group 12 — both fit.

Option (C): Hg, Cn.

Hg fits; Cn ( $Z=112$ ) is the group-12 element $[\text{Rn}]5f^{14}6d^{10}7s^2$ . Both fit.

Option (D): Cu, Zn.

Zn is $3d^{10}4s^2$ (fits), but Cu ( $Z=29$ ) is the classic exception: its ground state is $[\text{Ar}]3d^{10}4s^1$ , not $3d^9 4s^2$ , because a completely filled $d^{10}$ subshell is especially stable and pulls one electron out of $4s$ . So Cu is $(n-1)d^{10}ns^1$ and does not show the $ns^2$ configuration. …
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COMEDK 2023Set 2023-E1 markMCQ
$\mathrm{Mn}^{2+}$
›Reveal solutionSolution

The incorrect statement is (A): fluorine stabilises higher oxidation states because of the high lattice (and bond) enthalpy of its compounds, not the low lattice enthalpy. B, C and D are correct NCERT statements.

Option (A) — incorrect. Fluorine, being small and highly electronegative, forms fluorides with high lattice enthalpy (and high M–F bond enthalpy). It is this high lattice/bond enthalpy that lets fluorine stabilise the highest oxidation states of transition metals. The statement wrongly attributes it to "low lattice enthalpy."

Option (B) — correct. The irregular ionisation enthalpies in this series arise from the stability of the half-filled $d^5$ configurations produced ( $\mathrm{Mn^{2+}}$ , $\mathrm{Fe^{3+}}$ ), as noted in NCERT.

Option (C) — correct. Transition metals form alloys readily because their metallic radii are similar (within ~15%). …
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KCET 2019Set A-11 markMCQ
$Ce (Z = 58)$
›Reveal solutionSolution
The question tests your understanding of lanthanide chemistry, specifically the stability and redox behaviour of cerium. The incorrect statement is (A): $Ce^{4+}$ is an oxidising agent, not a reducing agent.

Concept & Intuition

Cerium is the first element in the lanthanide series (Z=58). Its ground-state electronic configuration is $[Xe]\,4f^1\,5d^1\,6s^2$ . The key to its chemistry lies in the stability of the empty $4f$ subshell ( $4f^0$ ) and the half-filled $4f$ subshell ( $4f^7$ ). For cerium, losing four electrons gives the $Ce^{4+}$ ion with a $[Xe]$ configuration — a noble gas core, which is exceptionally stable. This stability makes $Ce^{4+}$ a strong oxidising agent (it readily accepts electrons to go back to the more common $Ce^{3+}$ state). In contrast, $Ce^{3+}$ has a $[Xe]\,4f^1$ configuration and is the most stable oxidation state for cerium in aqueous solution.

Now let’s examine each statement.

Statement (A): $Ce^{4+}$ is a reducing agent.

A reducing agent is a substance that donates electrons (gets oxidised itself). $Ce^{4+}$ has a strong tendency to gain one electron and become $Ce^{3+}$ (the $4f^1$ configuration is more stable than $4f^0$ in most chemical environments). This means $Ce^{4+}$ is an oxidising agent, not a reducing agent. In fact, $Ce^{4+}$ is a well-known oxidising agent in analytical chemistry (e.g., in cerimetric titrations). So this statement is false.

Statement (B): Atomic size of Ce is more than that of Lu.

This is true. Across the lanthanide series (from Ce, Z=58, to Lu, Z=71), there is a steady decrease in atomic and ionic radii — the lanthanide contraction. The 4f electrons are poorly shielding, so as nuclear charge increases, the electron cloud is pulled inward. Ce is near the beginning of the series, Lu at the end, so Ce has a larger atomic radius than Lu.

Statement (C): Ce in +3 oxidation state is more stable than in +4. …
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Q.Why E⊖E^\ominusE⊖ values for Mn, Ni and Zn are more negative than expected?

The Core Idea: What E⊖E^\ominusE⊖ Actually Measures

Step-by-Step: Why Mn, Ni, and Zn Stand Out

1. The Expected Trend

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4. Ni: The d8d^8d8 Anomaly

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Q.Why $E^\ominus$ values for Mn, Ni and Zn are more negative than expected?

The Core Idea: What $E^\ominus$ Actually Measures

2. Mn: The Half-Filled $d^5$ Stability

4. Ni: The $d^8$ Anomaly

Unlock everything free for 14 days