The product of the oxidation of $I^{-}$ with $MnO_4^{-}$ in alkaline medium is : (A) $IO_4^{-}$ (B) $I_2$ (C) $IO^{-}$ (D) $IO_3^{-}$

The product of the oxidation of I⁻ with MnO₄⁻ in alkaline medium is : (A) IO₄⁻ (B) I₂ (C) IO⁻ (D) IO₃⁻

Question

I^{-}

CBSECBSE Class XII Board 2025MCQ· 1mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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: …

Start your 14-day free trial to unlock the full solution →

In alkaline medium, permanganate ( $MnO_4^-$ ) oxidises iodide ( $I^-$ ) to iodate ( $IO_3^-$ ), not to iodine or periodate. The balanced reaction shows $I^-$ loses 6 electrons to form $IO_3^-$ , while $MnO_4^-$ gains 3 electrons to form $MnO_2$ . The correct product is $IO_3^-$ , option (D).

Why the medium matters

The oxidation state of iodine in its products depends heavily on the pH of the solution. Permanganate is a powerful oxidising agent, but its reduction product changes with medium:

In acidic medium: $MnO_4^- \rightarrow Mn^{2+}$ (gains 5 electrons)
In neutral/alkaline medium: $MnO_4^- \rightarrow MnO_2$ (gains 3 electrons)

This difference in electron gain per mole of permanganate directly affects how far it can oxidise iodide. In alkaline medium, permanganate is a milder oxidising agent (gains only 3 electrons) compared to acidic medium (gains 5 electrons). Yet it still oxidises $I^-$ all the way to $IO_3^-$ , not stopping at $I_2$ .

Step-by-step reasoning

Identify the half-reactions

Iodide ( $I^-$ ) has oxidation state $-1$ . The possible products given are:
- $IO_4^-$ : iodine in $+7$ state
- $I_2$ : iodine in $0$ state
- $IO^-$ : iodine in $+1$ state (hypoiodite)
- $IO_3^-$ : iodine in $+5$ state (iodate)
In alkaline medium, permanganate reduces to $MnO_2$ (manganese in $+4$ state, from $+7$ in $MnO_4^-$ ).
Balance the oxidation half-reaction

Iodide going to iodate:

I^- \rightarrow IO_3^-

Balance oxygen with water (alkaline medium):

I^- + 3H_2O \rightarrow IO_3^- + 6H^+

Balance charge: left side has $-1$ , right side has $-1 + 6 = +5$ . Add 6 electrons to right:

I^- + 3H_2O \rightarrow IO_3^- + 6H^+ + 6e^-

In alkaline medium, add $OH^-$ to neutralise $H^+$ :

I^- + 6OH^- \rightarrow IO_3^- + 3H_2O + 6e^-

So each $I^-$ loses 6 electrons to become $IO_3^-$ .

Balance the reduction half-reaction

Permanganate to manganese dioxide in alkaline medium:

MnO_4^- \rightarrow MnO_2

Balance oxygen with water:

MnO_4^- + 2H_2O \rightarrow MnO_2 + 4OH^-

Balance charge: left $-1$ , right $-4$ . Add 3 electrons to left:

MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-

So each $MnO_4^-$ gains 3 electrons.

Combine the half-reactions

To equalise electrons: multiply reduction half by 2 (gives 6 electrons gained) and oxidation half by 1 (gives 6 electrons lost):

2MnO_4^- + 4H_2O + 6e^- \rightarrow 2MnO_2 + 8OH^-

I^- + 6OH^- \rightarrow IO_3^- + 3H_2O + 6e^-

Adding:

2MnO_4^- + I^- + 4H_2O + 6OH^- \rightarrow 2MnO_2 + IO_3^- + 3H_2O + 8OH^-

Cancel $3H_2O$ from both sides and $6OH^-$ from both sides: …

Start your 14-day free trial to unlock the full solution →

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

Showing the 12 most recent of 19 on this concept.

CBSE 2026Set 56/3/11 markMCQ
$Cr_2O_7^{2-}$
›Reveal solutionSolution
In alkaline medium, dichromate ( $Cr_2O_7^{2-}$ ) converts to chromate ( $CrO_4^{2-}$ ) without any change in oxidation state — it’s a simple acid-base equilibrium, not a redox reaction. The correct option is (B).

The key to this question lies in understanding that the conversion of dichromate to chromate is not a redox reaction — the oxidation state of chromium remains +6 throughout. Many students instinctively think of reduction to $Cr^{3+}$ because they associate dichromate with strong oxidizing behaviour, but that only happens in acidic medium. In alkaline conditions, the chemistry is entirely different.

Let’s walk through the reasoning step by step.

Recall the oxidation state of chromium in dichromate.

In $Cr_2O_7^{2-}$ , each oxygen is -2, so total from seven oxygens is -14. The ion has a -2 charge, so the sum of oxidation states of the two chromium atoms must be +12. Hence each Cr is in the +6 state.

Now consider the alkaline medium.

When you add a base (like NaOH) to a solution of $K_2Cr_2O_7$ , the dichromate ion reacts with hydroxide ions. The reaction is:

$Cr_2O_7^{2-} + 2OH^- \rightarrow 2CrO_4^{2-} + H_2O$
Notice that the oxidation state of Cr in $CrO_4^{2-}$ is also +6 (four oxygens at -2 give -8, charge -2, so Cr = +6). No electrons are transferred — this is an acid-base equilibrium, not a redox change.

Why does this happen? Dichromate exists in equilibrium with chromate, and the position depends on pH. In acidic solution, the equilibrium shifts toward dichromate; in alkaline solution, it shifts toward chromate. The reaction is:

$2CrO_4^{2-} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} + H_2O$
Adding $OH^-$ removes $H^+$ , pulling the equilibrium to the left — producing chromate. …
Start your 14-day free trial to see the full solution →
CBSE 2026Set 56/1/11 markMCQ
$Q. Assertion (A) : Actinoids show wide range of oxidation states. Reason (R) : Actinoids are radioactive in nature. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.$
›Reveal solutionSolution
The assertion that actinoids show a wide range of oxidation states is true, but the reason given — that they are radioactive — is not the correct explanation. The correct explanation lies in the small energy gap between 5f, 6d, and 7s orbitals, which allows many electrons to participate in bonding. So the answer is option (B).

The question tests your understanding of why actinoids (elements 90–103, from thorium to lawrencium) exhibit so many different oxidation states. Many students memorise that “actinoids show variable oxidation states” and also know they are radioactive, so they assume the second explains the first. That’s a trap.

Let’s break it down properly.

Is Assertion (A) true?

Yes. Actinoids display a remarkably wide range of oxidation states. For example, uranium shows +3, +4, +5, and +6; neptunium and plutonium go from +3 to +7. This is far more varied than most d-block elements. The reason is that the 5f, 6d, and 7s orbitals are very close in energy. Electrons from all three can be lost with relatively little energy cost, so many different oxidation numbers become accessible.

Is Reason (R) true?

Yes, actinoids are indeed radioactive. All actinoid nuclei are unstable and decay over time. So the reason statement is factually correct.

Does the radioactivity explain the wide range of oxidation states?

No. Radioactivity is a nuclear property — it depends on the instability of the nucleus (proton/neutron ratio, nuclear binding energy). Oxidation states are an electronic property — they depend on how easily electrons are lost from the outer orbitals. These two phenomena are completely independent.

Watch out

A common mistake is to think that because both statements are true, the reason must be the explanation. But correlation is not causation. Radioactivity does not cause variable oxidation states; the orbital energy structure does.

What actually causes the wide range of oxidation states in actinoids? …
Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following oxidation states is common for all lanthanoids?$
(a) +2
(b) +3
(c) +4
(d) +5
›Reveal solutionSolution

All lanthanoids show a characteristic +3 oxidation state because it corresponds to a stable, similar electronic configuration achieved after losing the two 6s and one 4f (or 5d) electron.

Lanthanoids have the general electronic configuration [Xe] 4f^(1-14) 5d^(0-1) 6s2. Removal of the two 6s electrons and one more electron (from 4f or 5d) gives the Ln3+ ion, which is the most stable and commonly observed oxidation state across the entire series, from Ce to Lu.

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set 56/4/11 markMCQ
$I^{-}$
›Reveal solutionSolution
In alkaline medium, permanganate ( $MnO_4^-$ ) oxidises iodide ( $I^-$ ) to iodate ( $IO_3^-$ ), not to iodine or periodate. The balanced reaction shows $I^-$ loses 6 electrons to form $IO_3^-$ , while $MnO_4^-$ gains 3 electrons to form $MnO_2$ . The correct product is $IO_3^-$ , option (D).

Why the medium matters

The oxidation state of iodine in its products depends heavily on the pH of the solution. Permanganate is a powerful oxidising agent, but its reduction product changes with medium:

In acidic medium: $MnO_4^- \rightarrow Mn^{2+}$ (gains 5 electrons)

In neutral/alkaline medium: $MnO_4^- \rightarrow MnO_2$ (gains 3 electrons)

This difference in electron gain per mole of permanganate directly affects how far it can oxidise iodide. In alkaline medium, permanganate is a milder oxidising agent (gains only 3 electrons) compared to acidic medium (gains 5 electrons). Yet it still oxidises $I^-$ all the way to $IO_3^-$ , not stopping at $I_2$ .

Step-by-step reasoning

Identify the half-reactions

Iodide ( $I^-$ ) has oxidation state $-1$ . The possible products given are:

$IO_4^-$ : iodine in $+7$ state

$I_2$ : iodine in $0$ state

$IO^-$ : iodine in $+1$ state (hypoiodite)

$IO_3^-$ : iodine in $+5$ state (iodate)

In alkaline medium, permanganate reduces to $MnO_2$ (manganese in $+4$ state, from $+7$ in $MnO_4^-$ ).

Balance the oxidation half-reaction

Iodide going to iodate:

$I^- \rightarrow IO_3^-$
Balance oxygen with water (alkaline medium):
$I^- + 3H_2O \rightarrow IO_3^- + 6H^+$
Balance charge: left side has $-1$ , right side has $-1 + 6 = +5$ . Add 6 electrons to right:
$I^- + 3H_2O \rightarrow IO_3^- + 6H^+ + 6e^-$
In alkaline medium, add $OH^-$ to neutralise $H^+$ :
$I^- + 6OH^- \rightarrow IO_3^- + 3H_2O + 6e^-$
So each $I^-$ loses 6 electrons to become $IO_3^-$ .

Balance the reduction half-reaction

Permanganate to manganese dioxide in alkaline medium:

$MnO_4^- \rightarrow MnO_2$
Balance oxygen with water:
$MnO_4^- + 2H_2O \rightarrow MnO_2 + 4OH^-$
Balance charge: left $-1$ , right $-4$ . Add 3 electrons to left:
$MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$
So each $MnO_4^-$ gains 3 electrons.

Combine the half-reactions

To equalise electrons: multiply reduction half by 2 (gives 6 electrons gained) and oxidation half by 1 (gives 6 electrons lost):

$2MnO_4^- + 4H_2O + 6e^- \rightarrow 2MnO_2 + 8OH^-$ $I^- + 6OH^- \rightarrow IO_3^- + 3H_2O + 6e^-$
Adding:
$2MnO_4^- + I^- + 4H_2O + 6OH^- \rightarrow 2MnO_2 + IO_3^- + 3H_2O + 8OH^-$
Cancel $3H_2O$ from both sides and $6OH^-$ from both sides: …
Start your 14-day free trial to see the full solution →
CBSE 2025Set 56/6/11 markMCQ
$Q. For the following question, two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A) : Actinoids show wide range of oxidation states. Reason (R) : Actinoids are radioactive in nature.$
›Reveal solutionSolution
The assertion that actinoids show a wide range of oxidation states is true, but the reason given — that they are radioactive — does not explain this property. The correct answer is (B).

The question tests your understanding of why actinoids exhibit variable oxidation states. The key is to separate two distinct facts: actinoids are radioactive, and they do show many oxidation states — but the radioactivity is not the cause of the oxidation state variability.

Let’s break this down.

Why do actinoids show a wide range of oxidation states?

The 5f, 6d, and 7s orbitals in actinoids are very close in energy. This means electrons can be removed from any of these orbitals with relatively little energy cost. As you move across the actinoid series, the 5f orbitals gradually become more stable, but early actinoids (like Th, Pa, U, Np, Pu) can lose anywhere from 3 to 7 electrons. For example, uranium shows +3, +4, +5, and +6; plutonium shows +3, +4, +5, +6, and +7. This is the real reason for the wide range — it’s an electronic structure effect, not a nuclear one.

What about radioactivity?

Yes, all actinoids are radioactive — their nuclei are unstable and decay over time. But radioactivity is a nuclear property, while oxidation states depend on electron configuration. A nucleus decaying does not directly change how many electrons an atom can lose or gain in a chemical reaction. So while both statements are factually true, the reason does not explain the assertion. …
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$Q. What is the common oxidation state of Lanthanoids?$
›Reveal solutionSolution

All lanthanoids overwhelmingly favour the +3 oxidation state, since their poorly-bonding 4f electrons are not readily involved, leaving the same outer 5d/6s electrons available across the series.

Across the entire lanthanide series, the +3 oxidation state is by far the most common and stable one, shown by essentially every lanthanoid. This is because the 4f electrons are deeply buried and well-shielded, taking little part in bonding, while the outer $5d^{0-1}6s^2$ electrons are readily lost to give the stable $Ln^{3+}$ ion. Occasional +2 or +4 states occur only for a few elements whe …

Start your 14-day free trial to see the full solution →
CBSE 2024Set A11 markMCQ
$Q. Which of the following pair of metal oxides are amphoteric?$
(a) $\text{V}_2\text{O}_5$ , $\text{Cr}_2\text{O}_3$
(b) $\text{Mn}_2\text{O}_7$ , $\text{CrO}_3$
(c) $\text{V}_2\text{O}_5$ , $\text{V}_2\text{O}_4$
(d) $\text{CrO}$ , $\text{V}_2\text{O}_5$
›Reveal solutionSolution

$\text{V}_2\text{O}_5$ and $\text{Cr}_2\text{O}_3$ are the amphoteric pair — option (a).

For transition-metal oxides, the character changes from basic (low oxidation state) through amphoteric to acidic (high oxidation state). $\text{Cr}_2\text{O}_3$ (Cr in +3) is amphoteric — it dissolves in acids to give $\text{Cr}^{3+}$ salts and in alkali to give chromite. $\text{V}_2\text{O}_5$ (V in +5) is chiefly acidic but is genuinely amphoteric, dissolving in both acids and alka …
Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. Which of the following oxidation state is common for all lanthanoids ?$
(a) +2
(b) +3
(c) +4
(d) +5
›Reveal solutionSolution

Every lanthanoid shows the +3 oxidation state as its characteristic and most stable state, even though a few also show +2 or +4 in special cases.

Lanthanoids (Ce to Lu) have the general electronic configuration $[Xe]4f^{1-14}5d^{0-1}6s^2$ . Losing the two 6s electrons and one 4f/5d electron gives the stable, half-filled/fully-filled-favouring $Ln^{3+}$ ion, which is why +3 is the predominant and universally shown oxidation state across the whole series. A handful of lanthanoids additi …

Start your 14-day free trial to see the full solution →
CBSE 2023Set 56/1/11 markMCQ
$Q. The most common and stable oxidation state of a Lanthanoid is : (A) + 2 (B) + 3 (C) + 4 (D) + 6$
›Reveal solutionSolution
Lanthanoids overwhelmingly prefer the +3 oxidation state due to the stability gained from losing the two 6s and one 5d/4f electron, achieving a configuration analogous to noble gases or half-filled/filled f-subshells. The answer is (B) +3.

Why Lanthanoids Love +3: Electronic Configuration and Stability

The lanthanoid series (elements 57–71: La through Lu) sits in the f-block, where the 4f orbitals are being progressively filled. To understand their oxidation state preference, we need to look at what electrons are available and what configurations become stable upon ionization.

A typical lanthanoid has the general electronic configuration:
$[\text{Xe}] \, 4f^{0-14} \, 5d^{0-1} \, 6s^2$
The 6s electrons are outermost and easiest to remove. The 5d and 4f orbitals are close in energy, so sometimes one electron occupies 5d instead of 4f. When a lanthanoid forms a cation, it loses electrons in a specific order: 6s electrons go first, then 5d, then 4f (because 4f is more tightly held, being an inner orbital).

Step-by-Step Reasoning

First ionization removes 6s electrons

All lanthanoids have two 6s electrons. Removing both gives a +2 state, but this is rarely the stopping point because the resulting ion still has relatively accessible 5d or 4f electrons.

Third electron removal: the key to +3 stability

After losing the two 6s electrons, removing one more electron (from 5d if occupied, otherwise from 4f) produces the +3 oxidation state. This configuration turns out to be remarkably stable across the entire series.

Why? The resulting $\text{Ln}^{3+}$ ion achieves one of several favorable electronic arrangements:

For La ( $4f^0$ ): $[\text{Xe}]$ — a noble gas configuration.

For Gd ( $4f^7$ ): half-filled f-subshell with all spins parallel (exchange energy stabilization).

For Lu ( $4f^{14}$ ): completely filled f-subshell.

For others: partially filled 4f with reasonable stability.

Why not +2?

The +2 state does exist for a few lanthanoids (Eu, Yb) where it leads to half-filled or filled f-subshells ( $4f^7$ for Eu²⁺, $4f^{14}$ for Yb²⁺), but these are exceptions, not the rule. Most lanthanoids find +2 too reducing and unstable in aqueous solution.

Why not +4 or higher?

Removing a fourth electron means breaking into the tightly held 4f subshell (which is shielded and contracted). The ionization energy jumps dramatically. Only Ce commonly shows +4 (because Ce⁴⁺ achieves $4f^0 = [\text{Xe}]$ ), and even that is a strong oxidizing agent. Higher states like +6 are virtually unknown in lanthanoids—the 4f electrons are too stable to remove. …
Start your 14-day free trial to see the full solution →
CBSE 2023Set 56/2/11 markMCQ
$[Fe(CO)_5]$
›Reveal solutionSolution

Carbonyl (CO) is a neutral ligand that does not contribute any charge. With five neutral CO ligands, the overall complex is neutral, so Fe must be in the 0 oxidation state. The correct option is (B).

Why this is a trick question — and how to see through it

Most students memorise that transition metals in coordination compounds usually show positive oxidation states like +2 or +3. Iron especially is famous for Fe(II) and Fe(III). So when you see [Fe(CO)5], the instinct is to guess +2 or +3. That instinct is wrong here — and the reason is beautiful.

The key is to ask: What charge does each ligand bring?

CO (carbonyl) is a neutral ligand. It donates a lone pair to the metal but carries no net charge. If every ligand is neutral, and the overall complex is neutral (no square brackets with a superscript charge), then the metal must be in the zero oxidation state.

This is not a rare exception — it is a whole class of compounds called metal carbonyls, where metals often exist in low or zero oxidation states. CO is a strong field ligand that stabilises these low states through back-bonding.

Step-by-step reasoning

1. Identify the charge on each ligand.

CO is carbon monoxide — a neutral molecule. In coordination chemistry, neutral ligands contribute 0 to the oxidation state calculation. Other examples: NH₃, H₂O, PPh₃.

2. Identify the overall charge on the complex.

The formula is written as [Fe(CO)5] — no superscript charge. That means the complex is neutral: overall charge = 0.

3. Set up the oxidation state equation.

Let the oxidation state of Fe be $x$ .

Each CO contributes 0. There are 5 CO ligands.

So:
$x + 5(0) = 0$
4. Solve for $x$ .
$x = 0$
That is the entire calculation — it takes one line once you know the rule.

Watch out

A common mistake is to treat CO as if it were a charged ligand like CN⁻ or Cl⁻. CO is not cyanide — it is neutral. Do not assign it a −1 charge. Also, do not confuse this with ferrocene or other organometallics where the ligand (like cyclopentadienyl) is anionic. …

Start your 14-day free trial to see the full solution →
CBSE 2023Set 56/2/11 markMCQ
$Q. Which of the following characteristics of transition metals is associated with their catalytic activity ? (A) Paramagnetic nature (B) Colour of hydrated ions (C) High enthalpy of atomisation (D) Variable oxidation states$
›Reveal solutionSolution
The catalytic activity of transition metals arises primarily from their ability to adopt variable oxidation states, which allows them to form intermediate complexes and lower activation energy. The correct option is (D).

Why this question tests a core idea

Catalysis is about providing an alternative reaction pathway with a lower activation energy. For a substance to be a good catalyst, it must be able to temporarily bind to reactants, change its own electronic state, and then release the products. Transition metals excel at this because they can change their oxidation state easily — often by ±1 — without breaking down. This flexibility lets them shuttle electrons to and from reactants, stabilising transition states that would otherwise be too high in energy.

The other options — paramagnetism, colour, and high enthalpy of atomisation — are important properties of transition metals, but they don't directly explain catalytic activity. Let's see why.

Step-by-step reasoning

Paramagnetic nature (A)

Paramagnetism arises from unpaired electrons. While many transition metal ions are paramagnetic, this property has no direct role in catalysis. A catalyst doesn't need unpaired electrons to speed up a reaction — it needs to form bonds with reactants and then break them. Paramagnetism is a consequence of electronic configuration, not a cause of catalytic behaviour.

Colour of hydrated ions (B)

The colour of transition metal complexes comes from $d$ – $d$ transitions — electrons jumping between split $d$ orbitals when they absorb visible light. This is fascinating, but it's a spectroscopic property. Colour tells us about the electronic structure of the ion, but it doesn't help the ion catalyse a reaction. A colourless catalyst can be just as effective.

High enthalpy of atomisation (C)

This refers to the energy required to convert a solid metal into isolated gaseous atoms. Transition metals have high enthalpies of atomisation because of strong metallic bonding (due to unpaired $d$ electrons contributing to bonding). This property is related to the strength of the metal lattice, not to its ability to change oxidation states during a catalytic cycle. In fact, a very high enthalpy of atomisation might make it harder for the metal to leave the lattice and participate in solution-phase catalysis.

Variable oxidation states (D) …
Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 markMCQ
$Q. Which of the following lanthanoid ions in solution is a good oxidizing agent ?$
(a) $Eu^{2+}$
(b) $Yb^{2+}$
(c) $Sm^{2+}$
(d) $Tb^{4+}$
›Reveal solutionSolution
+3 is the overwhelmingly preferred oxidation state across the whole lanthanide series, so an unusual +4 ion like Tb⁴⁺ tends to gain an electron and revert to +3 — making it a good oxidising agent.

Across the lanthanide series, +3 is by far the most stable and common oxidation state (arising from the overall energetics of the whole series, not just an individual ion's own f-subshell configuration). Ions that deviate from +3 — whether to +2 or +4 — tend to revert back to +3, and in doing so they act as either reducing or oxidising agents:

+2 lanthanide ions (Eu²⁺, Sm²⁺, Yb²⁺) tend to lose an electron to revert to +3 — they act as reducing agents. …
Start your 14-day free trial to see the full solution →

Q.The product of the oxidation of I−I^{-}I− with MnO4−MnO_4^{-}MnO4−​ in alkaline medium is : (A) IO4−IO_4^{-}IO4−​ (B) I2I_2I2​ (C) IO−IO^{-}IO− (D) IO3−IO_3^{-}IO3−​

Why the medium matters

Step-by-step reasoning

🎓Unlock everything free for 14 days

Q.The product of the oxidation of $I^{-}$ with $MnO_4^{-}$ in alkaline medium is : (A) $IO_4^{-}$ (B) $I_2$ (C) $IO^{-}$ (D) $IO_3^{-}$

Unlock everything free for 14 days