Which of the following is amphoteric oxide? $Mn_2O_7$, $CrO_3$, $Cr_2O_3$, $CrO$, $V_2O_5$, $V_2O_4$ (i) $V_2O_5$, $Cr_2O_3$ (ii) $Mn_2O_7$, $CrO_3$ (iii) $CrO$, $V_2O_5$ (iv) $V_2O_5$, $V_2O_4$

> [!TLDR] > Amphoteric oxides are those that react with both acids and bases. Among the given oxides, $V_2O_5$ and $Cr_2O_3$ are amphoteric, making option **(i)** correct. The key to identifying amphoteric oxides lies in the **oxidation state** of the metal. In transition metals, as the oxidation state increases, the oxide becomes more acidic. Low oxidation states give basic oxides, intermediate states give amphoteric oxides, and high states give acidic oxides. This trend mirrors the behaviour of non-metal oxides. Let’s examine each oxide step by step. 1. **$Mn_2O_7$** — Manganese is in the +7 oxidation state. This is the highest common state for Mn. The oxide is a green oily liquid that reacts violently with water to give permanganic acid ($HMnO_4$), a strong acid. It is **acidic**, not amphoteric. 2. **$CrO_3$** — Chromium is in the +6 state. This is a dark red solid that dissolves in water to form chromic acid ($H_2CrO_4$). It is a strong oxidizing agent and clearly **acidic**. (It is sometimes called chromic anhydride.) 3. **$Cr_2O_3$** — Chromium is in the +3 state. This is a green solid. It is insoluble in water but dissolves in both acids (giving $Cr^{3+}$ salts) and strong bases (giving chromite ions, $[Cr(OH)_4]^-$ or $CrO_2^-$). This dual behaviour makes it **amphoteric**. 4. **$CrO$** — Chromium is in the +2 state. This is a black solid that behaves as a **basic** oxide — it reacts with acids to form $Cr^{2+}$ salts but does not react with bases. 5. **$V_2O_5$** — Vanadium is in the +5 state. This is an orange-yellow solid. It is predominantly acidic (dissolves in bases to give vanadates), but it also reacts with strong acids (e.g., concentrated $HCl$) to form vanadyl salts ($VO_2^+$). Hence it is **amphoteric**, though weakly basic. …

Which of the following is amphoteric oxide? Mn₂O₇, CrO₃, Cr₂O₃, CrO, V₂O₅, V₂O₄ (i) V₂O₅, Cr₂O₃ (ii) Mn₂O₇, Cr

NCERT Exemplar · Q12

Mn_2O_7

(i)

V_2O_5

Cr_2O_3

(ii)

Mn_2O_7

CrO_3

(iii)

CrO

V_2O_5

(iv)

V_2O_5

V_2O_4

Sikkim CbseMCQ· 1mImportance★★★★★

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Concept understanding — Lanthanide Contraction

Lanthanide Contraction: The Intuition

Imagine you are walking through a dense forest. With every step forward, you push through thick undergrowth. The deeper you go, the more tired you become — each step feels a little harder, and you find yourself hunching forward, your shoulders pulling inward. That inward pull is exactly what happens inside the lanthanide atoms.

The lanthanides are the 14 elements from cerium ( $Ce$ , atomic number 58) to lutetium ( $Lu$ , atomic number 71). As you move from one element to the next, you add one proton to the nucleus and one electron to the atom. The new electron goes into a 4f orbital — a set of orbitals that are shaped like clover leaves and sit deep inside the atom, close to the nucleus.

Here is the key: 4f orbitals are poorly shielded. They do not spread out far from the nucleus, and they do not block the nuclear charge from pulling on the outer electrons. So when you add a proton, the nucleus gets stronger, and the 4f electrons do almost nothing to stop that extra pull. The result? The entire electron cloud — especially the outermost electrons — gets pulled inward. The atom shrinks.

Note

Shielding is the ability of inner electrons to "block" the outer electrons from feeling the full positive charge of the nucleus. Electrons in s and p orbitals shield well; 4f electrons shield very poorly.

The Precise Statement

Lanthanide contraction is the steady and significant decrease in the atomic and ionic radii of the lanthanide elements as atomic number increases from 58 (Ce) to 71 (Lu).

\text{Atomic radius} \propto \frac{1}{Z_{\text{eff}}}

The total contraction across the entire series is about 15–20 picometers — roughly 10–15% of the initial radius. That is a substantial shrinkage for a single row of the periodic table.

Why It Matters

This contraction has two enormous consequences in chemistry:

Similarity of post-lanthanide elements: After lutetium, the next elements are hafnium ( $Hf$ , 72), tantalum ( $Ta$ , 73), and tungsten ( $W$ , 74). Because the lanthanide contraction has made the atoms so small, these elements have almost identical atomic and ionic radii to their counterparts directly above them in the periodic table — zirconium ( $Zr$ ), niobium ( $Nb$ ), and molybdenum ( $Mo$ ). This is why zirconium and hafnium are chemically almost inseparable — they are the same size.
Difficulty in separating lanthanides: All lanthanide ions ( $Ln^{3+}$ ) have nearly identical chemical properties because their radii change so gradually. Separating them requires hundreds of repeated steps (ion-exchange chromatography, solvent extraction) — a painstaking process that was a major challenge in early nuclear chemistry.

Watch out

A common mistake is to think lanthanide contraction means the atoms get smaller because the 4f orbitals are "full" or because of some repulsion effect. It is purely due to poor shielding of the 4f electrons, which lets the nuclear charge pull everything inward.

The Numbers (for reference)

Element	Atomic Number	Ionic Radius ( $Ln^{3+}$ , pm)
Ce	58	103.4
Pr	59	101.3
Nd	60	99.5

Why this formula?

Lanthanide Contraction: Why It Happens

The Lanthanide Contraction is the steady decrease in atomic and ionic radii of the lanthanide elements (Ce to Lu) as atomic number increases. The key observation: the radii shrink by about 1–2 pm per element, despite adding electrons to the 4f subshell.

The Core Question

Why does adding electrons not increase the size, but instead decrease it?

The Formula That Governs It

The effective nuclear charge ( $Z_{\text{eff}}$ ) experienced by an electron is:

Z_{\text{eff}} = Z - S

Where:

$Z$ = atomic number (protons in nucleus)
$S$ = shielding constant (screening by inner electrons)

The key formula for the trend in ionic radii ( $r$ ) across the lanthanides is:

r \propto \frac{n^2}{Z_{\text{eff}}}

Where $n$ is the principal quantum number of the outermost electron (here, $n=6$ for the 6s orbital).

The Derivation: Step by Step

1. What happens when you add a proton and an electron?

Each lanthanide adds:

+1 proton to the nucleus (increases $Z$ by 1)
+1 electron to the 4f subshell

2. The 4f orbital is "penetrating" but poorly shielding

The 4f orbital has a radial distribution that peaks close to the nucleus (inside the 5s and 5p shells).
However, 4f electrons are very poor at shielding the outer 6s electrons from the nuclear charge.

Why?

The 4f orbital is diffuse and deeply buried — it does not effectively screen the outer electrons because:

Its shape (complex, multi-lobed) means it doesn't occupy the space between the nucleus and the 6s electrons efficiently.
The 4f electrons are inside the 5s/5p shells, so they don't block the nuclear pull on the 6s electrons.

3. The net effect on $Z_{\text{eff}}$

When you add one proton ( $\Delta Z = +1$ ) and one 4f electron ( $\Delta S \approx 0.85$ to 0.95), the change in effective nuclear charge is:

\Delta Z_{\text{eff}} \approx +1 - 0.85 = +0.15 \text{ to } +0.05

Result: $Z_{\text{eff}}$ increases slightly with each element.

4. How this shrinks the radius

From the formula $r \propto \frac{n^2}{Z_{\text{eff}}}$ :

$n$ (the principal quantum number of the 6s orbital) stays constant at 6.
$Z_{\text{eff}}$ increases.
Therefore, $r$ decreases. …

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Amphoteric oxides are those that react with both acids and bases. Among the given oxides, $V_2O_5$ and $Cr_2O_3$ are amphoteric, making option (i) correct.

The key to identifying amphoteric oxides lies in the oxidation state of the metal. In transition metals, as the oxidation state increases, the oxide becomes more acidic. Low oxidation states give basic oxides, intermediate states give amphoteric oxides, and high states give acidic oxides. This trend mirrors the behaviour of non-metal oxides.

Let’s examine each oxide step by step.

$Mn_2O_7$ — Manganese is in the +7 oxidation state. This is the highest common state for Mn. The oxide is a green oily liquid that reacts violently with water to give permanganic acid ( $HMnO_4$ ), a strong acid. It is acidic, not amphoteric.
$CrO_3$ — Chromium is in the +6 state. This is a dark red solid that dissolves in water to form chromic acid ( $H_2CrO_4$ ). It is a strong oxidizing agent and clearly acidic. (It is sometimes called chromic anhydride.)
$Cr_2O_3$ — Chromium is in the +3 state. This is a green solid. It is insoluble in water but dissolves in both acids (giving $Cr^{3+}$ salts) and strong bases (giving chromite ions, $[Cr(OH)_4]^-$ or $CrO_2^-$ ). This dual behaviour makes it amphoteric.
$CrO$ — Chromium is in the +2 state. This is a black solid that behaves as a basic oxide — it reacts with acids to form $Cr^{2+}$ salts but does not react with bases.
$V_2O_5$ — Vanadium is in the +5 state. This is an orange-yellow solid. It is predominantly acidic (dissolves in bases to give vanadates), but it also reacts with strong acids (e.g., concentrated $HCl$ ) to form vanadyl salts ( $VO_2^+$ ). Hence it is amphoteric, though weakly basic. …

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Common Mistakes Students Make on Amphoteric Oxides of Transition Metals

Mistake 1: Confusing Amphoteric with Acidic or Basic

The Error: Students often assume that all oxides of transition metals in high oxidation states are acidic, and those in low states are basic. They forget that intermediate oxidation states often give amphoteric behaviour.

Example from this question:

$Mn_2O_7$ (Mn in +7) → strongly acidic (not amphoteric)
$CrO$ (Cr in +2) → basic (not amphoteric)
$Cr_2O_3$ (Cr in +3) → amphoteric (correct)

How to Avoid: Memorise the trend:

Low oxidation state → Basic

Intermediate oxidation state → Amphoteric

High oxidation state → Acidic

For 3d series, remember:

$Cr_2O_3$ (+3) and $V_2O_5$ (+5) are classic amphoteric oxides.

Mistake 2: Thinking $V_2O_5$ is Only Acidic

The Error: Because vanadium is in +5 (high), students label $V_2O_5$ as purely acidic. But $V_2O_5$ reacts with both acids and bases — it is amphoteric.

How to Avoid: Recall the reaction:

With base: $V_2O_5 + 2OH^- \rightarrow 2VO_3^- + H_2O$ (acts as acid)
With acid: $V_2O_5 + 2H^+ \rightarrow 2VO_2^+ + H_2O$ (acts as base)

So $V_2O_5$ is amphoteric, not just acidic.

Mistake 3: Assuming $CrO_3$ is Amphoteric Because Chromium is a Transition Metal

The Error: $CrO_3$ (Cr in +6) is often mistaken as amphoteric. In reality, it is strongly acidic — it forms chromic acid ( $H_2CrO_4$ ) in water.

How to Avoid: Remember the oxidation state rule:

Cr in +6 → acidic oxide ( $CrO_3$ )
Cr in +3 → amphoteric ( $Cr_2O_3$ )
Cr in +2 → basic ( $CrO$ )

Mistake 4: Confusing $V_2O_4$ with $V_2O_5$

The Error: $V_2O_4$ (V in +4) is not amphoteric — it is basic. Students often group all vanadium oxides together. …

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Q.Which of the following is amphoteric oxide?
$Mn_2O_7$ , $CrO_3$ , $Cr_2O_3$ , $CrO$ , $V_2O_5$ , $V_2O_4$

Concept understanding — Lanthanide Contraction

Lanthanide Contraction: The Intuition

The Precise Statement

Why It Matters

The Numbers (for reference)

Why this formula?

Lanthanide Contraction: Why It Happens

The Core Question

The Formula That Governs It

The Derivation: Step by Step

1. What happens when you add a proton and an electron?

2. The 4f orbital is "penetrating" but poorly shielding

3. The net effect on $Z_{\text{eff}}$

4. How this shrinks the radius

Method: Periodic Trend of Oxide Acidity (Basic → Amphoteric → Acidic)

Steps:

Common Mistakes Students Make on Amphoteric Oxides of Transition Metals

Mistake 1: Confusing Amphoteric with Acidic or Basic

Mistake 2: Thinking $V_2O_5$ is Only Acidic

Mistake 3: Assuming $CrO_3$ is Amphoteric Because Chromium is a Transition Metal

Mistake 4: Confusing $V_2O_4$ with $V_2O_5$

Q.Which of the following is amphoteric oxide? Mn2O7Mn_2O_7Mn2​O7​, CrO3CrO_3CrO3​, Cr2O3Cr_2O_3Cr2​O3​, CrOCrOCrO, V2O5V_2O_5V2​O5​, V2O4V_2O_4V2​O4​

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Q.Which of the following is amphoteric oxide?
$Mn_2O_7$ , $CrO_3$ , $Cr_2O_3$ , $CrO$ , $V_2O_5$ , $V_2O_4$

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