Q.Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?

Concept and intuition:

Each mineral listed is a naturally occurring compound that contains a specific metal. To identify which pairs share the same metal, we need to recall the chemical composition of each ore. The trick is not to confuse similar-sounding names or to assume that minerals with “-ite” endings are related. Instead, focus on the metal element that is economically extracted from each ore.

Step-by-step reasoning:

1. Pair I: Kernite and Kaolinite

   • Kernite is a borate mineral with formula $\text{Na}_2\text{B}_4\text{O}_7 \cdot 4\text{H}_2\text{O}$ — it is an ore of boron, not a metal in the usual sense.
   • Kaolinite is a clay mineral, $\text{Al}_2\text{Si}_2\text{O}_5(\text{OH})_4$, which is an ore of aluminium.
   • These two contain different metals (boron vs. aluminium), so they are not ores of the same metal.

2. Pair II: Magnetite and Siderite

   • Magnetite is $\text{Fe}_3\text{O}_4$, an important ore of iron.
   • Siderite is $\text{FeCO}_3$, also an ore of iron.
   • Both yield iron, so this pair is correct.

3. Pair III: Zincite and Calamine

   • Zincite is $\text{ZnO}$, an ore of zinc.
   • Calamine is a historical name for zinc carbonate, $\text{ZnCO}_3$ (or sometimes zinc silicate), also an ore of zinc.
   • Both yield zinc, so this pair is correct.

4. Pair IV: Cuprite and Malachite

   • Cuprite is $\text{Cu}_2\text{O}$, an ore of copper.
   • Malachite is $\text{Cu}_2\text{CO}_3(\text{OH})_2$, also an ore of copper.
   • Both yield copper, so this pair is correct.

Thus, the correct pairs are II, III, and IV only.

The key concept here is lanthanoid contraction and its consequences, plus a comparison of oxidation state variability between the lanthanoid and actinoid series. Let’s break it down.

1. Understanding lanthanoid contraction

   As we move across the lanthanoid series (Ce to Lu), the 4f orbitals are filled. The 4f electrons are poor at shielding the nuclear charge, so the effective nuclear charge ($Z_{\text{eff}}$) increases steadily. This pulls the outer electron shells inward, causing a gradual decrease in atomic and ionic radii — this is lanthanoid contraction.

2. Effect on 4d and 5d series

   Because of lanthanoid contraction, the radii of the 5d series elements (e.g., Hf, Ta, W) are very close to those of their 4d counterparts (e.g., Zr, Nb, Mo). For instance, the atomic radius of Zr is about 160 pm and Hf is about 159 pm — nearly identical. This is a direct consequence of the contraction. So Statement I is correct.

3. Comparing oxidation states: lanthanoids vs. actinoids

   Lanthanoids typically show a +3 oxidation state as the most stable, with a few showing +2 or +4 (e.g., Eu²⁺, Ce⁴⁺). In contrast, actinoids exhibit a much wider range: from +3 to +6 (and even +7 for Np and Pu) because their 5f, 6d, and 7s electrons are closer in energy and more available for bonding. Therefore, actinoids exhibit more oxidation states than lanthanoids, making Statement II false.

4. Conclusion

   Only Statement I is correct; Statement II is incorrect.

The key to this problem is understanding the Lewis acid–base behaviour of xenon fluorides. Xenon hexafluoride ($\mathrm{XeF_6}$) is a strong fluoride-ion donor — it can give away an $\mathrm{F}^-$ to a suitable acceptor. Xenon difluoride ($\mathrm{XeF_2}$) and xenon tetrafluoride ($\mathrm{XeF_4}$) are weaker donors, but they can act as fluoride-ion donors to very strong Lewis acids like $\mathrm{PF_5}$ and $\mathrm{SbF_5}$. The products are ionic compounds where the xenon-containing species becomes a cation (if it donates $\mathrm{F}^-$) or an anion (if it accepts $\mathrm{F}^-$).

Let’s examine each reaction one by one.

1. Reaction I: $\mathrm{XeF_6} + \mathrm{NaF} \rightarrow \mathrm{Na}[\mathrm{XeF_7}]$

   Here, $\mathrm{NaF}$ provides $\mathrm{F}^-$ ions. $\mathrm{XeF_6}$ can accept an $\mathrm{F}^-$ because xenon in $\mathrm{XeF_6}$ has 12 valence electrons around it (expanded octet) and can accommodate one more pair to form $\mathrm{XeF_7}^-$. This is a well-known reaction — $\mathrm{XeF_6}$ behaves as a Lewis acid. The product $\mathrm{Na}[\mathrm{XeF_7}]$ is a stable salt. So reaction I is correct.

2. Reaction II: $\mathrm{XeF_2} + \mathrm{PF_5} \rightarrow [\mathrm{XeF}]^+ [\mathrm{PF_6}]^-$

   $\mathrm{PF_5}$ is a strong Lewis acid (it can accept $\mathrm{F}^-$ to form $\mathrm{PF_6}^-$). $\mathrm{XeF_2}$ can donate an $\mathrm{F}^-$ ion, leaving behind $\mathrm{XeF}^+$. This is a classic reaction — $\mathrm{XeF_2}$ acts as a fluoride-ion donor toward $\mathrm{PF_5}$, giving the ionic salt $[\mathrm{XeF}]^+[\mathrm{PF_6}]^-$. So reaction II is correct.

3. Reaction III: $\mathrm{XeF_4} + \mathrm{SbF_5} \rightarrow [\mathrm{XeF_3}]^+ [\mathrm{SbF_6}]^-$

   $\mathrm{SbF_5}$ is an even stronger Lewis acid than $\mathrm{PF_5}$. $\mathrm{XeF_4}$ can donate an $\mathrm{F}^-$ to $\mathrm{SbF_5}$, forming $\mathrm{SbF_6}^-$ and leaving $\mathrm{XeF_3}^+$. This reaction is indeed known — the product is $[\mathrm{XeF_3}]^+[\mathrm{SbF_6}]^-$. So reaction III is also correct.

Since all three reactions are correct, the answer should be option (D) — I, II, III.

The lanthanoids are the 14 elements from cerium (Z=58) to lutetium (Z=71) where the 4f subshell is progressively filled. Their chemistry is dominated by the +3 oxidation state, but some elements show +2 or +4 states when doing so leads to a particularly stable electronic configuration — either a half-filled 4f⁷ or a fully-filled 4f¹⁴ subshell. This stability principle is the single most powerful tool for predicting redox behaviour in this series.

Let’s examine each statement one by one.

1. Statement I: Ce⁴⁺ and Tb⁴⁺ act as oxidising agents

   Ce has the ground-state configuration [Xe]4f¹5d¹6s². Ce³⁺ is [Xe]4f¹, while Ce⁴⁺ is [Xe]4f⁰ — a noble gas configuration. That empty 4f subshell is exceptionally stable, so Ce⁴⁺ readily gains an electron to become Ce³⁺. In other words, Ce⁴⁺ is a strong oxidising agent.

   Tb has the configuration [Xe]4f⁹6s². Tb³⁺ is [Xe]4f⁸, but Tb⁴⁺ is [Xe]4f⁷ — a half-filled subshell. Half-filled stability makes Tb⁴⁺ eager to accept an electron and drop to Tb³⁺, so it too acts as an oxidising agent.

   Statement I is correct.

2. Statement II: Eu²⁺ and Yb²⁺ act as oxidising agents

   Eu has the configuration [Xe]4f⁷6s². Eu²⁺ is [Xe]4f⁷ — half-filled, very stable. To act as an oxidising agent, Eu²⁺ would need to gain an electron to become Eu⁺, which would disrupt the half-filled stability. That is energetically unfavourable. Instead, Eu²⁺ tends to lose an electron (be oxidised) to Eu³⁺, making it a reducing agent, not an oxidising one.

   Yb has the configuration [Xe]4f¹⁴6s². Yb²⁺ is [Xe]4f¹⁴ — fully filled, very stable. For the same reason, Yb²⁺ prefers to lose an electron to become Yb³⁺ rather than gain one. So Yb²⁺ is also a reducing agent.

   Statement II is false.

3. Statement III: Mischmetal is an alloy of 95% iron and 5% lanthanoid metal

   This is backwards. Mischmetal is an alloy of about 95% lanthanoid metals (mainly cerium, lanthanum, neodymium, and praseodymium) and about 5% iron, along with traces of other elements. It is used in lighter flints because the lanthanoids are pyrophoric — they spark when struck.

   Statement III is false.

4. Statement IV: La³⁺ and Ce⁴⁺ are diamagnetic in nature

   La has the configuration [Xe]5d¹6s². La³⁺ is [Xe] — no unpaired electrons, so it is diamagnetic.

   Ce⁴⁺ is [Xe]4f⁰, also with no unpaired electrons, so it too is diamagnetic.

   Statement IV is correct.

Only statements I and IV are correct.

Concept & Intuition

Diborane (B₂H₆) reacts violently with water to give boric acid (H₃BO₃) and hydrogen gas. The key is to understand the acid–base behaviour and solid‑state structure of boric acid. Many students mistakenly think H₃BO₃ is tribasic because it has three OH groups, but in water it acts as a Lewis acid, accepting OH⁻ to form [B(OH)₄]⁻, releasing only one H⁺ per molecule — hence it is monobasic. In the solid state, boric acid molecules are held together by hydrogen bonds into a layered structure, similar to graphite. It is only sparingly soluble in cold water (about 5 g/100 mL), not “highly soluble”.

Step‑by‑step reasoning

1. Identify the product X Diborane hydrolysis:

So X is boric acid (orthoboric acid), H₃BO₃.

2. Evaluate statement I: “It is a tribasic acid” A tribasic acid donates three protons (H⁺). Boric acid does not donate H⁺ directly; instead it accepts OH⁻ from water:

Only one H⁺ is produced per molecule. Thus it is monobasic, not tribasic.

Statement I is false.

3. Evaluate statement II: “It is a weak monobasic acid”

   The equilibrium above has $K_a \approx 5.8 \times 10^{-10}$, so it is indeed a weak acid and monobasic.

   Statement II is true.

4. Evaluate statement III: “It has a layer structure”

   In the solid state, H₃BO₃ molecules form hydrogen‑bonded sheets (each B is trigonal planar, and OH groups link to neighbouring molecules). This gives a layered structure (like graphite).

   Statement III is true.

5. Evaluate statement IV: “It is highly soluble in water”

   Solubility of boric acid in water at 20 °C is about 5 g/100 mL — moderate, not “highly soluble” (compare NaCl: 36 g/100 mL).

   Statement IV is false.

6. Match with options

   True statements: II and III.

   Option (A) lists II & III.

   Option (B) I & III, (C) I & IV, (D) II & IV — only (A) matches.

The question asks about the role of water in two different reactions — whether it gets oxidized (loses electrons, oxygen’s oxidation number increases) or reduced (gains electrons, oxygen’s oxidation number decreases, or hydrogen’s does). The unbalanced equations are a clue: we need to figure out what products form, then check the oxidation states.

Let’s take them one at a time.

1. Reaction (I): $\mathrm{H_2O(l) + Na(s) \rightarrow}$

   Sodium is a highly reactive metal. When it reacts with water, it displaces hydrogen. The products are sodium hydroxide and hydrogen gas:

   $\mathrm{2Na(s) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g)}$

   Now look at water. In $\mathrm{H_2O}$, hydrogen has an oxidation number of $+1$ and oxygen $-2$. In the product $\mathrm{H_2}$, hydrogen has an oxidation number of $0$. So hydrogen’s oxidation number decreases from $+1$ to $0$ — that’s a gain of electrons, i.e., reduction. Water is being reduced (specifically, the hydrogen in water is reduced to H₂). Oxygen stays at $-2$ in NaOH, so no change there.

2. Reaction (II): $\mathrm{H_2O(l) + F_2(g) \rightarrow}$

   Fluorine is the most electronegative element — it will oxidize water. The reaction produces oxygen and hydrogen fluoride:

   $\mathrm{2F_2(g) + 2H_2O(l) \rightarrow 4HF(aq) + O_2(g)}$

   In water, oxygen is $-2$. In $\mathrm{O_2}$, oxygen’s oxidation number is $0$. So oxygen’s oxidation number increases from $-2$ to $0$ — that’s a loss of electrons, i.e., oxidation. Water is being oxidized (the oxygen in water is oxidized to O₂). Hydrogen stays at $+1$ in HF, so no change there.

So, summarizing:

• In (I), water is reduced to H₂.
• In (II), water is oxidized to O₂.

This matches exactly with option (D).

Concept & Intuition

The key is to first identify the product of complete hydrolysis of XeF₆. Xenon hexafluoride reacts with water to replace all fluorine atoms with oxygen, ultimately forming xenon trioxide (XeO₃). This molecule is not flat; it has a trigonal pyramidal geometry due to a lone pair on xenon. Each Xe–O bond is a double bond (one σ and one π). Counting all σ and π bonds in the molecule gives the answer.

Step-by-step reasoning

1. Determine the product of complete hydrolysis

   XeF₆ + 3 H₂O → XeO₃ + 6 HF

   Complete hydrolysis means all six F atoms are replaced by oxygen atoms. The stable oxide formed is XeO₃ (xenon trioxide).

2. Draw the Lewis structure of XeO₃

   Xenon is in group 18, so it has 8 valence electrons. Each oxygen contributes 6, giving a total of 8 + 3×6 = 26 valence electrons.

   • Place Xe in the center, three O atoms around it.
   • Connect each O to Xe with a single bond (uses 6 electrons).
   • Distribute remaining 20 electrons to satisfy octets: each O gets 6 more (3 lone pairs), using 18 electrons.
   • That leaves 2 electrons, which become a lone pair on Xe.
   • Now Xe has only 6 electrons in bonds (three single bonds) plus a lone pair — that’s only 8 electrons, but Xe can expand its octet. To reduce formal charges, form double bonds: each Xe=O bond uses one lone pair from oxygen to make a π bond. This gives each O a formal charge of 0 and Xe a formal charge of 0. Final structure: Xe with one lone pair, three double bonds to O.

3. Identify bond types

   Each Xe=O double bond consists of:

   • 1 σ bond (the head-on overlap)
   • 1 π bond (the side-on overlap) There are three such double bonds.

4. Count total σ and π bonds

   • σ bonds: 3 (one per Xe–O bond)
   • π bonds: 3 (one per Xe–O bond) Total = 3 + 3 = 6.

5. Match with options

   The total number is 6, which corresponds to option (C).

The key concept here is periodic trends in oxide acidity. Across a period, as you move from left to right, the electronegativity of the element increases and the oxide becomes more acidic. Also, for a given element, higher oxidation states produce more acidic oxides. Here, all elements are in the third period (Al, Si, P, S), so we compare their oxides directly.

1. Identify the nature of each oxide

   • $\mathrm{Al_2O_3}$: Aluminum is a metal, so its oxide is amphoteric (can act as both acid and base), but it is predominantly basic in character.
   • $\mathrm{SiO_2}$: Silicon is a metalloid; its oxide is weakly acidic.
   • $\mathrm{P_2O_3}$: Phosphorus is a nonmetal; its oxide is acidic, but phosphorus here is in the +3 oxidation state (less acidic than its +5 oxide).
   • $\mathrm{SO_2}$: Sulfur is a nonmetal; its oxide is strongly acidic (sulfur in +4 state, but still more electronegative than P).

2. Rank by increasing acidic strength

   • The least acidic is $\mathrm{Al_2O_3}$ (amphoteric, leans basic).
   • Next is $\mathrm{SiO_2}$ (weak acid).
   • Then $\mathrm{P_2O_3}$ (moderate acid).
   • Most acidic is $\mathrm{SO_2}$ (strong acid). So the order is: $\mathrm{Al_2O_3} < \mathrm{SiO_2} < \mathrm{P_2O_3} < \mathrm{SO_2}$.

3. Match with the options

   • Option (C) gives exactly this order: $\mathrm{Al_2O_3} < \mathrm{SiO_2} < \mathrm{P_2O_3} < \mathrm{SO_2}$.

The question is about why atomic radii increase so little as we move from left to right across the third (5d) transition series — elements like Hf, Ta, W, Re, Os, Ir, Pt, Au. Normally, adding protons pulls the electron cloud inward, so radii shrink across a period. But here, the radii barely change at all. The reason lies in what happens before the 5d series begins.

Before the 5d orbitals start filling (at element 72, Hf), the 4f subshell is being filled across the lanthanide series (elements 58 to 71). The 4f electrons are very poor at shielding the nuclear charge — they are deeply buried and have a diffuse, "penetrating" shape. So each added proton in the nucleus pulls the outer electrons (including the 5d electrons) more strongly, causing a steady contraction across the lanthanides. This is the lanthanoid contraction.

By the time we reach Hf, the effective nuclear charge felt by the 5d electrons is already very high. Adding more protons across the 5d series increases it only slightly, because the 5d electrons themselves shield each other better. The result: the atomic radius hardly changes.

Now, the question asks: which orbitals are X and Y? X is the orbital being filled before the 5d series, causing the contraction. That is the 4f orbital. Y is the orbital being filled during the 5d series — that is the 5d orbital itself.

Let’s go through the options:

1. Option (A): X = 4f, Y = 5d — This matches exactly: 4f fills before 5d, causing lanthanoid contraction, and 5d fills during the series. This is correct.

2. Option (B): X = 5f, Y = 5d — The 5f orbitals fill in the actinide series, which comes after the 5d series, not before. So this is wrong.

3. Option (C): X = 5d, Y = 4f — This reverses the order. The 4f fills before 5d, not after. So this is incorrect.

4. Option (D): X = 4f, Y = 4d — The 4d series is the second transition series, not the third. The 4f filling affects the 5d series, not the 4d series. So this is wrong.

The Haber process is the industrial method for producing ammonia ($\mathrm{NH_3}$) from nitrogen ($\mathrm{N_2}$) and hydrogen ($\mathrm{H_2}$). The reaction is:

This is an exothermic, reversible reaction. The challenge is that nitrogen is extremely stable — its triple bond ($\mathrm{N \equiv N}$) is one of the strongest in chemistry. Breaking it requires a high activation energy, so without a catalyst, the reaction would be impractically slow even at high temperatures and pressures.

The catalyst lowers the activation energy by providing a surface where nitrogen and hydrogen molecules can adsorb and dissociate into atoms. The key is that the catalyst must be a transition metal (or its compound) that can bind nitrogen atoms moderately — strongly enough to weaken the triple bond, but not so strongly that the product ammonia cannot leave.

1. Why iron? Finely divided iron (often with small amounts of promoters like $\mathrm{K_2O}$ and $\mathrm{Al_2O_3}$) is the standard catalyst. Iron is relatively cheap, abundant, and effective at dissociating $\mathrm{N_2}$ at the temperatures used (around 400–500°C). The "finely divided" form maximizes surface area, which is crucial for heterogeneous catalysis.

2. Why not the others? Let's check each option:

   • (A) $\mathrm{Cr_2O_3}$: Chromium(III) oxide is used as a catalyst in some other processes (e.g., the dehydrogenation of alkanes), but not in the Haber process.
   • (B) $\mathrm{V_2O_5}$: Vanadium(V) oxide is the catalyst for the Contact process (manufacture of sulfuric acid), not for ammonia synthesis.
   • (D) $\mathrm{Mn_2O_3}$: Manganese(III) oxide has no significant role as a catalyst in the Haber process.

3. A common misconception: Some students confuse the catalyst for the Haber process with that for the Ostwald process (which uses platinum or rhodium to oxidize ammonia to nitric acid). The Haber process specifically uses iron.

The key here is to separate two different properties of $SF_6$: its chemical stability and its physical state. Many students confuse the two, thinking that being a gas somehow implies instability, or that a stable compound must be a solid. Let’s break it down.

Why is $SF_6$ highly stable?

Sulfur hexafluoride is exceptionally inert because of its molecular structure. Sulfur is in the +6 oxidation state, and the six fluorine atoms surround it octahedrally. The S–F bonds are very strong (bond energy ~330 kJ/mol), and the molecule is sterically protected — the fluorine atoms shield the sulfur atom from attack. Moreover, $SF_6$ is kinetically inert: it does not react with water, acids, or bases at room temperature, and it is non-flammable. This stability is a result of the bonding and geometry, not its physical state.

What about the reason?

The reason says "$SF_6$ is a gas". This is actually true — $SF_6$ is a colourless, odourless gas at room temperature (it sublimes at −64°C). But being a gas has nothing to do with its chemical stability. Many gases are highly reactive (e.g., $F_2$, $Cl_2$), and many stable compounds are solids (e.g., $NaCl$). So the reason is a true statement, but it does not explain the assertion.

Now let’s match this to the options:

1. Option (A) says both are true and (R) explains (A). That fails because (R) does not explain (A).
2. Option (B) says both are true but (R) is not the correct explanation. This would be correct if (R) were true — but wait, is (R) actually true? Yes, $SF_6$ is a gas. So (B) seems plausible. However, we must check the exact wording of the reason: "$SF_6$ is a gas". That is a factual statement, and it is true. So (A) is true, (R) is true, but (R) is not the correct explanation. That points to (B).
3. Option (C) says (A) is true but (R) is false. But (R) is not false — $SF_6$ is indeed a gas. So (C) is incorrect.
4. Option (D) says (A) is false but (R) is true. (A) is not false, so this is wrong.

So the correct option is (B) — both statements are true, but the reason does not explain the assertion.