Q.Suggest a list of the substances where carbon can exhibit oxidation states from –4 to +4 and nitrogen from –3 to +5.

The question tests your knowledge of qualitative analysis of organic compounds — specifically, how to identify the presence of nitrogen, sulphur, phosphorus, and halogens (like iodine) in a given organic substance. Each element, after being converted into an inorganic ion through sodium fusion (Lassaigne’s test), reacts with a specific reagent to produce a distinct colour or precipitate. The key is to remember which reagent pairs with which element, and why.

Let’s go through each element one by one.

1. Nitrogen (A) — When an organic compound containing nitrogen is fused with sodium, sodium cyanide (NaCN) forms. In Lassaigne’s test, this cyanide ion reacts with freshly prepared ferrous sulphate in acidic medium to form Prussian blue (ferric ferrocyanide). The reagent used is FeSO₄ followed by H⁺ (dilute H₂SO₄ or HCl). So A matches with III.

2. Sulphur (B) — Sodium fusion converts sulphur into sodium sulphide (Na₂S). This sulphide ion gives a violet colour with sodium nitroprusside, Na₂[Fe(CN)₅NO]. That’s a very specific and sensitive test. So B matches with IV.

3. Phosphorus (C) — Phosphorus in the organic compound is oxidised to phosphate during fusion (or by nitric acid in a separate test). The phosphate ion reacts with ammonium molybdate, (NH₄)₂MoO₄, in the presence of nitric acid to give a canary-yellow precipitate of ammonium phosphomolybdate. So C matches with I. …

Concept & Intuition

This question tests two classic inorganic reactions. For Statement I, white phosphorus ($P_4$) disproportionates in hot aqueous alkali: one phosphorus atom is oxidized and another is reduced. The sodium salt formed is sodium hypophosphite ($NaH_2PO_2$), where phosphorus has an oxidation state of +1. For Statement II, concentrated nitric acid is a strong oxidizing agent; sulfur is oxidized to sulfuric acid ($H_2SO_4$), which is a liquid, not a gas. The only gaseous product is nitrogen dioxide ($NO_2$), but the statement claims two gaseous products, which is false.

Step-by-step reasoning

1. Analyze Statement I: Oxidation state of phosphorus in the sodium salt from $P_4$ + alkali White phosphorus reacts with hot aqueous NaOH:

The salt is sodium hypophosphite ($NaH_2PO_2$). To find the oxidation state of P:

• Na is +1, each H is +1, O is –2.
• Let oxidation state of P be $x$.
• Equation: $+1 + 2(+1) + x + 2(-2) = 0 \Rightarrow 1 + 2 + x - 4 = 0 \Rightarrow x - 1 = 0 \Rightarrow x = +1$. So Statement I is correct.

2. Analyze Statement II: Sulfur with conc. $HNO_3$ …

Concept & Intuition

An amphoteric oxide can behave as an acid (reacting with a base) or as a base (reacting with an acid). This usually happens for oxides of elements near the “staircase” line between metals and nonmetals in the periodic table — especially for elements in intermediate oxidation states. For main-group elements, the higher oxides tend to be acidic, the lower ones basic, and the intermediate ones amphoteric. For p-block metals like Sn, Pb, and Ga, their oxides are often amphoteric. Nonmetal oxides (like CO, CO₂, B₂O₃) are acidic; basic oxides come from active metals.

Let’s check each oxide one by one.

1. CO (carbon monoxide) — A neutral oxide; it does not react with acids or bases under normal conditions. Not amphoteric.

2. B₂O₃ (boron trioxide) — A classic acidic oxide; it reacts with water to give boric acid and with bases to form borates. Does not act as a base. Not amphoteric.

3. SnO₂ (tin(IV) oxide) — Reacts with strong acids (e.g., HCl) to form tin(IV) salts and with strong bases (e.g., NaOH) to form stannates. Amphoteric.

4. PbO₂ (lead(IV) oxide) — Reacts with acids (e.g., HCl) to give lead(IV) salts (though it also oxidizes HCl to Cl₂) and with bases to form plumbates. Amphoteric.

5. Ga₂O₃ (gallium(III) oxide) — Gallium is in group 13, just below aluminum. Like Al₂O₃, Ga₂O₃ dissolves in both acids (giving Ga³⁺ salts) and bases (giving gallates). Amphoteric.

6. SnO (tin(II) oxide) — Reacts with acids to give Sn²⁺ salts and with bases to form stannites. Amphoteric.

7. PbO (lead(II) oxide) — Reacts with acids to give Pb²⁺ salts and with bases to form plumbites. Amphoteric. …

Understanding Oxide Classification

Oxides can be classified based on their acid-base behavior:

• Acidic oxides: React with bases to form salts (e.g., CO₂, SO₃)
• Basic oxides: React with acids to form salts (e.g., Na₂O, CaO)
• Amphoteric oxides: React with both acids AND bases (e.g., Al₂O₃, ZnO)
• Neutral oxides: Don't react with acids or bases (e.g., CO, N₂O)

The key trend: As we move down Group 14, the +2 oxidation state becomes more stable (inert pair effect), and lower oxidation state oxides tend to be more basic or amphoteric, while higher oxidation state oxides are more acidic.

Analyzing Each Option

Let me examine each set systematically:

1. Option (A): SnO and PbO classified as Neutral

Both tin(II) oxide and lead(II) oxide are amphoteric, not neutral:

• SnO reacts with acids: $\text{SnO} + 2\text{HCl} \rightarrow \text{SnCl}_2 + \text{H}_2\text{O}$
• SnO reacts with bases: $\text{SnO} + 2\text{NaOH} \rightarrow \text{Na}_2\text{SnO}_2 + \text{H}_2\text{O}$
• Similarly, PbO shows amphoteric behavior with both acids and bases

This classification is INCORRECT.

2. Option (B): SnO₂ and PbO₂ classified as Amphoteric

Both are indeed amphoteric:

• SnO₂ reacts with strong acids and strong bases
• PbO₂ similarly shows amphoteric character
• This is correct ✓

3. Option (C): SiO₂ and GeO₂ classified as Acidic

Both silicon dioxide and germanium dioxide are acidic:

• SiO₂ reacts with bases: $\text{SiO}_2 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SiO}_3 + \text{H}_2\text{O}$
• GeO₂ shows similar acidic behavior
• This is correct ✓

4. Option (D): CO₂ and GeO classified as Acidic …

Classify each oxide by the nature of the element (non-metal/metalloid oxides in higher states are acidic; metal oxides are basic; borderline metals give amphoteric oxides; a few non-metal oxides are neutral):

The key concept here is electronic configuration and stability. The second ionisation enthalpy is the energy needed to remove an electron from a singly charged positive ion (M⁺ → M²⁺). The trend across a period is not simply increasing, because the stability of the electron configuration in the +1 ion matters greatly. A half-filled or fully filled subshell is especially stable, so removing an electron from such a configuration costs more energy.

Let’s work through each element step by step.

1. Write the ground state configurations of the neutral atoms

   • Carbon (Z=6): $1s^2 2s^2 2p^2$
   • Nitrogen (Z=7): $1s^2 2s^2 2p^3$
   • Oxygen (Z=8): $1s^2 2s^2 2p^4$
   • Fluorine (Z=9): $1s^2 2s^2 2p^5$

2. Determine the configuration of the +1 ion (M⁺) after first ionisation

   The first electron removed is always from the outermost orbital.

   • C⁺: $1s^2 2s^2 2p^1$
   • N⁺: $1s^2 2s^2 2p^2$
   • O⁺: $1s^2 2s^2 2p^3$
   • F⁺: $1s^2 2s^2 2p^4$

3. Now consider the second ionisation: removing an electron from M⁺

   The energy required depends on the stability of the M⁺ configuration.

   • Carbon (C⁺ → C²⁺): C⁺ has $2p^1$. Removing that single p-electron is relatively easy — no special stability.
   • Nitrogen (N⁺ → N²⁺): N⁺ has $2p^2$. Removing one p-electron leaves $2p^1$, no special stability.
   • Oxygen (O⁺ → O²⁺): O⁺ has $2p^3$ — this is a half-filled p subshell, which is exceptionally stable. Removing an electron from a half-filled shell requires a lot of energy.
   • Fluorine (F⁺ → F²⁺): F⁺ has $2p^4$. Removing one p-electron leaves $2p^3$ (half-filled), which is stable. So the process is less difficult than for oxygen, because the product is stable, but the starting configuration is not as stable as oxygen’s half-filled one.

4. Compare the energies

   • Oxygen’s second ionisation is the highest because you are breaking a half-filled shell. …

The concept first

Two periodic ideas explain nearly everything odd about group 13:

1. Inert-pair effect. Going down the group the $ns^2$ electrons become poorer at participating in bonding (poor shielding by intervening $d$ and $f$ electrons ⇒ stronger effective nuclear pull). So the lower ($+1$) oxidation state becomes more stable as you descend, and $\ce{Tl^{+}}$ is more stable than $\ce{Tl^{3+}}$.
2. d-block contraction. Gallium comes straight after the first transition series; those $3d$ electrons shield badly, so Ga's electrons feel a bigger effective nuclear charge and Ga ends up slightly smaller than Al.

Statement-by-statement

I) $\text{Tl} > \text{In} > \text{Ga}$ for $+1$ stability — TRUE. Directly the inert-pair effect. ($\ce{TlCl}$ is stable; $\ce{GaCl}$ disproportionates.)

II) Boron has the lowest m.p./b.p. — FALSE. Boron is not a simple molecular non-metal; it forms a giant covalent network of $\ce{B12}$ icosahedra, so it melts around 2453 K — the highest in the group. (Gallium, ironically, melts in your hand at 303 K.) …

The key to this problem is recognizing the structure of peroxydisulphuric acid, often called Marshall's acid. Many students mistakenly assign oxidation states without accounting for the peroxide (–O–O–) linkage, which changes the usual rules. Let’s build the reasoning step by step.

1. Write the molecular formula and recall the structure Peroxydisulphuric acid is H₂S₂O₈. Its structure is not simply two SO₄ groups joined; instead, it has a peroxide bridge:

Each sulphur is bonded to two terminal oxygens (double bonds), one oxygen from the peroxide bridge, and one oxygen from an –OH group.

2. Determine the oxidation state of sulphur
   • Oxygen usually has oxidation state –2, except in peroxides where it is –1.
   • Here, the two oxygens in the –O–O– bridge are each –1.
   • The other six oxygens (two terminal double-bonded O per S, plus one –OH oxygen per S) are each –2.
   • Hydrogen in –OH is +1. Let the oxidation state of each sulphur be $x$. The molecule is neutral, so:

Both sulphur atoms have the same oxidation state: +6.

3. Count the number of S–OH bonds …

The key to solving this is understanding that oxidation states are assigned based on electronegativity and bonding context, not just counting bonds blindly. In carbon suboxide, the molecule is symmetric: two terminal carbons are each double-bonded to an oxygen (more electronegative) and to a central carbon. The central carbon is only bonded to other carbons (same electronegativity), so its oxidation state is 0. The terminal carbons each lose two electrons to oxygen, giving them +2.

Let’s work through it step by step.

1. Draw the structure. Carbon suboxide has the linear structure $\mathrm{O=C=C=C=O}$. Each carbon is sp-hybridized, and the molecule is cumulated (consecutive double bonds). The central carbon is bonded only to two other carbons; the terminal carbons are each bonded to one oxygen and one carbon.

2. Recall the rules for assigning oxidation states.

   • Oxygen is almost always –2 (except in peroxides, etc., which don’t apply here).
   • A bond between identical atoms (C–C) contributes 0 to the oxidation state of each.
   • For a bond between different atoms, all bonding electrons are assigned to the more electronegative atom.
   • The oxidation state of an atom = (number of valence electrons) – (number of electrons assigned to it after electronegativity adjustment).

3. Assign electrons for a terminal carbon.

   Consider the leftmost carbon: it is double-bonded to oxygen and double-bonded to the central carbon.

   • In the C=O bond, oxygen is more electronegative, so both pairs of electrons (4 electrons) go to oxygen. That means this carbon “loses” 4 electrons relative to its neutral count.
   • In the C=C bond with the central carbon, both atoms are carbon, so the electrons are shared equally: each carbon gets 2 electrons from that bond.
   • A neutral carbon has 4 valence electrons. After assigning: it keeps 2 from the C=C bond and 0 from the C=O bond, so it has 2 electrons assigned.
   • Oxidation state = 4 – 2 = +2.

4. Assign electrons for the central carbon. …

The key concept here is the disproportionation of nitrous acid (HNO₂). When sodium nitrite (NaNO₂) reacts with a strong acid like HCl, it first forms nitrous acid, which is unstable and decomposes into two different nitrogen oxides. Understanding which oxides form requires balancing the oxidation states of nitrogen.

1. Write the initial reaction. Sodium nitrite reacts with hydrochloric acid:

The nitrous acid (HNO₂) is the species that further reacts.

2. Recognize the instability of nitrous acid.

   Nitrous acid is not stable under acidic conditions and decomposes spontaneously. The decomposition involves a disproportionation reaction: nitrogen in HNO₂ has an oxidation state of +3, and it both increases and decreases to form two different oxides.

3. Determine the possible nitrogen oxides.

   Common nitrogen oxides are NO (oxidation state +2), NO₂ (+4), N₂O (+1), N₂O₃ (+3), and N₂O₅ (+5). Since HNO₂ is at +3, plausible products are NO (+2) and NO₂ (+4) — one reduced, one oxidized.

4. Balance the decomposition reaction.

   The classic decomposition of nitrous acid is:

   $2\,\mathrm{HNO_2} \rightarrow \mathrm{NO} + \mathrm{NO_2} + \mathrm{H_2O}$ …

The question asks about the oxidation states of lead in Pb$_3$O$_4$. A common trap is to calculate an average oxidation number and treat it as the actual state — but that misses the real chemistry. Pb$_3$O$_4$ is not a simple compound where all lead atoms are identical; it is a mixed oxide, also known as red lead or minium, with a definite structure.

The key insight: Pb$_3$O$_4$ can be thought of as 2PbO · PbO$_2$. That is, two lead atoms are in the +2 state (as in PbO) and one lead atom is in the +4 state (as in PbO$_2$). The formula Pb$_3$O$_4$ is just a convenient way to write the overall composition.

Let’s verify this systematically.

1. Assign variables for oxidation states. Let the oxidation state of Pb be $x$ and that of O be $-2$ (standard). For Pb$_3$O$_4$, the sum of oxidation states must be zero: $3x + 4(-2) = 0$.

2. Solve for $x$. $3x - 8 = 0$ gives $x = \frac{8}{3} \approx 2.67$. This is the average oxidation state of lead in the compound.

3. Interpret the average. An average of $+2.67$ does not mean any lead atom actually has that charge. It simply tells us that the mixture of +2 and +4 states yields this mean. If we have $a$ atoms of Pb(II) and $b$ atoms of Pb(IV), with $a+b=3$, then the total charge contributed by lead is $2a + 4b$. Setting this equal to the total negative charge from oxygen ($+8$) gives $2a + 4b = 8$.

4. Solve for $a$ and $b$. From $a+b=3$ and $2a+4b=8$: subtract twice the first equation from the second: $(2a+4b) - 2(a+b) = 8 - 6$, so $2b = 2$, hence $b=1$ and $a=2$. So there are exactly two Pb(II) atoms and one Pb(IV) atom per formula unit.