3.2.5 · D5p-Block
Question bank — Group 15 (Nitrogen family) — N₂ inertness; NH₃ synthesis (Haber); HNO₃ (Ostwald); oxides of N (N₂O, NO, NO₂, N₂O₄, N₂O₅)
Build the pictures first (so the traps make sense on this page)
Before the traps, four visual anchors. Every "why" answer below points back to one of these figures — you should not have to leave the page to get the intuition.
Figure 1 — bond-strength bars. The "159 kJ/mol" number is the single-bond dissociation energy (N–N): the energy to pull apart one N–N single bond, a standard tabulated bond enthalpy. The triple bond (N≡N) is 941 kJ/mol. The bars show why "three singles" can't beat one triple.
Figure 2 — Le Chatelier plot. NH₃ yield rises as pressure rises (fewer product moles) but falls as temperature rises (exothermic). The industrial point sits where the two pressures meet a workable rate. A catalyst does not move any curve — it only gets you onto the curve faster.
Figure 3 — odd electrons and the NO₂ ⇌ N₂O₄ dimer. NO and NO₂ each carry one unpaired electron (paramagnetic; the MO logic mirrors N2 molecule MO diagram). Two NO₂ pair those electrons into an N–N bond → diamagnetic N₂O₄. Crucially, dimerising loses entropy (, two molecules → one) and releases heat (), so via the term wins at high T, tearing the dimer back to brown NO₂.
Figure 4 — the kinetic wall. N₂'s reactions with H₂ are downhill in energy (thermodynamically fine) but blocked by a huge activation barrier (breaking N≡N). This is the picture behind every "inertness is kinetic, not thermodynamic" answer, and it explains oxidation-state bookkeeping used in Oxidation States and Redox.
True or false — justify
Reveal each with reasoning, not a bare verdict.
N₂ is unreactive because each nitrogen has a stable full octet.
False. O₂ also has full octets yet is reactive — the real cause is the kinetic barrier of the 941 kJ/mol triple bond (Figure 4), not the octet. See Oxidation States and Redox for why octet ≠ inertness.
N₂'s inertness is thermodynamic (its reactions are non-spontaneous).
False. Many N₂ reactions are thermodynamically favourable (e.g. forming NH₃, ΔH < 0, Figure 4); they're just kinetically blocked by the huge activation energy of breaking N≡N.
A catalyst in the Haber process increases the equilibrium yield of NH₃.
False. A catalyst speeds forward and reverse rates equally, so it reaches the same equilibrium faster but leaves unchanged; since percent conversion is fixed by (Figure 2), yield is unchanged.
Lowering the temperature always gives more ammonia, so we should run Haber as cold as possible.
False for practice. Low T does favour more NH₃ at equilibrium (exothermic — the yield curve of Figure 2), but the rate becomes uselessly slow; ~700 K is the compromise. See Le Chatelier Principle.
Increasing pressure shifts the Haber equilibrium toward NH₃.
True. There are 4 gas moles on the left and 2 on the right; higher pressure shifts toward the side with fewer gas moles (products), as the rising yield curve of Figure 2 shows.
Step 3 of Ostwald, , produces only nitric acid.
False. It is a disproportionation: one N goes up to +5 (HNO₃), another down to +2 (NO). The NO must be recycled or a third of the nitrogen is lost. See Disproportionation Reactions.
N₂O₄ is paramagnetic like NO₂.
False. NO₂ has an unpaired electron (paramagnetic); when two NO₂ dimerise, those odd electrons pair up into an N–N bond (Figure 3), making N₂O₄ diamagnetic.
NO is diamagnetic because it has an even total electron count.
False. NO has 11 valence electrons — an odd number — leaving one unpaired electron, so it is paramagnetic. Its MO reasoning parallels N2 molecule MO diagram.
Dilute and concentrated HNO₃ give the same nitrogen product with metals.
False. Dilute HNO₃ mostly yields NO, concentrated yields NO₂ — the more concentrated (stronger oxidiser) the acid, the higher the N oxidation state of the gas.
Aqua regia dissolves gold because HNO₃ alone is a strong enough oxidiser.
False. HNO₃ alone can't; the Cl⁻ from HCl complexes the gold (), removing product and driving the oxidation. See Aqua regia and Noble Metals.
Spot the error
The trap phrase is bolded inside each statement — say why it's wrong before revealing.
"P forms P–P single bonds in P₄ because phosphorus is more electronegative than N."
Electronegativity is not the reason (and N is actually more electronegative). The real cause is poor 3p–3p π overlap in the larger P atom, so single bonds beat a would-be triple bond — opposite of nitrogen (Figure 1).
"Since 3 × (N–N single bond) = 3 × 159 ≈ 477 kJ/mol is less than 941, three single bonds are more stable than one triple bond."
Backwards. Because 477 ≪ 941, the triple bond releases far more energy and is hugely more stable, not less — that is exactly what the taller 941 bar in Figure 1 shows.
"We use high pressure in Haber mainly to increase the reaction rate."
The primary reason is equilibrium shift (fewer product moles, Figure 2), not rate. Rate is handled by temperature and the Fe catalyst.
"In Ostwald Step 1, NH₃ is reduced to NO by oxygen."
It is oxidised, not reduced: nitrogen goes from −3 in NH₃ to +2 in NO.
"Without the Pt/Rh catalyst, NH₃ + O₂ would still cleanly give NO."
Without the catalyst NH₃ combusts to N₂ + H₂O; the catalyst is what steers the product to NO.
"N₂O₅ is the anhydride of HNO₂."
N₂O₅ (N at +5) is the anhydride of HNO₃; HNO₂ (N at +3) corresponds to N₂O₃.
"NO₂ is colourless and N₂O₄ is brown."
Reversed — NO₂ is brown (odd electron), N₂O₄ is colourless/pale (paired electrons, Figure 3).
"Raising temperature on favours N₂O₄."
Dimerisation is exothermic and entropy-lowering (), so the term dominates as T rises and heating favours NO₂ (the monomer), not N₂O₄ (Figure 3).
Why questions
Explain the mechanism, not just the fact — each answer points to a figure above.
Why is the N≡N bond so much stronger relative to a single N–N bond than, say, the C≡C is relative to C–C?
Nitrogen's small size gives excellent 2p–2p π overlap, so the two π bonds add a lot (the 941 vs 3×159 gap in Figure 1); the N–N single bond is also weakened by lone-pair/lone-pair repulsion, exaggerating the ratio.
Why does high pressure specifically work through the expression?
: scaling all partial pressures up puts on top but on the bottom, so the quotient drops below and the system moves right to restore it — the rising branch of Figure 2.
Why does NO₂ dimerise but NO essentially does not (at normal conditions)?
NO₂'s unpaired electron sits mostly on N and readily pairs via an N–N bond (Figure 3); NO's unpaired electron is delocalised in a π MO* across both atoms, making an N–N dimer bond weak, so NO stays monomeric.
Why is nitrous oxide (N₂O) chemically neutral rather than acidic?
It has no easily formed corresponding oxoacid and does not react with water to give H⁺ or OH⁻; N₂O and NO are neutral oxides, unlike the higher acidic oxides.
Why must the temperature in Ostwald Step 2 () be lowered after the hot Step 1?
NO oxidation to NO₂ is favoured at lower temperature, and, like all such / associations (Figure 3), cooling makes the penalty small so conversion proceeds.
Why can't we simply heat N₂ hard enough to react it easily instead of using a catalyst?
Enough heat would break N≡N but also shifts the exothermic NH₃ equilibrium backward (Figure 2) and wastes energy; a catalyst instead lowers the activation barrier of Figure 4 so moderate T suffices.
Edge cases
Boundary and degenerate scenarios you must not be surprised by.
What happens to the Haber yield if pressure is raised but temperature is also raised a lot?
The two effects oppose (Figure 2): high P helps yield, but the higher T (exothermic reaction) hurts it, so yield may barely improve while energy cost rises — a poor trade.
In the disproportionation , what is the fate of the NO if it is not recycled?
One-third of the nitrogen is lost as NO gas, cutting the theoretical HNO₃ yield; recycling NO back to Step 2 is what makes Ostwald efficient.
If NO₂ contains N at +4 and N₂O₄ also at +4, is any redox happening when they interconvert?
No. Dimerisation only pairs electrons and forms an N–N bond (Figure 3); oxidation states are unchanged, so it is not a redox process. See Oxidation States and Redox.
What is the limiting behaviour of the equilibrium as very low?
The term vanishes, so the exothermic dimerisation dominates and the mixture becomes almost entirely colourless N₂O₄ — the brown colour fades (Figure 3).
Zero-catalyst limit: what does Haber's rate approach even at high P and moderate T?
The rate approaches effectively zero on any useful timescale because the N≡N activation barrier is enormous (Figure 4) — pressure alone can't fix a kinetic wall.
For NH₃ vs PH₃, why does the "inert-because-strong-bond" logic not transfer down the group?
Nitrogen's strength story rests on strong 2p π bonding and H-bonding basicity; P's larger orbitals give weaker bonds and PH₃ is a weaker base, so trends invert. See Group 15 Hydrides PH3 vs NH3.
In the brown ring test, is the iron oxidised, reduced, or unchanged when forms?
NO binds as a ligand and the complex is written as Fe(II) with NO⁺; the key point is a coordination/redox interplay, not a simple spectator — the brown colour is a charge-transfer feature of the complex.
Connections
- N2 molecule MO diagram — the odd-electron / paramagnetism reasoning (Figure 3)
- Le Chatelier Principle — pressure and temperature trade-offs (Figure 2)
- Disproportionation Reactions — Ostwald Step 3 and NO₂ in water
- Oxidation States and Redox — every "which way is N going?" trap (Figure 4)
- Group 15 Hydrides PH3 vs NH3 — why nitrogen's rules don't descend
- Aqua regia and Noble Metals — the gold-dissolution trap