Group 15 (Nitrogen family) — N₂ inertness; NH₃ synthesis (Haber); HNO₃ (Ostwald); oxides of N (N₂O, NO, NO₂, N₂O₄, N₂O₅): Worked examples

The scenario matrix

Before solving anything, let us list every class of problem this topic can throw. Think of it like a grid: each row is a type of situation, and we must have at least one worked example hitting each.

#	Case class	What makes it tricky	Example
A	Oxidation state — normal molecule	neutral, O is $- 2$	Ex 1
B	Oxidation state — fractional / average	atoms in different environments	Ex 2
C	Oxidation state — element with a positive partner	O bonded to something less electronegative? sign flips	Ex 3 (peroxide/ion twist)
D	Le Chatelier — change pressure	count gas moles both sides	Ex 4
E	Le Chatelier — change temperature	sign of $Δ H$ decides direction	Ex 5
F	Le Chatelier — degenerate case (equal moles)	pressure does nothing	Ex 6
G	Stoichiometry / recycle (word problem)	Ostwald atom economy	Ex 7
H	Disproportionation identification	one element splits up and down	Ex 8
I	Paramagnetism / odd-electron count	count valence electrons	Ex 9
J	Exam twist — combine two ideas	dimerisation + Le Chatelier + colour	Ex 10

We now walk through all ten. Every numeric answer is machine-checked at the bottom.

Case A — oxidation state, ordinary molecule

Worked example Ex 1 — Find the oxidation state of N in

$H N O_{3}$ . Forecast: guess before reading. HNO₃ is "the top of the nitrogen ladder." What number do you expect?

Step 1 — Assign the atoms we already know. Why this step? We anchor to the two elements whose states almost never change: H is $+ 1$ , O is $- 2$ . In Oxidation States and Redox these are the standard reference values.

Step 2 — Let the unknown be $x$ and write the neutrality equation. Why this step? The molecule carries zero total charge, so the oxidation states must sum to $0$ . There is $1$ H, $1$ N, $3$ O: $(+ 1) + x + 3 (- 2) = 0$

Step 3 — Solve. Why this step? Pure algebra now. $1 + x - 6 = 0 \Rightarrow x = + 5$

Verify: $+ 5$ is nitrogen's maximum state (it has 5 valence electrons, all "given away" to oxygen). Matches the parent note: HNO₃ is the anhydride-acid of $N_{2} O_{5}$ , also $+ 5$ . ✔

Case B — fractional / average oxidation state

Case C — sign-flip twist (the peroxide-style trap)

Worked example Ex 3 — Find the oxidation state of N in the nitrite ion

$N O_{2}^{-}$ , and contrast with the trap "O is always $- 2$ ." Forecast: Is N in $N O_{2}^{-}$ the same $+ 4$ as in the neutral $N O_{2}$ molecule? Guess yes/no.

Step 1 — Include the ion's charge in the sum. Why this step? For an ion, oxidation states sum to the charge, not zero. Here charge $= - 1$ . Oxygen is still $- 2$ (there is no O–O bond, so no peroxide flip): $x + 2 (- 2) = - 1$

Step 2 — Solve. $x - 4 = - 1 \Rightarrow x = + 3$

Step 3 — Compare with neutral $N O_{2}$ . Why this step? In neutral $N O_{2}$ : $x + 2 (- 2) = 0 \Rightarrow x = + 4$ . So adding one electron (making the anion) drops N from $+ 4$ to $+ 3$ . The lesson: the total charge changes the answer — never blindly copy the neutral value.

[!mistake] The trap this case guards against "O is always $- 2$ " is usually true — but in peroxides ( $O_{2}^{2 -}$ ) O is $- 1$ , and in $O F_{2}$ it is $+ 2$ . Nitrogen oxides have no O–O bonds, so O safely stays $- 2$ here. Always ask "is oxygen bonded to another O or to F?" before trusting $- 2$ .

Verify: $N O_{2}^{-}$ (nitrite, N at $+ 3$ ) is the conjugate base of $H N O_{2}$ , whose N is also $+ 3$ . Consistent. ✔

Case D — Le Chatelier under pressure

Worked example Ex 4 — In the Haber equilibrium

$N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ , the total pressure is doubled at fixed temperature. Which way does it shift? Forecast: more NH₃ or less? (Recall Le Chatelier Principle.)

Step 1 — Count gas moles on each side. Why this step? Pressure changes push a system toward the side with fewer gas molecules, because fewer molecules take less volume and relieve the squeeze. Left $= 1 + 3 = 4$ mol; right $= 2$ mol.

Step 2 — Compare sides. Why this step? Right side has fewer moles ( $2 < 4$ ), so raising pressure favours the right → more NH₃.

Step 3 — Confirm with $K_{p}$ . Why this step? Let each partial pressure scale by factor $2$ under the squeeze. The reaction quotient $Q_{p} = \frac{p _{N H_{3}}^{2}}{p _{N_{2}} p _{H_{2}}^{3}} ⟶ \frac{( 2 p _{N H_{3}} ) ^{2}}{( 2 p _{N_{2}} ) ( 2 p _{H_{2}} ) ^{3}} = \frac{4}{2 \cdot 8} Q_{p} = \frac{1}{4} Q_{p} .$ $Q_{p}$ instantly drops below $K_{p}$ , so the system runs forward (right) to rebuild $Q_{p}$ up to $K_{p}$ .

Verify: The factor is $\frac{2 ^{2}}{2 ^{1} \cdot 2 ^{3}} = \frac{4}{16} = \frac{1}{4}$ , i.e. $Q$ falls to a quarter → definitely shifts right. Matches "high P helps Haber." ✔

Case E — Le Chatelier under temperature

Worked example Ex 5 — The Haber reaction has

$Δ H = - 92 kJ/mol$ . Temperature is raised. What happens to NH₃ yield? Forecast: yield up or down when we heat an exothermic reaction?

Step 1 — Treat heat as a "product" for an exothermic reaction. Why this step? $Δ H < 0$ means the forward reaction releases heat: $N_{2} + 3 H_{2} ⇌ 2 N H_{3} + heat .$ Adding heat (raising T) is like adding a product.

Step 2 — Apply Le Chatelier. Why this step? Add product ⇒ system shifts backward (left) to consume the added heat ⇒ less NH₃.

Step 3 — Reconcile with why industry still uses $\sim 700$ K. Why this step? Low T gives higher yield but the reaction crawls (rate too slow). So $700$ K is a compromise: slightly lower yield, but reached fast enough to be profitable, aided by the Fe catalyst.

Verify: Sign of $Δ H$ is negative ⇒ raising T lowers yield. Numerically the equilibrium constant obeys $ln K \propto - Δ H / (R T)$ ; with $Δ H < 0$ , increasing $T$ decreases $K$ . ✔

Case F — degenerate case: pressure does nothing

Worked example Ex 6 — For

$N O (g) + \frac{1}{2} O_{2} (g) ⇌ N O_{2} (g)$ vs. the model reaction $N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)$ , what does raising pressure do to the second one? Forecast: guess before counting.

Step 1 — Count moles for the second reaction. Why this step? Left $= 1 + 1 = 2$ mol gas; right $= 2$ mol gas. Equal.

Step 2 — Apply the rule. Why this step? Pressure shifts toward fewer moles — but here both sides are equal, so there is no "fewer" side. Pressure change ⇒ no shift at all. This is the degenerate / zero case.

Step 3 — Check the $Q_{p}$ algebra to be sure. Why this step? Scale every pressure by $k$ : $Q_{p} = \frac{p _{N O}^{2}}{p _{N_{2}} p _{O_{2}}} ⟶ \frac{( k p _{N O} ) ^{2}}{( k p _{N_{2}} ) ( k p _{O_{2}} )} = \frac{k ^{2}}{k ^{2}} Q_{p} = Q_{p} .$ $Q_{p}$ is unchanged ⇒ still equal to $K_{p}$ ⇒ no shift.

Verify: exponent on $k$ is $2 - (1 + 1) = 0$ , so $k^{0} = 1$ : pressure genuinely irrelevant. ✔

Case G — Ostwald recycle word problem

Figure — Group 15 (Nitrogen family) — N₂ inertness; NH₃ synthesis (Haber); HNO₃ (Ostwald); oxides of N (N₂O, NO, NO₂, N₂O₄, N₂O₅)

Worked example Ex 7 — In Step 3 of Ostwald,

$3 N O_{2} + H_{2} O \to 2 H N O_{3} + N O$ , you feed 300 mol of $N O_{2}$ . (a) How many mol of $H N O_{3}$ form in one pass? (b) How many mol of NO must be recycled? (c) If all recycled NO is re-oxidised and re-absorbed with 100% efficiency, what is the eventual mol of $H N O_{3}$ from the original 300 mol of $N O_{2}$ -equivalent nitrogen? Forecast: guess whether you eventually get 200 or 300 mol of HNO₃.

Step 1 — Use the stoichiometric ratios for one pass. Why this step? The balanced equation says $3 N O_{2} \to 2 H N O_{3} + 1 N O$ . So per 3 mol in: 2 mol acid, 1 mol NO out. Scale by $300/3 = 100$ : $H N O_{3} = 2 \times 100 = 200 mol, N O = 1 \times 100 = 100 mol .$ This answers (a) $= 200$ and (b) $= 100$ .

Step 2 — Track the nitrogen atoms (conservation). Why this step? Look at the figure: the $100$ mol NO goes back to Step 2, becomes $100$ mol $N O_{2}$ , and re-enters Step 3. Each recycle converts $\frac{2}{3}$ of the incoming N into acid and returns $\frac{1}{3}$ as NO. So the acid produced is a geometric series: $H N O_{3} = 300 (\frac{2}{3} + \frac{1}{3} \cdot \frac{2}{3} + (\frac{1}{3})^{2} \frac{2}{3} + \dots) = 300 \cdot \frac{2/3}{1 - 1/3} .$

Step 3 — Sum the geometric series. Why this step? Infinite geometric sum $\sum a r^{n} = \frac{a}{1 - r}$ with $a = \frac{2}{3}, r = \frac{1}{3}$ : $\frac{2/3}{2/3} = 1 \Rightarrow H N O_{3} = 300 \times 1 = 300 mol .$ With perfect recycling all nitrogen atoms end up as $H N O_{3}$ .

Verify: Atom economy demands it — nitrogen can't vanish, and with 100% recycle nothing leaves as NO, so 300 mol N → 300 mol HNO₃. First-pass numbers: $200$ acid $+ 100$ NO $= 300$ N accounted. ✔

Case H — spot the disproportionation

Worked example Ex 8 — In

$3 N O_{2} + H_{2} O \to 2 H N O_{3} + N O$ , prove this is a disproportionation and identify which atoms go up and which go down. (See Disproportionation Reactions.) Forecast: which single element is both oxidised and reduced here?

Step 1 — Assign N in every species. Why this step? Disproportionation means one element in one starting oxidation state ends up in two different states, one higher one lower.

$N O_{2}$ : $x + 2 (- 2) = 0 \Rightarrow x = + 4$ .
$H N O_{3}$ : (from Ex 1) $N = + 5$ .
$N O$ : $x + (- 2) = 0 \Rightarrow x = + 2$ .

Step 2 — Compare starting and ending states. Why this step? All the N started at $+ 4$ (in $N O_{2}$ ). Some ended higher at $+ 5$ (oxidised, in HNO₃) and some ended lower at $+ 2$ (reduced, in NO). Same element, split both ways ⇒ disproportionation. ✔

Step 3 — Balance electrons to check the $3 : 2 : 1$ ratio. Why this step? Two N atoms each lose $1 e^{-}$ ( $+ 4 \to + 5$ ): total $2 e^{-}$ lost. One N atom gains $2 e^{-}$ ( $+ 4 \to + 2$ ): $2 e^{-}$ gained. Electrons balance ⇒ $2$ oxidised : $1$ reduced ⇒ the $3 N O_{2}$ split as $2$ HNO₃ $+ 1$ NO.

Verify: Electrons lost $= 2 \times (5 - 4) = 2$ ; electrons gained $= 1 \times (4 - 2) = 2$ . Equal ⇒ balanced disproportionation. ✔

Case I — paramagnetism from odd electrons

Worked example Ex 9 — Is

$N O$ paramagnetic? How many unpaired electrons? (Compare with N2 molecule MO diagram.) Forecast: even or odd total valence electrons?

Step 1 — Count valence electrons. Why this step? Magnetism comes from unpaired electrons; if the total is odd, at least one electron must be unpaired. N contributes $5$ , O contributes $6$ : $5 + 6 = 11 valence electrons (odd) .$

Step 2 — Odd total ⇒ one unpaired electron. Why this step? Electrons pair up two-per-orbital; an odd count leaves exactly one lonely electron ⇒ NO is paramagnetic with $1$ unpaired electron.

Step 3 — Contrast with $N O_{2}$ and $N_{2} O_{4}$ . Why this step? $N O_{2}$ : $5 + 2 \times 6 = 17$ (odd) ⇒ also paramagnetic. Two $N O_{2}$ join their lonely electrons into an N–N bond ⇒ $N_{2} O_{4}$ : $2 \times 17 = 34$ (even) ⇒ diamagnetic. This is why $N O_{2}$ dimerises.

Verify: $11$ is odd ⇒ ≥1 unpaired ⇒ paramagnetic; $34$ is even ⇒ can be fully paired ⇒ diamagnetic. ✔

Case J — exam twist: colour + dimerisation + Le Chatelier

Worked example Ex 10 — A sealed tube of brown

$N O_{2} / N_{2} O_{4}$ at equilibrium, $2 N O_{2} (brown) ⇌ N_{2} O_{4} (colourless), Δ H = - 57 kJ/mol$ , is cooled in ice. Predict the colour change, and confirm with mole counting. Forecast: does the tube get darker brown or paler as you cool it?

Step 1 — Identify what changes with temperature. Why this step? The dimerisation is exothermic ( $Δ H < 0$ ), so heat is a product: $2 N O_{2} ⇌ N_{2} O_{4} + heat$ . Cooling removes heat ⇒ shifts forward (toward $N_{2} O_{4}$ ).

Step 2 — Translate the shift into colour. Why this step? $N O_{2}$ is the brown species; $N_{2} O_{4}$ is colourless. Shifting forward consumes brown $N O_{2}$ ⇒ the tube goes paler / colourless on cooling.

Step 3 — Cross-check with the pressure/mole argument. Why this step? Forward direction goes $2$ mol $\to 1$ mol (fewer gas molecules), so cooling and compressing both favour colourless $N_{2} O_{4}$ — two independent reasons pointing the same way, exactly as the parent note states.

Verify: $Δ H < 0$ + cooling ⇒ forward ⇒ more $N_{2} O_{4}$ ⇒ paler. Consistent with "low T / high P favours $N_{2} O_{4}$ ." ✔

Active Recall

Recall When does raising pressure NOT shift an equilibrium?

When both sides have the same number of gas moles (Ex 6: $N_{2} + O_{2} ⇌ 2 N O$ , $2$ vs $2$ ). Then $Q_{p}$ is unchanged by any pressure scaling.

Recall Why is

$N O_{2}$ paramagnetic but $N_{2} O_{4}$ is not? $N O_{2}$ has $17$ valence electrons (odd) → one unpaired electron. Two of them pair up their lonely electrons into an N–N bond to give $N_{2} O_{4}$ ( $34$ electrons, even, all paired) → diamagnetic.

Recall In Ostwald Step 3, what fraction of nitrogen becomes HNO₃ per pass, and eventually?

Per pass: $\frac{2}{3}$ becomes HNO₃, $\frac{1}{3}$ returns as NO. With perfect recycling (geometric series), eventually all nitrogen becomes HNO₃.

Connections

Parent topic (Hinglish)
Oxidation States and Redox — the rules used in Ex 1, 2, 3, 8
Le Chatelier Principle — Ex 4, 5, 6, 10
Disproportionation Reactions — Ex 8
N2 molecule MO diagram — Ex 9 paramagnetism logic
Aqua regia and Noble Metals — HNO₃ oxidising chemistry