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

Before any equation, we need two ideas in plain words.

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

Figure s01 — The oxidation ladder. Read the vertical axis as nitrogen's oxidation state. Watch the coloured dots climb from the teal $N H_{3}$ dot at $- 3$ , up through $N O$ ( $+ 2$ ) and $N O_{2}$ ( $+ 4$ ), to the plum $H N O_{3}$ dot at $+ 5$ . The orange arrows are the reactions; notice the biggest single jump is the very first one.

Step 1 — Locate the starting rung: N in $N H_{3}$ is $- 3$

WHAT. We compute nitrogen's oxidation state in ammonia so we know where the climb begins.

WHY. You cannot plan a climb without knowing the ground floor. Every later step is measured against this $- 3$ .

PICTURE. Look at $N H_{3}$ : one N surrounded by three H. Since N is more electronegative than H, each H hands its shared electron to N and is left at $+ 1$ ; the three H's carry $+ 3$ total. The molecule is neutral, so N must cancel them.

Figure s02 — Ammonia's bookkeeping. Observe the central teal N labelled $- 3$ bonded to three plum H's each labelled $+ 1$ . Trace each bond line: because N wins every tug-of-war, the three $+ 1$ 's must be cancelled by a $- 3$ on nitrogen. The equation under the molecule is that cancellation written out.

Step 2 — First climb: burn $N H_{3}$ over hot Pt/Rh to make $N O$ ( $- 3 \to + 2$ )

WHAT. Pass ammonia + oxygen over a platinum–rhodium gauze at ~500 K, $9 bar$ . Nitrogen jumps from $- 3$ to $+ 2$ .

WHY a catalyst, and not just a flame? Here is the key question a tool must answer: ammonia burning in air naturally goes all the way to harmless $N_{2}$ — useless to us. The Pt/Rh gauze is a steering wheel: it lowers the activation barrier for the path to $N O$ specifically, so nitrogen stops at the $+ 2$ rung instead of falling back to $0$ . Without it we lose our nitrogen to $N_{2}$ .

WHY these exact coefficients (mass + electron balance, not a guess). We do not memorise $5 O_{2}$ and $6 H_{2} O$ ; we derive them.

Check the climb: in $N O$ , oxygen is $- 2$ , molecule neutral, so N $= + 2$ . Nitrogen rose by $+ 2 - (- 3) = 5$ units. That's a big single jump — the catalyst earns its keep.

Figure s03 — The catalyst as a steering wheel. Follow two arrows leaving the teal $N H_{3}$ dot: the solid orange arrow (labelled "Pt/Rh catalyst") lands on the wanted $N O$ at $+ 2$ ; the dashed grey arrow (labelled "no catalyst") drops to wasted $N_{2}$ at $0$ . The picture's whole point: only the catalysed path reaches the rung we want.

Step 3 — Second climb: cool the gas, add more $O_{2}$ , get $N O_{2}$ ( $+ 2 \to + 4$ )

WHAT. Let the $N O$ cool and meet fresh oxygen. It grabs one more oxygen and becomes brown $N O_{2}$ .

WHY cool it? (a mini-derivation, not a slogan). Two facts about this reaction both point to low temperature, and we quantify each.

Figure s04 — Adding one oxygen. Watch the orange $N O$ dot ( $+ 2$ ) meet a faint extra O, cross the teal "cool" arrow, and arrive as the larger orange $N O_{2}$ dot ( $+ 4$ , brown gas). The balanced equation sits beneath. The key visual: one oxygen added = two rungs climbed.

Figure s05 — The $N O_{2} ⇌ N_{2} O_{4}$ equilibrium. See two brown $N O_{2}$ dots (each with a small unpaired-electron mark) join across a double arrow into one colourless $N_{2} O_{4}$ . The arrow bends toward $N_{2} O_{4}$ under "cool / high P" because both reduce gas moles — the same push that made Step 3 go.

Step 4 — The tricky rung: $N O_{2}$ + water splits two ways (disproportionation)

WHAT. Dissolve $N O_{2}$ (and its $N_{2} O_{4}$ ) in water. Instead of every nitrogen going up, the nitrogens split: some go up to $+ 5$ (the $H N O_{3}$ we want) and some fall down to $+ 2$ (back to $N O$ ).

WHY does it split — and why is that the right tool? This is a disproportionation: a single starting oxidation state ( $+ 4$ ) is unstable enough that it is energetically cheaper for the nitrogens to share out — some grabbing more oxygen, some giving it up — than for all of them to move the same way. Water is the medium that lets this rearrangement happen. No external oxidiser is needed; the $N O_{2}$ oxidises and reduces itself.

Formula Deriving the

$3 N O_{2} + H_{2} O \to 2 H N O_{3} + N O$ equation from scratch Electron balance sets the up:down ratio. Each atom going up loses $1$ electron ( $+ 4 \to + 5$ ); each atom going down gains $2$ electrons ( $+ 4 \to + 2$ ). To make electrons cancel: $(atoms up) \times 1 = (atoms down) \times 2 \Rightarrow up : down = 2 : 1$ So the smallest set is 2 up + 1 down = 3 N atoms, i.e. we must start from $3 N O_{2}$ . Mass balance fixes the water and confirms the integers. With $3 N O_{2}$ (giving $2 H N O_{3} + 1 N O$ ), count each element:

N: $3$ left, $2 + 1 = 3$ right. ✓
H: the $2 H N O_{3}$ need $2$ H, so we need exactly $1 H_{2} O$ (which supplies $2$ H). ✓
O: left $= 3 \times 2 + 1 = 7$ ; right $= 2 \times 3 + 1 = 7$ . ✓

Every element balances with the smallest whole numbers, so this is the minimal integer equation — the electron ledger and the mass ledger agree. $3 all at + 4 N O_{2} + fixes H, adds 1 O H_{2} O \to 2 N = + 5 (up) H N O_{3} + N = + 2 (down) N O$

Figure s06 — The disproportionation fork. From one orange " $3 N O_{2}$ , all $+ 4$ " dot, follow the plum arrow climbing to $2 H N O_{3}$ ( $+ 5$ , "lose 1e⁻ each") and the teal arrow dropping to $1 N O$ ( $+ 2$ , "gains 2e⁻"). The two-to-one splay of the arrows is exactly the $2 : 1$ ratio the electron balance forced.

Step 5 — Close the loop: recycle the escaped $N O$

WHAT. Take the $N O$ made in Step 4 and send it back to Step 3.

WHY. That $N O$ is nitrogen at $+ 2$ — exactly the rung Step 3 starts from. Feeding it back means no nitrogen is wasted; the process becomes a loop that converts almost all ammonia to nitric acid.

Figure s07 — The recycle loop. Three coloured dots (NO → NO₂ → HNO₃) are chained by solid arrows; a dashed teal arc loops the $+ 2$ NO byproduct from the HNO₃ end all the way back to the NO₂ stage. Watch that arc — it is the reason almost no nitrogen leaves the plant unused.

Step 6 — Edge case: what if you skip the catalyst? (thermodynamics vs kinetics)

WHAT. Consider removing the Pt/Rh gauze in Step 2.

WHY show it. To prove the catalyst isn't optional decoration — and to be honest about why the wrong product wins without it.

Figure s08 — Catalyst-free dead end. On an oxidation-state axis, the teal $N H_{3}$ dot at $- 3$ rises only to the grey $N_{2}$ dot at $0$ (dashed arrow), never reaching the plum $+ 5$ goal marked above. The picture makes the loss visceral: without the catalyst the ladder collapses at the ground floor.

The one-picture summary

Everything above is one climb with one recycle loop: $- 3 \to + 2 \to + 4 \to$ (fork) $+ 5$ product $+$ $+ 2$ recycled.

Figure s09 — The whole process in one frame. The main orange staircase climbs $N H_{3} (- 3) \to N O (+ 2) \to N O_{2} (+ 4) \to H N O_{3} (+ 5)$ ; a dashed teal branch drops from the $+ 4$ rung to $N O (+ 2)$ and a dotted teal arc carries it back up — the recycle loop. Everything on this page is compressed into these arrows.

Recall Feynman retelling — the whole walkthrough in plain words

Imagine nitrogen wearing outfits made of oxygen. In ammonia it wears nothing — it's at the bottom, outfit number $- 3$ . We want it in the fanciest outfit, $+ 5$ , which is nitric acid. So we dress it up in stages. First (Step 2) we heat ammonia with oxygen over a special metal grid — the grid is a bouncer that only lets nitrogen into the "+2 room" ( $N O$ ); without the bouncer nitrogen sneaks back to plain air ( $N_{2}$ , outfit $0$ ), which it actually prefers — so the bouncer wins by being fast, not by changing what nitrogen likes. Then (Step 3) we cool it (cooling helps because the reaction gives off heat and squishes three gas molecules into two) and give it another oxygen so it reaches $+ 4$ (brown $N O_{2}$ , which likes to pair up into colourless $N_{2} O_{4}$ ). Now the tricky bit (Step 4): when this hits water, the nitrogens can't all move together — three of them split, two climbing to $+ 5$ (our acid!) and one sliding down to $+ 2$ ( $N O$ again). Rather than waste that one, we send it back up the stairs (Step 5). Round and round, until almost all our nitrogen ends up as nitric acid.

Connections

Parent topic
Oxidation States and Redox — the bookkeeping behind every rung
Disproportionation Reactions — Step 4's two-way fork
Le Chatelier Principle — why we cool Step 3
N2 molecule MO diagram — why the catalyst-free path dead-ends at inert N₂