NCERT Solutions · Class 11 Chemistry Redox Reactions

Oxidation number is the charge an atom would carry if every bond were purely ionic — and in a neutral compound, every atom's oxidation number, multiplied by how many atoms of that kind are present, must sum to zero. So for each formula below, fix the oxidation numbers you already know from standard rules, let the underlined atom's oxidation number be \(\displaystyle x\), and solve.The fixed rules you need throughout: (a) \(\displaystyle NaH_2PO_4\), oxidation number of \(\displaystyle P\). \(\displaystyle Na = +1\), each \(\displaystyle H = +1\) (bonded to non-metals here, not a hydride), each \(\displaystyle O = -2\). \[(+1) + 2(+1) + x + 4(-2) = 0 \] \[1 + 2 + x - 8 = 0 \implies x = +5 \](b) \(\displaystyle NaHSO_4\), oxidation number of \(\displaystyle S\). \(\displaystyle Na = +1\), \(\displaystyle H = +1\), each \(\displaystyle O = -2\). \[(+1) + (+1) + x + 4(-2) = 0 \] \[1 + 1 + x - 8 = 0 \implies x = +6 \](c) \(\displaystyle H_4P_2O_7\), oxidation number of \(\displaystyle P\). Each \(\displaystyle H = +1\), each \(\displaystyle O = -2\); there are two \(\displaystyle P\) atoms, so their combined contribution is \(\displaystyle 2x\). \[4(+1) + 2x + 7(-2) = 0 \] \[4 + 2x - 14 = 0 \implies 2x = 10 \implies x = +5 \](d) \(\displaystyle K_2MnO_4\), oxidation number of \(\displaystyle Mn\). Each \(\displaystyle K = +1\), each \(\displaystyle O = -2\). \[2(+1) + x + 4(-2) = 0 \] \[2 + x - 8 = 0 \implies x = +6 \](e) \(\displaystyle CaO_2\), oxidation number of \(\displaystyle O\). This is calcium peroxide, not a normal oxide — the aside above about peroxides applies directly, so do not assume \(\displaystyle O = -2\) here. \(\displaystyle Ca = +2\), and there are two \(\displaystyle O\) atoms, each with unknown \(\displaystyle x\). \[(+2) + 2x = 0 \implies x = -1 \] That \(\displaystyle -1\) is exactly the peroxide signature: each oxygen shares one electron with the other oxygen instead of taking two full electrons from a metal.(f) \(\displaystyle NaBH_4\), oxidation number of \(\displaystyle B\). Here \(\displaystyle H\) is bonded to boron in a hydride-type anion, and boron is less electronegative than hydrogen, so each \(\displaystyle H = -1\) — the opposite of its usual sign. \(\displaystyle Na = +1\). \[(+1) + x + 4(-1) = 0 \] \[1 + x - 4 = 0 \implies x = +3 \](g) \(\displaystyle H_2S_2O_7\), oxidation number of \(\displaystyle S\). Each \(\displaystyle H = +1\), each \(\displaystyle O = -2\); two \(\displaystyle S\) atoms contribute \(\displaystyle 2x\). \[2(+1) + 2x + 7(-2) = 0 \] \[2 + 2x - 14 = 0 \implies 2x = 12 \implies x = +6 \](h) \(\displaystyle KAl(SO_4)_2 \cdot 12H_2O\), oxidation number of \(\displaystyle S\). \(\displaystyle K = +1\) and \(\displaystyle Al = +3\) are both fixed. There are two sulfate groups, so \(\displaystyle 2x\) for sulfur and \(\displaystyle 2 \times 4 = 8\) oxygens from the sulfate part, each \(\displaystyle -2\). The \(\displaystyle 12\,H_2O\) is neutral water of crystallisation, so it adds \(\displaystyle 0\) — do not try to fold its hydrogens or oxygens into the balance. \[(+1) + (+3) + 2x + 8(-2) + 0 = 0 \] \[1 + 3 + 2x - 16 = 0 \implies 2x = 12 \implies x = +6 \]Answer: (a) P = +$\displaystyle 5$ (b) S = +$\displaystyle 6$ (c) P = +$\displaystyle 5$ (d) Mn = +$\displaystyle 6$ (e) O = −$\displaystyle 1$ (f) B = +$\displaystyle 3$ (g) S = +$\displaystyle 6$ (h) S = +$\displaystyle 6$

The sum-of-oxidation-numbers rule only gives you an AVERAGE — real atoms of the same element inside one molecule can sit in different bonding environments and carry different actual values that average out to that number.Two tools are needed throughout:Rule $\displaystyle 1$ (whole-species check). For any species, \[\sum (\text{oxidation number} \times \text{number of atoms}) = \text{overall charge on the species} \] using the fixed values \(\displaystyle \mathrm{H} = +1\) (except metal hydrides) and \(\displaystyle \mathrm{O} = -2\) (except peroxides). This is fast, but it silently assumes every atom of the underlined element in the formula is identical — that is the assumption that breaks in several parts below.Rule $\displaystyle 2$ (bond-by-bond check, for when Rule $\displaystyle 1$'s assumption fails). Look at the actual bonds around one specific atom. For every bond to a more electronegative atom, add +$\displaystyle 1$ to the atom you're scoring (per bond, so a double bond counts twice); for every bond to a less electronegative atom (like H), add −$\displaystyle 1$; for a bond to an atom of the same element (like C–C), add 0. Order used here: \(\displaystyle \mathrm{O} > \mathrm{C} > \mathrm{H}\).(a) KI₃ — oxidation number of IPotassium is a Group $\displaystyle 1$ metal, always \(\displaystyle +1\). KI₃ is neutral, so by Rule $\displaystyle 1$: \[(+1) + 3x = 0 \implies x = -\tfrac{1}{3} \]A fraction cannot be the real oxidation number of a single iodine atom — oxidation numbers of actual atoms are whole numbers. This is the tell that Rule $\displaystyle 1$'s "all three I atoms are identical" assumption is wrong here. KI₃ is really \(\displaystyle \mathrm{K^+}\) paired with the triiodide ion, which is best pictured as an \(\displaystyle \mathrm{I_2}\) molecule held together with an iodide ion, \(\displaystyle [\mathrm{I-I-I}]^{-}\). In that picture two of the iodine atoms are still bonded only to each other, exactly as in elemental \(\displaystyle \mathrm{I_2}\) (oxidation number \(\displaystyle 0\)), and the third carries the ionic charge (oxidation number \(\displaystyle -1\)). Check: \(\displaystyle \dfrac{0+0+(-1)}{3} = -\dfrac{1}{3}\), matching the Rule-$\displaystyle 1$ average.(b) H₂S₄O₆ — oxidation number of SH₂S₄O₆ is neutral. By Rule $\displaystyle 1$, with \(\displaystyle \mathrm{H}=+1\) ($\displaystyle 2$ atoms) and \(\displaystyle \mathrm{O}=-2\) ($\displaystyle 6$ atoms): \[2(+1) + 4x + 6(-2) = 0 \implies 4x = 10 \implies x = +2.5 \]Again a non-integer, so the four sulfur atoms are not chemically identical. Tetrathionic acid's anion has the chain structure \(\displaystyle [\mathrm{O_3S{-}S{-}S{-}SO_3}]^{2-}\): the two end sulfur atoms are each bonded to three oxygens (in the sulfonate-type arrangement also seen in dithionate, \(\displaystyle \mathrm{S_2O_6^{2-}}\), where each S is \(\displaystyle +5\)), while the two middle sulfur atoms are bonded only to other sulfur atoms and carry no oxygen at all, so their oxidation number is \(\displaystyle 0\). Check: end sulfurs \(\displaystyle +5\) each, middle sulfurs \(\displaystyle 0\) each, average \(\displaystyle =\dfrac{5+5+0+0}{4} = 2.5\), matching Rule $\displaystyle 1$, and the ionic-charge check also comes out right: \(\displaystyle 2(+5)+2(0)+6(-2) = 10-12 = -2\), the charge on \(\displaystyle \mathrm{S_4O_6^{2-}}\).(c) Fe₃O₄ — oxidation number of FeFe₃O₄ is neutral, with \(\displaystyle \mathrm{O} = -2\) ($\displaystyle 4$ atoms): \[3x + 4(-2) = 0 \implies x = +\tfrac{8}{3} \approx +2.67 \]A pitfall worth flagging here: it's tempting to call this "iron($\displaystyle 8$/$\displaystyle 3$)," but iron never actually exists in a \(\displaystyle +8/3\) state. Fe₃O₄ (magnetite) is a mixed oxide, effectively \(\displaystyle \mathrm{FeO}\cdot\mathrm{Fe_2O_3}\) — one iron atom is genuinely \(\displaystyle \mathrm{Fe^{2+}}\) and the other two are genuinely \(\displaystyle \mathrm{Fe^{3+}}\) (confirmed by magnetic and structural measurements showing two distinct iron sites). Check: \(\displaystyle \dfrac{2+3+3}{3} = \dfrac{8}{3}\), the same average Rule $\displaystyle 1$ gave.(d) CH₃CH₂OH — oxidation number of both carbonsWhole-molecule average (Rule $\displaystyle 1$) for neutral \(\displaystyle \mathrm{C_2H_6O}\), with H = +$\displaystyle 1$ ($\displaystyle 6$ atoms) and O = −$\displaystyle 2$ ($\displaystyle 1$ atom): \[2x + 6(+1) + (-2) = 0 \implies 2x = -4 \implies x = -2 \]But the two carbons sit in different environments, so use Rule $\displaystyle 2$ on each: Average: \(\displaystyle \dfrac{-3+(-1)}{2} = -2\), matching Rule 1. The mistake to avoid is quoting "\(\displaystyle -2\)" as if it were the oxidation number of each carbon — it is the average of two genuinely different values, \(\displaystyle -3\) and \(\displaystyle -1\).(e) CH₃COOH — oxidation number of both carbonsWhole-molecule average (Rule $\displaystyle 1$) for neutral \(\displaystyle \mathrm{C_2H_4O_2}\), with H = +$\displaystyle 1$ ($\displaystyle 4$ atoms) and O = −$\displaystyle 2$ ($\displaystyle 2$ atoms): \[2x + 4(+1) + 2(-2) = 0 \implies 2x = 0 \implies x = 0 \]By Rule $\displaystyle 2$ on each carbon: Average: \(\displaystyle \dfrac{-3+(+3)}{2} = 0\), matching Rule $\displaystyle 1$ — the two carbons are equal and opposite, not both zero.Answer: (a) KI₃ — average \(\displaystyle -\tfrac13\) for I; actually two I atoms at $\displaystyle 0$ and one at −$\displaystyle 1$ (I₂ bound to I⁻). (b) H₂S₄O₆ — average +$\displaystyle 2.5$ for S; actually the two terminal S atoms are +$\displaystyle 5$ each and the two central S atoms are 0. (c) Fe₃O₄ — average +$\displaystyle 8$/$\displaystyle 3$ for Fe; actually one Fe²⁺ and two Fe³⁺ (mixed oxide FeO·Fe₂O₃). (d) CH₃CH₂OH — average −$\displaystyle 2$ for C; actually −$\displaystyle 3$ (CH₃ carbon) and −$\displaystyle 1$ (CH₂OH carbon). (e) CH₃COOH — average $\displaystyle 0$ for C; actually −$\displaystyle 3$ (CH₃ carbon) and +$\displaystyle 3$ (COOH carbon). In every case the sum-of-oxidation-numbers formula only returns the average across chemically different atoms of the same element; the real, integer values come from examining the actual bonds (or, for KI₃, the actual structure) atom by atom.

A reaction is redox only when oxidation numbers move in both directions at once — one element's number goes up (it is oxidized, it loses electrons) by exactly as much as another element's number goes down (it is reduced, it gains electrons). To justify each equation below, the oxidation number of every atom is tracked from reactant to product, and it is checked that electrons lost equal electrons gained. Three rules are used throughout: a free (uncombined) element such as \(\displaystyle \mathrm{H_2} \), \(\displaystyle \mathrm{F_2} \), or \(\displaystyle \mathrm{K}(s) \) is always $\displaystyle 0$; oxygen is \(\displaystyle -2 \) in every species here; and the oxidation numbers of all atoms in a neutral molecule or formula unit must add up to 0. For hydrogen there is one extra rule that trips people up, given in part (c) below: hydrogen is \(\displaystyle +1 \) when it is bonded to a non-metal or a metalloid, but \(\displaystyle -1 \) (a hydride ion) when it is bonded only to a metal.(a) \(\displaystyle \mathrm{CuO}(s) + \mathrm{H_2}(g) \rightarrow \mathrm{Cu}(s) + \mathrm{H_2O}(g) \)Atom count check: $\displaystyle 1$ Cu, $\displaystyle 1$ O, $\displaystyle 2$ H on each side — already balanced.In \(\displaystyle \mathrm{CuO} \), oxygen is \(\displaystyle -2 \) and the compound is neutral, so copper's number \(\displaystyle x \) satisfies \(\displaystyle x + (-2) = 0 \), giving \(\displaystyle x = +2 \). In \(\displaystyle \mathrm{H_2O} \), \(\displaystyle 2(+1) + (-2) = 0 \), so hydrogen is \(\displaystyle +1 \).\[\underset{+2}{\mathrm{Cu}} \rightarrow \underset{0}{\mathrm{Cu}} \quad (\text{gains } 2\, e^-,\ \text{reduced}) \qquad\qquad \underset{0}{\mathrm{H_2}} \rightarrow 2\,\underset{+1}{\mathrm{H}} \quad (\text{loses } 1\, e^- \text{ each, } 2\, e^- \text{ total, oxidized}) \]$\displaystyle 2$ electrons lost by hydrogen exactly match $\displaystyle 2$ electrons gained by copper. \(\displaystyle \mathrm{CuO} \) is the oxidizing agent (it gets reduced) and \(\displaystyle \mathrm{H_2} \) is the reducing agent (it gets oxidized), so this is a redox reaction.(b) \(\displaystyle \mathrm{Fe_2O_3}(s) + 3\mathrm{CO}(g) \rightarrow 2\mathrm{Fe}(s) + 3\mathrm{CO_2}(g) \)Atom count check: Fe $\displaystyle 2$ = $\displaystyle 2$, O \(\displaystyle 3+3=6 \) on the left and \(\displaystyle 3\times2=6 \) on the right, C \(\displaystyle 3=3\) — balanced.In \(\displaystyle \mathrm{Fe_2O_3} \): \(\displaystyle 2x + 3(-2) = 0 \Rightarrow x = +3 \). In \(\displaystyle \mathrm{CO} \): \(\displaystyle x + (-2) = 0 \Rightarrow x=+2 \). In \(\displaystyle \mathrm{CO_2} \): \(\displaystyle x + 2(-2) = 0 \Rightarrow x = +4 \).\[2\,\underset{+3}{\mathrm{Fe}} \rightarrow 2\,\underset{0}{\mathrm{Fe}} \quad (\text{gains } 3\,e^- \text{ each, } 6\,e^- \text{ total}) \qquad\qquad 3\,\underset{+2}{\mathrm{C}} \rightarrow 3\,\underset{+4}{\mathrm{C}} \quad (\text{loses } 2\,e^- \text{ each, } 6\,e^- \text{ total}) \]$\displaystyle 6$ electrons lost by carbon equal $\displaystyle 6$ gained by iron. Iron is reduced, carbon is oxidized — redox.(c) \(\displaystyle 4\mathrm{BCl_3}(g) + 3\mathrm{LiAlH_4}(s) \rightarrow 2\mathrm{B_2H_6}(g) + 3\mathrm{LiCl}(s) + 3\mathrm{AlCl_3}(s) \)Atom count check: B \(\displaystyle 4 = 2\times2 \); Cl \(\displaystyle 4\times3=12 \) on the left, \(\displaystyle 3\times1 + 3\times3 = 12 \) on the right; Li \(\displaystyle 3=3 \); Al \(\displaystyle 3=3 \); H \(\displaystyle 3\times4=12=2\times6 \) — balanced.This one is easy to get wrong because at first glance nothing looks like an oxidizer or a reducer. The atom that actually moves is hydrogen, and it moves between the two hydrogen conventions stated above.In \(\displaystyle \mathrm{LiAlH_4} \), hydrogen is bonded only to the metals lithium and aluminium, so it is the hydride ion, \(\displaystyle -1 \). Working out the metals from that: chlorine is always \(\displaystyle -1 \) here (bonded to a less electronegative partner in every chloride), so in \(\displaystyle \mathrm{LiCl} \), \(\displaystyle \mathrm{Li} = +1 \); in \(\displaystyle \mathrm{AlCl_3} \), \(\displaystyle \mathrm{Al} = +3 \). These match the values \(\displaystyle \mathrm{Li}=+1 \), \(\displaystyle \mathrm{Al}=+3 \) already carried inside \(\displaystyle \mathrm{LiAlH_4} \) — neither Li nor Al changes.In \(\displaystyle \mathrm{BCl_3} \), boron is bonded only to the more electronegative chlorine, so \(\displaystyle x + 3(-1) = 0 \Rightarrow \mathrm{B} = +3 \).In the product \(\displaystyle \mathrm{B_2H_6} \), hydrogen is now bonded to boron, a metalloid, not a metal — so by the rule above hydrogen here is \(\displaystyle +1 \), the opposite sign from the hydride ion it was in \(\displaystyle \mathrm{LiAlH_4} \). Electroneutrality of \(\displaystyle \mathrm{B_2H_6} \) then forces boron's number: \(\displaystyle 2x + 6(+1) = 0 \Rightarrow x = -3 \).\[4\,\underset{+3}{\mathrm{B}} \rightarrow 4\,\underset{-3}{\mathrm{B}} \quad (\text{gains } 6\,e^- \text{ each, } 24\,e^- \text{ total, reduced}) \qquad\qquad 12\,\underset{-1}{\mathrm{H}} \rightarrow 12\,\underset{+1}{\mathrm{H}} \quad (\text{loses } 2\,e^- \text{ each, } 24\,e^- \text{ total, oxidized}) \]$\displaystyle 24$ electrons lost by hydrogen exactly equal $\displaystyle 24$ gained by boron, while Li, Al and Cl are unchanged spectators. Boron is reduced and hydrogen is oxidized, so this is a redox reaction even though it looks, at a glance, like a plain substitution.(d) \(\displaystyle 2\mathrm{K}(s) + \mathrm{F_2}(g) \rightarrow 2\mathrm{K^+F^-}(s) \)Atom count check: K \(\displaystyle 2=2\), F \(\displaystyle 2=2\) — balanced.Both reactants are free elements, so \(\displaystyle \mathrm{K} = 0 \) and \(\displaystyle \mathrm{F} = 0 \) before reaction. The product is written explicitly as the ion pair, so the oxidation numbers are just the ionic charges: \(\displaystyle \mathrm{K^+} = +1 \), \(\displaystyle \mathrm{F^-} = -1 \).\[2\,\underset{0}{\mathrm{K}} \rightarrow 2\,\underset{+1}{\mathrm{K}} \quad (\text{loses } 1\,e^- \text{ each, } 2\,e^- \text{ total, oxidized}) \qquad\qquad \underset{0}{\mathrm{F_2}} \rightarrow 2\,\underset{-1}{\mathrm{F}} \quad (\text{gains } 1\,e^- \text{ each, } 2\,e^- \text{ total, reduced}) \]$\displaystyle 2$ electrons leave the two potassium atoms and land on the two fluorine atoms — a direct electron transfer, the clearest possible redox reaction, and the reason the product is written with explicit \(\displaystyle + \) and \(\displaystyle - \) charges rather than as a plain formula.(e) \(\displaystyle 4\mathrm{NH_3}(g) + 5\mathrm{O_2}(g) \rightarrow 4\mathrm{NO}(g) + 6\mathrm{H_2O}(g) \)Atom count check: N \(\displaystyle 4=4\); H \(\displaystyle 4\times3=12=6\times2\); O \(\displaystyle 5\times2=10\) on the left, \(\displaystyle 4\times1+6\times1=10\) on the right — balanced.In \(\displaystyle \mathrm{NH_3} \), hydrogen is bonded to the non-metal nitrogen, so hydrogen is \(\displaystyle +1 \); neutrality gives nitrogen \(\displaystyle x + 3(+1) = 0 \Rightarrow x = -3 \). In \(\displaystyle \mathrm{O_2} \), oxygen is a free element, \(\displaystyle 0 \). In \(\displaystyle \mathrm{NO} \), oxygen is \(\displaystyle -2 \), so nitrogen is \(\displaystyle x + (-2) = 0 \Rightarrow x = +2 \). In \(\displaystyle \mathrm{H_2O} \), hydrogen stays \(\displaystyle +1 \) and oxygen stays \(\displaystyle -2 \) — hydrogen does not change across this whole equation, which is the detail easiest to miss when scanning for what reacted.\[4\,\underset{-3}{\mathrm{N}} \rightarrow 4\,\underset{+2}{\mathrm{N}} \quad (\text{loses } 5\,e^- \text{ each, } 20\,e^- \text{ total, oxidized}) \qquad\qquad 10\,\underset{0}{\mathrm{O}} \rightarrow 10\,\underset{-2}{\mathrm{O}} \quad (\text{gains } 2\,e^- \text{ each, } 20\,e^- \text{ total, reduced}) \](The $\displaystyle 10$ oxygen atoms come from the $\displaystyle 5$ molecules of \(\displaystyle \mathrm{O_2} \).) $\displaystyle 20$ electrons lost by nitrogen equal $\displaystyle 20$ gained by oxygen. Nitrogen is oxidized and oxygen is reduced — redox, even though hydrogen is a bystander throughout.Answer: All five are redox reactions. (a) Cu is reduced (\(\displaystyle +2\to0\)), H is oxidized (\(\displaystyle 0\to+1\)). (b) Fe is reduced (\(\displaystyle +3\to0\)), C is oxidized (\(\displaystyle +2\to+4\)). (c) B is reduced (\(\displaystyle +3\to-3\)), H is oxidized (\(\displaystyle -1\to+1\)); Li, Al, Cl are unchanged. (d) K is oxidized (\(\displaystyle 0\to+1\)), F is reduced (\(\displaystyle 0\to-1\)). (e) N is oxidized (\(\displaystyle -3\to+2\)), O is reduced (\(\displaystyle 0\to-2\)); H is unchanged.

Assign an oxidation number to every atom, then watch which ones actually change — that is what "redox" means.The equation as given, \[\text{H}_2\text{O(s)} + \text{F}_2\text{(g)} \rightarrow \text{HF(g)} + \text{HOF(g)} \] is already balanced: count atoms on each side. Left side has $\displaystyle 2$ H, $\displaystyle 1$ O, $\displaystyle 2$ F. Right side: HF contributes $\displaystyle 1$ H and $\displaystyle 1$ F, HOF contributes $\displaystyle 1$ H, $\displaystyle 1$ O and $\displaystyle 1$ F, giving a total of $\displaystyle 2$ H, $\displaystyle 1$ O, $\displaystyle 2$ F. Every atom count matches, so no coefficients need adjusting.Step $\displaystyle 1$: Oxidation number of hydrogen. In \(\displaystyle \text{H}_2\text{O}\), oxygen is more electronegative than hydrogen, so H is assigned \(\displaystyle +1\). In \(\displaystyle \text{HF}\), fluorine is more electronegative than hydrogen, so H is again \(\displaystyle +1\). In \(\displaystyle \text{HOF}\) (hydrogen–oxygen–fluorine, bonded as H–O–F), H is bonded only to the more electronegative O, so H is once more \(\displaystyle +1\). Hydrogen stays at \(\displaystyle +1\) throughout — it does not change.Step $\displaystyle 2$: Oxidation number of oxygen. Oxygen is almost always \(\displaystyle -2\) unless bonded to fluorine (which is more electronegative than oxygen) or in a peroxide. In \(\displaystyle \text{H}_2\text{O}\), O is bonded to H only, so O \(\displaystyle = -2\). In \(\displaystyle \text{HOF}\), the oxygen atom is bonded to H (less electronegative than O) and to F (more electronegative than O). The bond to H contributes \(\displaystyle -1\) to O's count and the bond to F contributes \(\displaystyle +1\), giving O \(\displaystyle = (-1) + (+1) = -2\). So oxygen is \(\displaystyle -2\) on both sides too — no change here either.A quick check: the oxidation numbers in a neutral molecule must sum to zero. In HOF: H \(\displaystyle (+1)\) + O \(\displaystyle (-2)\) + F \(\displaystyle (?) = 0\), so F \(\displaystyle = +1\). That is consistent with assigning O \(\displaystyle =-2\) here.Step $\displaystyle 3$: Oxidation number of fluorine — this is the step people skip. In \(\displaystyle \text{F}_2\), fluorine is bonded only to another fluorine atom (identical electronegativity), so by definition an element in its free form has oxidation number \(\displaystyle 0\).On the product side there are two different fluorine environments, and treating them as the same is the mistake to avoid: Step $\displaystyle 4$: Compare oxidation numbers before and after.\[\underset{0}{\text{F}_2} \;\longrightarrow\; \underset{-1}{\text{F(in HF)}} \qquad \text{(gain of 1 electron per F atom — reduction)} \] \[\underset{0}{\text{F}_2} \;\longrightarrow\; \underset{+1}{\text{F(in HOF)}} \qquad \text{(loss of 1 electron per F atom — oxidation)} \]Of the two fluorine atoms that started in \(\displaystyle \text{F}_2\) at oxidation number \(\displaystyle 0\), one ends up at \(\displaystyle -1\) (in HF) and the other ends up at \(\displaystyle +1\) (in HOF). Since oxidation numbers change — one atom is reduced, the other is oxidized, and both changes come from the same element in the same starting substance — this satisfies the definition of a redox reaction. (This particular pattern, where one element is simultaneously oxidized and reduced starting from a single species, is called disproportionation — a subclass of redox reactions, not an exception to them.)Hydrogen \(\displaystyle (+1 \to +1)\) and oxygen \(\displaystyle (-2 \to -2)\) undergo no change in oxidation number, so they are spectators to the electron transfer; the entire redox action is carried by fluorine.Answer: Yes, it is a redox reaction. Oxidation numbers: H stays \(\displaystyle +1\) and O stays \(\displaystyle -2\) throughout, but fluorine changes from \(\displaystyle 0\) in \(\displaystyle \text{F}_2\) to \(\displaystyle -1\) in HF (reduction, gain of an electron) and to \(\displaystyle +1\) in HOF (oxidation, loss of an electron). Because the same element (F) is simultaneously oxidized and reduced, this is a redox reaction — specifically a disproportionation reaction of fluorine.

The all-O-is-\(\displaystyle -2\) rule breaks down the moment a molecule has an O–O bond — and \(\displaystyle \text{H}_2\text{SO}_5\) has exactly that.The rule you normally use is: in a neutral molecule the oxidation numbers of every atom, weighted by how many of that atom are present, add up to zero; in an ion they add up to the ion's charge. Hydrogen is taken as \(\displaystyle +1\) and oxygen as \(\displaystyle -2\) — but only when oxygen is behaving as an ordinary oxide ion. When two oxygens are joined to each other (a peroxide linkage, \(\displaystyle -\text{O}-\text{O}-\)), each of those two oxygens is \(\displaystyle -1\), not \(\displaystyle -2\), because the O–O bond is between two atoms of identical electronegativity and contributes nothing to either atom's oxidation number.Oxidation number of S in \(\displaystyle \text{H}_2\text{SO}_5\) — the naive calculation, and why it failsIf you (wrongly) treat all five oxygens as ordinary \(\displaystyle -2\) oxide oxygens:\[2(+1) + x + 5(-2) = 0 \implies x = +8 \]This is the fallacy the question is pointing at. Sulphur sits in group $\displaystyle 16$, and the highest oxidation number any group-$\displaystyle 16$ element can show is \(\displaystyle +6\) (equal to the number of valence electrons available to lose or share). An oxidation number of \(\displaystyle +8\) for sulphur is not possible — so a calculation that produces it is not describing the real molecule.The fix is the actual structure of \(\displaystyle \text{H}_2\text{SO}_5\) (peroxymonosulfuric acid, also called Caro's acid). Sulphur sits at the centre, bonded to: two oxygens joined to it by double bonds (\(\displaystyle \text{S}=\text{O}\)), one oxygen joined to it and to a hydrogen (\(\displaystyle \text{S}-\text{O}-\text{H}\)), and one oxygen that is joined to sulphur on one side and, instead of going straight to a second hydrogen, is joined to another oxygen which then carries the second hydrogen — i.e. a \(\displaystyle -\text{O}-\text{O}-\text{H}\) peroxo chain hanging off sulphur. So of the five oxygens, three are ordinary (each \(\displaystyle -2\)) and two form a peroxide bridge (each \(\displaystyle -1\)).Redo the sum with that correction:\[2(+1) + x + \underbrace{3(-2)}_{\text{ordinary O}} + \underbrace{2(-1)}_{\text{peroxide O}} = 0 \] \[2 + x - 6 - 2 = 0 \implies x = +6 \]Sulphur is \(\displaystyle +6\) — consistent with its maximum possible value. The aside worth remembering: whenever an oxoacid's "naive" oxidation number comes out higher than the element's group number, suspect a peroxide (or superoxide) linkage before assuming the arithmetic is wrong.Oxidation number of Cr in \(\displaystyle \text{Cr}_2\text{O}_7^{2-}\) (the dichromate ion)Here every oxygen genuinely is an ordinary oxide oxygen (\(\displaystyle -2\)) — there is no O–O bond in this ion, so no correction is needed. Let \(\displaystyle x\) be the oxidation number of chromium. The ion carries a charge of \(\displaystyle -2\), so the oxidation numbers must sum to \(\displaystyle -2\), not zero:\[2x + 7(-2) = -2 \] \[2x - 14 = -2 \implies 2x = 12 \implies x = +6 \]Chromium is \(\displaystyle +6\), which is its maximum oxidation state (it has $\displaystyle 6$ valence electrons in group $\displaystyle 6$) — so this result needs no fallacy-hunting; it is simply consistent.Structure of \(\displaystyle \text{Cr}_2\text{O}_7^{2-}\): picture two \(\displaystyle \text{CrO}_4\) tetrahedra (the same tetrahedral unit found in the yellow chromate ion, \(\displaystyle \text{CrO}_4^{2-}\)) sharing one corner oxygen between them. That single shared oxygen bridges the two chromium atoms (\(\displaystyle \text{Cr}-\text{O}-\text{Cr}\)); each chromium keeps three more oxygens of its own, not shared with the other chromium. This is exactly the geometry you'd get by fusing two chromate tetrahedra at one vertex and squeezing out a water molecule.Oxidation number of N in \(\displaystyle \text{NO}_3^-\) (the nitrate ion)Again, no O–O bonds, so oxygen is an ordinary \(\displaystyle -2\) here too. Let \(\displaystyle x\) be nitrogen's oxidation number; the ion's charge is \(\displaystyle -1\):\[x + 3(-2) = -1 \] \[x - 6 = -1 \implies x = +5 \]Nitrogen is \(\displaystyle +5\), which is its maximum oxidation state (group $\displaystyle 15$, five valence electrons) — again a consistent result, not a fallacy.Structure of \(\displaystyle \text{NO}_3^-\): nitrogen sits at the centre of a flat (planar) triangle with one oxygen at each corner — a trigonal planar ion. If you tried to draw one fixed Lewis structure, you'd draw one \(\displaystyle \text{N}=\text{O}\) double bond and two \(\displaystyle \text{N}-\text{O}\) single bonds (each of those two carrying part of the negative charge). But experimentally all three nitrogen–oxygen bonds are the same length, shorter than a single bond and longer than a true double bond. That happens because the ion is a resonance hybrid of three equivalent structures — the "extra" double-bond character is shared equally among all three N–O bonds rather than sitting fixed on one of them.Answer: In \(\displaystyle \text{H}_2\text{SO}_5\), sulphur is \(\displaystyle +6\) (a naive count assuming every oxygen is \(\displaystyle -2\) wrongly gives \(\displaystyle +8\), which is impossible for sulphur; the fallacy is resolved by recognising that two of the five oxygens form a peroxide, \(\displaystyle -\text{O}-\text{O}-\), bridge and are therefore \(\displaystyle -1\) each, not \(\displaystyle -2\)). In \(\displaystyle \text{Cr}_2\text{O}_7^{2-}\), chromium is \(\displaystyle +6\) — two \(\displaystyle \text{CrO}_4\) tetrahedra joined through one shared bridging oxygen. In \(\displaystyle \text{NO}_3^-\), nitrogen is \(\displaystyle +5\) — a trigonal planar ion whose three N–O bonds are made equivalent by resonance.

The Roman numeral in a compound's name is the metal ion's charge — cross-multiply that charge against the charge on the negative ion to get the formula, so that positive and negative charges add up to zero.For each compound, identify the metal cation (its charge is given by the Roman numeral) and the anion (its charge you must know), then use the crossover rule: the number of cations needed equals the anion's charge (dropping the sign), and the number of anions needed equals the cation's charge (dropping the sign). Reduce the resulting subscripts to the smallest whole-number ratio.(a) Mercury(II) chloride Mercury(II) means the mercury ion is \(\displaystyle \text{Hg}^{2+} \). Chloride is \(\displaystyle \text{Cl}^{-} \). To balance one \(\displaystyle 2+ \) charge, two \(\displaystyle 1- \) charges are needed: \[\text{Hg}^{2+} + 2\,\text{Cl}^{-} \rightarrow \text{HgCl}_2 \] Charge check: \(\displaystyle (+2) + 2(-1) = 0 \). Formula: \(\displaystyle \text{HgCl}_2 \) (mercury(II) chloride).(b) Nickel(II) sulphate Nickel(II) is \(\displaystyle \text{Ni}^{2+} \). Sulphate is the polyatomic ion \(\displaystyle \text{SO}_4^{2-} \) (one sulphur, four oxygens, overall charge \(\displaystyle 2- \)). The charges already match one-to-one, \(\displaystyle 2+ \) with \(\displaystyle 2- \), so no subscripts are needed on either ion: \[\text{Ni}^{2+} + \text{SO}_4^{2-} \rightarrow \text{NiSO}_4 \] Charge check: \(\displaystyle (+2) + (-2) = 0 \). Formula: \(\displaystyle \text{NiSO}_4 \) (nickel(II) sulphate).(c) Tin(IV) oxide Tin(IV) is \(\displaystyle \text{Sn}^{4+} \). Oxide is \(\displaystyle \text{O}^{2-} \). One \(\displaystyle \text{Sn}^{4+} \) needs two \(\displaystyle \text{O}^{2-}\) ions to cancel its charge (crossing over $\displaystyle 4$ and $\displaystyle 2$ gives a ratio $\displaystyle 4$:$\displaystyle 2$, which reduces to $\displaystyle 2$:$\displaystyle 1$ for oxide:tin, i.e. \(\displaystyle \text{SnO}_2 \) rather than the unreduced \(\displaystyle \text{Sn}_2\text{O}_4 \) — always reduce to lowest terms): \[\text{Sn}^{4+} + 2\,\text{O}^{2-} \rightarrow \text{SnO}_2 \] Charge check: \(\displaystyle (+4) + 2(-2) = 0 \). Formula: \(\displaystyle \text{SnO}_2 \) (tin(IV) oxide).(d) Thallium(I) sulphate Thallium(I) is \(\displaystyle \text{Tl}^{+} \), a singly charged cation — this is the step people slip on, because sulphate compounds are usually pictured with a $\displaystyle 2$+ metal (as in part b), but here the metal carries only a single positive charge, so it takes two thallium ions to balance one sulphate. Sulphate is \(\displaystyle \text{SO}_4^{2-} \). To cancel a \(\displaystyle 2-\) charge using \(\displaystyle 1+\) ions, two thallium ions are required: \[2\,\text{Tl}^{+} + \text{SO}_4^{2-} \rightarrow \text{Tl}_2\text{SO}_4 \] Charge check: \(\displaystyle 2(+1) + (-2) = 0 \). Formula: \(\displaystyle \text{Tl}_2\text{SO}_4 \) (thallium(I) sulphate).(e) Iron(III) sulphate Iron(III) is \(\displaystyle \text{Fe}^{3+} \). Sulphate is \(\displaystyle \text{SO}_4^{2-} \). The charges are $\displaystyle 3$ and $\displaystyle 2$, which do not cancel one-to-one, so cross-multiply: take as many iron ions as the sulphate's charge magnitude ($\displaystyle 2$) and as many sulphate ions as the iron's charge magnitude ($\displaystyle 3$), giving the smallest whole-number ratio \(\displaystyle 2:3 \): \[2\,\text{Fe}^{3+} + 3\,\text{SO}_4^{2-} \rightarrow \text{Fe}_2(\text{SO}_4)_3 \] Charge check: \(\displaystyle 2(+3) + 3(-2) = 6 - 6 = 0 \). The sulphate ion is enclosed in brackets before the subscript $\displaystyle 3$ because the subscript applies to the whole \(\displaystyle \text{SO}_4 \) group (one sulphur and four oxygens), not to a single oxygen — writing \(\displaystyle \text{Fe}_2\text{SO}_{4\,3}\) or \(\displaystyle \text{Fe}_2\text{SO}_{12}\) would misstate how many sulphur atoms are present. Formula: \(\displaystyle \text{Fe}_2(\text{SO}_4)_3 \) (iron(III) sulphate).(f) Chromium(III) oxide Chromium(III) is \(\displaystyle \text{Cr}^{3+} \). Oxide is \(\displaystyle \text{O}^{2-} \). Again the charges are $\displaystyle 3$ and $\displaystyle 2$, so cross-multiply: $\displaystyle 2$ chromium ions (to supply \(\displaystyle 2 \times 3 = 6 \) positive charge) and $\displaystyle 3$ oxide ions (to supply \(\displaystyle 3 \times 2 = 6 \) negative charge): \[2\,\text{Cr}^{3+} + 3\,\text{O}^{2-} \rightarrow \text{Cr}_2\text{O}_3 \] Charge check: \(\displaystyle 2(+3) + 3(-2) = 0 \). Formula: \(\displaystyle \text{Cr}_2\text{O}_3 \) (chromium(III) oxide).Answer: (a) \(\displaystyle \text{HgCl}_2 \) (b) \(\displaystyle \text{NiSO}_4 \) (c) \(\displaystyle \text{SnO}_2 \) (d) \(\displaystyle \text{Tl}_2\text{SO}_4 \) (e) \(\displaystyle \text{Fe}_2(\text{SO}_4)_3 \) (f) \(\displaystyle \text{Cr}_2\text{O}_3 \)

The oxidation number of every atom in a molecule must add up to zero — and once you fix hydrogen at \(\displaystyle +1\) and oxygen/chlorine at \(\displaystyle -2\)/\(\displaystyle -1\), you can hand-pick a short family of compounds that walks carbon, or nitrogen, through every single state in between.The rule that does all the work here: for a neutral molecule, \[\sum (\text{number of atoms of each element}) \times (\text{its oxidation number}) = 0. \] Two anchors are fixed because of electronegativity: hydrogen is less electronegative than C, N, O and Cl, so it is always taken as \(\displaystyle +1\); oxygen is more electronegative than everything on this list except fluorine, so it is \(\displaystyle -2\); chlorine, being more electronegative than carbon, is \(\displaystyle -1\). A bond between two atoms of the same element (a C–C or N–N bond) contributes zero, because the shared electron pair splits evenly between identical atoms — this is exactly the step people forget when a molecule has more than one carbon, and it's why ethane's two identical carbons don't each "steal" an oxidation number from the other.Carbon: −$\displaystyle 4$ to +$\displaystyle 4$Two matched series of chlorinated hydrocarbons cover all nine states, because swapping one H for one Cl always raises carbon's oxidation number by 2.Series $\displaystyle 1$ — one carbon (each swap changes the single carbon's state by $\displaystyle 2$, so this series lands on every even value): Series $\displaystyle 2$ — two carbons, one Cl added to each carbon per step (this raises the shared, symmetric oxidation number by $\displaystyle 2$ each time, landing on every odd value): Nitrogen: −$\displaystyle 3$ to +$\displaystyle 5$Nitrogen's own oxide/hydride ladder does the same job in one family each side of zero. Every entry above was checked the same way — write the neutral-molecule sum, plug in H = \(\displaystyle +1\), O = \(\displaystyle -2\), Cl = \(\displaystyle -1\), and treat any bond between two atoms of the same element as contributing \(\displaystyle 0\) — so the list is not just plausible, each oxidation number is forced by that one equation.Answer: Carbon (−$\displaystyle 4$ → +$\displaystyle 4$): \(\displaystyle CH_4\) (−$\displaystyle 4$), \(\displaystyle C_2H_6\) (−$\displaystyle 3$), \(\displaystyle CH_3Cl\) (−$\displaystyle 2$), \(\displaystyle C_2H_4Cl_2\) (−$\displaystyle 1$), \(\displaystyle CH_2Cl_2\) ($\displaystyle 0$), \(\displaystyle C_2H_2Cl_4\) (+$\displaystyle 1$), \(\displaystyle CHCl_3\) (+$\displaystyle 2$), \(\displaystyle C_2Cl_6\) (+$\displaystyle 3$), \(\displaystyle CCl_4\) (+$\displaystyle 4$). Nitrogen (−$\displaystyle 3$ → +$\displaystyle 5$): \(\displaystyle NH_3\) (−$\displaystyle 3$), \(\displaystyle N_2H_4\) (−$\displaystyle 2$), \(\displaystyle NH_2OH\) (−$\displaystyle 1$), \(\displaystyle N_2\) ($\displaystyle 0$), \(\displaystyle N_2O\) (+$\displaystyle 1$), \(\displaystyle NO\) (+$\displaystyle 2$), \(\displaystyle N_2O_3\) (+$\displaystyle 3$), \(\displaystyle NO_2\) (+$\displaystyle 4$), \(\displaystyle N_2O_5\) (+$\displaystyle 5$).

A species can be BOTH an oxidising and a reducing agent only if the reactive atom's oxidation number sits strictly between the lowest and highest values that element ever reaches; if the atom already sits at one of those extremes, it can only move the other way.To act as an oxidising agent an atom's oxidation number must be able to fall (it gets reduced, pulling electrons from something else). To act as a reducing agent the same atom's oxidation number must be able to rise (it gets oxidised, giving electrons away). An atom already at the maximum oxidation number the element ever shows has nowhere higher to go, so it can only fall — oxidant only. An atom already at the minimum has nowhere lower to go, so it can only rise — reductant only. An atom sitting in between can go either way. Work out where S in \(\displaystyle \text{SO}_2 \), O in \(\displaystyle \text{H}_2\text{O}_2 \), O in \(\displaystyle \text{O}_3 \), and N in \(\displaystyle \text{HNO}_3 \) actually sit.Oxidation number is found from the rule that the oxidation numbers in a neutral molecule add to zero, taking O as \(\displaystyle -2\) and H as \(\displaystyle +1\) (aside: peroxide oxygen is the one standard exception to "O is always \(\displaystyle -2\)" — that exception is exactly what this question turns on).Sulphur in \(\displaystyle \text{SO}_2 \). Let \(\displaystyle x\) be sulphur's oxidation number: \[x + 2(-2) = 0 \implies x = +4 \] Sulphur's range runs from \(\displaystyle -2\) (in \(\displaystyle \text{H}_2\text{S}\)) up to \(\displaystyle +6\) (in \(\displaystyle \text{H}_2\text{SO}_4\)). \(\displaystyle +4\) sits strictly inside that range, so S in \(\displaystyle \text{SO}_2\) can move either way.Oxidised, \(\displaystyle +4 \to +6\) — \(\displaystyle \text{SO}_2\) acting as a reducing agent: \[2\text{KMnO}_4 + 5\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 2\text{H}_2\text{SO}_4 \] Balance it by electrons: Mn falls \(\displaystyle +7 \to +2\) ($\displaystyle 5$-electron gain per atom, $\displaystyle 10$ electrons for the two Mn); S rises \(\displaystyle +4 \to +6\) ($\displaystyle 2$-electron loss per atom), so $\displaystyle 5$ sulphur atoms are needed to supply the $\displaystyle 10$ electrons the two manganese atoms take — that fixes the $\displaystyle 5$ : $\displaystyle 2$ ratio, and the rest ($\displaystyle 2$ K, $\displaystyle 20$ O, $\displaystyle 4$ H on each side) falls into place automatically. The products are potassium sulphate, manganese(II) sulphate, and sulphuric acid.Reduced, \(\displaystyle +4 \to 0\) — \(\displaystyle \text{SO}_2\) acting as an oxidising agent: \[2\text{H}_2\text{S} + \text{SO}_2 \rightarrow 3\text{S} + 2\text{H}_2\text{O} \] Sulphur in \(\displaystyle \text{H}_2\text{S}\) (\(\displaystyle -2\)) is oxidised to elemental sulphur (\(\displaystyle 0\)), a $\displaystyle 2$-electron loss per atom; sulphur in \(\displaystyle \text{SO}_2\) (\(\displaystyle +4\)) is reduced to elemental sulphur (\(\displaystyle 0\)), a $\displaystyle 4$-electron gain per atom. Two \(\displaystyle \text{H}_2\text{S}\) (losing \(\displaystyle 2\times2=4\) electrons) exactly supply what one \(\displaystyle \text{SO}_2\) needs, and both sulphurs end up as the same product, solid sulphur — $\displaystyle 2$ S + $\displaystyle 1$ S in gives $\displaystyle 3$ S out, and $\displaystyle 4$ H in gives $\displaystyle 2$ water molecules out.Oxygen in \(\displaystyle \text{H}_2\text{O}_2 \). Let \(\displaystyle y\) be the oxidation number of each O atom: \[2(+1) + 2y = 0 \implies y = -1 \] Oxygen's range runs from \(\displaystyle -2\) (in water or any oxide) up to \(\displaystyle 0\) (in \(\displaystyle \text{O}_2\)). \(\displaystyle -1\) sits strictly inside, so \(\displaystyle \text{H}_2\text{O}_2\) can also go either way.Reduced, \(\displaystyle -1 \to -2\) — \(\displaystyle \text{H}_2\text{O}_2\) acting as an oxidising agent: \[\text{PbS} + 4\text{H}_2\text{O}_2 \rightarrow \text{PbSO}_4 + 4\text{H}_2\text{O} \] Sulphur in lead sulphide (\(\displaystyle -2\)) is oxidised to sulphur in sulphate (\(\displaystyle +6\)), an $\displaystyle 8$-electron loss; peroxide oxygen falls \(\displaystyle -1 \to -2\), a $\displaystyle 2$-electron gain per \(\displaystyle \text{H}_2\text{O}_2\) molecule, so $\displaystyle 4$ molecules are needed to absorb the $\displaystyle 8$ electrons sulphur gives up. This is the reaction that turns blackened lead sulphide back into white lead sulphate — the standard way old paintings are restored.Oxidised, \(\displaystyle -1 \to 0\) — \(\displaystyle \text{H}_2\text{O}_2\) acting as a reducing agent: \[2\text{KMnO}_4 + 5\text{H}_2\text{O}_2 + 3\text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 8\text{H}_2\text{O} + 5\text{O}_2 \] Mn falls \(\displaystyle +7 \to +2\) ($\displaystyle 5$-electron gain per atom, $\displaystyle 10$ for two Mn); peroxide oxygen rises \(\displaystyle -1 \to 0\) as oxygen gas, a $\displaystyle 2$-electron loss per \(\displaystyle \text{H}_2\text{O}_2\), so $\displaystyle 5$ molecules supply the $\displaystyle 10$ electrons the manganese atoms take. (Aside: this is also why \(\displaystyle \text{H}_2\text{O}_2\) instantly turns acidified purple \(\displaystyle \text{KMnO}_4\) colourless — a standard test for peroxide.)Oxygen in \(\displaystyle \text{O}_3 \). Ozone is an element, so — exactly like \(\displaystyle \text{O}_2\) — every oxygen atom in it has oxidation number $\displaystyle 0$ by definition. That is the top of oxygen's range, not the middle: oxygen has no everyday oxidation state above $\displaystyle 0$ (it only goes positive bonded to fluorine, which is not the case here). An oxygen atom in \(\displaystyle \text{O}_3\) therefore has nowhere to go but down — it can be reduced but it cannot be oxidised. Ozone can only be an oxidising agent. \[2\text{KI} + \text{O}_3 + \text{H}_2\text{O} \rightarrow 2\text{KOH} + \text{I}_2 + \text{O}_2 \] Iodide is oxidised, \(\displaystyle -1 \to 0\), each iodine losing $\displaystyle 1$ electron, so the two iodide ions together lose $\displaystyle 2$ electrons. One of the three oxygen atoms of ozone is reduced, \(\displaystyle 0 \to -2\) (ending up in potassium hydroxide), a $\displaystyle 2$-electron gain, while the other two oxygen atoms leave unchanged as oxygen gas — which is why the equation needs only one \(\displaystyle \text{O}_3\) and produces exactly one \(\displaystyle \text{O}_2\) alongside it. This is the starch–iodide test used to detect ozone.Nitrogen in \(\displaystyle \text{HNO}_3 \). Let \(\displaystyle z\) be nitrogen's oxidation number: \[(+1) + z + 3(-2) = 0 \implies z = +5 \] Nitrogen has $\displaystyle 5$ valence electrons, so \(\displaystyle +5\) — every valence electron given up — is the ceiling nitrogen can reach; N in \(\displaystyle \text{HNO}_3\) is already sitting at that ceiling. It has nowhere higher to go, so it cannot be oxidised, only reduced (to \(\displaystyle +4\) in \(\displaystyle \text{NO}_2\), \(\displaystyle +2\) in \(\displaystyle \text{NO}\), or lower, depending on concentration and what it reacts with). Nitric acid can only be an oxidising agent. \[\text{Cu} + 4\text{HNO}_3(\text{conc.}) \rightarrow \text{Cu}(\text{NO}_3)_2 + 2\text{NO}_2 + 2\text{H}_2\text{O} \] Copper is oxidised, \(\displaystyle 0 \to +2\), losing $\displaystyle 2$ electrons; nitrogen is reduced, \(\displaystyle +5 \to +4\) (as nitrogen dioxide gas), a $\displaystyle 1$-electron gain per N atom, so two nitrogen atoms are reduced for every copper atom oxidised, matching the $\displaystyle 2$ electrons copper releases. (Aside: the other two nitrate ions among the "$\displaystyle 4$ \(\displaystyle \text{HNO}_3\)" are spectators — they end up unchanged as \(\displaystyle \text{NO}_3^-\) in \(\displaystyle \text{Cu(NO}_3)_2\); don't read "$\displaystyle 4$ \(\displaystyle \text{HNO}_3\)" as "$\displaystyle 4$ nitrogens reduced.")Answer: In \(\displaystyle \text{SO}_2\) (S at \(\displaystyle +4\), sulphur's range being \(\displaystyle -2\) to \(\displaystyle +6\)) and in \(\displaystyle \text{H}_2\text{O}_2\) (O at \(\displaystyle -1\), oxygen's range being \(\displaystyle -2\) to \(\displaystyle 0\)), the reactive atom sits at an intermediate oxidation state, so it can rise (giving reducing behaviour, e.g. \(\displaystyle \text{SO}_2 \to \text{SO}_4^{2-}\) or \(\displaystyle \text{H}_2\text{O}_2 \to \text{O}_2\)) or fall (giving oxidising behaviour, e.g. \(\displaystyle \text{SO}_2 \to \text{S}\) or \(\displaystyle \text{H}_2\text{O}_2 \to \text{H}_2\text{O}\)) — both are seen. In \(\displaystyle \text{O}_3\) the oxygen is already at $\displaystyle 0$, its highest attainable state, and in \(\displaystyle \text{HNO}_3\) the nitrogen is already at \(\displaystyle +5\), its highest possible state; both atoms can only fall (be reduced), never rise, so ozone and nitric acid can act only as oxidising agents.

A balanced equation only guarantees that mass is conserved — it says nothing about which atom of which reactant ends up in which product. That second question needs an isotope tracer, and the "expanded" equations are written specifically to record what the tracer experiments found.Reaction (a): photosynthesisName every formula first: \(\displaystyle \text{CO}_2 \) is carbon dioxide, \(\displaystyle \text{H}_2\text{O} \) is water, \(\displaystyle \text{C}_6\text{H}_{12}\text{O}_6 \) is glucose, and \(\displaystyle \text{O}_2 \) is dioxygen (oxygen gas).The textbook's compact form, \[6\text{CO}_2(g) + 6\text{H}_2\text{O}(l) \rightarrow \text{C}_6\text{H}_{12}\text{O}_6(aq) + 6\text{O}_2(g), \] is balanced ($\displaystyle 6$ C, $\displaystyle 12$ H, $\displaystyle 18$ O on each side), but balancing alone cannot tell you where the oxygen atoms in the released \(\displaystyle \text{O}_2 \) came from — \(\displaystyle \text{CO}_2 \) and \(\displaystyle \text{H}_2\text{O} \) both supply oxygen, and once you write a single pooled count of O atoms you have thrown away the information about which molecule each one started in. Going purely by the compact equation, a student could just as easily guess (wrongly) that the \(\displaystyle \text{O}_2 \) is released by breaking the C–O bonds of \(\displaystyle \text{CO}_2 \), by analogy with respiration running backward.That guess was tested directly: plants were fed water in which the oxygen atom was replaced by the heavy isotope \(\displaystyle ^{18}\text{O} \) (i.e. \(\displaystyle \text{H}_2\,^{18}\text{O} \)), together with ordinary \(\displaystyle \text{CO}_2 \). The \(\displaystyle \text{O}_2 \) gas that came off was enriched in \(\displaystyle ^{18}\text{O} \) — proof that the evolved oxygen originates in water, not in carbon dioxide.Now write the equation so that this fact is visible instead of hidden: \[6\text{CO}_2(g) + 12\text{H}_2\text{O}(l) \rightarrow \text{C}_6\text{H}_{12}\text{O}_6(aq) + 6\text{H}_2\text{O}(l) + 6\text{O}_2(g) \] Check that it still balances the same way as before, just with a bigger water count: C: \(\displaystyle 6=6\). H: left \(\displaystyle 12\times2=24\); right \(\displaystyle 12+6\times2=24\). O: left \(\displaystyle 6\times2+12\times1=24\); right \(\displaystyle 6+6\times1+6\times2=6+6+12=24\). All three atoms balance, and if you cancel the $\displaystyle 6$ water molecules that appear on both sides you get back exactly the compact equation — so this is not a different reaction, only a different bookkeeping of the same one.The reason for that bookkeeping: all $\displaystyle 12$ oxygen atoms needed to make the $\displaystyle 6$ \(\displaystyle \text{O}_2 \) molecules are drawn from the $\displaystyle 12$ water molecules on the left (matching the \(\displaystyle ^{18}\text{O} \) result), while the $\displaystyle 6$ oxygen atoms of \(\displaystyle \text{CO}_2 \) supply the oxygen that ends up in glucose and in the $\displaystyle 6$ water molecules regenerated as a by-product on the right. Writing "$\displaystyle 12$ \(\displaystyle \text{H}_2\text{O} \) in, $\displaystyle 6$ \(\displaystyle \text{H}_2\text{O} \) back out" is the only way to show, atom for atom, that water is consumed twice as fast as the net equation implies and that none of its oxygen survives into the glucose.Reaction (b): ozone and hydrogen peroxideName the formulas: \(\displaystyle \text{O}_3 \) is ozone, \(\displaystyle \text{H}_2\text{O}_2 \) is hydrogen peroxide.\[\text{O}_3(g) + \text{H}_2\text{O}_2(l) \rightarrow \text{H}_2\text{O}(l) + \text{O}_2(g) + \text{O}_2(g) \] Atom count: O, left \(\displaystyle 3+2=5\), right \(\displaystyle 1+2+2=5\); H, left \(\displaystyle 2\), right \(\displaystyle 2\). Balanced — and it is the same statement as "\(\displaystyle \rightarrow \text{H}_2\text{O} + 2\text{O}_2 \)", just with the coefficient "$\displaystyle 2$" split into two separate \(\displaystyle \text{O}_2 \) terms.Splitting it matters because the two \(\displaystyle \text{O}_2 \) molecules are not chemically equivalent, which oxidation numbers make precise. In \(\displaystyle \text{O}_3 \), an element in its own free form, oxygen is at oxidation number \(\displaystyle 0\). The trap here is treating every \(\displaystyle \text{O–O}\) bonded species like a normal oxide: oxygen in \(\displaystyle \text{H}_2\text{O}_2 \) is \(\displaystyle -1\), not \(\displaystyle -2\), because it is a peroxide linkage — two oxygens sharing the reduction between them rather than one oxygen fully reduced. In the products, oxygen is \(\displaystyle -2\) in \(\displaystyle \text{H}_2\text{O} \) and back to \(\displaystyle 0\) in \(\displaystyle \text{O}_2 \).Follow the electrons: the two oxygen atoms of \(\displaystyle \text{H}_2\text{O}_2 \) go from \(\displaystyle -1\) to \(\displaystyle 0\) as they become one \(\displaystyle \text{O}_2 \) molecule — each loses one electron, an oxidation, \(\displaystyle 2\) electrons lost in total. One "odd" oxygen atom of \(\displaystyle \text{O}_3 \) goes from \(\displaystyle 0\) to \(\displaystyle -2\) as it becomes the oxygen of \(\displaystyle \text{H}_2\text{O} \) — a reduction, gaining \(\displaystyle 2\) electrons. The other two oxygen atoms of \(\displaystyle \text{O}_3 \) stay at \(\displaystyle 0\) and simply pair off as the second \(\displaystyle \text{O}_2 \) molecule, untouched. The \(\displaystyle 2\) electrons lost by peroxide exactly equal the \(\displaystyle 2\) electrons gained by ozone's odd atom, so the redox bookkeeping closes. This is exactly what "\(\displaystyle \text{H}_2\text{O} + \text{O}_2 + \text{O}_2 \)" is recording: one \(\displaystyle \text{O}_2 \) is untouched ozone oxygen, the other \(\displaystyle \text{O}_2 \) is oxidised peroxide oxygen, and the water's single oxygen atom is the one atom of ozone that got reduced — none of that distinction survives if you write it as the pooled term "\(\displaystyle 2\text{O}_2\)".The technique to find the actual pathBoth explanations above rest on the same experimental method: isotopic labelling. A reactant is prepared with an oxygen atom replaced by a distinguishable isotope — most often the stable heavy isotope \(\displaystyle ^{18}\text{O} \) (occasionally a radioactive isotope, tracked with a Geiger counter instead) — and the products are then analysed, by mass spectrometry for a stable isotope, to see which product molecule carries the label. Labelling the water in reaction (a) showed the \(\displaystyle ^{18}\text{O} \) coming out in \(\displaystyle \text{O}_2 \), proving water is the oxygen source; labelling the ozone in reaction (b) (using \(\displaystyle ^{18}\text{O}_3 \)) shows the label appearing in the water product and only in one of the two \(\displaystyle \text{O}_2 \) fractions, proving ozone's odd atom becomes the water oxygen while the other \(\displaystyle \text{O}_2 \) is peroxide-derived. A balanced equation alone can never distinguish these possibilities, because balancing only counts atoms, not their identity — isotopic labelling is what lets you follow an individual atom's identity through the reaction.**Answer: The compact equations are correctly balanced but treat every oxygen atom as interchangeable, so they cannot show which reactant supplies which product atom. The expanded forms, \(\displaystyle 6\text{CO}_2+12\text{H}_2\text{O}\rightarrow \text{C}_6\text{H}_{12}\text{O}_6+6\text{H}_2\text{O}+6\text{O}_2\) and \(\displaystyle \text{O}_3+\text{H}_2\text{O}_2\rightarrow \text{H}_2\text{O}+\text{O}_2+\text{O}_2\), are the same mass-balanced reactions written to reflect the actual atom path established by experiment: in (a) all the oxygen released as \(\displaystyle \text{O}_2\) comes from water, not \(\displaystyle \text{CO}_2\), so $\displaystyle 12$ (not $\displaystyle 6$) water molecules must appear, with $\displaystyle 6$ regenerated as product; in (b) one \(\displaystyle \text{O}_2\) is unreacted ozone oxygen while the other \(\displaystyle \text{O}_2\) is oxidised (\(\displaystyle -1\to0\)) hydrogen-peroxide oxygen, and the lone water molecule carries the ozone oxygen atom that was reduced (\(\displaystyle 0\to-2\)). The technique used to establish and verify such a path is isotopic labelling — replacing an oxygen atom with the traceable isotope \(\displaystyle ^{18}\text{O} \) in one reactant and locating it in the products by mass spectrometry.

A compound turns into a strong oxidising agent when its central atom is sitting in an oxidation state it does not want to hold — it grabs an electron from anything nearby just to fall back to the state it actually prefers.Step $\displaystyle 1$ — find the oxidation number of silver in \(\displaystyle AgF_2\).Fluorine is the most electronegative element there is, so in every single compound it is assigned the oxidation number \(\displaystyle -1\) — there is no atom that can pull electron density away from it, so this value never changes. Let the oxidation number of silver be \(\displaystyle x\). Since \(\displaystyle AgF_2\) is an electrically neutral compound, the oxidation numbers of all atoms must add up to zero: \[x + 2(-1) = 0 \;\implies\; x = +2 \] So in \(\displaystyle AgF_2\), silver is present as \(\displaystyle \mathrm{Ag^{2+}}\), i.e. in the \(\displaystyle +2\) oxidation state.Aside: the step most students skip here is not the arithmetic — it's asking afterward whether that number is actually a comfortable oxidation state for the element, rather than stopping once a number has been produced.Step $\displaystyle 2$ — check whether \(\displaystyle +2\) is comfortable for silver.Silver's ground-state electron configuration is \(\displaystyle [Kr]\,4d^{10}\,5s^{1}\). Removing the single outer \(\displaystyle 5s\) electron gives \(\displaystyle \mathrm{Ag^{+}}\) with configuration \(\displaystyle 4d^{10}\) — a completely filled d-subshell, which is an especially stable arrangement. This is exactly why \(\displaystyle +1\) is silver's normal, everyday oxidation state, seen in familiar compounds such as \(\displaystyle AgNO_3\), \(\displaystyle AgCl\), and \(\displaystyle Ag_2O\).To reach \(\displaystyle \mathrm{Ag^{2+}}\), a second electron has to be pulled out of that already-filled, stable \(\displaystyle 4d^{10}\) shell, leaving \(\displaystyle 4d^{9}\) — a broken, unfilled shell, which is far less stable. Silver has no reason to stay in this state; it "wants" to pick up one electron and drop straight back down to the comfortable \(\displaystyle 4d^{10}\), \(\displaystyle \mathrm{Ag^{+}}\) configuration.Step $\displaystyle 3$ — write the electron-gain (reduction) half-reaction.\[Ag^{2+} + e^{-} \; \rightarrow \; Ag^{+} \]Check the balance: one silver atom appears on each side, and the charges match too — on the left, \(\displaystyle +2\) combined with the electron's \(\displaystyle -1\) gives \(\displaystyle +1\); on the right, the charge is \(\displaystyle +1\). Both atoms and charge balance, so nothing further needs adjusting.Step $\displaystyle 4$ — connect this to oxidising power.An oxidising agent is, by definition, a species that gets reduced by taking an electron away from something else. Because the \(\displaystyle \mathrm{Ag^{2+}}\) inside \(\displaystyle AgF_2\) is so strongly driven to complete exactly this reduction — gain one electron and revert to the stable \(\displaystyle \mathrm{Ag^{+}}\) state — it seizes an electron from whatever it is placed in contact with, oxidising that other substance in the process. This is also precisely why \(\displaystyle AgF_2\) itself is so hard to keep as a stable compound in the first place: the same pull of silver back toward \(\displaystyle +1\) that makes the compound decompose is the pull that makes it snatch electrons so aggressively — i.e. act as a very strong oxidising agent — in the moments it does exist.Answer: In \(\displaystyle AgF_2\), silver is forced into the unusual \(\displaystyle +2\) oxidation state (since F is always \(\displaystyle -1\) and the compound is neutral, \(\displaystyle x + 2(-1) = 0 \Rightarrow x = +2\)). Silver's stable, preferred state is \(\displaystyle +1\), with the extra-stable, fully filled \(\displaystyle 4d^{10}\) configuration, so \(\displaystyle \mathrm{Ag^{2+}}\) (configuration \(\displaystyle 4d^{9}\)) strongly tends to gain one electron and fall back to \(\displaystyle \mathrm{Ag^{+}}\): \(\displaystyle Ag^{2+} + e^{-} \rightarrow Ag^{+}\). This strong drive to pick up an extra electron is exactly what makes \(\displaystyle AgF_2\), wherever it manages to form, act as a very strong oxidising agent.