What are the oxidation states of oxygen in $KO_2$, of oxygen in $H_2O_2$, and of sulphur in $H_2SO_4$ respectively? | NeetCrackers

Home › Chemistry › Q

ChemistryRedox Reactions

What are the oxidation states of oxygen in $KO_2$, of oxygen in $H_2O_2$, and of sulphur in $H_2SO_4$ respectively?

Options

1

$-\frac{1}{2}$, $-1$, $+6$

2

$-1$, $-\frac{1}{2}$, $+6$

3

$-\frac{1}{2}$, $-1$, $-6$

4

$+\frac{1}{2}$, $-1$, $+6$

Correct Answer

$-\frac{1}{2}$, $-1$, $+6$

Solution

1

$KO_2$ (superoxide): K = +1, let O = $x$

$$+1 + 2x = 0 \implies x = -\frac{1}{2}$$

$H_2O_2$ (peroxide): H = +1, let O = $x$

$$2(+1) + 2x = 0 \implies x = -1$$

2

$H_2SO_4$: H = +1, O = -2, let S = $x$

$$2(+1) + x + 4(-2) = 0 \implies 2 + x - 8 = 0 \implies x = +6$$

$KO_2$: O = $-\frac{1}{2}$ (superoxide) | $H_2O_2$: O = $-1$ (peroxide) | $H_2SO_4$: S = $+6$

Theory: Redox Reactions

1. Oxidation State Rules

Oxidation state (OS): hypothetical charge on an atom if all bonds were ionic. Rules in order of priority: (1) Free element: OS = 0. (2) Monoatomic ion: OS = charge. (3) F in compounds: always -1. (4) O in compounds: usually -2 (except peroxides -1, superoxides -1/2, ozonides -1/3, OF2: +2, O2F2: +1). (5) H in compounds: +1 (except metal hydrides: -1, e.g., NaH, CaH2). (6) In ionic compounds: OS = ionic charge. (7) Sum of OS = 0 for neutral molecule; = charge for ions. Electronegativity determines OS: more electronegative atom gets negative OS in a bond.

2. Special Oxidation States of Oxygen

Oxide (O²⁻): OS = -2. Most common. Examples: Na2O, MgO, CaO, Fe2O3. Peroxide (O2²⁻): OS = -1. Examples: H2O2, Na2O2, BaO2. Characterised by O-O single bond. Superoxide (O2⁻): OS = -1/2. Examples: KO2, NaO2, CsO2. Paramagnetic (1 unpaired electron). Ozonide (O3⁻): OS = -1/3. Examples: KO3. Positive oxidation states: OF2 (fluorine more electronegative): O = +2. O2F2: O = +1. These are the only cases where O has positive OS. Important for NEET: memorise the special cases — peroxides and superoxides are frequently tested in oxidation state questions.

3. Balancing Redox Reactions

Two methods: Oxidation number method: (1) Identify atoms changing OS. (2) Find increase and decrease in OS. (3) Balance electrons by multiplying half-reactions. (4) Balance other atoms (H, O by adding H2O and H⁺). Half-reaction (ion-electron) method: (1) Separate into oxidation and reduction half-reactions. (2) Balance atoms (O by H2O, H by H⁺ in acid; OH⁻ and H2O in base). (3) Balance charge by adding electrons. (4) Multiply to equalise electrons. (5) Add half-reactions. Important redox reagents for NEET: KMnO4 (n=5 in acid, n=3 in neutral, n=1 in base). K2Cr2O7 (n=6). Na2S2O3 (n=1). SnCl2 (n=2). FeSO4 (n=1). H2O2 (n=2, both oxidising and reducing agent).

4. Disproportionation and Comproportionation

Disproportionation: same element simultaneously oxidised AND reduced. Example: Cl2 + NaOH → NaCl + NaOCl + H2O. Cl goes from 0 to -1 (NaCl) and +1 (NaOCl). H2O2 → H2O + O2 (O goes from -1 to -2 and 0). 2NO2 + H2O → HNO2 + HNO3 (N goes from +4 to +3 and +5). White phosphorus (P4) in NaOH → PH3 + NaH2PO2. Comproportionation (inverse of disproportionation): two different oxidation states of same element combine to give intermediate. Cu⁰ + Cu²⁺ → 2Cu⁺. SnO + SnO2 → Sn2O3 at high T. These are important conceptual reactions for NEET.

5. Electrochemical Series and Redox Potential

Standard reduction potential (E°) measures tendency to be reduced. More positive E° = stronger oxidising agent. More negative E° = stronger reducing agent. Series (high to low E°): F2 (+2.87V) > MnO4⁻/Mn²⁺ (+1.51V) > Cl2 (+1.36V) > Cr2O7²⁻ (+1.33V) > O2 (+1.23V) > Br2 (+1.09V) > Fe³⁺/Fe²⁺ (+0.77V) > I2 (+0.54V) > Cu²⁺/Cu (+0.34V) > H⁺/H2 (0.00V) > Pb²⁺ (-0.13V) > Ni²⁺ (-0.25V) > Fe²⁺ (-0.44V) > Zn²⁺ (-0.76V) > Al³⁺ (-1.66V) > Mg²⁺ (-2.37V) > Na⁺ (-2.71V) > Li⁺ (-3.04V). A metal higher in the series displaces metals below it from their salt solutions.

6. Redox Titrations

Permanganate titrations: KMnO4 (purple) acts as self-indicator. In acid (H2SO4): MnO4⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H2O. Equivalent weight = M/5 = 158/5 = 31.6 g/eq. Used to estimate: Fe²⁺ (iron in ore), oxalic acid (C2O4²⁻, n=2), H2O2 (n=2). Dichromate titrations: K2Cr2O7 + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H2O. n=6. Equivalent weight = M/6 = 294/6 = 49 g/eq. Indicator: diphenylamine or ferroin. Iodometric: I2/Na2S2O3. Cu²⁺ + I⁻ → CuI + I2 (n=1 for Cu). I2 + S2O3²⁻ → I⁻ + S4O6²⁻ (n=1 for S2O3). Cerimetry: Ce⁴⁺/Ce³⁺ used instead of KMnO4 (more stable, sharper endpoint).

7. Industrial Redox Processes

Manufacture of H2SO4 (Contact process): S → SO2 → SO3 (V2O5 catalyst) → H2SO4. S oxidised from 0 to +4 to +6. Manufacture of HNO3 (Ostwald process): NH3 → NO (Pt/Rh catalyst, 900°C) → NO2 → HNO3. N goes from -3 to +2 to +4 to +5. Chlor-alkali process: 2NaCl + 2H2O → Cl2 + H2 + 2NaOH (electrolysis). Cl- oxidised to Cl2 at anode; H+ reduced to H2 at cathode. Thermite process: Al + Fe2O3 → Al2O3 + Fe. Al oxidised (0 to +3), Fe reduced (+3 to 0). Used in welding. Photography: AgBr + hv → Ag + Br (photoreduction of Ag⁺). Development: hydroquinone reduces remaining Ag⁺. Fixing: Na2S2O3 dissolves unexposed AgBr.

8. Oxidising and Reducing Agents

Oxidising agents (accept electrons, get reduced): F2, Cl2, Br2, I2 (halogens, in that order). KMnO4, K2Cr2O7 (strong, in acid). HNO3 (conc. and dilute, different products). H2SO4 (conc.). O3, H2O2 (in acidic conditions). MnO2. Reducing agents (donate electrons, get oxidised): metals (Li, Na, K, Mg, Al, Zn, Fe). H2, CO. H2S, HI, HBr, HCl (in that order, I⁻ strongest). Na2S2O3, SnCl2, FeSO4, oxalic acid. Dual nature (both oxidising and reducing): H2O2 (reducing with KMnO4, oxidising with KI). SO2 (reducing with KMnO4, oxidising with H2S). Cl2 (oxidising usually, but reducing with F2). NO2 (varies). Understanding which agent is stronger determines direction of reaction.

Frequently Asked Questions

1. Why is oxidation state of O -1/2 in KO2 and not -1? ⌄

KO2 is potassium superoxide containing the O2⁻ (superoxide) ion. In O2⁻: total charge = -1, two oxygen atoms share this charge equally, so each O has OS = -1/2. This is a fractional oxidation state — formally each O is -1/2. Compare: in O2²⁻ (peroxide like in H2O2): total charge = -2, each O = -1. In O²⁻ (regular oxide): charge = -2, one O = -2. The fractional value is perfectly valid for OS calculations. Similarly, in Fe3O4 (magnetite): Fe has OS = +8/3 (fractional, because it contains both Fe²⁺ and Fe³⁺ in 1:2 ratio).

2. What compounds have oxygen in positive oxidation state? ⌄

Oxygen has positive OS only when bonded to the more electronegative fluorine (EN of F = 4.0, EN of O = 3.5). OF2: OS of O = +2 (F = -1, 2F = -2, so O = +2). O2F2: OS of O = +1 (F = -1, total charge = 0: 2O + 2(-1) = 0, O = +1). These are the only known stable compounds where O has positive OS. Note: ozone O3 has OS of O = 0 (free element). In H2O2, O2, and most oxides: O has -1, -2 OS. Memorise for NEET: F-O compounds give O positive OS; all other compounds give O zero or negative OS.

3. How to identify oxidising vs reducing agents in a reaction? ⌄

In any redox reaction: species that gets oxidised (OS increases) = reducing agent (it reduces the other). Species that gets reduced (OS decreases) = oxidising agent (it oxidises the other). Example: in 2H2S + SO2 → 3S + 2H2O: H2S: S goes from -2 to 0 (oxidised → H2S is reducing agent). SO2: S goes from +4 to 0 (reduced → SO2 is oxidising agent). For NEET: always identify which element changes OS in each reactant. The one going to higher OS = reducing agent; going to lower OS = oxidising agent.

4. What is the oxidation state of S in different sulphur compounds? ⌄

Important oxidation states of sulphur: H2S: S = -2 (lowest). S: 0 (elemental). SO2: S = +4. H2SO3: S = +4. SO3: S = +6. H2SO4: S = +6 (highest in stable compound). Na2S2O3 (thiosulphate): S = +2 average (one S = 0, one S = +4? No: each S = +2). S2O3²⁻: sum = -2, so 2x = -2+(-2×3 from O)... careful: 2x + 3(-2) = -2, 2x = -2+6 = 4, x = +2. Na2S2O5 (pyrosulphite): S = +4. K2S2O8 (peroxydisulphate): S = +7 (contains peroxo linkage, O-O bond). Sulphur shows all even OS from -2 to +6. This versatility makes S chemistry very rich and important in NEET.

5. How does the concept of OS relate to electron gain and loss? ⌄

Oxidation state is a bookkeeping tool that tracks electrons. Increase in OS = loss of electrons = oxidation. Decrease in OS = gain of electrons = reduction. In KMnO4 acting as oxidising agent in acid: Mn goes from +7 to +2 (gains 5 electrons per Mn). In Fe²⁺ acting as reducing agent: Fe goes from +2 to +3 (loses 1 electron per Fe). Balancing: 1 mol KMnO4 (n=5) reacts with 5 mol FeSO4 (n=1 each). This electron bookkeeping is why equivalent weights and n-factors are important: they tell us how many moles of electrons each species transfers.

Previous Questions

Standard enthalpy formation Ba2+ aq BaCl2 aq -206.9 Cl- -39.7 kcal/mol

Chemistry · Answer: -128.5 kcal/mol

Equal number of atoms 12g carbon 212g Na2CO3 Na2O H2 A B C only

Chemistry · Answer: A, B and C only

Match purification techniques simple fractional reduced pressure steam distillation

Chemistry · Answer: A-IV, B-III, C-I, D-II

First order reaction k=0.03s-1 concentration 7.2 to 0.9 mol/L time 69.3s

Chemistry · Answer: 69.3 s

Coordination complexes wavelength absorbed order B A D C crystal field splitting

Chemistry · Answer: B < A < D < C