QCC Notes
CLASS 11 · CHEMISTRYJEE MAIN × NEETहिंदी
§ 7.2NCERT Class 11 · Chemistry · Chapter 7

Oxidation Number Rules and Structural Applications

QCC Notes · Class 11 Chemistry NCERT-aligned Updated 26 Sep 2026 11 min read

While the electron-transfer model cleanly describes pure ionic reactions, many redox reactions involve covalent molecules where electrons are not fully transferred but merely redistributed. To track electron density shifts systematically, chemists developed the operational concept of Oxidation Number (Oxidation State). In this module, we examine the rigorous NCERT assignment rules, resolve the famous paradox of fractional oxidation states through structural chemistry, and apply Stock notation.

1. Definition and Core Concepts

The oxidation number of an element in a compound is defined as the electrical charge that an atom of that element appears to have when all other atoms bonded to it are hypothetically removed as ions, assigning shared electron pairs entirely to the more electronegative partner.

Valency vs Oxidation Number

  • Valency: Combining capacity of an element. It is always a whole number (1, 2, 3...) and carries no plus or minus sign.
  • Oxidation Number: Formal charge assigned based on electronegativity. It can be positive, negative, zero, or fractional.
  • Example: In CH₄, valency of C is 4 and oxidation number is -4. In CH₂Cl₂, valency is 4 and oxidation number is 0. In CCl₄, valency is 4 and oxidation number is +4.

Fundamental Axiom

In homonuclear covalent bonds (between identical atoms, e.g., C–C, Cl–Cl), the bonding electrons are shared equally; neither atom is assigned a charge (contribution = 0).

In heteronuclear bonds (e.g., C–H, C–O), the bonding pair is allocated completely to the more electronegative atom (EN order: F > O > N ≈ Cl > Br > I > S > C > H > Metals).

2. NCERT Rules for Assigning Oxidation Numbers

To compute oxidation states without drawing complete Lewis structures for every molecule, IUPAC and NCERT establish seven universal rules:

  1. Free / Elemental State: The oxidation number of an element in its elementary, uncombined form is always zero.
    Examples: H₂, O₂, Cl₂, O₃, P₄, S₈, Na, Mg, Fe, C(graphite) → all have oxidation number = 0.
  2. Monatomic Ions: The oxidation number equals the actual electrical charge on the ion.
    Examples: Na+ (+1), Mg²+ (+2), Al³+ (+3), Fe³+ (+3), Cl− (-1), O²− (-2), N³− (-3).
  3. Oxygen Atom:
    • In the vast majority of compounds (oxides), oxygen is -2 (e.g., H₂O, CO₂, H₂SO₄).
    • In peroxides (containing the −O–O− linkage), oxygen is -1 (e.g., H₂O₂, Na₂O₂, BaO₂).
    • In superoxides (containing the O₂− ion), oxygen is -1/2 (e.g., KO₂, RbO₂, CsO₂).
    • In oxygen fluorides, because fluorine is more electronegative than oxygen: in OF₂, oxygen is +2; in O₂F₂, oxygen is +1.
  4. Hydrogen Atom:
    • In most compounds bonded to non-metals, hydrogen is +1 (e.g., H₂O, HCl, NH₃, CH₄).
    • In binary metal hydrides (Group 1, Group 2 metals where H forms ionic hydride ion H−), hydrogen is -1 (e.g., LiH, NaH, CaH₂).
  5. Halogens (Group 17):
    • Fluorine is the most electronegative element; it has an oxidation state of -1 in all its compounds without exception.
    • Other halogens (Cl, Br, I) have an oxidation state of -1 in simple halides (e.g., NaCl, KBr). However, when bonded to more electronegative atoms (O or F) in oxoacids, oxoanions, and interhalogens, they display positive oxidation numbers up to +7 (e.g., HClO (+1), KClO₃ (+5), HClO₄ (+7), IF₇ (+7)).
  6. Group 1 & 2 Metals:
    Alkali metals (Li, Na, K, Rb, Cs) are strictly +1; Alkaline earth metals (Be, Mg, Ca, Sr, Ba) are strictly +2; Aluminium is always +3 in its compounds.
  7. Sum of Oxidation Numbers:
    • In a neutral chemical compound, the algebraic sum of oxidation numbers of all constituent atoms is zero.
    • In a polyatomic ion, the algebraic sum of oxidation numbers equals the net charge on the ion. (e.g., in SO₄²−: S + 4(-2) = -2 ⇒ S = +6).

3. The Paradox of Fractional Oxidation States & Structural Reality

Applying the algebraic rules mechanically occasionally produces fractional oxidation numbers. However, an atom cannot lose or gain a fraction of an electron. Fractional oxidation numbers are merely statistical averages of atoms residing in chemically non-equivalent structural environments.

+-----------------------------------------------------------------------------------+ | STRUCTURAL FORMULAE OF FAMOUS "PARADOXICAL" MOLECULES | +-----------------------------------------------------------------------------------+ (a) Chromium Peroxide (CrO5) — "Butterfly Structure": O (-2, Oxo) || Cr (+6, Center) / (-1) O ---- O (-1) Peroxo ring 1 (-1) O ---- O (-1) Peroxo ring 2 Algebraic formula gave +10 (Wrong!). True structural state = +6. (b) Carbon Suboxide (C3O2): (+2) (0) (+2) O = C = C* = C = O Terminal carbons bonded to O → +2 each Central carbon C* bonded only to C → 0 Average Oxidation State = (+2 + 0 + 2) / 3 = +4/3. (c) Tetrathionate Ion (S4O6^2-): O O || || (-1) O - S(+5) - S(0) - S(0) - S(+5) - O (-1) || || O O Two terminal sulfur atoms: +5 each; Two central bridging sulfur atoms: 0 each. Average Oxidation State = (+5 + 0 + 0 + +5) / 4 = +10/4 = +2.5. (d) Tribromooctaoxide (Br3O8): O O O || || || O = Br(+6)- Br(+4)- Br(+6) = O || || || O O O Two terminal Br atoms: +6 each; Central Br atom: +4. Average Oxidation State = (+6 + 4 + +6) / 3 = +16/3. +-----------------------------------------------------------------------------------+
Figure 7.3: Structural resolution of oxidation states in CrO5, C3O2, S4O6 2-, and Br3O8.
Molecule / Ion Apparent Algebraic Formula Structural Reality & Linkages Individual True Oxidation Numbers Average State
CrO₅ (Chromium peroxide) Cr + 5(-2) = 0 ⇒ +10 (Impossible) Butterfly structure: 1 oxo (=O), 2 peroxo (−O–O−) rings Cr = +6; 1 O at -2; 4 O at -1 Cr = +6
H₂SO₅ (Caro's acid) 2(+1) + S + 5(-2) = 0 ⇒ +8 (Impossible) Peroxomonosulfuric acid: contains 1 peroxo (−O–O−) bond S = +6; 3 O at -2; 2 O at -1 S = +6
H₂S₂O₈ (Marshall's acid) 2(+1) + 2S + 8(-2) = 0 ⇒ +7 (Impossible) Peroxodisulfuric acid: contains 1 peroxo bridge between two −SO₃H Both S = +6; 6 O at -2; 2 O at -1 S = +6
C₃O₂ (Carbon suboxide) 3C + 2(-2) = 0 ⇒ C = +4/3 Linear O=C=C*=C=O structure Terminal C = +2; Central C* = 0 Average = +4/3
S₄O₆²− (Tetrathionate) 4S + 6(-2) = -2 ⇒ S = +2.5 −O₃S–S–S–SO₃− chain Two terminal S = +5; Two central S = 0 Average = +2.5
Fe₃O₄ (Magnetite) 3Fe + 4(-2) = 0 ⇒ Fe = +8/3 Mixed oxide: FeO · Fe₂O₃ (spinel lattice) One Fe²+ (+2); Two Fe³+ (+3) Average = +8/3

4. Stock Notation and Nomenclature

In 1919, German chemist Alfred Stock proposed representing the oxidation state of an element in a compound by Roman numerals placed in parentheses directly after the name or symbol of the element.

Stock Formula Representation:

  • FeSO₄ → Fe(II)SO₄ [Iron(II) sulphate]
  • Fe₂(SO₄)₃ → Fe₂(III)(SO₄)₃ [Iron(III) sulphate]
  • Cu₂O → Cu₂(I)O [Copper(I) oxide]
  • CuO → Cu(II)O [Copper(II) oxide]
  • MnO₂ → Mn(IV)O₂ [Manganese(IV) oxide]

Stock Coordination / Complex Compounds:

  • HAuCl₄ → H[Au(III)Cl₄] [Hydrogen tetrachloroaurate(III)]
  • K₂Cr₂O₇ → Potassium dichromate(VI)
  • KMnO₄ → Potassium manganate(VII)
  • K₄[Fe(CN)₆] → Potassium hexacyanoferrate(II)
  • K₃[Fe(CN)₆] → Potassium hexacyanoferrate(III)

5. Solved Examples & Numerical Applications

Example 1: Determine the oxidation number of the underlined element in each of the following species: (a) KMnO₄, (b) K₂Cr₂O₇, (c) NaH₂PO₄, (d) NH₄NO₃.
Solution:

(a) In KMnO₄: Potassium is +1, Oxygen is -2. Let Mn = x.

(+1) + x + 4(-2) = 0 ⇒ x − 7 = 0 ⇒ Mn = +7.

(b) In K₂Cr₂O₇: Potassium is +1, Oxygen is -2. Let Cr = x.

2(+1) + 2x + 7(-2) = 0 ⇒ 2 + 2x − 14 = 0 ⇒ 2x = 12 ⇒ Cr = +6.

(c) In NaH₂PO₄: Na = +1, H = +1, O = -2. Let P = x.

(+1) + 2(+1) + x + 4(-2) = 0 ⇒ 3 + x − 8 = 0 ⇒ P = +5.

(d) In NH₄NO₃: This is an ionic compound composed of NH₄+ and NO₃−. The two nitrogen atoms have completely different oxidation states:

  • In NH₄+: x + 4(+1) = +1 ⇒ N = -3.
  • In NO₃−: y + 3(-2) = -1 ⇒ N = +5.

Assigning an average of +1 would mask the fundamental reality of the molecule.

Example 2: A student calculates the oxidation number of sulfur in sodium thiosulphate (Na₂S₂O₃) as +2. Draw its Lewis structure and state the actual individual oxidation numbers of the two sulfur atoms.
Solution:

Step 1: Algebraic Average:

2(+1) + 2(S) + 3(-2) = 0 ⇒ 2 + 2S − 6 = 0 ⇒ 2S = 4 ⇒ S = +2 (Average).

Step 2: Structural Analysis:

The thiosulphate ion [S₂O₃]²− is structurally derived from sulphate [SO₄]²− by replacing one peripheral oxygen atom with a terminal sulfur atom. The central sulfur atom is bonded to three oxygen atoms and one terminal sulfur atom via a coordinate or double bond:

[O₃S − S]²−

Since the terminal sulfur is more electronegative than the central sulfur (or by treating the terminal sulfur as sulfide-like S²−), the terminal sulfur has an oxidation number of -2 (or 0 in coordinate bonding view), and the central sulfur has an oxidation number of +6 (or +4).

In standard NCERT treatment: Terminal S = -2, Central S = +6; Average = (+6 + (-2)) / 2 = +2.

Example 3: Bleaching powder has the chemical formula CaOCl₂ (or Ca(OCl)Cl). What are the oxidation states of the two chlorine atoms present in it?
Solution:

Step 1: Analyze constituent ions:

Bleaching powder is a mixed salt consisting of calcium ions (Ca²+), chloride ions (Cl−), and hypochlorite ions (OCl−):

Ca²+ [Cl]− [OCl]−

Step 2: Calculate individual chlorine states:

  • For the simple chloride ion (Cl−): Oxidation state = -1.
  • For the hypochlorite ion (OCl−): O is -2, let Cl = x ⇒ (-2) + x = -1 ⇒ x = +1.

Thus, the two chlorine atoms in bleaching powder exist in -1 and +1 oxidation states (Average = 0).

Example 4: What is the maximum and minimum oxidation state that an element of Group 15 (e.g., Nitrogen or Phosphorus) and Group 16 (e.g., Sulfur) can display? Explain why HNO₃ can act only as an oxidant, while HNO₂ can act as both an oxidant and a reductant.
Solution:

Step 1: Determine periodic group limits:

  • Group 15 (ns² np³): Maximum = +5 (loss of all 5 valence electrons); Minimum = 5 − 8 = -3 (gain of 3 electrons to complete octet).
  • Group 16 (ns² np⁴): Maximum = +6 (loss of 6 electrons); Minimum = 6 − 8 = -2 (gain of 2 electrons).

Step 2: Analyze HNO₃ vs HNO₂:

  • In HNO₃ (Nitric acid), nitrogen is in its maximum oxidation state of +5. It cannot increase its oxidation number further (cannot be oxidized). It can only decrease its oxidation state (gain electrons ⇒ undergo reduction). Therefore, HNO₃ acts exclusively as an oxidizing agent.
  • In HNO₂ (Nitrous acid), nitrogen is in an intermediate oxidation state of +3. It can be oxidized to +5 (acting as a reducing agent) or reduced to +2 (NO), +1 (N₂O), 0 (N₂), or -3 (NH₃) (acting as an oxidizing agent). Therefore, HNO₂ acts as both an oxidant and a reductant.

Frequently Asked Questions (Class 11 & JEE/NEET)

Q1. Why can an element never exhibit an oxidation state exceeding its group number?
The maximum oxidation state of any representative (main group) element is strictly limited by the total number of valence electrons present in its outermost shell (Group valence). Removing core electrons from an inner noble gas shell requires astronomical amounts of ionization energy that can never be compensated by chemical bond formation. (For Group 17, max = +7; Group 16, max = +6; Group 15, max = +5).
Q2. What is the oxidation number of oxygen in KO₂ and OF₂?
In potassium superoxide (KO₂), potassium is an alkali metal (+1). By rule of neutrality, 2(O) + 1 = 0 ⇒ O = -1/2. In oxygen difluoride (OF₂), fluorine is more electronegative (-1). Thus, O + 2(-1) = 0 ⇒ O = +2.
Q3. Why does hydrogen exhibit an oxidation number of -1 in NaH but +1 in HCl?
In NaH, sodium is an electropositive metal (electronegativity ~0.9) while hydrogen is more electronegative (~2.1), so hydrogen completely captures the electron to form the hydride anion (H⁻, state -1). In HCl, chlorine is far more electronegative (~3.0) than hydrogen, pulling electron density away and leaving hydrogen with state +1.
Q4. How does oxidation state correlate with acidic character of oxides?
As the oxidation state of a central non-metal or transition metal increases, its size shrinks and its effective nuclear charge/polarizing power increases, drawing electron density strongly from attached oxygen atoms and facilitating the release of H⁺ ions. Consequently, higher oxidation states yield more acidic oxides. For example, MnO (+2, basic) < Mn₂O₃ (+3, weakly basic) < MnO₂ (+4, amphoteric) < Mn₂O₇ (+7, strongly acidic).
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