Q.Assign oxidation number to the following underlined elements: a) KMnO4 (Mn underlined) b) H2O2 (one O underlined) c) H4P2O7 (one O underlined)

Showing the 12 most recent of 49 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.The oxidation numbers of boron in NaBH4 and Cr in K2Cr2O7 are

   (a) +4, +3

   (b) +3, +6

   (c) -3, +6

   (d) -4, +12

   ›Reveal solutionSolution

   Use the standard oxidation-number rules: Na = +1, H = -1 in metal hydrides, K = +1, O = -2, and the sum of oxidation numbers in a neutral compound equals zero.

   For NaBH4:

   Na = +1 (alkali metal, always +1).

   H = -1 (H bonded to a less electronegative metal takes -1, as in metal hydrides).

   Let oxidation number of B = x.

   Sum = 0 (neutral compound):

   (+1) + x + 4(-1) = 0

   1 + x - 4 = 0

   x = +3

   So B is +3 in NaBH4.

   For K2Cr2O7:

   K = +1 each, so 2 K contribute +2. …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Oxidation number of oxygen in H2O2 is -

   (a) -1

   (b) -2

   (c) -1/2

   (d) +1

   ›Reveal solutionSolution

   Oxygen has oxidation number -1 in H2O2 because of the O-O peroxide bond.

   To find oxidation number of O in H2O2: the molecule has formula H2O2, structure H-O-O-H. Each H contributes +1 (standard oxidation number of hydrogen bonded to a non-metal). Let oxidation number of each O be x. Since the molecule is neutral overall: 2(+1) + 2(x) = 0, so 2x = -2, x = -1.

   …

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3. CBSE 2026Set ANNUAL1 mark
   Q.Oxidation number of Fe in Fe2O3 is ______.

   ›Reveal solutionSolution

   Iron has oxidation number +3 in Fe2O3.

   Let the oxidation number of Fe be x. Oxygen is assigned its standard oxidation number, -2 (no peroxide or superoxide linkage here). Fe2O3 is electrically neutral, so: 2(x) + 3 …

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4. CBSE 2026Set ANNUAL1 mark
   Q.Oxidation number of chlorine (Cl) in ClO3⁻ is .................

   ›Reveal solutionSolution

   Solving $x + 3(-2) = -1$ gives the oxidation number of chlorine as $+5$.

   Let the oxidation number of Cl be $x$. Oxygen is assigned $-2$ in this oxoanion. The sum of oxidation numbers equals the overall ionic charge: …

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5. CBSE 2026Set sz1 markMCQ
   Q.Select the correct one: The oxidation state of chromium in chromium trioxide is

   (a) +3

   (b) +4

   (c) +5

   (d) +6

   ›Reveal solutionSolution

   Using the standard oxidation-state rule for oxygen (-2) and balancing to zero net charge for the neutral molecule CrO3 gives Cr = +6.

   Let the oxidation state of chromium in CrO3 be x.

   Oxygen is assigned an oxidation state of -2 (standard rule, since CrO3 is not a peroxide or superoxide).

   CrO3 is a neutral molecule, so the sum of oxidation states must equal zero:

   x + 3(-2) = 0

   x - 6 = 0

   x = +6

   …

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6. CBSE 2026Set ANNUAL1 markMCQ
   Q.In which of the following oxidation number of oxygen is maximum ?

   (a) H2O2

   (b) K2O

   (c) KO2

   (d) O2F2

   ›Reveal solutionSolution

   Oxygen shows +1 in O2F2, its maximum among the given species.

   Assign oxidation numbers of oxygen:

   • H2O2: O = −1 (peroxide).
   • K2O: O = −2 (normal oxide).
   • KO2: O = −1/2 (superoxide). …

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7. CBSE 2026Set ANNUAL1 markMCQ
   Q.In which of the following compounds oxidation number of Cl is + 5 ?

   (a) HClO4

   (b) HClO2

   (c) HClO3

   (d) HClO

   ›Reveal solutionSolution

   Cl is +5 in HClO3.

   Using H = +1, O = −2 and net charge 0:

   • HClO: 1 + Cl − 2 = 0 → Cl = +1.
   • HClO2: 1 + Cl − 4 = 0 → Cl = +3.
   • HClO3: 1 + Cl − 6 = 0 → Cl = +5. …

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8. CBSE 2026Set ANNUAL1 markMCQ
   Q.The oxidation number of Sulphur in Na2S4O6 is :

   (1) 2.5

   (2) 1.5

   (3) 2

   (4) 3

   ›Reveal solutionSolution

   The average oxidation number of sulphur in Na2S4O6 is +2.5 — option (1).

   For the neutral compound Na2S4O6, the sum of all oxidation numbers is zero. Taking Na = +1 and O = -2, and letting the (average) oxidation number of S be x:

   2(+1) + 4(x) + 6(-2) = 0

   2 + 4x - 12 = 0

   4x = 10

   x = +2.5

   …

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9. CBSE 2025Set ANNUAL1 markMCQ
   Q.Oxidation state of Cl in CaOCl2 is

   (a) 0

   (b) +1

   (c) -1

   (d) +1, -1

   ›Reveal solutionSolution

   CaOCl2 (bleaching powder) contains chlorine in two oxidation states at once: +1 and -1.

   Bleaching powder, CaOCl2, is best represented as the mixed salt Ca(OCl)Cl, containing both a hypochlorite ion (OCl-, in which Cl has oxidation state +1) and a chloride ion (Cl-, oxidation state -1). This dual oxidation state is a direct consequence of how bleaching powder is manufactured: by passing chlorine gas (oxidation state 0) over s …

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10. CBSE 2025Set ANNUAL1 markMCQ
    Q.What is the oxidation state of S in H2SO4?

    (a) +4

    (b) +5

    (c) +6

    (d) +8

    ›Reveal solutionSolution

    Using the standard oxidation-state rules (H = +1, O = -2) and the fact that the molecule is neutral overall, solving for S gives +6.

    Oxidation state rules used here:

    • Hydrogen is almost always +1 in compounds (except metal hydrides)
    • Oxygen is almost always -2 in compounds (except peroxides/superoxides)
    • The sum of all oxidation states in a neutral molecule must equal zero

    For H2SO4 (2 H atoms, 1 S atom, 4 O atoms), let x = oxidation state of S:

    2(+1) + 1(x) + 4(-2) = 0

    2 + x - 8 = 0

    x - 6 = 0

    x = +6

    …

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11. CBSE 2025Set ANNUAL1 mark
    Q.Fill in the blank: Oxidation number of Sulphur (S) in H2S2O7 is ___________.

    ›Reveal solutionSolution

    Using the standard oxidation states H = +1, O = -2 and solving for S in the neutral molecule gives S = +6 (same as in H$_2$SO$_4$, since H$_2$S$_2$O$_7$ is oleum/pyrosulfuric acid built from two SO$_4$ units).

    $H_2S_2O_7$ is a neutral molecule, so the sum of all oxidation numbers must equal 0.

    Let the oxidation number of each S atom be $x$ (both S atoms are equivalent by symmetry in pyrosulfuric acid).

    $2(+1) + 2(x) + 7(-2) = 0$

    $2 + 2x - 14 = 0$

    $2x = 12$

    $x = +6$

    …

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12. CBSE 2025Set ANNUAL1 markMCQ
    Q.When methane is burnt in O2 to produce CO2 and H2O the oxidation number of carbon changes by?

    (a) +4

    (b) +8

    (c) Zero

    (d) +2

    ›Reveal solutionSolution

    Carbon's oxidation state goes from -4 in methane to +4 in carbon dioxide during combustion, a change of +8.

    In CH4, each H is assigned +1 (by convention). Let oxidation number of C = x.

    x + 4(+1) = 0 (CH4 is neutral)

    x = -4

    In CO2, oxygen is assigned -2 (standard). Let oxidation number of C = y.

    y + 2(-2) = 0

    y = +4

    …

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