What is the oxidation number of chlorine in Cl₂O?
Let Cl = x. O = -2. Equation: 2x + (-2) = 0, 2x = 2, x = +1.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Applications of Redox Reactions and Disproportionation
Latest questions in this category.
Let Cl = x. O = -2. Equation: 2x + (-2) = 0, 2x = 2, x = +1.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Applications of Redox Reactions and Disproportionation
P (0) is oxidized to +3 in H₃PO₃ and reduced to -3 in PH₃; P reduces itself in this disproportionation.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Applications of Redox Reactions and Disproportionation
In Mg(OH)2 + 2HCl → MgCl₂ + 2H₂O, no element changes oxidation state (Mg: +2, H: +1, O: -2, Cl: -1).
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Applications of Redox Reactions and Disproportionation
S (-2 in H₂S) becomes 0 in S, losing electrons.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Applications of Redox Reactions and Disproportionation
TeO₂ (Te: +4 to 0) gains electrons, reduced by H₂ (0 to +1).
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Applications of Redox Reactions and Disproportionation
In OF₂, O is less electronegative than F (-1). Equation: x + 2(-1) = 0, x = +2.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Applications of Redox Reactions and Disproportionation
Reduction involves electron gain. In 2FeCl₃ + SnCl₂ → 2FeCl₂ + SnCl₄, Fe (+3) gains an electron to become Fe (+2).
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Applications of Redox Reactions and Disproportionation
In CuO + H₂ → Cu + H₂O, Cu (+2 to 0) is reduced, and H (0 to +1) is oxidized.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Applications of Redox Reactions and Disproportionation
The oxidizing agent accepts electrons. Cl₂ (0) gains electrons to become Cl− (-1), oxidizing Na.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Applications of Redox Reactions and Disproportionation
In 3Br₂ + 6NaOH → 5NaBr + NaBrO₃ + 3H₂O, Br (0) is oxidized to +5 in NaBrO₃ and reduced to -1 in NaBr.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Applications of Redox Reactions and Disproportionation
Let the oxidation number of S be x. H = +1, O = -2. Equation: 2(+1) + x + 4(-2) = 0. Solving: 2 + x - 8 = 0, x = +6.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Applications of Redox Reactions and Disproportionation
In 3S + 6NaOH → 2Na₂S + Na₂SO₃ + 3H₂O, S (0) is oxidized to +4 in Na₂SO₃ and reduced to -2 in Na₂S.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Applications of Redox Reactions and Disproportionation