CBSE Class 11 Chemistry Chapter 8: Redox Reactions NCERT Solutions

To assign oxidation numbers, we use the following general rules:

The oxidation number of an element in its free state is zero.

The oxidation number of alkali metals (Group 1) in compounds is +1.

The oxidation number of alkaline earth metals (Group 2) in compounds is +2.

The oxidation number of Hydrogen is +1 when bonded to non-metals and -1 when bonded to metals.

The oxidation number of Oxygen is usually -2, but it is -1 in peroxides, -1/2 in superoxides, and positive when bonded to fluorine.

The sum of oxidation numbers in a neutral compound is zero.

The sum of oxidation numbers in a polyatomic ion is equal to the charge of the ion.

Let's calculate the oxidation number for the underlined elements:

1. For $Na\backslash underline\{H\_2PO\_4\}$:

   Let the oxidation number of Phosphorus (P) be x.\nWe know the oxidation numbers: Na = +1, H = +1, O = -2.\nApplying the rule for neutral compounds: $$1(+1)\; +\; 2(+1)\; +\; 1(x)\; +\; 4(-2)\; =\; 0$$ $$1\; +\; 2\; +\; x\; -\; 8\; =\; 0$$ $$3\; +\; x\; -\; 8\; =\; 0$$ $$x\; -\; 5\; =\; 0$$ $$x\; =\; +5$$\nTherefore, the oxidation number of Phosphorus (P) is +5.

2. For $Na\backslash underline\{HSO\_4\}$:

   Let the oxidation number of Sulfur (S) be x.\nWe know the oxidation numbers: Na = +1, H = +1, O = -2.\nApplying the rule for neutral compounds:

3. For $H\_4P\_2O\_7\}$:

   Let the oxidation number of Phosphorus (P) be x.\nWe know the oxidation numbers: H = +1, O = -2.\nApplying the rule for neutral compounds: $$4(+1)\; +\; 2(x)\; +\; 7(-2)\; =\; 0$$ $$4\; +\; 2x\; -\; 14\; =\; 0$$ $$2x\; -\; 10\; =\; 0$$ $$2x\; =\; +10$$ $$x\; =\; +5$$\nTherefore, the oxidation number of Phosphorus (P) is +5.

4. For $K\_2\backslash underline\{MnO\_4\}$:

   Let the oxidation number of Manganese (Mn) be x.\nWe know the oxidation numbers: K = +1, O = -2.\nApplying the rule for neutral compounds:

5. For $Ca\backslash underline\{O\_2\}$:

   Let the oxidation number of Oxygen (O) be x.\nWe know the oxidation number: Ca = +2.\nThis is a peroxide (CaO₂), where oxygen has an oxidation state of -1.\nApplying the rule for neutral compounds: $$(+2)\; +\; 2(x)\; =\; 0$$ $$2\; +\; 2x\; =\; 0$$ $$2x\; =\; -2$$ $$x\; =\; -1$$\nTherefore, the oxidation number of Oxygen (O) is -1.

6. For $Na\backslash underline\{BH\_4\}$:

   Let the oxidation number of Boron (B) be x.\nWe know the oxidation numbers: Na = +1. In metal hydrides like NaBH₄, Hydrogen acts as a hydride ion (H⁻), so its oxidation number is -1.\nApplying the rule for neutral compounds: $$1(+1)\; +\; 1(x)\; +\; 4(-1)\; =\; 0$$ $$1\; +\; x\; -\; 4\; =\; 0$$ $$x\; -\; 3\; =\; 0$$ $$x\; =\; +3$$\nTherefore, the oxidation number of Boron (B) is +3.

7. For $H\_2\backslash underline\{S\_2O\_7\}$