CBSE Class 11 Chemistry Chapter 8: Redox Reactions NCERT Solutions

To assign oxidation numbers, we use the general rules and set the sum of oxidation numbers in a neutral compound to zero.

1. For $Na\backslash underline\{H\_2PO\_4\}$: Let the oxidation number of Phosphorus (P) be x. We know the standard oxidation numbers: Na = +1, H = +1, O = -2. The sum of oxidation numbers in $NaH\_2PO\_4$ is 0. So, $1(+1)\; +\; 2(+1)\; +\; 1(x)\; +\; 4(-2)\; =\; 0$ $1\; +\; 2\; +\; x\; -\; 8\; =\; 0$ $3\; +\; x\; -\; 8\; =\; 0$ $x\; -\; 5\; =\; 0\; \backslash Rightarrow\; x\; =\; +5$ Therefore, the oxidation number of P is +5.
2. For $Na\backslash underline\{HSO\_4\}$: Let the oxidation number of Sulfur (S) be x. Standard oxidation numbers: Na = +1, H = +1, O = -2. The sum of oxidation numbers in $NaHSO\_4$ is 0. So, $1(+1)\; +\; 1(+1)\; +\; 1(x)\; +\; 4(-2)\; =\; 0$ $1\; +\; 1\; +\; x\; -\; 8\; =\; 0$ $2\; +\; x\; -\; 8\; =\; 0$ $x\; -\; 6\; =\; 0\; \backslash Rightarrow\; x\; =\; +6$ Therefore, the oxidation number of S is +6.
3. For $H\_4\backslash underline\{P\_2O\_7\}$: Let the oxidation number of Phosphorus (P) be x. Standard oxidation numbers: H = +1, O = -2. The sum of oxidation numbers in $H\_4P\_2O\_7$ is 0. So, $4(+1)\; +\; 2(x)\; +\; 7(-2)\; =\; 0$ $4\; +\; 2x\; -\; 14\; =\; 0$ $2x\; -\; 10\; =\; 0\; \backslash Rightarrow\; 2x\; =\; +10\; \backslash Rightarrow\; x\; =\; +5$ Therefore, the oxidation number of P is +5.
4. For $K\_2\backslash underline\{MnO\_4\}$: Let the oxidation number of Manganese (Mn) be x. Standard oxidation numbers: K = +1, O = -2. The sum of oxidation numbers in $K\_2MnO\_4$ is 0. So, $2(+1)\; +\; 1(x)\; +\; 4(-2)\; =\; 0$ $2\; +\; x\; -\; 8\; =\; 0$ $x\; -\; 6\; =\; 0\; \backslash Rightarrow\; x\; =\; +6$ Therefore, the oxidation number of Mn is +6.
5. For $\backslash underline\{CaO\_2\}$: Let the oxidation number of Oxygen (O) be x. Standard oxidation number: Ca = +2. The sum of oxidation numbers in $CaO\_2$ is 0. So, $(+2)\; +\; 2(x)\; =\; 0$ $2\; +\; 2x\; =\; 0\; \backslash Rightarrow\; 2x\; =\; -2\; \backslash Rightarrow\; x\; =\; -1$ Therefore, the oxidation number of O is -1. This indicates it is a peroxide.
6. For $Na\backslash underline\{BH\_4\}$: Let the oxidation number of Boron (B) be x. Standard oxidation numbers: Na = +1. In metal hydrides, Hydrogen usually has an oxidation number of -1. The sum of oxidation numbers in $NaBH\_4$ is 0. So, $1(+1)\; +\; 1(x)\; +\; 4(-1)\; =\; 0$ $1\; +\; x\; -\; 4\; =\; 0$ $x\; -\; 3\; =\; 0\; \backslash Rightarrow\; x\; =\; +3$ Therefore, the oxidation number of B is +3.
7. For $H\_2\backslash underline\{S\_2O\_7\}$: Let the oxidation number of Sulfur (S) be x. Standard oxidation numbers: H = +1, O = -2. The sum of oxidation numbers in $H\_2S\_2O\_7$ is 0. So, $2(+1)\; +\; 2(x)\; +\; 7(-2)\; =\; 0$ $2\; +\; 2x\; -\; 14\; =\; 0$ $2x\; -\; 12\; =\; 0\; \backslash Rightarrow\; 2x\; =\; +12\; \backslash Rightarrow\; x\; =\; +6$ Therefore, the oxidation number of S is +6.
8. For $KAl(SO\_4)\_2.12H\_2O$: This is a hydrated salt. We can consider the oxidation states of the elements within the ions and the water molecules separately. For the $SO\_4$ group, let the oxidation number of S be y. Oxygen is -2, and the sulfate ion has a charge of -2. $y\; +\; 4(-2)\; =\; -2\; \backslash Rightarrow\; y\; -\; 8\; =\; -2\; \backslash Rightarrow\; y\; =\; +6$. So, S has an oxidation number of +6. For the water molecule ($H\_2O$), Hydrogen is +1 and Oxygen is -2. The oxidation numbers of K and Al are +1 and +3 respectively, as they are in Group 1 and Group 13. The question asks for the underlined element, but no element is underlined in this part. Assuming it refers to S within the sulfate group: The oxidation number of S in $KAl(SO\_4)\_2.12H\_2O$ is +6.