Question

Определить С.О. (Степень окисления) элементов по формуле. а) SO2 SO3 K2S в) CL2 LiCl Cl2O3 KCLI4 б) NO N2O NO2 Na3Iv

Answer 1

Не так немного.SO2 S+2, SO3 S+3, J2S K+1 S-2. Cl2 0, LiCl Li+1 CL-1

Answer 2

Determining Oxidation States of Elements in Compounds

Let's determine the oxidation states (S.O.) of the elements in the given compounds.

# a) SO2, SO3, K2S

- SO2: - The oxidation state of oxygen is usually -2, so the total oxidation state for the compound is 0. - Using algebraic manipulation, the oxidation state of sulfur in SO2 can be calculated as follows: - Let the oxidation state of sulfur be x. - The sum of the oxidation states of all atoms in the compound equals the charge of the compound, which is 0. - Therefore, 2(-2) + x = 0 - Solving for x, we get x = +4. - Therefore, the oxidation state of sulfur in SO2 is +4.

- SO3: - Similar to SO2, the oxidation state of oxygen is -2, and the total oxidation state for the compound is 0. - Using the same approach as before: - Let the oxidation state of sulfur be x. - The sum of the oxidation states of all atoms in the compound equals the charge of the compound, which is 0. - Therefore, 3(-2) + x = 0 - Solving for x, we get x = +6. - Therefore, the oxidation state of sulfur in SO3 is +6.

- K2S: - The oxidation state of potassium (K) in compounds is usually +1, and the oxidation state of sulfur in compounds is typically -2. - Using the same approach as before: - Let the oxidation state of sulfur be x. - The sum of the oxidation states of all atoms in the compound equals the charge of the compound, which is 0. - Therefore, 2(+1) + x = 0 - Solving for x, we get x = -2. - Therefore, the oxidation state of sulfur in K2S is -2.

# б) Cl2, LiCl, Cl2O3, KClO4

- Cl2: - Chlorine exists as diatomic molecules (Cl2) with an oxidation state of 0.

- LiCl: - The oxidation state of lithium (Li) in compounds is usually +1, and the oxidation state of chlorine in compounds is typically -1. - Therefore, the oxidation state of chlorine in LiCl is -1.

- Cl2O3: - The sum of the oxidation states of all atoms in the compound equals the charge of the compound, which is 0. - Let the oxidation state of chlorine be x. - Therefore, 3(-2) + 2x = 0 - Solving for x, we get x = +3. - Therefore, the oxidation state of chlorine in Cl2O3 is +3.

- KClO4: - The sum of the oxidation states of all atoms in the compound equals the charge of the compound, which is 0. - Let the oxidation state of chlorine be x. - Therefore, 4(-2) + 1 + x = 0 - Solving for x, we get x = +7. - Therefore, the oxidation state of chlorine in KClO4 is +7.

# в) NO, N2O, NO2, Na3Iv

- NO: - The sum of the oxidation states of all atoms in the compound equals the charge of the compound, which is 0. - Let the oxidation state of nitrogen be x. - Therefore, -2 + x = 0 - Solving for x, we get x = +2. - Therefore, the oxidation state of nitrogen in NO is +2.

- N2O: - The sum of the oxidation states of all atoms in the compound equals the charge of the compound, which is 0. - Let the oxidation state of nitrogen be x. - Therefore, 2(-2) + x = 0 - Solving for x, we get x = +4. - Therefore, the oxidation state of nitrogen in N2O is +4.

- NO2: - The sum of the oxidation states of all atoms in the compound equals the charge of the compound, which is 0. - Let the oxidation state of nitrogen be x. - Therefore, 2(-2) + x = 0 -