Question

use the periodic table to give the possible oxidation states for each element. place lower numbers first (-1 is lower than 1). fr 1 pb 2 and 3 s 4, 5, 6, and 7 f 8 a. 0 b. 1 c. -1 d. 2 e. -2 f. 3 g. -3 h. 4 i. -4 j. 5 k. -5 l. 6 m. -6

Explanation:

Step1: Determine Fr oxidation state

Francium (Fr) is in Group 1 of the periodic table. Elements in Group 1 have a + 1 oxidation state in most compounds. So the oxidation state of Fr is 1.

Step2: Determine Pb oxidation state

Lead (Pb) is a post - transition metal. It commonly has oxidation states of +2 and +4. So the oxidation states of Pb are 2 and 4.

Step3: Determine S oxidation state

Sulfur (S) can have oxidation states of - 2 in sulfides, +4 in sulfur dioxide ($SO_2$) and sulfites ($SO_{3}^{2 - }$), +6 in sulfuric acid ($H_2SO_4$) and sulfates ($SO_{4}^{2 - }$). So the oxidation states of S are -2, 4, 6.

Step4: Determine F oxidation state

Fluorine (F) is the most electronegative element. It always has an oxidation state of - 1 in compounds. So the oxidation state of F is -1.

Answer:

Fr: B. 1
Pb: D. 2, H. 4
S: E. -2, H. 4, L. 6
F: C. -1