Match the species in List I with the oxidation state of Xe in List II. List I: P XeF2; Q XeF4; R XeOF4; S XeO4. List II: 1 = +2; 2 = +4; 3 = +6; 4 = +8.

JEE Main levelSingle correct optionDifficulty 2 of 3ph130

Match the species in List I with the oxidation state of Xe in List II. List I: P $XeF X_{2}$ ; Q $XeF X_{4}$ ; R $XeOF X_{4}$ ; S $XeO X_{4}$ . List II: 1 = +2; 2 = +4; 3 = +6; 4 = +8.

XeF2: two F at

- 1

gives +2. XeF4 gives +4. In XeOF4, one O (

- 2

) and four F (

- 4

) total

- 6

, so Xe = +6. In XeO4, 4 O(

- 2

) means Xe = +8.