Question

1. Write the oxidation number for each atom: a. CrO42- b. HC104 d. Sg c. NF3 e. P2O5 g. CHC13

Write the oxidation number for each atom

Please help me solve these 6. I don't know how to do them.

Answer 1

In generally i) The sum of the oxidation states of all the atoms of ions in a neutral compound is equal to zero. " The sum of the oxidation states of all the atoms in a charged compound is equal to charge on that compound (1) The oxidation state of State is equal to an element, in an elementary zero. Ex: S, Al, c is zero. In generally for I Group elements oxidation state = +1. I Group elements oxidation state = +2. orygen 3-2, In peroxides of Halogens (C1, F, I, BX) = -1. rules to write oxidation numbers. use these For halogens exception: (CI, B8, I) halogen bonded withog (CI, B8, I) f i.e. If CI, BX, I bonded with 'o' or 'F' widation number Varie. *'F' is a1' in any compound.

@ crata Applicable dis rule and In generally ' oxidation number=-=; :: cF+ 4(-2) = -2 cv = 8-2 = +6 --+6 and 05-21 ordation numbers. Hcy In generally M₃ + 1 Cl3-4 cai is bonded with o) But maximum oxygen oxidation number is = -2' only. :. Find the cl' olidation number (+1) + Cl + 4(-2) = 0 ( i rule) C1 - 8 - +7 maximum positive oxidation number for di is t 1 oridation numbers >> H= +1 Cl = +7 0=-2

© NE In generally F = -1 N+36-1) = 0 N = +3 + N F + 3) -1 cidation numbers @ So 5o = 's' oxidation number is z ero. This is element so zero. It is an elemental sulfur. In generally O's-2 2p+ s(-2) = 0 = 2P = 10 = p=105 : oxidation numbers >> P = +5 03-2

A CHCl In generally cha-1 H +4 NÓ . C+1+36-1) =0 = c = + " oxidation numbers H₃ +1 a -1 Cyta