Respuesta :

These are four questons and four answers:

Answers:

  • 1)  7⁺
  • 2) 1⁻
  • 3) 3⁺
  • 4) 5⁺

Explanation:

Question 1) Cl₂O₇:

a) Net charge of the compound: 0

b) Rule: oxygen works with oxidation state +2, except with peroxides.

d) Rule: balance of charges: ∑ of the charges = net charge

Call X the oxidation number of Cl:

  • 2×X + 7 (-2) = 0
  • 2X - 14 = 0
  • 2X = +14
  • X = +14 /2 = + 7

Conclusion: the oxidation number of Cl in Cl₂O₇ is 7⁺.

Question 2) AlCl₄⁻

a) Net charge of the ion: - 1

b) Rule: common oxidation number of Al in compounds: +3

c) Rule: balance of charges: ∑ charges = net charge = - 1

  • 1 (+3) + 4X = - 1
  • +3 + 4X = - 1
  • 4X = - 1 - 3
  • 4X = - 4
  • X = - 1

Conclusion: the oxidation number of Cl in AlCl₄⁻ is 1 ⁻.

Question 3) Ba(ClO₂)₂

a) Net charge of the compound: 0

b) Rule: common oxidation number of BA in compounds: +2

c) Rule: common oxidation number of O in compounds (except in peroxides): -2

d) Rule: balance of charges: ∑ charges = net charge = 0

  • +2 + 2X + 4 (-2) = 0
  • 2X +2 - 8 = 0
  • 2X - 6 = 0
  • 2X = +6
  • X = + 3

Conclusion: the oxidation number of Cl in Ba(ClO₂)₂  is 3⁺.

Question 4) CIF₄⁺

a) Net charge of the ion: + 1

b) Rule: common oxidation number of F : - 1 (it is the most electronegative)

c) Rule: balance of charges: ∑ charges = net charge = + 1

  • X + 4(-1) = +1
  • X - 4 = +1
  • X = +1 + 4
  • X = + 5

Conclusion: the oxidation number of Cl in ClF₄⁺ is 5⁺.