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⁺.