Question

Find the oxidation number of oxygen in KO₃ and Na₂O₂.​

Answer 1

Answer:

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Explanation:

KO3

KO3Suppose oxidation number of O = x

KO3Suppose oxidation number of O = x⇒1 + 3x = 0

KO3Suppose oxidation number of O = x⇒1 + 3x = 03x = −1

KO3Suppose oxidation number of O = x⇒1 + 3x = 03x = −1x = - 1/3 = - 0.33

KO3Suppose oxidation number of O = x⇒1 + 3x = 03x = −1x = - 1/3 = - 0.33

KO3Suppose oxidation number of O = x⇒1 + 3x = 03x = −1x = - 1/3 = - 0.33Na2O2

KO3Suppose oxidation number of O = x⇒1 + 3x = 03x = −1x = - 1/3 = - 0.33Na2O2Suppose oxidation number of O = x

KO3Suppose oxidation number of O = x⇒1 + 3x = 03x = −1x = - 1/3 = - 0.33Na2O2Suppose oxidation number of O = x2 × 1 + 2x = 0

KO3Suppose oxidation number of O = x⇒1 + 3x = 03x = −1x = - 1/3 = - 0.33Na2O2Suppose oxidation number of O = x2 × 1 + 2x = 02 + 2x = 0

KO3Suppose oxidation number of O = x⇒1 + 3x = 03x = −1x = - 1/3 = - 0.33Na2O2Suppose oxidation number of O = x2 × 1 + 2x = 02 + 2x = 02x = −2⇒x = - 2/2

KO3Suppose oxidation number of O = x⇒1 + 3x = 03x = −1x = - 1/3 = - 0.33Na2O2Suppose oxidation number of O = x2 × 1 + 2x = 02 + 2x = 02x = −2⇒x = - 2/2x = -1

Answer 2

Explanation:

Answer

KO

3

Suppose oxidation number of O=x, ⇒1+3x=0

3x=−1

x=−

3

1

=−0.33

Na

2

O

2

Suppose oxidation number of O=x

2×1+2x=0

2+2x=0

2x=−2⇒x=−

2

2

x=−1