Question

Assign oxidation numbers to each element in (a) \(\mathrm{HIO}_{3}\) (b) \(\mathrm{NaMnO}_{4}\) (c) \(\mathrm{SnO}_{2}\) (d) NOF (e) \(\mathrm{NaO}_{2}\)

Short answer

Answer: The oxidation numbers for each element are as follows: - HIO3: H(+1), I(+5), O(-2) - NaMnO4: Na(+1), Mn(+7), O(-2) - SnO2: Sn(+4), O(-2) - NOF: N(+3), O(-2), F(-1) - NaO2: Na(+1), O(-1)

Step-by-step solution

HIO3 - Assigning Oxidation Numbers

In HIO3, we have the following elements: Hydrogen (H), Iodine (I), and Oxygen (O). According to the rules above: - Hydrogen (H): Oxidation number is +1. - Oxygen (O): Oxidation number is -2. - Iodine (I): We can find its oxidation number by setting up an equation, because the sum of oxidation numbers in a compound is equal to zero. \(+1 + x + 3(-2)=0\) \(x = 1 - 6 = +5\) HIO3: H(+1), I(+5), O(-2)

NaMnO4 - Assigning Oxidation Numbers

In NaMnO4, we have the following elements: Sodium (Na), Manganese (Mn), and Oxygen (O). According to the rules above: - Sodium (Na): Oxidation number is +1 (monatomic ion) - Oxygen (O): Oxidation number is -2. - Manganese (Mn): We can find its oxidation number by setting up an equation, because the sum of oxidation numbers in a compound is equal to zero. \(+1 + x + 4(-2)=0\) \(x = 1 - 8 = +7\) NaMnO4: Na(+1), Mn(+7), O(-2)

SnO2 - Assigning Oxidation Numbers

In SnO2, we have the following elements: Tin (Sn) and Oxygen (O). According to the rules above: - Oxygen (O): Oxidation number is -2. - Tin (Sn): We can find its oxidation number by setting up an equation, because the sum of oxidation numbers in a compound is equal to zero. \(x + 2(-2)=0\) \(x = 4\) SnO2: Sn(+4), O(-2)

NOF - Assigning Oxidation Numbers

In NOF, we have the following elements: Nitrogen (N), Oxygen (O), and Fluorine (F). - Oxygen (O): Oxidation number is -2. - Fluorine (F): Oxidation number is -1 (halogen) - Nitrogen (N): We can find its oxidation number by setting up an equation, because the sum of oxidation numbers in a compound is equal to zero. \(x - 2 - 1=0\) \(x = 3\) NOF: N(+3), O(-2), F(-1)

NaO2 - Assigning Oxidation Numbers

In NaO2, we have the following elements: Sodium (Na) and Oxygen (O). In this compound, we have an oxygen peroxide (O2) group. Based on the rules above: - Sodium (Na): Oxidation number is +1 (monatomic ion) - Oxygen (O): In a peroxide, the oxidation number is -1. NaO2: Na(+1), O(-1)