Certainly! Let's analyze the oxidation numbers for the compounds [tex]\( \text{MgCl}_2 \)[/tex] and [tex]\( \text{MnO}_4^- \)[/tex]. Here's a step-by-step solution for each compound:
### Problem 30: [tex]\( \text{MgCl}_2 \)[/tex]
1. Identify the oxidation states of individual elements:
- Magnesium (Mg) is an alkaline earth metal, and typically, the oxidation number of Mg in compounds is [tex]\( +2 \)[/tex].
- Chlorine (Cl) usually has an oxidation number of [tex]\( -1 \)[/tex] in its anionic form.
2. Determine the total oxidation number for the compound:
- The formula of the compound is [tex]\( \text{MgCl}_2 \)[/tex].
- Therefore, there is 1 magnesium atom and 2 chlorine atoms in the compound.
3. Calculate the sum of the oxidation numbers:
- Oxidation number of Mg: [tex]\( +2 \)[/tex]
- Oxidation number of Cl: [tex]\( -1 \)[/tex]
- For 2 chlorine atoms: [tex]\( 2 \times (-1) = -2 \)[/tex]
4. Sum up the oxidation numbers:
[tex]\[
\text{Oxidation number of Mg} + \text{(Oxidation number of Cl)} \times 2 = +2 + (-2) = 0
\][/tex]
Therefore, the sum of the oxidation numbers in [tex]\( \text{MgCl}_2 \)[/tex] is 0, as expected for a neutral compound.
### Problem 31: [tex]\( \text{MnO}_4^- \)[/tex] (Permanganate Ion)
1. Identify the oxidation states of individual elements:
- Oxygen (O) typically has an oxidation number of [tex]\( -2 \)[/tex] in most compounds.
- We need to determine the oxidation number of manganese (Mn) in the permanganate ion ( [tex]\( \text{MnO}_4^- \)[/tex] ).
2. Determine the total oxidation number for the ion:
- The permanganate ion has a charge of [tex]\( -1 \)[/tex].
- There is 1 manganese atom and 4 oxygen atoms in the ion.
3. Set up the equation for the sum of oxidation numbers:
- Let the oxidation number of Mn be [tex]\( x \)[/tex].
- Oxidation number of O: [tex]\( -2 \)[/tex]
- For 4 oxygen atoms: [tex]\( 4 \times (-2) = -8 \)[/tex]
4. Calculate the oxidation number of Mn:
[tex]\[
x + 4 \times (-2) = -1
\][/tex]
[tex]\[
x - 8 = -1
\][/tex]
[tex]\[
x = +7
\][/tex]
5. Check the sum of the oxidation numbers:
[tex]\[
\text{Oxidation number of Mn} + \text{(Oxidation number of O)} \times 4 = +7 + (-8) = -1
\][/tex]
Therefore, the sum of the oxidation numbers in [tex]\( \text{MnO}_4^- \)[/tex] is [tex]\( -1 \)[/tex], which matches the charge of the ion.
