What is the oxidation state of iron?
Fe(NH4)2(SO4)2
Group of answer choices
Fe2+
Fe3+
Fe
Fe4+
The correct answer and explanation is :
The correct oxidation state of iron (Fe) in the compound Fe(NH4)2(SO4)2 is Fe2+.
Explanation:
To determine the oxidation state of iron in Fe(NH4)2(SO4)2, we need to consider the other components in the compound and apply the rules for oxidation states.
- Ammonium ion (NH4+):
The ammonium ion NH4+ has a charge of +1. Since there are two ammonium ions in the compound, the total contribution to the charge from the ammonium ions is +2. - Sulfate ion (SO4^2-):
The sulfate ion has a charge of -2. Since there are two sulfate ions in the compound, the total contribution to the charge from the sulfate ions is -4. - The compound as a whole is neutral:
The entire compound Fe(NH4)2(SO4)2 must have a neutral charge because no charge is indicated for the overall molecule. Therefore, the sum of the oxidation states of all components must equal zero.
Let’s now break down the equation:
- The total charge from the ammonium ions is +2 (from two NH4+ ions).
- The total charge from the sulfate ions is -4 (from two SO4^2- ions).
- Let the oxidation state of iron be ( x ).
Now, using the equation:
[
x + 2(+1) + 2(-2) = 0
]
[
x + 2 – 4 = 0
]
[
x – 2 = 0
]
[
x = +2
]
Thus, the oxidation state of iron (Fe) in Fe(NH4)2(SO4)2 is Fe2+.
Conclusion:
The oxidation state of iron (Fe) in the compound Fe(NH4)2(SO4)2 is Fe2+.