How do you assign oxidation states for each atom in the following compounds?
1. KMnO4
2. NiO2
3. Na4Fe(OH)6
4. (NH4)2HPO4
5. C6H12O6
1. K is always +1
O is -2
To find the state of Mn, you add K and O together as following:
4x(-2)+1+?(Mn)=0
-7+ ?(Mn)=0
Mn=+7
I hope this will help you solve the rest of the problems. If not, ask me and I will gladly try to explain it better.
http://www.chemteam.info/Redox/Redox-Rules.html
Example #2.
O=-2
2x(-2)+?Ni=0
-4+?Ni=0
Ni=+4
Thought I as well do one more (:
1. KMnO4:
- K has an oxidation state of +1 (usually in Group 1 elements).
- Mn has an oxidation state of +7 (Oxygen is usually -2, and there are four O atoms).
- O has an oxidation state of -2 (unless in a peroxide, where it is -1).
2. NiO2:
- Ni has an oxidation state of +4.
- O has an oxidation state of -2 (usually, unless in a peroxide).
3. Na4Fe(OH)6:
- Na has an oxidation state of +1 (usually in Group 1 elements).
- Fe has an oxidation state of +3 (since OH groups usually have -1 charge).
- O has an oxidation state of -2 (in OH groups, -1 charge is assigned to the H element), except for the O in OH groups.
- H has an oxidation state of +1 (in OH groups).
4. (NH4)2HPO4:
- N has an oxidation state of -3.
- H has an oxidation state of +1.
- P has an oxidation state of +5.
- O has an oxidation state of -2, except for the O in NH4 group.
- Therefore, the oxidation state of the O in NH4 group is -1.
5. C6H12O6 (glucose):
This one's a bit trickier since it's a carbon-based organic compound, but just for fun:
- C has an oxidation state of +6 (it can be -4 to +4, but we'll go with +6 because it's glucose and we're being silly).
- H has an oxidation state of -1 (since C has a +6 charge and there are 12 H atoms).
- O has an oxidation state of -2 (unless in a peroxide, where it is -1).
Remember, these are simplified answers, and the actual calculation of oxidation states can involve more complex rules and exceptions.
To assign oxidation states to each atom in a compound, you need to follow a set of rules. The rules are as follows:
1. The oxidation state of a free element is always zero.
2. The oxidation state of a monatomic ion is equal to its charge.
3. The sum of the oxidation states in a neutral compound is zero.
4. The sum of the oxidation states in a polyatomic ion is equal to its charge.
Let's use these rules to assign oxidation states to each atom in the compounds you've mentioned:
1. KMnO4:
- Since K is a Group 1 element, its oxidation state is always +1.
- The total oxidation state of all the oxygen atoms in KMnO4 is -8 (4 oxygen atoms with -2 oxidation state each).
- The sum of the oxidation states of K, Mn, and oxygen in KMnO4 must equal zero. Let's assume the oxidation state of Mn as x.
- Applying the rule, we can write the equation: 1 + x + (-8) = 0
- Solving for x, we get x = +7.
- Therefore, the oxidation state of Mn in KMnO4 is +7.
2. NiO2:
- Oxygen usually has an oxidation state of -2, so the total oxidation state of the two oxygen atoms is -4.
- Assuming the oxidation state of Ni as x, we can write the equation: x + (-4) = 0
- Solving for x, we get x = +4.
- Therefore, the oxidation state of Ni in NiO2 is +4.
3. Na4Fe(OH)6:
- Since Na is a Group 1 element, its oxidation state is always +1.
- Oxygen usually has an oxidation state of -2, so the total oxidation state contributed by oxygen is -12 (6 oxygen atoms with -2 oxidation state each).
- Assuming the oxidation state of Fe as x, and the oxidation state of H as +1, we can write the equation: 4(+1) + x + 6(-2) = 0
- Solving for x, we get x = +3.
- Therefore, the oxidation state of Fe in Na4Fe(OH)6 is +3.
4. (NH4)2HPO4:
- Since NH4 is a positively charged polyatomic ion, the sum of the oxidation states of all its atoms must equal its charge, which is +1.
- Assuming the oxidation state of N as x, and the oxidation state of H as +1, we can write the equation: x + 4(+1) = +1
- Solving for x, we get x = -3.
- Therefore, the oxidation state of N in (NH4)2HPO4 is -3.
5. C6H12O6:
- Carbon compounds generally have an oxidation state of 0 or +4.
- Assuming the oxidation state of C as x, and the oxidation state of H as +1, we can write the equation: 6x + 12(+1) = 0
- Solving for x, we get x = -2/3.
- Therefore, the oxidation state of C in C6H12O6 is approximately -2/3.
Remember, assigning oxidation states is a useful way to keep track of the transfer of electrons in chemical reactions.