What is the oxidation state of an individual sulfur atom in baso4?
Answers
Is +3 and O2 is -2 so when combining something like S8 + O2 you ger sulfur dioxide (SOz) but when balanced it is S2O3
the oxidation state of an individual sulfur atom in SO₃⁻² is +4.
Explanation:
1) Fundamental considerations: There are important rules that you need to know to find the oxidation states of neutral compounds or ions.
In a neutral compound the sum of the oxidation states is equal to zero. In an ion, the sum of the oxidation states is equal to the overall ionic charge.Except in peroxides, oxygen (O) atoms bonded to other atoms have oxidation state -2.Those are the relevant rules for this question.
2) Procedure:
The ion for which you have to fiind the oxidation state of an indivitual sulfur atom is SO₃⁻².Let x be the oxidation state of the sulfur (S) atom.Oxidation state of O = -2Total oxidation number for 3 atoms of O: 3 × (-2) = - 6Overall ionic charge of SO₃⁻²: - 2Sum of the oxidation states: x + (-6) = overal charge⇒ x + (-6) = -2Solve for x: x = - 2 + 6 = + 44) Conclusion: the oxidation state of an individual sulfur atom in SO₃⁻² is +4.
The oxidation state of single oxygen atom is -2 since it is more electronegative and like to gain electrons.
The sum of the oxidation states of each atom = charge of compound
Let’s say the oxidation state of Sulfur is x.
Hence,
x +4 * (-2) = -2
x – 8 = -2
x = +6
The oxidation state of individual Sulfur atom is +6
The oxidation state of an individual sulfur atom in BaSO4 = 6+
Explanation
In a molecule the oxidation number is the sum of oxidation numbers of its constituent atoms.
for our case it is the oxidation number of BaSO4 = Ba + S +O=0
Since the oxidation number of S is is unknown let be represented by YThe oxidation state of Ba= 2+ and O=2-
Therefore the oxidation state of Y is calculates as below= 2+ +y + (4 x2-)=0
= 2+ + y +8- =0
like terms together
=Y +2 -8=0
=y-6=0
take -6 to the other side
y= 6+
This does not make any since to me