DIRECTIONS: Assign the correct oxidation number of the individual atom in the compound or ion below. ANSWER IT ALL BELOW.
- S in S8
- Cl in CaCl2
- I in IO3
- C in HCO3
- S in FE2(SO2)3
- S in SO42-
- S in Na2SO4
- As in K3AsO4
- Cr in Cr2O72-
- N in NH4+
NOT A GRADED QUESTION. IT IS A RESEARCH QUESTION.
The Correct Answer and Explanation is:
Here are the oxidation numbers for each specified atom, along with a 300-word explanation:
Answers:
- S in S₈: 0
- Cl in CaCl₂: –1
- I in IO₃⁻: +5
- C in HCO₃⁻: +4
- S in Fe₂(SO₂)₃: +4
- S in SO₄²⁻: +6
- S in Na₂SO₄: +6
- As in K₃AsO₄: +5
- Cr in Cr₂O₇²⁻: +6
- N in NH₄⁺: –3
Explanation
Oxidation numbers help keep track of electrons in chemical reactions, especially redox (oxidation-reduction) reactions. They are assigned based on a set of standard rules:
- Pure elements, like S in S₈, have oxidation numbers of 0.
- In ionic compounds like CaCl₂, calcium (a Group 2 metal) is +2, and each Cl is –1.
- In polyatomic ions, we assign known oxidation numbers first (e.g., O is usually –2) and solve for the unknown:
- For IO₃⁻, 3(–2) + x = –1 → x = +5 for I.
- In HCO₃⁻, H is +1, O is –2, and solving gives C as +4.
- For SO₂²⁻ in Fe₂(SO₂)₃, assume O is –2, giving S as +4.
- SO₄²⁻ has S as +6 because 4(–2) + x = –2 → x = +6.
- In Na₂SO₄, Na is +1, and since there are two Na⁺, the SO₄²⁻ must balance as –2, confirming S is still +6.
- In K₃AsO₄, K is +1, and O is –2; solving gives As = +5.
- Cr₂O₇²⁻ has total charge –2; 7 O = –14, so 2Cr must be +12 → Cr = +6.
- In NH₄⁺, H is +1; 4(+1) + x = +1 → x = –3 for N.
Understanding oxidation states helps chemists identify electron flow, predict reactions, and balance redox equations efficiently.
