Determine the correct oxidation numbers for all three elements in Na2C2O4.

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Determine the correct oxidation numbers for all three elements in Na2C2O4. (Na is a Group I element) (The rules for assigning Oxidation States are given as below). 1) The oxidation numbers of the atoms in a neutral molecule must add up to zero. 2) Group I = +1, Group II = +2, Group III = -3 3) Fluorine = -1 (always) 4) Halogens (except in compounds with oxygen and other halogens) = -1 5) Hydrogen = +1 (except in metal hydrides e.g. LiH, Rule 2) 6) Oxygen = -2 (except in compounds with fluorine, Rule 3, and compounds with O-O bonds, Rules 2 & 4) A. Na: +1, C: +3, O: -2 B. Na: +1, C: +4, O: -2 C. Na: +1, C: +2, O: -2 D. Na: +1, C: +6, O: -2 E. Na: +2, C: +4, O: -2

The Correct Answer and Explanation is:

The correct answer is A. Na: +1, C: +3, O: -2.

Here’s how the oxidation states are determined:

Step 1: Assign Oxidation States Based on Rules

  1. Sodium (Na):
    • Sodium is a Group I element, and according to Rule 2, elements in Group I have an oxidation number of +1. Therefore, Na is assigned an oxidation state of +1.
  2. Oxygen (O):
    • Oxygen typically has an oxidation number of -2, as per Rule 6, unless it is part of a peroxide or compound with fluorine. Since oxygen is in the oxalate ion (C₂O₄²⁻), we apply the rule and assign -2 to each oxygen atom.
  3. Carbon (C):
    • The molecule is sodium oxalate (Na₂C₂O₄), which contains two carbon atoms. To find the oxidation number of carbon, we use the fact that the sum of all oxidation numbers in a neutral compound must equal zero.
    • We have 2 sodium atoms with an oxidation number of +1 each, and 4 oxygen atoms with an oxidation number of -2 each. So, we can set up an equation to solve for the oxidation state of carbon (C): 2(Na)+2(C)+4(O)=02(\text{Na}) + 2(\text{C}) + 4(\text{O}) = 02(Na)+2(C)+4(O)=0 Substituting known values: 2(+1)+2(C)+4(−2)=02(+1) + 2(\text{C}) + 4(-2) = 02(+1)+2(C)+4(−2)=0 2+2(C)−8=02 + 2(\text{C}) – 8 = 02+2(C)−8=0 2(C)−6=02(\text{C}) – 6 = 02(C)−6=0 2(C)=62(\text{C}) = 62(C)=6 C=+3\text{C} = +3C=+3 Thus, the oxidation state of carbon in sodium oxalate is +3.

Step 2: Verify the Sum of Oxidation Numbers

  • The sum of the oxidation states should be zero for a neutral compound:
    • Sodium (Na): 2 × (+1) = +2
    • Carbon (C): 2 × (+3) = +6
    • Oxygen (O): 4 × (-2) = -8
    • Total: +2 + 6 + (-8) = 0

Therefore, the oxidation numbers are:

  • Na: +1
  • C: +3
  • O: -2

Thus, the correct answer is A.

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