Question

What is the oxidation number of carbon in Na2C2O4?

Answer 1

`C^"+3"`

Explanation:

You need to do a little algebra on this one.

You have the substance

`Na_2C_2O_4` = zero oxidation state

Converting this into numbers based on their known oxidation states, you will have:

Na = 2 (+1) = +2
C = 2 (x) = 2x, where x is the unknown
O = 4 (-2) = -8

Hence,

(+2) + 2x + (-8) = 0

2x + (-6) = 0

2x = +6

x = +3

Therefore, the oxidation state of one Carbon atom in the substance `Na_2C_2O_4` is +3.

Answer 2

The oxidation number of each carbon atom in `"Na"_2"C"_2"O"_4"` is +3.

Explanation:

`"Na"_2"C"_2"O"_4"` is the compound sodium oxalate.

The sum of the oxidation numbers of the elements in a compound is zero.

The oxidation number for sodium is pretty much always `+1`.

Since there are two sodium atoms, the total oxidation number for sodium is `+2`.

The oxidation number for oxygen is `-2`, except in peroxides. Oxalate is not a peroxide, so the oxidation number here is still `-2`.

Since there are four oxygen atoms, the total oxidation number for the oxygen atoms is `-8`.

The sum of the oxidation numbers for sodium and oxygen is `+2 - 8 = -6`. Therefore, the total oxidation number for carbon must be `+6` in order for the sum of the oxidation numbers to equal zero.

Divide `+6` by two to get the oxidation number of each carbon atom, which is `bb(+3)`.

Oxidation numbers for all elements in the compound sodium oxalate:

`stackrel(2 xx +1)("Na"_2)" "stackrel(2 xx +3)("C"_2)" "stackrel(4 xx -2)("O"_4)`

`2xx(+1)+2 xx (+3)+4xx(-2) = 0`