What is the oxidation number of carbon in Na2C2O4?

2 Answers
Oct 25, 2015

Answer:

#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:

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#