What is the oxidation number of carbon in Na2C2O4?
2 Answers
Explanation:
You need to do a little algebra on this one.
You have the substance
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
The oxidation number of each carbon atom in
Explanation:
The sum of the oxidation numbers of the elements in a compound is zero.
The oxidation number for sodium is pretty much always
Since there are two sodium atoms, the total oxidation number for sodium is
#+2# .
The oxidation number for oxygen is
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#