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Grade: 12
        
the difference in oxidation no of two types of s-atoms in na2s4o6 is
5 months ago

Answers : (1)

Arun
11590 Points
							
Average oxidation number of S : 

Na₂S₄O₆ is a neutral compound, and thus the sum of oxidation number is 0. 
(Total oxidation numbers of Na) + (Total oxidation numbers of S) + (Total oxidations of O) = 0 
(+1) × 2 + (Average oxidation number of S) × 4 + (-2) × 6 = 0 
(Average oxidation number of S) × 4 = 10 
Average oxidation number of S = 2.5 


==== 
Oxidation numbers of individual S atoms : 

 

Consider the 2 S atoms in red. 
Each of the S atoms in red is attached to 2 S atoms. 
Hence, oxidation number of each S atom in red = 0 

Consider the part of -SO₃⁻. 
Since the part has a change of -1, the total oxidation numbers = 0 
(Total oxidation numbers of O) + (Oxidation number of S in blue) = -1 
(-2) × 3 + (Oxidation number of S in blue) = -1 
Hence, oxidation number of S in blue = +5 

Answers : 
There are two types of S atoms having different oxidation numbers. 
The oxidation of each S atom attached to 2 other S atoms = 0 
The oxidation of each S atom attached to 1 S atom and 3 O atoms = +5
 
 
5 months ago
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