0

Guest

Courses New

the difference in oxidation no of two types of s-atoms in na2s4o6 is

the difference in oxidation no of two types of s-atoms in na2s4o6 is

Grade:12

1 Answers

Arun
25750 Points
6 years ago
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
 
 

Think You Can Provide A Better Answer ?

Other Related Questions on Physical Chemistry

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More