00:01
Here in this problem we have to assign oxidation state of sulfur in each of the given compounds.
00:09
Here first one, oxidation number of sulfur will be zero.
00:14
Oxidation number of sulfur is zero air because it is in the elemental form.
00:23
Then h2 -so -4.
00:26
Here, oxidation number of hydrogen here is positive 1, and oxidation number of oxygen is negative 2 in its compound except in peroxide.
00:39
And let the oxidation number for sulfur be x.
00:43
Therefore, we can write two hydrogen atoms, so two times one plus 1 -1 -1 -s, so just x and four oxygen so four times negative two is equal to zero why zero because some of oxidation numbers of all the atoms in a species is equal to the total charge of the species here it is a neutral compound h2 so forth so charge total charge will be zero so two plus x minus 8 is equal to 0 so x is equal to positive 6 therefore oxidation number of sulfur here is plus 6 then third one is sodium hydrogen sulfate here sodium is the alkali metal and alkalisal is assigned oxidation number of positive one.
01:45
Same is true for hydrogen.
01:48
And oxygen is assigned negative to as oxidation number in its compound except in peroxide.
01:56
And let the oxidation number of sulfur be x.
01:59
So we can write for sodium 1 plus for hydrogen 1, 1 hydrogen atom 1.
02:08
And for sulfur x and 4 oxygen atoms 4 times.
02:12
Negative 2 is equal to 0...