Question

Find the oxidation number of sulphur in $S_{2}O_{7}^{2-}$.

Answer 1

· Diatomic molecules like ${ H }_{ 2 },\quad { Cl }_{ 2 }........$ have zero oxidation state.

· The "oxidation number" of a "mono atomic" ion is equal to the "ion's charge".

· The "sum" of all "oxidation numbers" in a "neutral compound" is zero.

· Alkaline metal have +1 oxidation number and "alkaline earth metals" have +2 oxidation number.

· Oxygen have  -2 oxidation state But the oxidation number of oxygen in peroxide form is -1.  

· Oxidation number is Hydrogen is -1 but in binary metal hydride oxidation number of hydrogen is +1.

· Oxidation number of halide are always -1, unless they are in combination with an oxygen or fluorine.

${ S }_{ 2 }{ O }_{ 7 }^{ 2- }$

Let the oxidation state of S be x.

The oxidation state of oxygen is -2

Therefore,

$2(x)\quad +\quad 7(-2)\quad =\quad -2$

$2x\quad -\quad 14\quad =\quad -2$

$2x\quad =\quad 12$

$x\quad =\quad 6$

Therefore, the "oxidation state" of "sulphur" is +6