Question

The average oxidation state of sulphur in Na2S406
is​

Answer 1

Answer:

Oxidation state of oxygen is -2 so, the oxidation state of  oxygen atoms will be -2× 6 = -12. Let the oxidation state of sulphur is x. Calculate oxidation number of sulphur from is as follows. Therefore, the oxidation number of sulphur in is 2.5.

Answer 2

$\tt\huge\bigstar\boxed {\pink {\sf solution:-}}\bigstar$

sodium

the oxidation number of sodium is 1

alkali earth metals have oxidation state of 1 as they have valency of 1.

sulphur

since we know that sulphur has a variable oxidation state, we'll assume the oxidation number of sulphur to be 'x'

oxygen

the oxidation state of oxygen is:

$\tt\implies 2\: in\: most \:of\: the\: cases$

$\tt\implies 1 \:in \:peroxides\:(like h2o2)$

$\tt\implies 1/2\: in \:superoxide$

since, this is a complex compound we will take the state of oxygen to be (-2).

______________________________

we know that:

sum of oxidation numbers in a compound is equal to the net charge of the compound.

so,

$\tt (1×2 )+ (4×x )+ (6×-2)=0$

$\tt (2)+(4x)-12=0$

$\tt x=\dfrac {10}{4 }$

thus the average oxidation state of sulphur is equal to 10/4 or 2.5