Question

find out the oxidation state of sulphur in so3 and h2so4
find out the oxidation state of CR in K2 cr2 O7 and SO2

please answer this question and you will receive 100 points​
its urgent

Answer 1

1.The Oxidation states in SO3(g) are:

Sulfur (+6) & Oxygen (-2), because SO3(g) has no charge.

However in

SO3- 2

(aq) the Oxidation states are: Sulfur (+4) & Oxygen (-2).

Don't get the two confused, they may both be written without the charge,

•Thus,SO3 is (aq) it will have a charge of -2

2.O.N of Oxygen= -2

O.N of Hydrogen= +1

O.N of Sulphur= x(say)

Now,

2(1+x+4−2)=0

x−6=0

x=6

•Thus, O.N. of Suplhur is 6

Answer 2

Answer:

the oxidation state of sulphur is +6 in both the cases

and oxidation state of Cr in K2Cr2O7 is +6

Explanation:

oxidation state of S in SO3 let b x

and oxidation state of O2 is -2

then,

x+3×(-2)=0

x+(-6)=0

x-6=0

x=+6

Now in H2SO4 let the oxidation state of sulphur be x

and oxidation state of hydrogen and oxygen is +1 and -2

then,

2×(+1)+x+4×(-2)=0

+2+x+(-8)=0

+2+x-8=0

+2+x=+8

x=+8-2

x=+6

Now the oxidation state of Cr in K2Cr2O7 let b x

and the oxidation state of K and O2 is +1 and -2

then,

2×(+1)+2x+7×(-2)=0

+2+2x-14=0

2x-12=0

2x=+12

x=+12÷2

x=+6

HOPE THUS WILL HELP U ....

THANKS DEAD