Question

9 Calculate oxidation number of underlined elements in following molecule: a) KCr,0 b) Na2S2O3​

Answer 1

Answer:

a) Let the Ox. no. of Cr in K2Cr2O7 be x.

We know that, Ox. no. of K=+1

Ox. no. of O=−2

So, 2(Ox.no.K)+2(Ox.no.Cr)+7(Ox.no.O)=0

2(+1)+2(x)+7(−2)=0

or +2+2x−14=0

or 2x=+14−2=+12

or x=+12/2

x=+6.

Hence, oxidation number of Cr in K2Cr2O7 is +6.

b)Na2S2O3:2(+1)+ 2(oxidation no of  S) +3(−2)=0

2(Oxidation no of S) = 4 

(Oxidation no of S) = 2