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  • Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5 Cr2O7^2- and NO3^-

8.5 Calculate the oxidation number of sulphur, chromium, and nitrogen in  H_{2}SO_{5}, Cr_{2}O_{7}^{2-}and NO_{3}^{-} . Suggest the structure of these compounds. Count for the fallacy. 

Answers (1)

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Answer-

(i)H_{2}SO_{5}  let the oxidation number of sulphur be x

So, 

\\2*1+x+5(-2)= 0\\ x= +8

There is a fallacy Sulphur cannot have +8 oxidation state because it has maximum +6 oxidation number, not more than that.The structure of H_{2}SO_{5} is shown as follows:

2(H)+1(S)+3(O)+2(O \:\:in \:peroxy\: linkage)

\Rightarrow 2(+1) + 1(x) + 3(-2) + 2(-1) = 0

\Rightarrow x = +6( Answer))

 

(ii) Cr_{2}O_{7}^{2-}

let the oxidation number of chromium be x

now 

\\2x+7*(-2)=-2\\ x= (-2+14)/2\\ x= +6

There is no fallacy here

(iii) NO_{3}^{-}

let assume oxidation number of N is x

Now, \\x+(-2)*3= -1\\(x-6)=-1\\x= +5

here is  no fallacy about the O.N of N in NO_{3}^{-}

Posted by

Gautam harsolia

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