Question

Question 8.5 Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5,Cr2O7(2-) and NO3(-). Suggest structure of these compounds. Count for the fallacy.

Class XI Redox Reactions Page 272

Answer 1

(i)H2SO5,
let oxidation number of S is x.
+2 + x +5(-2) = 0
+2 + x - 10 = 0
x = +8 ,but oxidation number of S can't be 8 or more than 8 because S has only 6 valance electrons. This fallacy is removed by calculating oxidation of S by chemical bonding method.
(see attachment ).

in H2SO5 two oxygen atoms are in -1 oxidation state.

Let see the oxidation number of S is x.
2(+1) + x + 3(-2)+2(-1) =0
+2 + x -6 -2 = 0
x = +6
Hence, oxidation number of S is +6.

(ij)Cr2O7^2-
Let the oxidation number of Cr is x.
2x +7(-2) = -2
2x = 12 =>x = +6
Let us consider the structure of Cr2O7^2- ion.(see attachment)

here we see that

now, 2x + 4(-2)+2(-1)+(-2)=0
2x - 12 =0 => x = +6
hence, oxidation number of Cr has same . So, for this molecule there is no fallacy.

(iii)NO3^-
let oxidation number of N is x
x + 3(-2) = -1
x = +5 ,
so, oxidation number of N is +5 ,
Now we see structure of NO3- ion
(see attachment)

here we see that , oxidation number of two O-atoms are -2 and one O-atom is -1
e.g., -1 + x + 2(-2) = 0
x = +5 , so, oxidation number of N is same . Hence, there is no fallacy for this molecule.

Answer 2

Explanation:

1.) H₂SO₅

Oxidation number of H = +1

Oxidation number of O = -2

let the oxidation number of S be x

2(+1)+x+5(-2)=0

2+x-10=0

x=8

oxidation number of S is 8  is impossible because the maximum O.N. of S cannot be more than six since, it has only six electrons in the valence shell. This fallacy is overcome if we calculate the O.N. of S by chemical bonding method. The structure of H2SO5 is

                     $O^{-2}$

                     ║      

$H^{+1}$----$O^{-2}$----$S^{+6}$----O----O----$H^{+1}$

                     ║                

                     $O^{-2}$

2(+1)+x+3(-2)+2(-1) (here O---O shares one electron so it has -1 Oxi. No.)=0

2+x-6-2=0

x = +6

__________________________________________________________

2.) $Cr_{2}$$O_{7} ^{-2}$

let oxidation number of Cr be x

2(x)+7(-2)= -2

2x-14 = -2

x = +6

           $O^{-2}$                 $O^{-2}$

           ║                     ║

$O^{-1}$----$Cr^{+6}$----$O^{-2}$----$Cr^{+6}$----$O^{-1}$

           ║                     ║

           $O^{-2}$                 $O^{-2}$

for checking

2x+4(-2)+2(-1)-2=0

2x-8-2-2 = 0

x = +6

__________________________________________________________

3.) NO₃⁻

let oxidation number of N be x

x+3(-2) = -1

x-6 = -1

x = +5

$O^{-2}$

║

$N^{+5}$----$O^{-1}$

║

$O^{-2}$

for checking

x+2(-2)-1 = 0

x-4-1 = 0

x = +5