Ncert Exercises & Solutions

8.5 Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O72– and NO3–. Suggest structure of these compounds. Count for the fallacy.

$(i) \overset{+ 1}{H} \overset{x}{S} {\overset{- 2}{O}}_{5}$
$2 (+ 1) + 1 (x) + 5 (- 2) = 0$
$\Rightarrow 2 + x - 10 = 0$
$\Rightarrow x = + 8$

However, the O.N. of S cannot be +8. S has six valence electrons. Therefore, the O.N. of S cannot be more than +6.
The structure of $H_{2} {SO}_{5}$ is shown as follows:

$Now, 2 (+ 1) + 1 (x) + 3 (- 2) + 2 (- 1) = 0$
$\Rightarrow 2 + x - 6 - 2 = 0$
$\Rightarrow x = + 6$
$Therefore, the O . N . of S is + 6 .$
$(ii) \overset{x}{C} r_{2} \overset{2 -}{O_{7}^{2 -}}$
$2 (x) + 7 (- 2) = - 2$
$\Rightarrow 2 x - 14 = - 2$
$\Rightarrow x = + 6$
$Here, there is no fallacy about the O . N . of Cr in {Cr}_{2} O_{7}^{2 -} .$
$The structure of {Cr}_{2} O_{7}^{2 -} is shown as follows :$

Here, each of the two Cr atoms exhibits the O.N. of +6.

$(iii) \overset{x}{N} \overset{2 -}{O_{3}^{-}}$
$1 (x) + 3 (- 2) = - 1$
$\Rightarrow x - 6 = - 1$
$\Rightarrow x = + 5$
$H e r e, t h e r e i s n o f a l l a c y a b o u t t h e O . N . o f N i n N O_{3}^{-} .$
$T h e s t r u c t u r e o f N O_{3}^{-} i s s h o w n a s f o l l o w s :$

The N atom exhibits the O.N. of +5.

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