Ncert Exercises & Solutions

23. Calculate the oxidation number of each sulphur atom in the following compounds.

(a) ${Na}_{2} S_{2} O_{3}$

(b) ${Na}_{2} S_{4} O_{6}$

(d) ${Na}_{2} {SO}_{4}$

The oxidation number of each sulphur atom in the following compounds are given below

(a)

{Na}_{2} S_{2} O_{3}

Let us consider the structure of

{Na}_{2} S_{2} O_{3}

There is a coordinate bond between two sulphur atoms. The oxidation number of acceptor S-atom is -2. Let, the oxidation number of other S-atom be x.

\underset{For Na}{2 (+ 1)} + \underset{For O - atoms}{3 \times (- 2)} + x + \underset{For coordinate S - atom}{1 (- 2) = 0}

x = + 6

Therefore, the two sulphur atoms in

{Na}_{2} S_{2} O_{3}

have -2 and +6 oxidation number.

(b)

{Na}_{2} S_{4} O_{6}

Let us consider the structure of

{Na}_{2} S_{4} O_{6}

In this structure, two central sulphur atoms have zero oxidation number because electron pair forming the S-S bond remain in the centre. Let, the oxidation number of (remaining S-atoms) S-atom be x.

\underset{For Na}{2 (+ 1)} + \underset{For O}{6 (- 2)} + 2 x + 2 (0) = 0

2 - 12 + 2 x = 0 or x = + \frac{10}{2} = + 5

Therefore, the two central S-atoms have zero oxidation state and two terminal S-atoms have +5 oxidation state each.

(c)

{Na}_{2} {SO}_{3}

Let the oxidation number of S in

{Na}_{2} {SO}_{3}

be x.

2 (+ 1) + x + 3 (- 2) = 0 or x = + 4

(d)

{Na}_{2} {SO}_{4}

Let the oxidation number of S be x.

2 (+ 1) + x + 4 (- 2) = 0 or x = + 6

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