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# Help me solve this - Electrochemistry - BITSAT

The oxidation number of sulphur in $S_8,S_2F_2,H_2S$ are

• Option 1)

0,1,-2

• Option 2)

2,1,-2

• Option 3)

0,1,2

• Option 4)

-2,1,-2

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As we learnt in

Oxidation Number -

Oxidation number denotes the oxidation state of an element in a compound ascertained according to a set of rules formulated on the basis that electron pair in a covalent bond belogs entirely to more electronegative elements.

-

Rules for Oxidation Number -

In all its compound, flourine has an oxidation number of -1. Other halogen (Cl, Br and I) also have oxidation number of -1, when they occur as halide ions in their compound.

- wherein

e.g. Cl, Br and I when combined with O2  have +ve oxidation number.

Rules for Oxidation Number -

The oxidation number of hydrogen is +1, except when it is bonded to metals in binary compounds (that is compound containing two elements)

- wherein

e.g. LiH,NaH and CaH2 its oxidation number is -1.

Lets take the species one by one

$S_{8}$                          O.N=0

$S_{2}F_{2}$                    O.N=+1

$H_{2}S$                     O.N= - 2

$\therefore$ Solution is 1

Option 1)

0,1,-2

Correct

Option 2)

2,1,-2

Incorrect

Option 3)

0,1,2

Incorrect

Option 4)

-2,1,-2

Incorrect

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