Question

Oxidation number of S in (CH3) 2S

Answer 1

Answer:

S is double-bonded to O (— 2 electrons) and is bonded to 2 C (+2 electrons). +2 + -2 = 0. S has a 0 Oxidation number in (CH3)2SO.

Answer 2

This is an incomplete question, here is a complete question.

Oxidation number of S in (CH₃)₂SO ?

The oxidation number of sulfur (S) is, (0)

Explanation :

Oxidation number : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

When the atoms are present in their elemental state then the oxidation number will be zero.

Rules for Oxidation Numbers :

The oxidation number of a free element is always zero.

The oxidation number of a monatomic ion equals the charge of the ion.

The oxidation number of  Hydrogen (H)  is +1, but it is -1 in when combined with less electronegative elements.

The oxidation number of  oxygen (O)  in compounds is usually -2, but it is -1 in peroxides.

The oxidation number of a Group 1 element in a compound is +1.

The oxidation number of a Group 2 element in a compound is +2.

The oxidation number of a Group 17 element in a binary compound is -1.

The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.

The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

The given compound is, $(CH_3)_2SO$

Let the oxidation state of 'S' be, 'x'

In this molecule, oxygen is double bonded with sulfur.

So, Hydrogen has the oxidation number +1, oxygen has (-2), and carbon has (-4)

$2[-4+3(+1)]+x+(-2)=0\\\\x=0$

Therefore, the oxidation number of sulfur (S) is, (0)

Learn more about : oxidation number

https://brainly.com/question/1542641

https://brainly.com/question/5418214

#learnwithbrainly