Determine the oxidation number (oxidation state) of each element in CH 4 . C : H : Determine the oxidation number (oxidation state) of each element in SO 2 . S : O : Determine the oxidation number (oxidation state) of each element in BF 3 . B : F :

1. In CH₄ molecule:

The oxidation number of H = (+ 1)

The oxidation number of C = (- 4)

2. In SO₂ molecule:

The oxidation number of O = (- 2)

the oxidation number of S = (+ 4)

3. In BF₃ molecule:

The oxidation number of F = (- 1)

the oxidation number of B = (+ 3)

Explanation:

Oxidation number represents the oxidation state of an element in a given chemical compound, which can be zero, negative or positive.

The overall charge on a given chemical molecule is equal to sum of the oxidation numbers of all the elements in that molecule.

Now, calculating the oxidation number-

1. In CH₄ molecule:

As we know, the oxidation number of H = (+ 1)

and the overall charge on the molecule = 0

Let the oxidation number of C be x

∴ x + (+ 1) × 4 = 0

⇒ x = (- 4)

Therefore, the oxidation number of C = (- 4)

2. In SO₂ molecule:

As we know, the oxidation number of O = (- 2)

and the overall charge on the molecule = 0

Let the oxidation number of S be x

∴ x + (- 2) × 2 = 0

⇒ x = (+ 4)

Therefore, the oxidation number of S = (+ 4)

3. In BF₃ molecule:

As we know, the oxidation number of F = (- 1)

and the overall charge on the molecule = 0

Let the oxidation number of B be x

∴ x + (- 1) × 3 = 0

⇒ x = (+ 3)

Therefore, the oxidation number of B = (+ 3)

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of c is as shown here in the image.