Question

Match List-I (oxidation number) with List-ll (the element) and select the correct answer using the codes given below the lists.

List-I	List-II
A. 2	1. Oxidation number of Mn in MnO2
B. 3	2. Oxidation number of S in H2SO4
C. 4	3. Oxidation number of Ca in CaO
D. 6	4. Oxidation number of Al in NaAIH4

Answer 1

Correct option is (a)

The oxidation number of any compound is always zero. This means that sum of the oxidation number of all the elements present in the compound is zero. By this rule, we can now find the oxidation number in given options here 

(1) Suppose, the oxidation number of Mn in MnO₂ is x. MnO₂, contains two oxygen ions, each with an oxidation number (-2).

⇒ x + 2 x (-2) = 0

⇒ x - 4 = 0

⇒ x = 4

(2) Suppose, the oxidation number of S in H₂SO₄ is x H₂SO₄ contains two hydrogen ion, each with an oxidation number (+ 1) and oxygen (-2).

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

⇒ 2 + x - 8 = 0

⇒ x = 6

(3) The oxidation number of Ca in CaO when oxygen has oxidation number (-2).

⇒ x - 2 = 0

⇒ x = 2

(4) Suppose, the oxidation number of Al in NaAIH₄ is x. NaAIH₄ contains one Na which has oxidation number (+1) 

and four H which has oxidation number (-1).

⇒ (+ 1) + x + 4(-1) = 0

⇒ 1 + x - 4 = 0

⇒ x = 3