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Answer 1

QUESTION ( For the benefit of readers )

The brown ring complex compound is formulated as $[Fe(H_2O)_5NO]SO_4$ . The oxidation state of iron is :

A) 1

B) 2

C) 3

D) 0

ANSWER :

OPTION A

EXPLANATION :

The oxidation number of a particular element can be calculated by taking the oxidation number of the whole compound as 0 .

The oxidation number of an element which is in free state is always zero .

The oxidation number of Group 1 elements is taken as + 1 and those of Group 17 is taken as - 1 .

Let the oxidation number of iron be x .

We know that the oxidation number of $H_2O$ is 0 since it is in free state .

The oxidation number of $NO$ will be + 1 .

The oxidation number of $SO_4$ will be - 2 as the radical carries negative charge .

So we must add the oxidation numbers to get 0 as the total oxidation numbers of the compound is 0 .

x + 5 × 0 + ( + 1 ) + ( - 2 ) = 0

⇒ x + 0 + 1 - 2 = 0

⇒ x - 1 = 0

⇒ x = + 1

The oxidation number is + 1 .

Answer 2

Explanation:

!HELLO

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