Question

What is the oxidation number of Br in the compound \[RbBr{{O}_{4}}\]?
(A) -1
(B) +7
(C) +1
(D) +4

Answer 1

Hint: Solve this question by summing the oxidation number of the compound since in the given compound is neutral and has net charge zero. Assign a random variable ‘x’ for calculating oxidation number of Br.

Complete step by step answer:
The compound \[RbBr{{O}_{4}}\]is called Rubidium Perbromate. This compound is neutral and therefore we can say it has a net charge = 0.
Rubidium (Rb) is an alkali metal or a group I element. Therefore, its oxidation number will be = +1.
Oxygen has an oxidation state of -2, since it has 6 electrons in its valence shell.
Let the oxidation number of Bromine be ‘x’.
Equating oxidation state of each element with the oxidation state of the compound,
(+1) + (x) + 4(-2) = 0
x = +7
Therefore, the answer is – option (b) – oxidation number of Br in the compound \[RbBr{{O}_{4}}\]is +7.

Additional Information: The oxidation number of an atom or element is a number that indicates the total number of electrons lost or gained by it.

Note: There are certain rules for calculating oxidation number –
- Oxidation number of any free element is zero.
- Oxidation number for a monatomic ion is equal to the net charge on it.
- Hydrogen, in general, has an oxidation state equal to +1. However, it is -1 in forms of a - compound with an element with lesser electronegativity.
- Alkali metals have oxidation number = +1
- Alkali earth metals have oxidation number = +2
- Halogens have oxidation number = -1
- Oxygen has an oxidation number = -2, but in case of peroxides, it is -1.