Question

Oxidation state of $Cr$ in $K_{2}C{r}_2{O}_7$ is

• A. +6
• B. -7
• C. +2
• D. -2

Answer 1

The correct option is A

+6

The oxidation number of oxygen is -2 in most of its compounds , except in peroxides like $H_2{O}_2,Ba{O}_2$ etc. where it is -1

1.Another interesting exception is found in the compound $O{F}_2$ (oxygen difluoride) where the oxidation number of oxygen is +2

2.This is due to the fact that fluorine being the most electronegative element known has always an oxidation number of -1.

3. In compounds formed by union of metals with non-metals, the metal atoms will have positive oxidation numbers and the non-metals will have negative oxidation numbers.

4. The oxidation number of alkali metals ($Li,Na,K$ etc.) is always +1 and those of alkaline earth metals ($Be,Mg,Ca$ etc) is +2.

Therefore, O.N. (O) = -2

O.N. (K) = +1 (alkali metal)

$K_{2}C{r}_2{O}_7$$⟹$ 2$\times$(1) +2$\times$n + 7$\times$(-2) = 0

$⟹$ n =6