Question

When ${\text{K}}_{\text{2}}\text{C}{\text{r}}_{\text{2}}{\text{O}}_{\text{7}}$ is converted into ${\text{K}}_{\text{2}}\text{Cr}{\text{O}}_4$, the change in oxidation number of Cr is

• A. 0
• B. 6
• C. 4
• D. 3

Answer 1

The correct option is A 0

For ${\text{K}}_{\text{2}}\text{C}{\text{r}}_{\text{2}}{\text{O}}_{\text{7}}$

we consider oxidation no. of Cr as x , we will have

2(1) + 2(x) + 7 × (-2) = 0.….(as oxidation no. of alkali metals is 1 and of oxygen is -2 .Also we know charge on overall atom is 0)

Hence we get x = 6

Oxidation no. Of Cr is 6

for ${\text{K}}_2\text{C}{\text{r}}_{\text{4}}$ similarly oxidation no. is 6

Hence change is 0

(A) is correct Option.