Question
The oxidation state of $$Cr$$ in $${K_2}C{r_2}{O_7}$$ is
A.
$$+5$$
B.
$$+3$$
C.
$$+6$$
D.
$$+7$$
Answer :
$$+6$$
Solution :
Let the oxidation state of $$Cr$$ is $$x$$
$$\eqalign{
& {K_2}C{r_2}{O_7} \cr
& \therefore \,\,2\left( { + 1} \right) + 2x + 7\left( { - 2} \right) = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 + 2x - 14 = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x - 12 = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x = 12 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \frac{{12}}{2} = + 6 \cr} $$