Calculate the oxidation number of underlined elements in the following species.
`ul(C)O_2 , ul(Cr_2)O_7^(2-) , ul(Pb_3) O_4 , ul(P)O_4^(3-)`

Explanation

1. C in `CO_2 ` Let oxidation number of C be x. Oxidation number of each O atom = -2. Sum of oxidation number of all atoms = x+2 (-2) `implies` x - 4.
As it is neutral molecule, the sum must be equal to zero.
`:.` x-4=0 (or) x= +4
2. `Cr " in " Cr_2O_7^(2-)`Let oxidation number of Cr = x. Oxidation number of each oxygen atom = -2. Sum of oxidation number of all atoms
`2 x + 7(-2) = 2 x -14`
Sum of oxidation number must be equal to the charge on the ion.
Thus , 2x -14 =-2
` 2x = + 12 `
`x = 12//2`
`x = 6`