An element `A` in a compound `ABD` has oxidation number `A^(n-)`. It is oxidised by `Cr_(2)O_(7)^(2-)` in acid medium. In the experiment `1.68xx10^(-3)` moles of `K_(2)Cr_(2)O_(7)` were used for `3.26xx10^(-3)` moles of `ABD`. The new oxidation number of `A` after oxidation is:

Answer

1

Explanation

`A^(2+) to A^(a+) + (a+2)e^(-)`
`6e^(-) + (Cr^(6+))_(2) to 2Cr^(3+)`
Also Meq. Of A = Meq. Of `K_(2)Cr_(2)O_(7)`
`=3.26 xx 10^(-3) (a+2) = 1.68 xx 10^(-3) xx 6`
`therefore a+2 =3, rArr a=3-2=1`