Question

An element A in a compound ABD has oxidation number A ^n-. It is oxidised by $C{r}_2{O}_7^{-2}$ in acidic medium. In the experiment 1.68$\times$${10}^{-3}$ x moles of$K_{2}C{r}_2{O}_7$ were used for 3.26$\times$${10}^{-3}$ moles of ABD. The new oxidation number of A after oxidation is:

• A. 3
• B. 3-n
• C. n-3
• D. +n

Answer 1

The correct option is B

3-n

First - let us write down the reaction:
$C{r}_2{O}_7^{2-}+A^{n-}BD⟶C{r}^{3+}+A^{-n+x}$

From stoichiometry, we have:
Equivalents of $C{r}_2{O}_7^{2-}$ = equivalents of ABD
$1.68\times {10}^{-3}=3.26\times {10}^{-3}\times x$
Solving, we get $x=3$.
Hence, the new oxidation number is: 3-n