Question

An element A in a compound ABD has oxidation number $A^{n-}$. It is oxidised by $Cr_{2}O^{2-}_{7}$ in acid medium. In the experiment $1.68 \times10^{-3}$ moles of $K_{2}Cr_{2} O_{7}$ was 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

Answer: (B)

Meq. of $K_{2} Cr_{2} O_{7} =$ Meq. of ABD
n-factor of $K_{2}Cr_{2}O_{7}$ , in acidic medium $= 6$
$6 \times 1.68 \times10^{-3} = x \times 3.26 \times 10^{-3}$
$x = 3$
$\Rightarrow $ New oxidation state of $A^{-n}$ will be = $-n + 3$