Question

\(KMnO_4\), acts as an oxidising agent in acidic medium. 'X' is the difference between the oxidation states of Mn in reactant and product. 'Y' is the number of 'd' electrons present in the brown red precipitate formed at the end of the acetate ion test with neutral ferric chloride. The value of X + Y is _________.

Answer 1

Answer is: 10

\(\underset{\text { (O.A) }}{\mathrm{KMnO}_{4}} \xrightarrow{\text { Acidic medium }} \mathrm{Mn}^{2+}\)

X is difference in oxidation state.

\( 7-2=5\)

So  \(\mathrm{X}=5\)

\(6 \mathrm{CH}_{3} \mathrm{COO}^{\ominus}+\mathrm{Fe}^{3+}+\mathrm{H}_{2} \mathrm{O} \rightarrow\left[\mathrm{Fe}_{3}\left(\mathrm{OH}_{2}\right)\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{6}\right]^{\oplus}+2 \mathrm{H}^{\oplus}\)

\( \left[\mathrm{Fe}_{3}(\mathrm{OH})_{2}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{6}\right]^{\oplus}+4 \mathrm{H}_{2} \mathrm{O} \rightarrow \underset{\text { Brown red ppt }}{\left[\mathrm{Fe}(\mathrm{OH})_{2}\left(\mathrm{CH}_{3} \mathrm{COO}\right]\right.}+\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{H}^{\oplus}\)

\( \mathrm{Fe}^{3+} \Rightarrow 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{0} \ \text{contains 5 d electrons}\)

\( So \ \mathrm{Y}=5\)

\( X+Y=5+5=10\)