Question

The reaction of $K_3\left[Fe\right(CN{)}_6]$ with freshly prepared $FeS{O}_4$ solution produces a dark blue precipitate called Turnbull’s blue. Reaction of $K_4\left[Fe\right(CN{)}_6]$ with the $FeS{O}_4$ solution in complete absence of air produces a white precipitate $X$, which turns blue in air. Mixing the $FeS{O}_4$ solution with $NaN{O}_3,$ followed by a slow addition of concentrated $H_{2}S{O}_4$ through the side of the test tube produces a brown ring.

Among the following, the brown ring is due to the formation of

• A. $\left[Fe\right(NO{)}_2(S{O}_4{)}_2{]}^{2-}$
• B. $\left[Fe\right(NO{)}_2(H_{2}O{)}_4{]}^{3+}$
• C. $\left[Fe\right(NO{)}_4\left(S{O}_4{)}_2\right]$
• D. $\left[Fe\right(NO\left)\right(H_{2}O{)}_5{]}^{2+}$

Answer 1

The correct option is D $\left[Fe\right(NO\left)\right(H_{2}O{)}_5{]}^{2+}$

$N{O}_3^{-}+3F{e}^{2+}+4H^{+}\to NO+3F{e}^{3+}+2H_{2}O$
$\left[Fe\right(H_{2}O{)}_6{]}^{2+}+NO\to \left[Fe\right(H_{2}O{)}_{5}NO{]}^{2+}+H_{2}O$