Question
The brown ring test for $$NO_2^ - $$ and $$NO_3^ - $$ is due to the formation of complex ion with a formula
A.
$${\left[ {Fe{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }}$$
B.
$${\left[ {Fe\left( {NO} \right){{\left( {CN} \right)}_5}} \right]^{2 + }}$$
C.
$${\left[ {Fe{{\left( {{H_2}O} \right)}_5}NO} \right]^{2 + }}$$
D.
$${\left[ {Fe\left( {{H_2}O} \right){{\left( {NO} \right)}_5}} \right]^{2 + }}$$
Answer :
$${\left[ {Fe{{\left( {{H_2}O} \right)}_5}NO} \right]^{2 + }}$$
Solution :
$${\left[ {Fe{{\left( {{H_2}O} \right)}_5}NO} \right]^{2 + }}\,ion\,\,{\text{is formed}}$$