
The ion or group detected by $\left[ {{K_2}Hg{I_4}} \right]$ is:
A.NO
B.$C{l^ - }$
C.$N{H^{ - 2}}$
D.$N{H_3}$
Answer
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Hint: $\left[ {{K_2}Hg{I_4}} \right]$ is the reagent used for the detection of group 0 cation. Identify the cation that belongs to group 0. $\left[ {{K_2}Hg{I_4}} \right]$ is dissolved in KOH and is used to perform Nessler’s test. A brown colouration or a precipitate of basic mercury (II) amido-iodine is formed.
Complete step by step answer:
Potassium tetraiodomercurate(II), $\left[ {{K_2}Hg{I_4}} \right]$, is an inorganic compound consisting of potassium cations and the tetraiodomercurate(II) anion.
the 0.09 mol/L solution of $\left[ {{K_2}Hg{I_4}} \right]$ in 2.5 mol/L KOH is called Nessler’s reagent.
It is used to detect the presence of ammonia ($N{H_3}$).
It is named after the German chemist Julius Nessler who devised this method of colorimetric measure of ammonia.
$\left[ {{K_2}Hg{I_4}} \right]$ solution has a pale yellow colour and becomes deep yellow in the presence of ammonia. At higher concentrations, a brown precipitate may form.
The reaction can be summarized as follows:
$2{K_2}Hg{I_4} + N{H_3} + 3KOH \to HgO.Hg(N{H_2})I + 7KI + 2{H_2}O$
The active species is ${\left[ {Hg{I_4}} \right]^{2 - }}$. A brown precipitate of basic mercury(II) amido-iodine ($HgO.Hg(N{H_2})I$) is formed.
Nitric oxide can be determined using a chemiluminescent reaction involving ozone. A sample containing nitric oxide is mixed with a large quantity of ozone. The nitric oxide reacts with the ozone to produce oxygen and nitrogen dioxide, accompanied by the emission of light (chemiluminescence) which can be measured with a photodetector.
\[NO + {O_3} \to {\text{ }}N{O_2} + {O_2} + h\nu \]
$C{l^ - }$ can be detected by heating it with $Mn{O_2}$ and concentrated ${H_2}S{O_4}$. A greenish-yellow gas ($C{l_2}$) is formed.
$N{H^{ - 2}}$ cannot form ammonia during Nessler's test. Hence it cannot be detected.
The correct answer is (D) $N{H_3}$.
Note: The brown precipitate formed in the reaction of ammonia with $\left[ {{K_2}Hg{I_4}} \right]$ is also called as the iodide of Millon’s base.
Salts containing $N{H_4}^ + $ ion can also be detected by Nessler's reagent.
Ammonia gas evolved by the action of sodium hydroxide and ammonium salts reacts with hydrochloric acid to give ammonium chloride, which is visible as dense white fume. On passing the gas through Nessler’s reagent, a brown colouration or a precipitate of basic mercury (II) amido-iodine is formed.
Complete step by step answer:
Potassium tetraiodomercurate(II), $\left[ {{K_2}Hg{I_4}} \right]$, is an inorganic compound consisting of potassium cations and the tetraiodomercurate(II) anion.
the 0.09 mol/L solution of $\left[ {{K_2}Hg{I_4}} \right]$ in 2.5 mol/L KOH is called Nessler’s reagent.
It is used to detect the presence of ammonia ($N{H_3}$).
It is named after the German chemist Julius Nessler who devised this method of colorimetric measure of ammonia.
$\left[ {{K_2}Hg{I_4}} \right]$ solution has a pale yellow colour and becomes deep yellow in the presence of ammonia. At higher concentrations, a brown precipitate may form.
The reaction can be summarized as follows:
$2{K_2}Hg{I_4} + N{H_3} + 3KOH \to HgO.Hg(N{H_2})I + 7KI + 2{H_2}O$
The active species is ${\left[ {Hg{I_4}} \right]^{2 - }}$. A brown precipitate of basic mercury(II) amido-iodine ($HgO.Hg(N{H_2})I$) is formed.
Nitric oxide can be determined using a chemiluminescent reaction involving ozone. A sample containing nitric oxide is mixed with a large quantity of ozone. The nitric oxide reacts with the ozone to produce oxygen and nitrogen dioxide, accompanied by the emission of light (chemiluminescence) which can be measured with a photodetector.
\[NO + {O_3} \to {\text{ }}N{O_2} + {O_2} + h\nu \]
$C{l^ - }$ can be detected by heating it with $Mn{O_2}$ and concentrated ${H_2}S{O_4}$. A greenish-yellow gas ($C{l_2}$) is formed.
$N{H^{ - 2}}$ cannot form ammonia during Nessler's test. Hence it cannot be detected.
The correct answer is (D) $N{H_3}$.
Note: The brown precipitate formed in the reaction of ammonia with $\left[ {{K_2}Hg{I_4}} \right]$ is also called as the iodide of Millon’s base.
Salts containing $N{H_4}^ + $ ion can also be detected by Nessler's reagent.
Ammonia gas evolved by the action of sodium hydroxide and ammonium salts reacts with hydrochloric acid to give ammonium chloride, which is visible as dense white fume. On passing the gas through Nessler’s reagent, a brown colouration or a precipitate of basic mercury (II) amido-iodine is formed.
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