Question

The reaction which produces yellow precipitate is

• A. $\underset{\mathrm{Sodium}\mathrm{hydroxide}}{\mathrm{NaOH}\left(\mathrm{aq}\right)}+\underset{\mathrm{Hydrochloric}\mathrm{acid}}{\mathrm{HCl}\left(\mathrm{aq}\right)}\to \underset{\mathrm{Sodium}\mathrm{chloride}}{\mathrm{NaCl}\left(\mathrm{s}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$
• B. $\underset{\mathrm{Sodium}\mathrm{hydroxide}}{2\mathrm{NaOH}\left(\mathrm{aq}\right)}+\underset{\mathrm{Barium}\mathrm{chloride}}{{\mathrm{BaCl}}_2\left(\mathrm{aq}\right)}\to \underset{\mathrm{Barium}\mathrm{hydroxide}}{\mathrm{Ba}{\left(\mathrm{OH}\right)}_2\left(\mathrm{s}\right)}+\underset{\mathrm{Sodium}\mathrm{chloride}}{2\mathrm{NaCl}\left(\mathrm{aq}\right)}$
• C. $\underset{\mathrm{Lead}\mathrm{nitrate}}{\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{aq}\right)}+\underset{\mathrm{Potassium}\mathrm{iodide}}{2\mathrm{KI}\left(\mathrm{aq}\right)}\to \underset{\mathrm{Potassium}\mathrm{nitrate}}{2{\mathrm{KNO}}_3\left(\mathrm{aq}\right)}+\underset{\mathrm{Lead}\mathrm{iodide}}{{\mathrm{PbI}}_2\left(\mathrm{s}\right)}$
• D. $\underset{\mathrm{Copper}\mathrm{oxide}}{\mathrm{CuO}\left(\mathrm{s}\right)}+\underset{\mathrm{Magnesium}}{\mathrm{Mg}\left(\mathrm{s}\right)}\to \underset{\mathrm{Copper}}{\mathrm{Cu}\left(\mathrm{s}\right)}+\underset{\mathrm{Magnesium}\mathrm{oxide}}{\mathrm{MgO}\left(\mathrm{s}\right)}$
• E. $\underset{\mathrm{Iron}}{4\mathrm{Fe}\left(\mathrm{s}\right)}+\underset{\mathrm{Oxygen}}{3{\mathrm{O}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Iron}\mathrm{oxide}}{2{\mathrm{Fe}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)}$

Answer 1

The correct option is C

$\underset{\mathrm{Lead}\mathrm{nitrate}}{\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{aq}\right)}+\underset{\mathrm{Potassium}\mathrm{iodide}}{2\mathrm{KI}\left(\mathrm{aq}\right)}\to \underset{\mathrm{Potassium}\mathrm{nitrate}}{2{\mathrm{KNO}}_3\left(\mathrm{aq}\right)}+\underset{\mathrm{Lead}\mathrm{iodide}}{{\mathrm{PbI}}_2\left(\mathrm{s}\right)}$

The explanation for the correct option:

Option (C): $\underset{\mathrm{Lead}\mathrm{nitrate}}{\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{aq}\right)}+\underset{\mathrm{Potassium}\mathrm{iodide}}{2\mathrm{KI}\left(\mathrm{aq}\right)}\to \underset{\mathrm{Potassium}\mathrm{nitrate}}{2{\mathrm{KNO}}_3\left(\mathrm{aq}\right)}+\underset{\mathrm{Lead}\mathrm{iodide}}{{\mathrm{PbI}}_2\left(\mathrm{s}\right)}$

1. Precipitation reaction is a reaction that results in the formation of insoluble solid material.
2. Iodide ion displaces nitrate from Lead nitrate and nitrate displaces Iodine from Potassium iodide when aqueous Lead nitrate is introduced to Potassium iodide. As a result, two new products are formed Potassium nitrate and Lead iodide.
3. It can also be considered a double displacement reaction because two ions interchange their positions.
4. The precipitate formed is the compound Lead iodide (${\mathrm{PbI}}_2$). It is yellow in color at room temperature and becomes red or orange when heated.
5. The presence of yellow Lead iodide particles in the reaction mixture shows that a chemical transition has occurred.
6. Hence, it is the correct option.

The explanation for the incorrect option:

Option (A): $\underset{\mathrm{Sodium}\mathrm{hydroxide}}{\mathrm{NaOH}\left(\mathrm{aq}\right)}+\underset{\mathrm{Hydrochloric}\mathrm{acid}}{\mathrm{HCl}\left(\mathrm{aq}\right)}\to \underset{\mathrm{Sodium}\mathrm{chloride}}{\mathrm{NaCl}\left(\mathrm{s}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

1. When Sodium hydroxide is reacted with acid it undergoes a neutralization reaction i.e. formation of salt along with water.
2. Here Sodium hydroxide reacts with Hydrochloric acid to give Sodium chloride and water.
3. It is highly soluble in water and hence doesn't precipitate.

Option (B): $\underset{\mathrm{Sodium}\mathrm{hydroxide}}{2\mathrm{NaOH}\left(\mathrm{aq}\right)}+\underset{\mathrm{Barium}\mathrm{chloride}}{{\mathrm{BaCl}}_2\left(\mathrm{aq}\right)}\to \underset{\mathrm{Barium}\mathrm{hydroxide}}{\mathrm{Ba}{\left(\mathrm{OH}\right)}_2\left(\mathrm{s}\right)}+\underset{\mathrm{Sodium}\mathrm{chloride}}{2\mathrm{NaCl}\left(\mathrm{aq}\right)}$

1. Reaction of Sodium hydroxide and Barium chloride is an example of a double displacement reaction. The hydroxide ion gets displaced by the chloride ion in Sodium hydroxide and the Chloride ion gets displaced with the hydroxide ion.
2. It results in the formation of Barium hydroxide and Sodium chloride, among which Barium hydroxide is a white color precipitate.

Option (D): $\underset{\mathrm{Copper}\mathrm{oxide}}{\mathrm{CuO}\left(\mathrm{s}\right)}+\underset{\mathrm{Magnesium}}{\mathrm{Mg}\left(\mathrm{s}\right)}\to \underset{\mathrm{Copper}}{\mathrm{Cu}\left(\mathrm{s}\right)}+\underset{\mathrm{Magnesium}\mathrm{oxide}}{\mathrm{MgO}\left(\mathrm{s}\right)}$

1. The is a redox reaction between Copper oxide and Magnesium, which demonstrates magnesium's ability to remove oxygen from Copper oxide i.e. Magnesium is more reactive than Copper
2. Magnesium oxide, often known as magnesia, is a hygroscopic solid mineral that is white in color.

Option (E): $\underset{\mathrm{Iron}}{4\mathrm{Fe}\left(\mathrm{s}\right)}+\underset{\mathrm{Oxygen}}{3{\mathrm{O}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Iron}\mathrm{oxide}}{2{\mathrm{Fe}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)}$

1. Metal oxides are created when metals react with Oxygen. The rate of the reaction is determined by the metal's reactivity.
2. Iron ox­ides are formed when the iron is burned. Iron ox­ides can differ depending on the re­ac­tion con­di­tions and the pro­por­tions be­tween the two par­tic­i­pants.
3. Iron ox­ide ${\mathrm{Fe}}_2{\mathrm{O}}_3$ is an or­ange-red powder that develops when iron in the air is oxidized.

Therefore, the Reaction which produces yellow precipitate is option (C) $\underset{\mathrm{Lead}\mathrm{nitrate}}{\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{aq}\right)}+\underset{\mathrm{Potassium}\mathrm{iodide}}{2\mathrm{KI}\left(\mathrm{aq}\right)}\to \underset{\mathrm{Potassium}\mathrm{nitrate}}{2{\mathrm{KNO}}_3\left(\mathrm{aq}\right)}+\underset{\mathrm{Lead}\mathrm{iodide}}{{\mathrm{PbI}}_2\left(\mathrm{s}\right)}$