Problem 91 - A Complete Resource Book in Chemistry for JEE Main

Question

A metal gives two chlorides \(A\) and \(B\). A gives black precipitate with \(\mathrm{NH}_{2}\) and \(\mathrm{B}\) gives white. With \(\mathrm{KI}, \mathrm{B}\) gives a yellow precipitate, which is soluble in excess of KI. A and B are respectively (a) \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}, \mathrm{HgCl}_{2}\) (b) \(\mathrm{HgCl}, \mathrm{Hg}_{2} \mathrm{Cl}_{2}\) (c) \(\mathrm{ZnCl}_{2}, \mathrm{HgCl}_{2}\) (d) \(\mathrm{HgCl}_{2}, \mathrm{ZnCl}_{2}\)

Step-by-Step Solution

Answer

(a) \( \mathrm{Hg}_{2} \mathrm{Cl}_{2}, \mathrm{HgCl}_{2} \)

Step 1: Understanding the Reaction with NH2

We need to determine which metal chloride gives a black precipitate with \( \mathrm{NH}_{2} \). According to the question, chloride \( A \) forms a black precipitate. It is known that \( \mathrm{HgCl}_{2} \) forms a black precipitate when it reacts with \( \mathrm{NH}_{4OH}\) due to the formation of mercury amidochloride, indicating that \( A \) could be \( \mathrm{Hg}_{2} \mathrm{Cl}_{2} \).

Step 2: Analyzing Reaction with KI

Chloride \( B \) gives a yellow precipitate with \( \mathrm{KI} \), which is soluble in excess. \(\mathrm{HgCl}_{2} \) reacts with \( \mathrm{KI} \) to form yellow mercuric iodide (\mathrm{HgI}_{2}\ ), which is indeed soluble in excess KI. This supports \( B \) being \( \mathrm{HgCl}_{2} \).

Step 3: Confirming All Evidence

Putting the observations together, we have \( A \) as \( \mathrm{Hg}_{2}\mathrm{Cl}_{2} \) since it forms a black precipitate with \( \mathrm{NH}_{2} \). \( B \) is confirmed to be \( \mathrm{HgCl}_{2} \) as it forms a yellow precipitate with \( \mathrm{KI} \) that dissolves in excess.

Key Concepts

Precipitation Reactions

Precipitation reactions are a fascinating and important area in inorganic chemistry. These reactions involve the formation of a solid, known as a precipitate, when two solutions are mixed. The precipitate forms because the ionic product of the resulting compound exceeds its solubility product, causing the ions to come out of solution and form a solid.

There are a few key points to remember about precipitation reactions:

• They occur when two aqueous solutions combine to form an insoluble compound.
• The solid formed is called a precipitate and can be removed by filtration.
• Typical reactions include mixing silver nitrate with sodium chloride to form silver chloride, a common example.

In our exercise, the reactions are used to identify unknown compounds based on the color and solubility of the precipitate formed. For example, one compound formed a black precipitate, allowing us to infer the metal chloride involved was likely mercury (Hg) containing, given its known reactions. Understanding these principles gives insight into predicting the behavior of ionic compounds in solution.

Coordination Compounds

Coordination compounds, a core topic in inorganic chemistry, are composed of a central metal atom surrounded by molecules or ions known as ligands. These compounds have unique properties and complex structures, as the ligands attach to the metal through coordinate covalent bonds.

Key characteristics of coordination compounds include:

• The central metal atom, often a transition metal, can form several bonds.
• Ligands donate a pair of electrons to form the bond, stabilizing the metal.
• These compounds exhibit varied geometry, such as octahedral or tetrahedral.

In coordination chemistry, the behavior of metal atoms like mercury in coordination compounds helps us understand the formation of stable complexes in solution. The specificity of mercury to form certain colored precipitates and complexes, like the yellow mercuric iodide with KI, highlights its role in forming diverse and identifiable coordination compounds.

HgCl2 Reactions

Mercury(II) chloride, or HgCl2, plays a prominent role in various reactions due to its unique chemistry. It serves as a versatile compound, possessing abilities to form distinct precipitates with numerous reagents.

Considerations when dealing with HgCl2 reactions include:

• HgCl2 reacts with ammonia (NH3) or ammonium hydroxide (NH4OH) to form black mercury amidochloride, a diagnostic feature in analysis.
• When treated with potassium iodide (KI), it forms a yellow precipitate known as mercuric iodide (HgI2), which is soluble in excess KI.
• This solubility behavior is crucial in determining the presence of HgCl2 in a mixture of substances.

In the exercise provided, identifying HgCl2 based on its predictable reaction patterns with NH3 and KI helped in solving the problem. Recognizing how these interactions proceed allows chemists to unravel complex reaction scenarios and deduce the identities of involved chemicals effectively.