Problem 158 - A Complete Resource Book in Chemistry for JEE Main

Question

The oxidation state of \(\mathrm{Cr}\) in \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]^{+} \quad[\mathbf{2 0 0 5}]\) (a) 0 (b) \(+1\) (c) \(+2\) (d) \(+3\)

Step-by-Step Solution

Answer

The oxidation state of \( \mathrm{Cr} \) is \(+3\).

Step 1: Understanding the Composition

The compound \( \left[ \mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]^{+} \) is a complex ion consisting of \( \mathrm{Cr} \), four \( \mathrm{NH}_3 \) ligands, and two \( \mathrm{Cl}^- \) ions. In this structure, \( \mathrm{NH}_3 \) is a neutral ligand, meaning it does not contribute to the overall charge.

Step 2: Assign Oxidation Numbers

In the complex ion, the chloride ion \( \mathrm{Cl}^- \) has a charge of \(-1\). Since there are two chloride ions, their total charge is \(-2\). The ion is represented as \( \left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]^{+} \), indicating it has an overall charge of \(+1\).

Step 3: Calculating the Oxidation State of Chromium

Let the oxidation state of \( \mathrm{Cr} \) be \( x \). The sum of all charges must equal the overall charge of the ion. We set up the equation: \[x + 0 \times 4 + (-1) \times 2 = +1.\] Simplifying gives \( x - 2 = +1 \).

Step 4: Solve for x

Rearrange the equation \( x - 2 = +1 \) to find \( x \). Add 2 to both sides, yielding \( x = +3 \). Therefore, the oxidation state of \( \mathrm{Cr} \) is \(+3\).

Key Concepts

Transition Metals

Transition metals are a group of elements found in the d-block of the periodic table. They are known for their ability to form various oxidation states and complex ions. This versatility is due to the presence of partially filled d-orbitals. Transition metals include chromium (Cr), iron (Fe), and copper (Cu), among others.

Transition metals have unique characteristics:

• Different oxidation states due to the involvement of d-electrons in bonding.
• Formation of colored compounds because the d-d electronic transitions often fall within visible light.
• Ability to form complex ions due to their variable oxidation states and small atomic radii, which allow them to form stable coordination compounds.

Chromium, a transition metal, is an excellent example of these properties. It can exist in different oxidation states, such as Cr(II), Cr(III), and Cr(VI), which are commonly used in industrial and chemical processes.

Coordination Compounds

Coordination compounds are specific types of complex ions that form when a metal ion binds to molecules or ions called ligands. These structures involve coordinate covalent bonds, where the ligands donate pairs of electrons to the metal ion.

In the given compound \([\text{Cr}(\text{NH}_3)_4 \text{Cl}_2]^+\), the chromium (Cr) acts as the central metal ion. It is surrounded by four ammonia (NH₃) molecules and two chloride (Cl^-) ions:

• Ligands: Neutral or charged species that donate electron pairs. In this case, \(\text{NH}_3\), a neutral ligand, contributes no charge.
• Geometry: The arrangement of these ligands around the central metal ion influences the geometry. The ammonia and chloride ions coordinate with the metal to form a stable octahedral structure.

Coordination compounds are significant in various chemical reactions and their properties can be altered by changing the ligands or metal ions.

Oxidation Number Calculation

The oxidation number is a concept used to keep track of electrons in chemical reactions. It indicates the charge an atom would have if the compound was composed of ions. Calculating the oxidation number involves several steps.

In the exercise, we determine the oxidation state of chromium in the complex ion \([\text{Cr}(\text{NH}_3)_4 \text{Cl}_2]^+\):

• Identify the charges of each component: ammonia (\(\text{NH}_3\)) is neutral, and each chloride ion (\(\text{Cl}^-\)) has a charge of \(-1\).
• Sum these charges with the oxidation state of the chromium ion:\[ x + 0 \times 4 + (-1) \times 2 = +1 \]
• Simplify the equation to find the oxidation state of Cr: \( x - 2 = +1 \). Solving for \( x \) gives \( x = +3 \).

This calculated oxidation number helps in understanding redox reactions and stabilization of the complex, making it a fundamental concept in chemistry.